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21473
|
\section{Size of Tree is One Less than Order/Necessary Condition/Induction Step/Proof 1}
Tags: Size of Tree is One Less than Order, Tree Theory, Trees
\begin{theorem}
Let the following hold:
:A tree of order $k$ is of size $k - 1$.
Then this holds:
:A tree of order $k + 1$ is of size $k$.
\end{theorem}
\begin{proof}
Let $T_{k + 1}$ be an arbitrary tree with $k + 1$ nodes.
Take any node $v$ of $T_{k+1}$ of degree $1$.
Such a node exists from Finite Tree has Leaf Nodes.
Consider $T_k$, the subgraph of $T_{k + 1}$ created by removing $v$ and the edge connecting it to the rest of the tree.
By Connected Subgraph of Tree is Tree, $T_k$ is itself a tree.
The order of $T_k$ is $k$, and it has one less edge than $T_{k + 1}$ by definition.
By hypothesis, $T_k$ has $k - 1$ edges.
So $T_{k + 1}$ must have $k$ edges.
{{qed}}
Category:Size of Tree is One Less than Order
\end{proof}
|
21474
|
\section{Size of Tree is One Less than Order/Necessary Condition/Induction Step/Proof 2}
Tags: Size of Tree is One Less than Order, Tree Theory, Trees
\begin{theorem}
Let the following hold:
:For all $j \le k$, a tree of order $j$ is of size $j-1$.
Then this holds:
:A tree of order $k+1$ is of size $k$.
\end{theorem}
\begin{proof}
Let $T_{k + 1}$ be any tree with $k + 1$ nodes.
Remove any edge $e$ of $T_{k + 1}$.
By definition of tree $T_{k + 1}$ has no circuits
Therefore from Condition for Edge to be Bridge it follows that $e$ must be a bridge.
So removing $e$ disconnects $T_{k + 1}$ into two trees $T_1$ and $T_2$, with $k_1$ and $k_2$ nodes, where $k_1 + k_2 = k + 1$.
By hypothesis, $T_1$ and $T_2$ have $k_1 - 1$ and $k_2 - 1$ edges.
Putting the edge $e$ back again, it can be seen that $T_{k + 1}$ has $\paren {k_1 - 1} + \paren {k_2 - 1} + 1 = k$ edges.
Therefore a tree of order $k+1$ is of size $k$.
{{qed}}
\end{proof}
|
21475
|
\section{Size of Tree is One Less than Order/Sufficient Condition}
Tags: Size of Tree is One Less than Order, Tree Theory, Trees
\begin{theorem}
Let $T$ be a connected simple graph of order $n$.
Let the size of $T$ be $n-1$.
Then $T$ is a tree.
\end{theorem}
\begin{proof}
By definition, the order of a tree is how many nodes it has, and its size is how many edges it has.
Suppose $T$ is a connected simple graph of order $n$ with $n - 1$ edges.
We need to show that $T$ is a tree.
{{AimForCont}} $T$ is not a tree.
Then it contains a circuit.
It follows from Condition for Edge to be Bridge that there is at least one edge in $T$ which is not a bridge.
So we can remove this edge and obtain a graph $T'$ which is connected and has $n$ nodes and $n - 2$ edges.
Let us try and construct a connected graph with $n$ nodes and $n - 2$ edges.
We start with the edgeless graph $N_n$, and add edges till the graph is connected.
We pick any two vertices of $N_n$, label them $u_1$ and $u_2$ for convenience, and use one edge to connect them, labelling that edge $e_1$.
We pick any other vertex, label it $u_3$, and use one edge to connect it to either $u_1$ or $u_2$, labelling that edge $e_2$.
We pick any other vertex, label it $u_4$, and use one edge to connect it to either $u_1, u_2$ or $u_3$, labelling that edge $e_3$.
We continue in this way, until we pick a vertex, label it $u_{n - 1}$, and use one edge to connect it to either $u_1, u_2, \ldots, u_{n - 2}$, labelling that edge $e_{n - 2}$.
That was the last of our edges, and the last vertex still has not been connected.
Therefore a graph with $n$ vertices and $n-2$ edges that such a graph ''cannot'' be connected.
Therefore we cannot remove any edge from $T$ without leaving it disconnected.
Therefore all the edges in $T$ are bridges.
Hence $T$ can contain no circuits.
Hence, by Proof by Contradiction, $T$ must be a tree.
{{qed}}
\end{proof}
|
21476
|
\section{Size of y-1 lt n and Size of y+1 gt 1 over n}
Tags: Unions of Families, Intersections of Families
\begin{theorem}
Let $T_n \subseteq \R$ be the subset of the set of real numbers $\R$ defined as:
:$T_n = \set {y: \size {y - 1} < n \land \size {y + 1} > \dfrac 1 n}$
Then:
:$T_n = \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}$
\end{theorem}
\begin{proof}
:500pxthumbright
First note that:
{{begin-eqn}}
{{eqn | l = T_n
| r = \set {y: \size {y - 1} < n \land \size {y + 1} > \dfrac 1 n}
| c =
}}
{{eqn | r = \set {y: \size {y - 1} < n} \cap \set {y: \size {y + 1} > \dfrac 1 n}
| c =
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | l = \set {y: \size {y - 1} < n}
| r = \openint {1 - n} {1 + n}
| c = Open Interval Defined by Absolute Value
}}
{{eqn | l = \set {y: \size {y + 1} > \dfrac 1 n}
| r = \R \setminus \closedint {-1 - \dfrac 1 n} {-1 + \dfrac 1 n}
| c = Complement of Closed Interval Defined by Absolute Value
}}
{{eqn | ll= \leadsto
| l = \set {y: \size {y - 1} < n} \cap \set {y: \size {y + 1} > \dfrac 1 n}
| r = \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}
| c =
}}
{{end-eqn}}
{{qed}}
Category:Unions of Families
Category:Intersections of Families
\end{proof}
|
21477
|
\section{Skewness in terms of Non-Central Moments}
Tags: Skewness
\begin{theorem}
Let $X$ be a random variable with mean $\mu$ and standard deviation $\sigma$.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \expect {\paren {\dfrac {X - \mu} \sigma}^3}
| c = {{Defof|Skewness}}
}}
{{eqn | r = \frac {\expect {X^3 - 3 \mu X^2 + 3 \mu^2 X - \mu^3} } {\sigma^3}
| c = Linearity of Expectation Function, Cube of Difference
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \expect {X^2} + 3 \mu^2 \expect X - \mu^3} {\sigma^3}
| c = Linearity of Expectation Function
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \paren {\expect {X^2} - \mu \expect X} - \mu^3} {\sigma^3}
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \paren {\expect {X^2} - \paren {\expect X}^2} - \mu^3} {\sigma^3}
| c = $\mu = \expect X$
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}
| c = Variance as Expectation of Square minus Square of Expectation
}}
{{end-eqn}}
{{qed}}
Category:Skewness
\end{proof}
|
21478
|
\section{Skewness of Bernoulli Distribution}
Tags: Skewness, Bernoulli Distribution
\begin{theorem}
Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac {1 - 2 p} {\sqrt {p q} }$
where $q = 1 - p$.
\end{theorem}
\begin{proof}
From Skewness in terms of Non-Central Moments:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where $\mu$ is the mean of $X$, and $\sigma$ the standard deviation.
We have, by Expectation of Bernoulli Distribution:
:$\mu = p$
By Variance of Bernoulli Distribution, we also have:
:$\var X = \sigma^2 = p \paren {1 - p}$
so:
:$\sigma = \sqrt {p \paren {1 - p} }$
By Raw Moment of Bernoulli Distribution, we have:
:$\expect {X^3} = p$
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {p - 3 p^2 \paren {1 - p} - p^3} {p^{3/2} \paren {1 - p}^{3/2} }
}}
{{eqn | r = \frac {p \paren {1 - p^2} - 3 p^2 \paren {1 - p} } {p^{3/2} \paren {1 - p}^{3/2} }
}}
{{eqn | r = \frac {p \paren {1 - p} \paren {1 + p} - 3 p^2 \paren {1 - p} } {p^{3/2} \paren {1 - p}^{3/2} }
| c = Difference of Two Squares
}}
{{eqn | r = \frac {p \paren {1 - p} \paren {1 + p - 3p} } {p^{3/2} \paren {1 - p}^{3/2} }
}}
{{eqn | r = \frac {1 - 2 p} {\sqrt {p \paren {1 - p} } }
}}
{{eqn | r = \frac {1 - 2 p} {\sqrt {p q} }
| c = $q = 1 - p$
}}
{{end-eqn}}
{{qed}}
Category:Bernoulli Distribution
Category:Skewness
\end{proof}
|
21479
|
\section{Skewness of Beta Distribution}
Tags: Beta Distribution, Skewness
\begin{theorem}
Let $X \sim \BetaDist \alpha \beta$ for some $\alpha, \beta > 0$, where $\operatorname {Beta}$ denotes the Beta distribution.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac {2 \paren {\beta - \alpha} \sqrt {\alpha + \beta + 1} } {\paren {\alpha + \beta + 2} \sqrt {\alpha \beta} }$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}
| c = Skewness in terms of Non-Central Moments
}}
{{eqn | r = \paren {\paren {\prod_{r \mathop = 0}^2 \frac {\alpha + r} {\alpha + \beta + r} } - \frac {3 \alpha} {\alpha + \beta} \cdot \frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } - \paren {\frac \alpha {\alpha + \beta} }^3} \paren {\frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } }^{-\frac 3 2}
| c = Raw Moment of Beta Distribution, Expectation of Beta Distribution, Variance of Beta Distribution
}}
{{eqn | r = \paren {\frac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + \beta}^2 - 3 \alpha^2 \beta \paren {\alpha + \beta + 2} - \alpha^3 \paren {\alpha + \beta + 1} \paren {\alpha + \beta + 2} } {\paren {\alpha + \beta}^3 \paren {\alpha + \beta + 1} \paren {\alpha + \beta + 2} } } \paren {\frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } }^{-\frac 3 2}
}}
{{eqn | r = \paren {\frac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + \beta}^2 - 3 \alpha^2 \beta \paren {\alpha + \beta + 2} - \alpha^3 \paren {\alpha + \beta + 1} \paren {\alpha + \beta + 2} } {\alpha + \beta + 2} } \sqrt {\frac {\alpha + \beta + 1} {\alpha^3 \beta^3} }
}}
{{eqn | r = \paren {\frac {2 \alpha \beta \paren {\beta - \alpha} } {\alpha + \beta + 2} } \sqrt {\frac {\alpha + \beta + 1} {\alpha^3 \beta^3} }
}}
{{eqn | r = \frac {2 \paren {\beta - \alpha} \sqrt {\alpha + \beta + 1} } {\paren {\alpha + \beta + 2} \sqrt {\alpha \beta} }
}}
{{end-eqn}}
The fifth step is justified by:
{{begin-eqn}}
{{eqn | o =
| r = \alpha \paren {\alpha + 1} \paren {\alpha + 2} \paren {\alpha + \beta}^2 - 3 \alpha^2 \beta \paren {\alpha + \beta + 2} - \alpha^3 \paren {\alpha + \beta + 1} \paren {\alpha + \beta + 2}
}}
{{eqn | r = \alpha^5 + 3 \alpha^4 + 2 \alpha^3 + 2 \alpha^4 \beta + 6 \alpha^3 \beta + 4 \alpha^2 \beta + \alpha^3 \beta^2 + 3 \alpha^2 \beta^2 + 2 \alpha \beta^2 - 3 \alpha^3 \beta - 3 \alpha^2 \beta^2 - 6 \alpha^2 \beta - \alpha^5 - 2 \alpha^4 \beta - \alpha^3 \beta^2 - 3 \alpha^4 - 3 \alpha^3 \beta - 2 \alpha^3
}}
{{eqn | r = - 2 \alpha^2 \beta + 2 \alpha \beta^2
}}
{{eqn | r = 2 \alpha \beta \paren {\beta - \alpha}
}}
{{end-eqn}}
{{qed}}
Category:Skewness
Category:Beta Distribution
\end{proof}
|
21480
|
\section{Skewness of Chi-Squared Distribution}
Tags: Chi-Squared Distribution, Skewness
\begin{theorem}
Let $n$ be a strictly positive integer.
Let $X \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \sqrt{\dfrac 8 n}$
\end{theorem}
\begin{proof}
From Skewness in terms of Non-Central Moments, we have:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where:
:$\mu$ is the expectation of $X$.
:$\sigma$ is the standard deviation of $X$.
By Expectation of Chi-Squared Distribution we have:
:$\mu = n$
By Variance of Chi-Squared Distribution we have:
:$\sigma = \sqrt {2 n}$
We also have:
{{begin-eqn}}
{{eqn | l = \expect {X^3}
| r = \prod_{k \mathop = 0}^2 \paren {n + 2 k}
| c = Raw Moment of Chi-Squared Distribution
}}
{{eqn | r = n \paren {n + 2} \paren {n + 4}
}}
{{eqn | r = n^3 + 6 n^2 + 8 n
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {n^3 + 6 n^2 + 8 n - 6 n^2 - n^3} {\paren {\sqrt {2 n} }^3}
}}
{{eqn | r = \frac {\sqrt {\paren {8 n}^2} } {\sqrt {\paren {2 n}^3} }
}}
{{eqn | r = \frac {\sqrt {64 n^2} } {\sqrt {8 n^3} }
}}
{{eqn | r = \sqrt {\frac 8 n}
}}
{{end-eqn}}
{{qed}}
Category:Skewness
Category:Chi-Squared Distribution
\end{proof}
|
21481
|
\section{Skewness of Continuous Uniform Distribution}
Tags: Skewness, Continuous Uniform Distribution
\begin{theorem}
Let $X$ be a continuous random variable which is uniformly distributed on a closed real interval $\closedint a b$.
Then the skewness $\gamma_1$ of $X$ is equal to $0$.
\end{theorem}
\begin{proof}
From the definition of skewness:
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
where:
:$\mu$ is the mean of $X$
:$\sigma$ is the standard deviation of $X$.
From the definition of the continuous uniform distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {b - a}$
So, from Expectation of Function of Continuous Random Variable:
:$\ds \gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3} = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \mu}^3 \rd x$
Then:
{{begin-eqn}}
{{eqn | l = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \mu}^3 \rd x
| r = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \frac {a + b} 2}^3 \rd x
| c = Expectation of Continuous Uniform Distribution
}}
{{eqn | r = \frac 1 {\paren {b - a} \sigma^3} \int_{a - \frac {a + b} 2}^{b - \frac {a + b} 2} u^3 \rd u
| c = substituting $u = x - \dfrac {a + b} 2$
}}
{{eqn | r = \frac 1 {\paren {b - a} \sigma^3} \int_{-\frac {b - a} 2}^{\frac {b - a} 2} u^3 \rd u
}}
{{end-eqn}}
We have:
:$\paren {-u}^3 = -u^3$
so we can see the integrand is odd.
So, by Definite Integral of Odd Function:
:$\ds \gamma_1 = \frac 1 {\paren {b - a} \sigma^3} \int_{-\frac {b - a} 2}^{\frac {b - a} 2} u^3 \rd u = 0$
{{qed}}
Category:Skewness
Category:Continuous Uniform Distribution
\end{proof}
|
21482
|
\section{Skewness of Exponential Distribution}
Tags: Exponential Distribution, Skewness
\begin{theorem}
Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$
Then the skewness $\gamma_1$ of $X$ is equal to $2$.
\end{theorem}
\begin{proof}
From Skewness in terms of Non-Central Moments, we have:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where:
:$\mu$ is the expectation of $X$.
:$\sigma$ is the standard deviation of $X$.
By Expectation of Exponential Distribution we have:
:$\mu = \beta$
By Variance of Exponential Distribution we have:
:$\sigma = \beta$
By Raw Moment of Exponential Distribution we also have:
:$\expect {X^3} = 3! \beta^3 = 6 \beta^3$
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {6 \beta^3 - 3\beta^3 - \beta^3} {\beta^3}
}}
{{eqn | r = \frac {2 \beta^3} {\beta^3}
}}
{{eqn | r = 2
}}
{{end-eqn}}
{{qed}}
Category:Skewness
Category:Exponential Distribution
\end{proof}
|
21483
|
\section{Skewness of F-Distribution}
Tags: F-Distribution, Skewness
\begin{theorem}
Let $n, m$ be strictly positive integers.
Let $X \sim F_{n, m}$ where $F_{n, m}$ is the F-distribution with $\tuple {n, m}$ degrees of freedom.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac {2 \paren {m + 2 n - 2} } {m - 6} \sqrt {\dfrac {2 \paren {m - 4} } {n \paren {m + n - 2} } }$
for $m > 6$, and does not exist otherwise.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Skewness
Category:F-Distribution
\end{proof}
|
21484
|
\section{Skewness of Gamma Distribution}
Tags: Skewness of Gamma Distribution, Skewness, Gamma Distribution
\begin{theorem}
Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac 2 {\sqrt \alpha}$
\end{theorem}
\begin{proof}
From the definition of skewness, we have:
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
where:
:$\mu$ is the expectation of $X$.
:$\sigma$ is the standard deviation of $X$.
By Expectation of Gamma Distribution, we have:
:$\mu = \dfrac \alpha \beta$
By Variance of Gamma Distribution, we have:
:$\sigma = \dfrac {\sqrt \alpha} \beta$
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {\expect {X^3 - 3 X^2 \mu + 3 X \mu^2 - \mu^3} } {\sigma^3}
| c = Cube of Difference
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \expect {X^2} + 3 \mu^2 \expect {X} - \mu^3} {\sigma^3}
| c = Linearity of Expectation Function
}}
{{eqn | r = \frac {\expect {X^3} - 3 \paren {\dfrac \alpha \beta} \paren {\dfrac {\alpha \paren {\alpha + 1} } {\beta^2} } + 3 \paren {\dfrac \alpha \beta}^2 \paren {\dfrac \alpha \beta} - \paren {\dfrac \alpha \beta}^3} {\paren {\dfrac {\alpha \sqrt \alpha} {\beta^3} } }
| c = Variance of Gamma Distribution
}}
{{end-eqn}}
To calculate $\gamma_1$, we must calculate $\expect {X^3}$.
From Moment in terms of Moment Generating Function:
:$\expect {X^n} = \map { {M_X}^{\paren n} } 0$
where $M_X$ is the moment generating function of $X$.
From Moment Generating Function of Gamma Distribution: Third Moment:
:$\map { {M_X}'''} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\paren {\beta - t}^{\alpha + 3} }$
Setting $t = 0$:
{{begin-eqn}}
{{eqn | l = \expect {X^3}
| r = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\paren {\beta - 0}^{\alpha + 3} }
}}
{{eqn | r = \dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3}
| c =
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {\expect {X^3} - 3 \paren {\dfrac \alpha \beta} \paren {\dfrac {\alpha \paren {\alpha + 1} } {\beta^2} } + 3 \paren {\dfrac \alpha \beta}^2 \paren {\dfrac \alpha \beta} - \paren {\dfrac \alpha \beta}^3} {\paren {\dfrac {\alpha \sqrt \alpha} {\beta^3} } }
}}
{{eqn | r = \frac {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} - 3 \paren {\dfrac \alpha \beta} \paren {\dfrac {\alpha \paren {\alpha + 1} } {\beta^2} } + 3 \paren {\dfrac \alpha \beta}^2 \paren {\dfrac \alpha \beta} - \paren {\dfrac \alpha \beta}^3} {\paren {\dfrac {\alpha \sqrt \alpha} {\beta^3} } }
}}
{{eqn | r = \frac {\alpha^3 + 3 \alpha^2 + 2 \alpha - 3 \paren {\alpha^3 + \alpha^2} + 3 \alpha^3 - \alpha^3} {\paren {\alpha \sqrt \alpha } }
| c = $\beta^3$ cancels
}}
{{eqn | r = \dfrac 2 {\sqrt \alpha}
}}
{{end-eqn}}
{{qed}}
Category:Gamma Distribution
Category:Skewness
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Prime.mover
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\end{proof}
|
21485
|
\section{Skewness of Gaussian Distribution}
Tags: Skewness of Gaussian Distribution, Skewness, Gaussian Distribution
\begin{theorem}
Let $X$ be a continuous random variable with a Gaussian distribution with parameters $\mu$ and $\sigma^2$ for some $\mu \in \R$ and $\sigma \in \R_{> 0}$.
Then the skewness $\gamma_1$ of $X$ is equal to $0$.
\end{theorem}
\begin{proof}
From the definition of skewness:
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
From the definition of the Gaussian distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {\sigma \sqrt{2 \pi} } \, \map \exp {-\dfrac { \paren {x - \mu}^2} {2 \sigma^2} }$
So, from Expectation of Function of Continuous Random Variable:
:$\displaystyle \gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3} = \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac {x - \mu} \sigma}^3 \map \exp {-\dfrac { \paren {x - \mu}^2} {2 \sigma^2} } \rd x$
Making a substitution of $u = x - \mu$:
:$\displaystyle \gamma_1 = \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac u \sigma}^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} } \rd u$
We have that:
:$\paren {-u}^3 \map \exp {-\dfrac {\paren {-u}^2} {2 \sigma^2} } = -u^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} }$
So we can see that the integrand is odd.
So, by Definite Integral of Odd Function:
:$\displaystyle \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac u \sigma}^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} } \rd u = 0$
giving:
:$\gamma_1 = 0$
{{qed}}
Category:Skewness
Category:Gaussian Distribution
515913
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\end{proof}
|
21486
|
\section{Skewness of Geometric Distribution}
Tags: Skewness, Skewness of Geometric Distribution, Geometric Distribution
\begin{theorem}
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Geometric Distribution
Category:Skewness
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wikitext
text/x-wiki
\end{proof}
|
21487
|
\section{Skewness of Geometric Distribution/Formulation 1}
Tags: Skewness, Skewness of Geometric Distribution
\begin{theorem}
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = \paren {1 - p} p^k$
Then the skewness of $X$ is given by:
:$\gamma_1 = \dfrac {1 + p} {\sqrt {p} }$
\end{theorem}
\begin{proof}
From the definition of skewness, we have:
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
where:
:$\mu$ is the expectation of $X$.
:$\sigma$ is the standard deviation of $X$.
By Expectation of Geometric Distribution: Formulation 1, we have:
:$\mu = \dfrac p {1 - p}$
By Variance of Geometric Distribution: Formulation 1, we have:
:$\sigma = \dfrac {\sqrt p } {1 - p}$
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {\expect {X^3 - 3 X^2 \mu + 3 X \mu^2 - \mu^3} } {\sigma^3}
| c = Cube of Difference
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \expect {X^2} + 3 \mu^2 \expect {X} - \mu^3} {\sigma^3}
| c = Linearity of Expectation Function
}}
{{eqn | r = \frac {\expect {X^3} - 3 \paren {\dfrac p {1 - p} } \expect {X^2} + 2 \paren {\dfrac {p^3 } {\paren {1 - p}^3} } } {\paren {\dfrac {p \sqrt p} {\paren {1 - p}^3} } }
| c =
}}
{{end-eqn}}
To calculate $\gamma_1$, we must calculate both $\expect {X^2}$ and $\expect {X^3}$.
From Moment in terms of Moment Generating Function:
:$\expect {X^n} = \map { {M_X}^{\paren n} } 0$
where $M_X$ is the moment generating function of $X$.
From Moment Generating Function of Geometric Distribution: Second Moment:
:$\map { {M_X}''} t = p \paren {1 - p} e^t \paren {\dfrac {1 + p e^t} {\paren {1 - p e^t}^3} }$
From Moment Generating Function of Geometric Distribution: Third Moment:
:$\map { {M_X}'''} t = p \paren {1 - p } e^t \paren {\dfrac {1 + 4p e^t + p^2 e^{2t} } {\paren {1 - p e^t}^4 } }$
Setting $t = 0$ and from Exponential of Zero, we have:
:$\map {M_X''} 0 = \dfrac {p \paren {1 + p} } {\paren {1 - p}^2 }$
:$\map {M_X'''} 0 = \dfrac {p \paren {1 + 4 p + p^2} } {\paren {1 - p}^3 }$
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {\expect {X^3} - 3 \paren {\dfrac p {1 - p} } \expect {X^2} + 2 \paren {\dfrac {p^3 } {\paren {1 - p}^3} } } {\paren {\dfrac {p \sqrt p} {\paren {1 - p}^3} } }
}}
{{eqn | r = \frac {\dfrac {p \paren {1 + 4 p + p^2} } {\paren {1 - p}^3 } - 3 \paren {\dfrac p {1 - p} } \dfrac {p \paren {1 + p} } {\paren {1 - p}^2 } + 2 \paren {\dfrac {p^3 } {\paren {1 - p}^3} } } {\paren {\dfrac {p \sqrt p} {\paren {1 - p}^3} } }
}}
{{eqn | r = \frac {\paren {1 + 4 p + p^2} - 3 p \paren {1 + p} + 2 p^2 } { \sqrt p }
| c = $\dfrac p {\paren {1 - p}^3}$ cancels
}}
{{eqn | r = \dfrac {1 + p} {\sqrt {p} }
}}
{{end-eqn}}
{{qed}}
Category:Skewness
Category:Skewness of Geometric Distribution
\end{proof}
|
21488
|
\section{Skewness of Geometric Distribution/Formulation 2}
Tags: Skewness, Skewness of Geometric Distribution
\begin{theorem}
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$ for some $0 < p < 1$.
:$\map X \Omega = \set {0, 1, 2, \ldots} = \N$
:$\map \Pr {X = k} = p \paren {1 - p}^k$
Then the skewness of $X$ is given by:
:$\gamma_1 = \dfrac {2 - p} {\sqrt {1 - p} }$
\end{theorem}
\begin{proof}
From the definition of skewness, we have:
:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
where:
:$\mu$ is the expectation of $X$.
:$\sigma$ is the standard deviation of $X$.
By Expectation of Geometric Distribution: Formulation 2, we have:
:$\mu = \dfrac {1 - p} p$
By Variance of Geometric Distribution: Formulation 2, we have:
:$\sigma = \dfrac {\sqrt {1 - p} } p$
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {\expect {X^3 - 3 X^2 \mu + 3 X \mu^2 - \mu^3} } {\sigma^3}
| c = Cube of Difference
}}
{{eqn | r = \frac {\expect {X^3} - 3 \mu \expect {X^2} + 3 \mu^2 \expect {X} - \mu^3} {\sigma^3}
| c = Linearity of Expectation Function
}}
{{eqn | r = \frac {\expect {X^3} - 3 \paren {\dfrac {1 - p} p } \expect {X^2} + 2 \paren {\dfrac {\paren {1 - p} \paren {1 - p}^2 } {p^3} } } {\paren {\dfrac {\paren {1 - p} \sqrt {1 - p} } {p^3} } }
| c =
}}
{{end-eqn}}
To calculate $\gamma_1$, we must calculate both $\expect {X^2}$ and $\expect {X^3}$.
From Moment in terms of Moment Generating Function:
:$\expect {X^n} = \map { {M_X}^{\paren n} } 0$
where $M_X$ is the moment generating function of $X$.
From Moment Generating Function of Geometric Distribution: Second Moment:
:$\map { {M_X}''} t = p \paren {1 - p} e^t \paren {\dfrac {1 + \paren {1 - p} e^t } {\paren {1 - \paren {1 - p} e^t}^3 } }$
From Moment Generating Function of Geometric Distribution: Third Moment:
:$\map { {M_X}'''} t = p \paren {1 - p} e^t \paren {\dfrac {1 + 4 \paren {1 - p} e^t + \paren {1 - p}^2 e^{2t} } {\paren {1 - \paren {1 - p} e^t}^4 } }$
Setting $t = 0$ and from Exponential of Zero, we have:
:$\map {M_X''} 0 = \dfrac {\paren {1 - p} \paren {2 - p} } {p^2 }$
:$\map {M_X'''} 0 = \dfrac {\paren {1 - p} \paren {6 - 6 p + p^2} } {p^3 }$
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {\expect {X^3} - 3 \paren {\dfrac {1 - p} p } \expect {X^2} + 2 \paren {\dfrac {\paren {1 - p} \paren {1 - p}^2 } {p^3} } } {\paren {\dfrac {\paren {1 - p} \sqrt {1 - p} } {p^3} } }
}}
{{eqn | r = \frac {\dfrac {\paren {1 - p} \paren {6 - 6 p + p^2} } {p^3 } - 3 \paren {\dfrac {1 - p} p } \dfrac {\paren {1 - p} \paren {2 - p} } {p^2 } + 2 \paren {\dfrac {\paren {1 - p} \paren {1 - p}^2 } {p^3} } } {\paren {\dfrac {\paren {1 - p} \sqrt {1 - p} } {p^3} } }
}}
{{eqn | r = \frac {\paren {6 - 6 p + p^2} - 3 \paren {1 - p} \paren {2 - p} + 2 \paren {1 - p}^2} { \sqrt {1 - p } }
| c = $\dfrac {1 - p} {p^3}$ cancels
}}
{{eqn | r = \frac {6 - 6 p + p^2 - 6 + 9 p - 3 p^2 + 2 - 4 p + 2 p^2} { \sqrt {1 - p } }
| c =
}}
{{eqn | r = \dfrac {2 - p} {\sqrt {1 - p} }
}}
{{end-eqn}}
{{qed}}
Category:Skewness
Category:Skewness of Geometric Distribution
\end{proof}
|
21489
|
\section{Skewness of Hat-Check Distribution}
Tags: Hat-Check Distribution, Skewness
\begin{theorem}
Let $X$ be a discrete random variable with a Hat-Check distribution with parameter $n$. ($n \gt 2$)
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = -1$
\end{theorem}
\begin{proof}
From Skewness in terms of Non-Central Moments:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where $\mu$ is the mean of $X$, and $\sigma$ the standard deviation.
We have, by Expectation of Hat-Check Distribution:
:$\expect X = n - 1$
By Variance of Hat-Check Distribution:
:$\var X = \sigma^2 = 1$
so:
:$\sigma = \sqrt 1 = 1$
To now calculate $\gamma_1$, we must calculate $\expect {X^3}$.
{{begin-eqn}}
{{eqn | l = \expect {X^3}
| r = \sum_{k \mathop = 0}^n {k^3 \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} }
| c = {{Defof|Hat-Check Distribution}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n {k^3 \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} }
| c = as the $k = 0$ term vanishes
}}
{{eqn | r = \sum_{y \mathop = n - 1}^0 \paren {n - y }^3 \dfrac 1 {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!}
| c = Let $y = n - k$
}}
{{eqn | r = n^3 \sum_{y \mathop = 0}^{n - 1} \dfrac 1 {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{y \mathop = 0}^{n - 1} \dfrac y {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!} + 3n \sum_{y \mathop = 0}^{n - 1} \dfrac {y^2} {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!} - \sum_{y \mathop = 0}^{n - 1} \dfrac {y^3} {y!} \sum_{s \mathop = 0}^{n - y} \dfrac {\paren {-1}^s} {s!}
| c =
}}
{{eqn | r = n^3 \sum_{k \mathop = 1}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{k \mathop = 1}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 3n \sum_{k \mathop = 1}^n \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 1}^n \dfrac {\paren {n - k}^3 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
| c = Let $y = n - k$
}}
{{eqn | r = n^3 \sum_{k \mathop = 1}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{k \mathop = 1}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 3n \sum_{k \mathop = 1}^n \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 1}^n \dfrac {\paren {n - k}^3 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 4 \dfrac {n^3} {n!} - 4 \dfrac {n^3} {n!}
| c = adding $0$
}}
{{eqn | r = n^3 \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{k \mathop = 0}^n \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 3n \sum_{k \mathop = 0}^n \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 0}^n \dfrac {\paren {n - k}^3 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
| c =
}}
{{eqn | r = n^3 \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{k \mathop = 0}^{n - 1} \dfrac {n - k} {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 3n \sum_{k \mathop = 0}^{n - 1} \dfrac {\paren {n - k}^2 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 0}^{n - 1} \dfrac {\paren {n - k}^3 } {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
| c = as the $k = n$ term vanishes
}}
{{eqn | r = n^3 \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - 3n^2 \sum_{k \mathop = 0}^{n - 1} \dfrac 1 {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} + 3n \sum_{k \mathop = 0}^{n - 1} \dfrac {\paren {n - k} } {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!} - \sum_{k \mathop = 0}^{n - 1} \dfrac {\paren {n - k}^2 } {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
| c = canceling terms
}}
{{end-eqn}}
To help complete the sum above, recall that:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
| r = 1
| c = Hat-Check Distribution Gives Rise to Probability Mass Function
}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \dfrac 1 {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
| r = 1
| c = Hat-Check Distribution Gives Rise to Probability Mass Function
}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \dfrac k {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
| r = n - 2
| c = Expectation of Hat-Check Distribution
}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \dfrac {k^2} {\paren {n - 1 - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
| r = \paren{n - 2}^2 + 1
| c = Variance of Hat-Check Distribution
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | r = n^3 - 3n^2 + 3n^2 - 3n \paren {n - 2} - n^2 + 2n \paren {n - 2} - \paren {n^2 - 4n + 5}
| c =
}}
{{eqn | r = n^3 - 3n^2 + 3n^2 - 3n^2 - n^2 + 2n^2 - n^2 + 6n - 4n + 4n - 5
| c =
}}
{{eqn | r = n^3 - 3n^2 + 6n - 5
| c =
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {\paren {n^3 - 3n^2 + 6n - 5 } - 3 \paren {n - 1} 1 - \paren {n - 1}^3} {1^{3/2} }
}}
{{eqn | r = \paren {n^3 - 3n^2 + 6n - 5 } - 3n + 3 - \paren {n^3 - 3n^2 + 3n - 1}
}}
{{eqn | r = -1
}}
{{end-eqn}}
{{qed}}
Category:Skewness
Category:Hat-Check Distribution
\end{proof}
|
21490
|
\section{Skewness of Poisson Distribution}
Tags: Skewness, Poisson Distribution
\begin{theorem}
Let $X$ be a discrete random variable with a Poisson distribution with parameter $\lambda$.
Then the skewness $\gamma_1$ of $X$ is given by:
:$\gamma_1 = \dfrac 1 {\sqrt \lambda}$
\end{theorem}
\begin{proof}
From Skewness in terms of Non-Central Moments:
:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
where $\mu$ is the mean of $X$, and $\sigma$ the standard deviation.
We have, by Expectation of Poisson Distribution:
:$\expect X = \lambda$
By Variance of Poisson Distribution:
:$\var X = \sigma^2 = \lambda$
so:
:$\sigma = \sqrt \lambda$
To now calculate $\gamma_1$, we must calculate $\expect {X^3}$.
We find this using the moment generating function of $X$, $M_X$.
By Moment Generating Function of Poisson Distribution, this is given by:
:$\ds \map {M_X} t = e^{\lambda \paren {e^t - 1} }$
From Moment in terms of Moment Generating Function:
:$\expect {X^3} = \map {M_X'''} 0$
In Variance of Poisson Distribution: Proof 3, it is shown that:
:$\ds \map {M_X''} t = \lambda \paren {\lambda e^t + 1} e^{\lambda \paren {e^t - 1} + t}$
So:
{{begin-eqn}}
{{eqn | l = \map {M_X'''} t
| r = \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1} \frac \d {\d t} \paren {\lambda \paren {e^t - 1} + t} \frac \d {\d \paren {\lambda \paren {e^t - 1} + t} } \paren {e^{\lambda \paren {e^t - 1} + t} }
| c = Product Rule for Derivatives, Exponential of Sum, Chain Rule for Derivatives
}}
{{eqn | r = \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1}^2 e^{\lambda \paren {e^t - 1} + t}
| c = Derivative of Exponential Function, Derivative of Power
}}
{{end-eqn}}
Setting $t = 0$:
{{begin-eqn}}
{{eqn | l = \expect {X^3}
| r = \lambda^2 e^{\lambda \paren {e^0 - 1} + 0} + \lambda \paren {\lambda e^0 + 1}^2 e^{\lambda \paren {e^0 - 1} + 0}
}}
{{eqn | r = \lambda^2 + \lambda \paren {\lambda + 1}^2
| c = Exponential of Zero
}}
{{eqn | r = \lambda^3 + 3 \lambda^2 + \lambda
| c = Square of Sum
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \gamma_1
| r = \frac {\lambda^3 + 3 \lambda^2 + \lambda - 3 \lambda^2 - \lambda^3} {\lambda^{3/2} }
}}
{{eqn | r = \frac \lambda {\lambda^{3/2} }
}}
{{eqn | r = \frac 1 {\sqrt \lambda}
}}
{{end-eqn}}
{{qed}}
Category:Skewness
Category:Poisson Distribution
\end{proof}
|
21491
|
\section{Skolem's Paradox}
Tags: Set Theory, Mathematical Logic
\begin{theorem}
Let $\LL$ be a countable first-order language.
Let $T$ be an $\LL$-theory which axiomatizes some version of set theory (for example, ZFC).
There is a countable model of $T$.
\end{theorem}
\begin{proof}
This is a straightforward application of the downward Löwenheim-Skolem Theorem.
{{qed}}
{{refactor|From here on down, decide where it belongs and how it is to be presented. At the moment it is too much like an encyclopedia article to be compatible with {{ProofWiki}}'s dictionary style.}}
\end{proof}
|
21492
|
\section{Slice Category is Category}
Tags: Slice Categories
\begin{theorem}
Let $\mathbf C$ be a metacategory.
Let $C \in \mathbf C_0$ be a object of $\mathbf C$.
Let $\mathbf C / C$ be the slice category of $\mathbf C$ over $C$.
Then $\mathbf C / C$ is a metacategory.
\end{theorem}
\begin{proof}
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
Suppose that $a: f \to g$ and $b: g \to h$ are morphisms in $\mathbf C / C$.
To show that $b \circ a: f \to h$ is a morphism as well, compute:
{{begin-eqn}}
{{eqn|l = \left({b \circ a}\right) \circ f
|r = b \circ \left({a \circ f}\right)
|c = composition in $\mathbf C$ is associative
}}
{{eqn|r = b \circ g
|c = $a$ is a morphism
}}
{{eqn|r = h
|c = $b$ is a morphism
}}
{{end-eqn}}
Hence $b \circ a: f \to h$ is a morphism.
For $(C2)$, observe that for $a: f \to g$, with objects $f: X \to C$ and $g: Y \to C$, we have:
{{begin-eqn}}
{{eqn|l = a \circ \operatorname{id}_f
|r = a \circ \operatorname{id}_X
|c = Definition of $\operatorname{id}_f$
}}
{{eqn|r = a
|c = Since $\mathbf C$ is a metacategory
}}
{{eqn|r = \operatorname{id}_Y \circ a
}}
{{eqn|r = \operatorname{id}_g \circ a
}}
{{end-eqn}}
Hence $(C2)$ is shown to hold.
Since the composition in $\mathbf C / C$ is inherited from $\mathbf C$, satisfaction of the associative property $(C3)$ is also inherited.
Hence $\mathbf C / C$ is a metacategory.
{{qed}}
Category:Slice Categories
\end{proof}
|
21493
|
\section{Slice Category of Order Category}
Tags: Slice Categories, Poset Categories, Order Categories, Preorder Categories
\begin{theorem}
Let $\mathbf P$ be a order category, and denote its ordering by $\preceq$.
Let $p \in \mathbf P_0$ be an object of $\mathbf P$.
Then:
:$\mathbf P \mathbin / p \cong p^\preceq$
where:
:$\mathbf P \mathbin / p$ is the slice of $\mathbf P$ over $p$
:$p^\preceq$ is the order category defined by the weak lower closure of $p$.
\end{theorem}
\begin{proof}
The objects of $\mathbf P \mathbin / p$ are morphisms $q \to p$ of $\mathbf P$.
The morphisms are $q \to r$ fitting into a commutative diagram:
::$\begin{xy}\xymatrix@C=1em{
q
\ar[rr]
\ar[rd]
& &
r
\ar[ld]
\\ &
p
}\end{xy}$
Define a functor $U: \mathbf P \mathbin / p \to \mathbf P$ by:
:$U \left({q \to p}\right) := q$
:$U \left({q \to r}\right) := q \to r$
Because there is at most one morphism $q \to p$ for each $q$, $U$ is injective on objects.
That $U$ is faithful is trivial.
Hence by Functor is Embedding iff Faithful and Injective on Objects, $U$ is an embedding.
That is, $\mathbf P \mathbin / p$ is isomorphic to the image of $U$.
Now the objects of the image of $U$ in $\mathbf P$ are those $q$ such that $q \preceq p$.
That is, the image of $U$ has as objects $p^\preceq$.
By Commutative Diagram in Preorder Category, every morphism $q \to r$ in $p^\preceq$ fits into a commutative triangle as above.
Thus all morphisms of $p^\preceq$ are in the image of $U$, so that $p^\preceq$ is precisely the image of $U$.
Hence the result.
{{qed}}
\end{proof}
|
21494
|
\section{Slope of Curve at Point equals Derivative}
Tags: Differential Calculus, Analytic Geometry
\begin{theorem}
Let $\CC$ be a curve embedded in the Cartesian plane described using the equation:
:$y = \map f x$
where $f$ is a real function.
Let there exist a unique tangent $\TT$ to $\CC$ at a point $P = \tuple {x_0, y_0}$ on $\CC$.
Then the slope of $\CC$ at $P$ is equal to the derivative of $f$ at $P$.
\end{theorem}
\begin{proof}
We have been given that there exists a unique tangent $\TT$ to $\CC$ at $P$.
By definition of tangent, $\TT$ has a slope $M$ given by:
:$m = \ds \lim_{h \mathop \to 0} \frac {\map f {x_0 + h} - \map f {x_0} } h$
This is the definition of the derivative of $f$ at $P$.
{{qed}}
\end{proof}
|
21495
|
\section{Slope of Normal is Minus Reciprocal of Tangent}
Tags: Analytic Geometry
\begin{theorem}
Let $C$ be a curve defined by a real function which is differentiable.
Let $P$ be a point on $C$.
Let the curvature of $C$ at $P$ be non-zero.
Let $r$ be the slope of the tangent to $C$ at $P$.
Let $s$ be the slope of the normal to $C$ at $P$.
Then:
:$r = -\dfrac 1 s$
\end{theorem}
\begin{proof}
By definition, the normal to $C$ at $P$ is defined as being perpendicular to the tangent at $P$ and in the same plane as $P$.
The result follows from Condition for Straight Lines in Plane to be Perpendicular.
{{qed}}
Category:Analytic Geometry
\end{proof}
|
21496
|
\section{Slope of Orthogonal Curves}
Tags: Analytic Geometry
\begin{theorem}
Let $C_1$ and $C_2$ be curves in a cartesian plane.
Let $C_1$ and $C_2$ intersect each other at $P$.
Let the slope of $C_1$ and $C_2$ at $P$ be $m_1$ and $m_2$.
Then $C_1$ and $C_2$ are orthogonal {{iff}}:
:$m_1 = -\dfrac 1 {m_2}$
\end{theorem}
\begin{proof}
Let the slopes of $C_1$ and $C_2$ at $P$ be defined by the vectors $\mathbf v_1$ and $\mathbf v_2$ represented as column matrices:
:$\mathbf v_1 = \begin{bmatrix} x_1 \\ y_1 \end{bmatrix} , \mathbf v_2 = \begin{bmatrix} x_2 \\ y_2 \end{bmatrix}$
By Non-Zero Vectors Orthogonal iff Perpendicular:
:$\mathbf v_1 \cdot \mathbf v_2 = 0$ {{iff}} $C_1$ is orthogonal to $C_2$
where $\mathbf v_1 \cdot \mathbf v_2$ denotes the dot product of $C_1$ and $C_2$.
Thus:
{{begin-eqn}}
{{eqn | l = \mathbf v_1 \cdot \mathbf v_2
| r = 0
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x_1 x_2 + y_1 y_2
| r = 0
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \frac {y_1} {x_1} + \frac {x_2} {y_2}
| r = 0
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \frac {x_1} {y_1}
| r = -\frac 1 {\paren {\dfrac {y_2} {x_2} } }
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = m_1
| r = -\frac 1 {m_2}
| c =
}}
{{end-eqn}}
{{qed}}
Category:Analytic Geometry
\end{proof}
|
21497
|
\section{Slope of Secant}
Tags: Analytic Geometry
\begin{theorem}
Let $f: \R \to \R$ be a real function.
Let the graph of $f$ be depicted on a Cartesian plane.
:400px
Let $AB$ be a secant of $f$ where:
:$A = \tuple {x, \map f x}$
:$A = \tuple {x + h, \map f {x + h} }$
Then the slope of $AB$ is given by:
:$\dfrac {\map f {x + h} - \map f x} h$
\end{theorem}
\begin{proof}
The slope of $AB$ is defined as the change in $y$ divided by the change in $x$.
Between $A$ and $B$:
:the change in $x$ is $\paren {x + h} - x = h$
:the change in $y$ is $\map f {x + h} - \map f x$.
Hence the result.
{{qed}}
\end{proof}
|
21498
|
\section{Slope of Straight Line joining Points in Cartesian Plane}
Tags: Straight Lines, Slope
\begin{theorem}
Let $p_1 := \tuple {x_1, y_1}$ and $p_2 := \tuple {x_2, y_2}$ be points in a cartesian plane.
Let $\LL$ be the straight line passing through $p_1$ and $p_2$.
Then the slope of $\LL$ is given by:
:$\tan \theta = \dfrac {y_2 - y_1} {x_2 - x_1}$
where $\theta$ is the angle made by $\LL$ with the $x$-axis.
\end{theorem}
\begin{proof}
:500px
The slope of a straight line is defined as the change in $y$ divided by the change in $x$.
The change in $y$ from $p_1$ to $p_2$ is $y_2 - y_1$.
The change in $x$ from $p_1$ to $p_2$ is $x_2 - x_1$.
By definition of tangent of $\theta$:
:$\tan \theta = \dfrac {y_2 - y_1} {x_2 - x_1}$
Hence the result.
{{qed}}
\end{proof}
|
21499
|
\section{Smaller Elements of Pythagorean Triple not both Odd}
Tags: Pythagorean Triples
\begin{theorem}
Let $\left({x, y, z}\right)$ be a Pythagorean triple, i.e. integers such that $x^2 + y^2 = z^2$.
Then $x$ and $y$ cannot both be odd.
\end{theorem}
\begin{proof}
{{AimForCont}} $x$ and $y$ are both odd such that:
:$\exists z \in \Z: x^2 + y^2 = z^2$
Then:
:$x^2 + y^2 \equiv 1 + 1 \equiv 2 \pmod 4$
But from Square Modulo 4:
:$z^2 \equiv 0 \pmod 4$ or $z^2 \equiv 1 \pmod 4$
Thus $x^2 + y^2$ can not be square.
It follows by Proof by Contradiction that $x$ and $y$ cannot both be odd.
{{qed}}
Category:Pythagorean Triples
\end{proof}
|
21500
|
\section{Smaller Number of Wythoff Pair is Smallest Number not yet in Sequence}
Tags: Beatty Sequences
\begin{theorem}
Consider the sequence of Wythoff pairs arranged in sequential order:
:$\tuple {0, 0}, \tuple {1, 2}, \tuple {3, 5}, \tuple {4, 7}, \tuple {6, 10}, \tuple {8, 13}, \ldots$
The first coordinate of each Wythoff pair is the smallest positive integer which has not yet appeared in the sequence.
\end{theorem}
\begin{proof}
From Sequence of Wythoff Pairs contains all Positive Integers exactly Once Each, every positive integer can be found in the sequence of Wythoff pairs.
From Difference between Terms of Wythoff Pair, the first coordinate is the smaller of the coordinates of the Wythoff pair.
So consider a given Wythoff pair.
Let $p$ be the smallest positive integer which has not yet appeared in the sequence.
It has to appear somewhere.
The terms of both the lower and upper Wythoff sequences are in ascending order.
$n$ must appear in the next Wythoff pair, otherwise a larger positive integer will appear before it when it eventually does appear.
$n$ cannot appear as the second coordinate, or it will be less than whatever number appears as the first coordinate.
Hence $n$ appears as the first coordinate of the next Wythoff pair.
{{qed}}
\end{proof}
|
21501
|
\section{Smaller of Thabit Pair is Tetrahedral}
Tags: Amicable Pairs, Amicable Numbers, Tetrahedral Numbers
\begin{theorem}
Let $\tuple {m_1, m_2}$ be a Thabit pair such that $m_1 < m_2$.
Then $m_1$ is a tetrahedral number.
\end{theorem}
\begin{proof}
By Thabit's Rule:
{{begin-eqn}}
{{eqn | l = m_1
| r = 2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}
| c =
}}
{{eqn | l = m_2
| r = 2^n \times \paren {9 \times 2^{2 n - 1} - 1}
| c =
}}
{{end-eqn}}
for some $n \in \Z_{\ge 0}$.
We have that:
{{begin-eqn}}
{{eqn | l = \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}
| r = 9 \times 2^{2 n - 1} - 3 \times 2^n - 3 \times 2^{n - 1} + 1
| c =
}}
{{eqn | o = <
| r = 9 \times 2^{2 n - 1} - 1
| c =
}}
{{end-eqn}}
and so $m_1 < m_2$.
Then:
{{begin-eqn}}
{{eqn | l = m_1
| r = 2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1}
| c =
}}
{{eqn | r = \frac {2^n \times \paren {3 \times 2^{n - 1} - 1} \times \paren {3 \times 2^n - 1} \times 6} 6
| c =
}}
{{eqn | r = \frac {\paren {3 \times 2^n} \times \paren {2 \times 3 \times 2^{n - 1} - 2} \times \paren {3 \times 2^n - 1} } 6
| c =
}}
{{eqn | r = \frac {\paren {3 \times 2^n} \times \paren {3 \times 2^n - 1} \times \paren {3 \times 2^n - 2} } 6
| c =
}}
{{eqn | r = \frac {k \times \paren {k + 1} \times \paren {k + 2} } 6
| c = where $k = 3 \times 2^n - 2$
}}
{{end-eqn}}
From Closed Form for Tetrahedral Numbers, this is the $k$th tetrahedral number where $k = 3 \times 2^n - 2$.
{{qed}}
\end{proof}
|
21502
|
\section{Smallest 10 Primes in Arithmetic Sequence}
Tags: Prime Numbers, Arithmetic Progressions, Arithmetic Sequences
\begin{theorem}
The smallest $10$ primes in arithmetic sequence are:
:$199 + 210 n$
for $n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.
These are also the smallest $8$ and $9$ primes in arithmetic sequence.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 199 + 0 \times 210
| r = 199
| c = which is the $46$th prime
}}
{{eqn | l = 199 + 1 \times 210
| r = 409
| c = which is the $80$th prime
}}
{{eqn | l = 199 + 2 \times 210
| r = 619
| c = which is the $114$th prime
}}
{{eqn | l = 199 + 3 \times 210
| r = 829
| c = which is the $145$th prime
}}
{{eqn | l = 199 + 4 \times 210
| r = 1039
| c = which is the $175$th prime
}}
{{eqn | l = 199 + 5 \times 210
| r = 1249
| c = which is the $204$th prime
}}
{{eqn | l = 199 + 6 \times 210
| r = 1459
| c = which is the $232$nd prime
}}
{{eqn | l = 199 + 7 \times 210
| r = 1669
| c = which is the $263$rd prime
}}
{{eqn | l = 199 + 8 \times 210
| r = 1879
| c = which is the $289$th prime
}}
{{eqn | l = 199 + 9 \times 210
| r = 2089
| c = which is the $316$th prime
}}
{{end-eqn}}
{{OEIS|A033168}}
But note that $199 + 10 \times 210 = 2299 = 11^2 \times 19$ and so is not prime.
Now we show that this is the smallest $10$ primes in arithmetic sequence.
By Divisibility of Common Difference of Arithmetic Sequence of Primes, the common difference $d$ of any $10$ primes in arithmetic sequence must be divisible by all primes less that $10$.
That is, the common difference is a multiple of $2 \times 3 \times 5 \times 7 = 210$.
Suppose $d = 210$.
Then the first term $p_1$ of the arithmetic sequence cannot be $11$, as $11 + 210 = 221 = 13 \times 17$.
Note that $210 \equiv 1 \pmod {11}$.
Hence if the $p_1$ gives a remainder $r > 1$ upon divison by $11$, the $\paren {12 - r}^{th}$ term of the sequence is equivalent to:
:$r + \paren {12 - r - 1} \times 210 \equiv 0 \pmod {11}$
and is divisible by $11$, which is impossible as all terms are primes.
Therefore $p_1 \equiv 1 \pmod {11}$.
The first few primes of this form are:
:$23, 67, 89, 199, \dots$
and we eliminate the first three candidates because:
:$23 + 5 \times 210 = 1073 = 29 \times 37$
:$67 + 3 \times 210 = 697 = 17 \times 41$
:$89 + 1 \times 210 = 229 = 13 \times 23$
which are not primes, and we have checked that $199 + 210 n$ are indeed $10$ primes in arithmetic sequence.
Now suppose $d > 210$.
Then $d \ge 420$ and the last term is greater than $420 \times 9 = 3780$.
This shows that the one we found here is the smallest $10$ primes in arithmetic sequence.
{{qed|lemma}}
\end{proof}
|
21503
|
\section{Smallest 17 Primes in Arithmetic Sequence}
Tags: Prime Numbers, Arithmetic Progressions, Arithmetic Sequences
\begin{theorem}
The smallest $17$ primes in arithmetic sequence are:
:$3\,430\,751\,869 + 87\,297\,210 n$
for $n = 0, 1, \ldots, 16$.
\end{theorem}
\begin{proof}
First we note that:
:$3\,430\,751\,869 - 87\,297\,210 = 3\,343\,454\,659 = 17\,203 \times 194\,353$
and so this arithmetic sequence of primes does not extend to $n < 0$.
{{begin-eqn}}
{{eqn | l = 3\,430\,751\,869 + 0 \times 87\,297\,210
| r = 3\,430\,751\,869
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 1 \times 87\,297\,210
| r = 3\,518\,049\,079
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 2 \times 87\,297\,210
| r = 3\,605\,346\,289
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 3 \times 87\,297\,210
| r = 3\,692\,643\,499
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 4 \times 87\,297\,210
| r = 3\,779\,940\,709
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 5 \times 87\,297\,210
| r = 3\,867\,237\,919
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 6 \times 87\,297\,210
| r = 3\,954\,535\,129
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 7 \times 87\,297\,210
| r = 4\,041\,832\,339
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 8 \times 87\,297\,210
| r = 4\,129\,129\,549
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 9 \times 87\,297\,210
| r = 4\,216\,426\,759
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 10 \times 87\,297\,210
| r = 4\,303\,723\,969
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 11 \times 87\,297\,210
| r = 4\,391\,021\,179
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 12 \times 87\,297\,210
| r = 4\,478\,318\,389
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 13 \times 87\,297\,210
| r = 4\,565\,615\,599
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 14 \times 87\,297\,210
| r = 4\,652\,912\,809
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 15 \times 87\,297\,210
| r = 4\,740\,210\,019
| c = which is prime
}}
{{eqn | l = 3\,430\,751\,869 + 16 \times 87\,297\,210
| r = 4\,827\,507\,229
| c = which is prime
}}
{{end-eqn}}
But note that $3\,430\,751\,869 + 17 \times 87\,297\,210 = 4\,914\,804\,439 = 41 \times 97 \times 1 235807$ and so is not prime.
{{ProofWanted|It remains to be shown that there are no smaller such APs}}
\end{proof}
|
21504
|
\section{Smallest 18 Primes in Arithmetic Sequence}
Tags: Prime Numbers, Arithmetic Progressions, Arithmetic Sequences
\begin{theorem}
The smallest $18$ primes in arithmetic sequence are:
:$107\,928\,278\,317 + 9\,922\,782\,870 n$
for $n = 0, 1, \ldots, 16$.
\end{theorem}
\begin{proof}
First we note that:
:$107\,928\,278\,317 - 9\,922\,782\,870 = 98\,005\,495\,447 = 29 \times 149 \times 22\,681\,207$
and so this arithmetic sequence of primes does not extend to $n < 0$.
{{begin-eqn}}
{{eqn | l = 107\,928\,278\,317 + 0 \times 9\,922\,782\,870
| r = 107\,928\,278\,317
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 1 \times 9\,922\,782\,870
| r = 117\,851\,061\,187
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 2 \times 9\,922\,782\,870
| r = 127\,773\,844\,057
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 3 \times 9\,922\,782\,870
| r = 137\,696\,626\,927
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 4 \times 9\,922\,782\,870
| r = 147\,619\,409\,797
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 5 \times 9\,922\,782\,870
| r = 157\,542\,192\,667
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 6 \times 9\,922\,782\,870
| r = 167\,464\,975\,537
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 7 \times 9\,922\,782\,870
| r = 177\,387\,758\,407
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 8 \times 9\,922\,782\,870
| r = 187\,310\,541\,277
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 9 \times 9\,922\,782\,870
| r = 197\,233\,324\,147
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 10 \times 9\,922\,782\,870
| r = 207\,156\,107\,017
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 11 \times 9\,922\,782\,870
| r = 217\,078\,889\,887
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 12 \times 9\,922\,782\,870
| r = 227\,001\,672\,757
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 13 \times 9\,922\,782\,870
| r = 236\,924\,455\,627
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 14 \times 9\,922\,782\,870
| r = 246\,847\,238\,497
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 15 \times 9\,922\,782\,870
| r = 256\,770\,021\,367
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 16 \times 9\,922\,782\,870
| r = 266\,692\,804\,237
| c = which is prime
}}
{{eqn | l = 107\,928\,278\,317 + 17 \times 9\,922\,782\,870
| r = 276\,615\,587\,107
| c = which is prime
}}
{{end-eqn}}
But note that $107\,928\,278\,317 + 18 \times 9\,922\,782\,870 = 286\,538\,369\,977 = 23 \times 181 \times 68\,829\,779$ and so is not prime.
{{ProofWanted|It remains to be shown that there are no smaller such APs}}
\end{proof}
|
21505
|
\section{Smallest 22 Primes in Arithmetic Sequence}
Tags: Prime Numbers, Arithmetic Progressions, Arithmetic Sequences
\begin{theorem}
The smallest $22$ primes in arithmetic sequence are:
:$11 \, 410 \, 337 \, 850 \, 553 + 4 \, 609 \, 098 \, 694 \, 200 n$
for $n = 0, 1, \ldots, 21$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 0 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 11 \, 410 \, 337 \, 850 \, 553
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 1 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 16 \, 019 \, 436 \, 544 \, 753
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 2 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 20 \, 628 \, 535 \, 238 \, 953
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 3 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 25 \, 237 \, 633 \, 933 \, 153
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 4 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 29 \, 846 \, 732 \, 627 \, 353
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 5 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 34 \, 455 \, 831 \, 321 \, 553
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 6 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 39 \, 064 \, 930 \, 015 \, 753
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 7 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 43 \, 674 \, 028 \, 709 \, 953
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 8 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 48 \, 283 \, 127 \, 404 \, 153
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 9 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 52 \, 892 \, 226 \, 098 \, 353
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 10 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 57 \, 501 \, 324 \, 792 \, 553
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 11 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 62 \, 110 \, 423 \, 486 \, 753
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 12 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 66 \, 719 \, 522 \, 180 \, 953
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 13 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 71 \, 328 \, 620 \, 875 \, 153
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 14 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 75 \, 937 \, 719 \, 569 \, 353
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 15 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 80 \, 546 \, 818 \, 263 \, 553
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 16 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 85 \, 155 \, 916 \, 957 \, 753
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 17 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 89 \, 765 \, 015 \, 651 \, 953
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 18 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 94 \, 374 \, 114 \, 346 \, 153
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 19 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 98 \, 983 \, 213 \, 040 \, 353
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 20 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 103 \, 592 \, 311 \, 734 \, 553
| c = which is prime
}}
{{eqn | l = 11 \, 410 \, 337 \, 850 \, 553 + 21 \times 4 \, 609 \, 098 \, 694 \, 200
| r = 108 \, 201 \, 410 \, 428 \, 753
| c = which is prime
}}
{{end-eqn}}
But note that $11 \, 410 \, 337 \, 850 \, 553 + 22 \times 4 \, 609 \, 098 \, 694 \, 200 = 112 \, 810 \, 509 \, 122 \, 953 = 61 \times 107 \times 1907 \times 9063277$ and so is not prime.
{{ProofWanted|It remains to be shown that there are no smaller such APs}}
\end{proof}
|
21506
|
\section{Smallest 3-Digit Permutable Prime}
Tags: Prime Numbers, 113, Permutable Primes
\begin{theorem}
The smallest $3$-digit permutable prime is $113$.
\end{theorem}
\begin{proof}
:$113$ is prime.
:$131$ is prime.
:$311$ is prime.
Consider the $3$-digit primes smaller than $113$:
:$101, 103, 107, 109$
They all contain a zero.
Thus, for each of these, at least one permutation ends in a zero.
Hence it is divisible by $10$ and so is not prime.
{{qed}}
\end{proof}
|
21507
|
\section{Smallest 5 Consecutive Primes in Arithmetic Sequence}
Tags: Prime Numbers, Arithmetic Progressions, Arithmetic Sequences
\begin{theorem}
The smallest $5$ consecutive primes in arithmetic sequence are:
:$9 \, 843 \, 019 + 30 n$
for $n = 0, 1, 2, 3, 4$.
Note that while there are many longer arithmetic sequences of far smaller primes, those primes are not consecutive.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 9 \, 843 \, 019 + 0 \times 30
| r = 9 \, 843 \, 019
| c = which is the $654 \, 926$th prime
}}
{{eqn | l = 9 \, 843 \, 019 + 1 \times 30
| r = 9 \, 843 \, 049
| c = which is the $654 \, 927$th prime
}}
{{eqn | l = 9 \, 843 \, 019 + 2 \times 30
| r = 9 \, 843 \, 079
| c = which is the $654 \, 928$th prime
}}
{{eqn | l = 9 \, 843 \, 019 + 3 \times 30
| r = 9 \, 843 \, 109
| c = which is the $654 \, 929$th prime
}}
{{eqn | l = 9 \, 843 \, 019 + 4 \times 30
| r = 9 \, 843 \, 139
| c = which is the $654 \, 930$th prime
}}
{{end-eqn}}
But note that $9 \, 843 \, 019 + 5 \times 30 = 9 \, 843 \, 169 = 7^2 \times 200 \, 881$ and so is not prime.
Inspection of tables of primes (or a computer search) will reveal that this is the smallest such sequence.
{{qed}}
\end{proof}
|
21508
|
\section{Smallest Adjacent Happy Numbers}
Tags: Happy Numbers
\begin{theorem}
The smallest adjacent happy numbers are $31$ and $32$.
\end{theorem}
\begin{proof}
This can be determined by testing all the positive integers in succession for happiness.
Checking $31$ and $32$:
{{begin-eqn}}
{{eqn | l = 31
| o = \to
| r = 3^2 + 1^2 = 9 + 1 = 10
| c =
}}
{{eqn | o = \to
| r = 1^2 = 1
| c = and so $31$ is happy.
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 32
| o = \to
| r = 3^2 + 2^2 = 9 + 4 = 13
| c =
}}
{{eqn | o = \to
| r = 1^2 + 3^2 = 1 + 9 = 10
}}
{{eqn | o = \to
| r = 1^2 = 1
| c = and so $32$ is happy.
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21509
|
\section{Smallest Consecutive Even Nontotients}
Tags: Nontotients
\begin{theorem}
The smallest pair of consecutive even nontotients is $74$ and $76$.
\end{theorem}
\begin{proof}
From the sequence of nontotients:
{{:Definition:Nontotient/Sequence}}
Hence, by inspection, it can be seen that $74$ and $76$ are the smallest such pair.
{{qed}}
\end{proof}
|
21510
|
\section{Smallest Consecutive Even Numbers such that Added to Divisor Count are Equal}
Tags: Divisor Counting Function, Tau Function, Even Integers, Definition:Divisor Counting Function
\begin{theorem}
$30$ is the smallest positive even integer $n$ with the property:
{{begin-eqn}}
{{eqn | l = n + \map {\sigma_0} n
| r = m
| c =
}}
{{eqn | l = \paren {n + 2} + \map {\sigma_0} {n + 2}
| r = m
| c =
}}
{{eqn | l = \paren {n + 4} + \map {\sigma_0} {n + 4}
| r = m
| c =
}}
{{end-eqn}}
where:
:$m \in \Z_{>0}$ is some positive integer
:$\map {\sigma_0} n$ is the divisor counting function: the number of divisors of $n$.
In this case, where $n = 30$, we have that $m = 38$.
\end{theorem}
\begin{proof}
From Divisor Counting Function from Prime Decomposition, we have:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
where the prime decomposition of $n$ is:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
{{begin-eqn}}
{{eqn | l = 2 + \map {\sigma_0} 2
| r = 2 + 2
| rr= = 4
| c = as $2 = 2^1$
}}
{{eqn | l = 4 + \map {\sigma_0} 4
| r = 4 + 7
| rr= = 7
| c = as $4 = 2^2$
}}
{{eqn | l = 6 + \map {\sigma_0} 6
| r = 6 + 4
| rr= = 10
| c = as $6 = 2^1 3^1$
}}
{{eqn | l = 8 + \map {\sigma_0} 8
| r = 8 + 4
| rr= = 12
| c = as $8 = 2^3$
}}
{{eqn | l = 10 + \map {\sigma_0} {10}
| r = 10 + 4
| rr= = 14
| c = as $10 = 2^1 5^1$
}}
{{eqn | l = 12 + \map {\sigma_0} {12}
| r = 12 + 6
| rr= = 18
| c = as $12 = 2^2 3^1$
}}
{{eqn | l = 14 + \map {\sigma_0} {14}
| r = 14 + 4
| rr= = 18
| c = as $14 = 2^1 7^1$
}}
{{eqn | l = 16 + \map {\sigma_0} {16}
| r = 16 + 5
| rr= = 21
| c = as $16 = 2^4$
}}
{{eqn | l = 18 + \map {\sigma_0} {18}
| r = 18 + 6
| rr= = 24
| c = as $18 = 2^1 3^2$
}}
{{eqn | l = 20 + \map {\sigma_0} {20}
| r = 20 + 6
| rr= = 26
| c = as $20 = 2^2 5^1$
}}
{{eqn | l = 22 + \map {\sigma_0} {22}
| r = 22 + 4
| rr= = 26
| c = as $22 = 2^1 11^1$
}}
{{eqn | l = 24 + \map {\sigma_0} {24}
| r = 24 + 8
| rr= = 32
| c = as $24 = 2^3 3^1$
}}
{{eqn | l = 26 + \map {\sigma_0} {26}
| r = 26 + 4
| rr= = 30
| c = as $26 = 2^1 13^1$
}}
{{eqn | l = 28 + \map {\sigma_0} {28}
| r = 28 + 6
| rr= = 34
| c = as $28 = 2^2 7^1$
}}
{{eqn | l = 30 + \map {\sigma_0} {30}
| r = 30 + 8
| rr= = 38
| c = as $30 = 2^1 3^1 5^1$
}}
{{eqn | l = 32 + \map {\sigma_0} {32}
| r = 32 + 6
| rr= = 38
| c = as $32 = 2^5$
}}
{{eqn | l = 34 + \map {\sigma_0} {34}
| r = 34 + 4
| rr= = 38
| c = as $34 = 2^1 17^1$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21511
|
\section{Smallest Cube whose Sum of Divisors is Cube}
Tags: Cube Numbers, Sigma Function, Divisor Sum Function
\begin{theorem}
The smallest cube $N$ such that $\map {\sigma_1} N$ is also a cube is:
:$27 \, 418 \, 521 \, 963 \, 671 \, 501 \, 273 \, 905 \, 190 \, 135 \, 082 \, 692 \, 041 \, 730 \, 405 \, 303 \, 870 \, 249 \, 023 \, 209$
where $\map {\sigma_1} N$ denotes the divisor sum of $N$.
\end{theorem}
\begin{proof}
We have that:
{{begin-eqn}}
{{eqn | l = N
| r = 27 \, 418 \, 521 \, 963 \, 671 \, 501 \, 273 \, 905 \, 190 \, 135 \, 082 \, 692 \, 041 \, 730 \, 405 \, 303 \, 870 \, 249 \, 023 \, 209
| c =
}}
{{eqn | r = 3^9 \times 7^3 \times 11^3 \times 13^3 \times 17^3 \times 41^3 \times 43^3 \times 47^3 \times 443^3 \times 499^3 \times 3583^3
| c =
}}
{{eqn | r = 30 \, 154 \, 214 \, 043 \, 975 \, 990 \, 969^3
| c =
}}
{{end-eqn}}
Then from {{DSFLink|27,418, 521,963, 671,501, 273,905, 190,135, 082,692, 041,730, 405,303, 870,249, 023,209|27 \, 418 \, 521 \, 963 \, 671 \, 501 \, 273 \, 905 \, 190 \, 135 \, 082 \, 692 \, 041 \, 730 \, 405 \, 303 \, 870 \, 249 \, 023 \, 209}}:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} N
| r = 65 \, 400 \, 948 \, 817 \, 364 \, 742 \, 403 \, 487 \, 616 \, 930 \, 512 \, 213 \, 536 \, 407 \, 552 \, 000 \, 000 \, 000 \, 000 \, 000
| c =
}}
{{eqn | r = 2^{39} \times 3^6 \times 5^{15} \times 7^3 \times 11^3 \times 13^3 \times 17^3 \times 29^3 \times 37^3 \times 61^3 \times 157^3
| c =
}}
{{eqn | r = 40 \, 289 \, 760 \, 243 \, 532 \, 800 \, 000^3
| c =
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that this is the smallest such number.}}
\end{proof}
|
21512
|
\section{Smallest Cunningham Chain of the First Kind of Length 12}
Tags: Cunningham Chains
\begin{theorem}
The smallest Cunningham chain of the first kind of length $12$ is:
:$554 \, 688 \, 278 \, 429$, $1 \, 109 \, 376 \, 556 \, 859$, $2 \, 218 \, 753 \, 113 \, 719$, $4 \, 437 \, 506 \, 227 \, 439$,
::$8 \, 875 \, 012 \, 454 \, 879$, $17 \, 750 \, 024 \, 909 \, 759$, $35 \, 500 \, 049 \, 819 \, 519$, $71 \, 000 \, 099 \, 639 \, 039$,
::$142 \, 000 \, 199 \, 278 \, 079$, $284 \, 000 \, 398 \, 556 \, 159$, $568 \, 000 \, 797 \, 112 \, 319$, $1 \, 136 \, 001 \, 594 \, 224 \, 639$
\end{theorem}
\begin{proof}
Let $C$ denote the sequence in question.
We have that $554 \, 688 \, 278 \, 429$ is prime.
First note that:
:$\dfrac {554 \, 688 \, 278 \, 429 - 1} 2 = 277 \, 344 \, 139 \, 214 = 2 \times 138 \, 672 \, 069 \, 607$
and so is not prime.
Thus $554 \, 688 \, 278 \, 429$ is not a safe prime, and thus fulfils the requirement for $C$ to be a Cunningham chain of the first kind.
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = 2 \times 554 \, 688 \, 278 \, 429 + 1
| r = 1 \, 109 \, 376 \, 556 \, 859
| c = which is prime
}}
{{eqn | n = 2
| l = 2 \times 1 \, 109 \, 376 \, 556 \, 859 + 1
| r = 2 \, 218 \, 753 \, 113 \, 719
| c = which is prime
}}
{{eqn | n = 3
| l = 2 \times 2 \, 218 \, 753 \, 113 \, 719 + 1
| r = 4 \, 437 \, 506 \, 227 \, 439
| c = which is prime
}}
{{eqn | n = 4
| l = 2 \times 4 \, 437 \, 506 \, 227 \, 439 + 1
| r = 8 \, 875 \, 012 \, 454 \, 879
| c = which is prime
}}
{{eqn | n = 5
| l = 2 \times 8 \, 875 \, 012 \, 454 \, 879 + 1
| r = 17 \, 750 \, 024 \, 909 \, 759
| c = which is prime
}}
{{eqn | n = 6
| l = 2 \times 17 \, 750 \, 024 \, 909 \, 759 + 1
| r = 35 \, 500 \, 049 \, 819 \, 519
| c = which is prime
}}
{{eqn | n = 7
| l = 2 \times 35 \, 500 \, 049 \, 819 \, 519 + 1
| r = 71 \, 000 \, 099 \, 639 \, 039
| c = which is prime
}}
{{eqn | n = 8
| l = 2 \times 71 \, 000 \, 099 \, 639 \, 039 + 1
| r = 142 \, 000 \, 199 \, 278 \, 079
| c = which is prime
}}
{{eqn | n = 9
| l = 2 \times 142 \, 000 \, 199 \, 278 \, 079 + 1
| r = 284 \, 000 \, 398 \, 556 \, 159
| c = which is prime
}}
{{eqn | n = 10
| l = 2 \times 284 \, 000 \, 398 \, 556 \, 159 + 1
| r = 568 \, 000 \, 797 \, 112 \, 319
| c = which is prime
}}
{{eqn | n = 11
| l = 2 \times 568 \, 000 \, 797 \, 112 \, 319 + 1
| r = 1 \, 136 \, 001 \, 594 \, 224 \, 639
| c = which is prime
}}
{{eqn | n = 12
| l = 2 \times 1 \, 136 \, 001 \, 594 \, 224 \, 639 + 1
| r = 2 \, 272 \, 003 \, 188 \, 449 \, 279
| c = which is $19 \times 119 \, 579 \, 115 \, 181 \, 541$ prime
}}
{{end-eqn}}
Establishing that this is indeed the smallest such Cunningham chain of the first kind of length $12$ can be done by a computer search.
{{qed}}
\end{proof}
|
21513
|
\section{Smallest Cunningham Chain of the First Kind of Length 6}
Tags: Cunningham Chains, Sophie Germain Primes
\begin{theorem}
The smallest Cunningham chain of the first kind of length $6$ is:
:$\tuple {89, 179, 359, 719, 1439, 2879}$
\end{theorem}
\begin{proof}
By definition, a Cunningham chain of the first kind is a sequence of prime numbers $\tuple {p_1, p_2, \ldots, p_n}$ such that:
: $p_{k + 1} = 2 p_k + 1$
: $\dfrac {p_1 - 1} 2$ is not prime
: $2 p_n + 1$ is not prime.
Thus each term except the last is a Sophie Germain prime.
{{:Definition:Sophie Germain Prime/Sequence}}
Let $P: \Z \to \Z$ be the mapping defined as:
:$\map P n = 2 n + 1$
Applying $P$ iteratively to each of the smallest Sophie Germain primes in turn:
{{begin-eqn}}
{{eqn | l = \map P 2
| r = 5
| c =
}}
{{eqn | l = \map P 5
| r = 11
| c =
}}
{{eqn | l = \map P {11}
| r = 23
| c =
}}
{{eqn | l = \map P {23}
| r = 47
| c =
}}
{{eqn | l = \map P {47}
| r = 95
| c = which is not prime
}}
{{end-eqn}}
Thus $\tuple {2, 5, 11, 23, 47}$ is a Cunningham chain of the first kind of length $5$.
{{begin-eqn}}
{{eqn | l = \map P 3
| r = 7
| c =
}}
{{eqn | l = \map P 7
| r = 15
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map P {29}
| r = 59
| c =
}}
{{eqn | l = \map P {59}
| r = 119
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map P {41}
| r = 83
| c =
}}
{{eqn | l = \map P {83}
| r = 167
| c =
}}
{{eqn | l = \map P {167}
| r = 335
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map P {53}
| r = 107
| c =
}}
{{eqn | l = \map P {107}
| r = 215
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map P {83}
| r = 167
| c =
}}
{{eqn | l = \map P {167}
| r = 335
| c = which is not prime
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map P {89}
| r = 179
| c =
}}
{{eqn | l = \map P {179}
| r = 359
| c =
}}
{{eqn | l = \map P {359}
| r = 719
| c =
}}
{{eqn | l = \map P {719}
| r = 1439
| c =
}}
{{eqn | l = \map P {1439}
| r = 2879
| c =
}}
{{eqn | l = \map P {2879}
| r = 5759
| c = which is not prime.
}}
{{end-eqn}}
It is noted that $\dfrac {89 - 1} 2 = 44$ which is not prime.
Hence the sequence of $6$:
:$\tuple {89, 179, 359, 719, 1439, 2879}$
{{qed}}
\end{proof}
|
21514
|
\section{Smallest Cunningham Chain of the First Kind of Length 7}
Tags: Cunningham Chains
\begin{theorem}
The smallest Cunningham chain of the first kind of length $7$ is:
:$\left({1 \, 122 \, 659, 2 \, 245 \, 319, 4 \, 490 \, 639, 8 \, 981 \, 279, 17 \, 962 \, 559, 35 \, 925 \, 119, 71 \, 850 \, 239}\right)$
\end{theorem}
\begin{proof}
Let $C$ denote the sequence in question.
We have that:
:$\dfrac {1 \, 122 \, 659 - 1} 2 = 561 \, 329 = 83 \times 6763$
and so is not prime.
Thus $1 \, 122 \, 659$ is not a safe prime, as is required for $C$ to be a Cunningham chain of the first kind.
Then:
{{begin-eqn}}
{{eqn | l = 2 \times 561 \, 329 + 1
| r = 1 \, 122 \, 659
| c = which is the $87 \, 359$th prime
}}
{{eqn | l = 2 \times 1 \, 122 \, 659 + 1
| r = 2 \, 245 \, 319
| c = which is the $165 \, 760$th prime
}}
{{eqn | l = 2 \times 2 \, 245 \, 319 + 1
| r = 4 \, 490 \, 639
| c = which is the $315 \, 347$th prime
}}
{{eqn | l = 2 \times 4 \, 490 \, 639 + 1
| r = 8 \, 981 \, 279
| c = which is the $601 \, 286$th prime
}}
{{eqn | l = 2 \times 8 \, 981 \, 279 + 1
| r = 17 \, 962 \, 559
| c = which is the $1 \, 149 \, 096$th prime
}}
{{eqn | l = 2 \times 17 \, 962 \, 559 + 1
| r = 35 \, 925 \, 119
| c = which is the $2 \, 199 \, 933$rd prime
}}
{{eqn | l = 2 \times 35 \, 925 \, 119 + 1
| r = 71 \, 850 \, 239
| c = which is the $4 \, 220 \, 407$th prime
}}
{{eqn | l = 2 \times 71 \, 850 \, 239 + 1
| r = 143 \, 700 \, 479
| c =
}}
{{eqn | r = 13 \times 47 \times 479 \times 491
| c = and so is not prime
}}
{{end-eqn}}
Establishing that this is indeed the smallest such Cunningham chain of the first kind of length $7$ can be done by a computer search.
{{qed}}
\end{proof}
|
21515
|
\section{Smallest Cunningham Chain of the Second Kind of Length 13}
Tags: Cunningham Chains
\begin{theorem}
The smallest Cunningham chain of the second kind of length $13$ is:
:$758 \, 083 \, 947 \, 856 \, 951$, $1 \, 516 \, 167 \, 895 \, 713 \, 901$, $3 \, 032 \, 335 \, 791 \, 427 \, 801$, $6 \, 064 \, 671 \, 582 \, 855 \, 601$, $12 \, 129 \, 343 \, 165 \, 711 \, 201$, $24 \, 258 \, 686 \, 331 \, 422 \, 401$, $48 \, 517 \, 372 \, 662 \, 844 \, 801$, $97 \, 034 \, 745 \, 325 \, 689 \, 601$, $194 \, 069 \, 490 \, 651 \, 379 \, 201$, $388 \, 138 \, 981 \, 302 \, 758 \, 401$, $776 \, 277 \, 962 \, 605 \, 516 \, 801$, $1 \, 552 \, 555 \, 925 \, 211 \, 033 \, 601$, $3 \, 105 \, 111 \, 850 \, 422 \, 067 \, 201$
\end{theorem}
\begin{proof}
Let $C$ denote the sequence in question.
We have that $758 \, 083 \, 947 \, 856 \, 951$ is prime.
First note that:
:$\dfrac {758 \, 083 \, 947 \, 856 \, 951 + 1} 2 = 379 \, 041 \, 973 \, 928 \, 476 = 2^2 \times 94 \, 760 \, 493 \, 482 \, 119$
and so is not prime.
Thus $758 \, 083 \, 947 \, 856 \, 951$ fulfils the requirement for $C$ to be a Cunningham chain of the second kind.
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = 2 \times 758 \, 083 \, 947 \, 856 \, 951 - 1
| r = 1 \, 516 \, 167 \, 895 \, 713 \, 901
| c = which is prime
}}
{{eqn | n = 2
| l = 2 \times 1 \, 516 \, 167 \, 895 \, 713 \, 901 - 1
| r = 3 \, 032 \, 335 \, 791 \, 427 \, 801
| c = which is prime
}}
{{eqn | n = 3
| l = 2 \times 3 \, 032 \, 335 \, 791 \, 427 \, 801 - 1
| r = 6 \, 064 \, 671 \, 582 \, 855 \, 601
| c = which is prime
}}
{{eqn | n = 4
| l = 2 \times 6 \, 064 \, 671 \, 582 \, 855 \, 601 - 1
| r = 12 \, 129 \, 343 \, 165 \, 711 \, 201
| c = which is prime
}}
{{eqn | n = 5
| l = 2 \times 12 \, 129 \, 343 \, 165 \, 711 \, 201 - 1
| r = 24 \, 258 \, 686 \, 331 \, 422 \, 401
| c = which is prime
}}
{{eqn | n = 6
| l = 2 \times 24 \, 258 \, 686 \, 331 \, 422 \, 401 - 1
| r = 48 \, 517 \, 372 \, 662 \, 844 \, 801
| c = which is prime
}}
{{eqn | n = 7
| l = 2 \times 48 \, 517 \, 372 \, 662 \, 844 \, 801 - 1
| r = 97 \, 034 \, 745 \, 325 \, 689 \, 601
| c = which is prime
}}
{{eqn | n = 8
| l = 2 \times 97 \, 034 \, 745 \, 325 \, 689 \, 601 - 1
| r = 194 \, 069 \, 490 \, 651 \, 379 \, 201
| c = which is prime
}}
{{eqn | n = 9
| l = 2 \times 194 \, 069 \, 490 \, 651 \, 379 \, 201 - 1
| r = 388 \, 138 \, 981 \, 302 \, 758 \, 401
| c = which is prime
}}
{{eqn | n = 10
| l = 2 \times 388 \, 138 \, 981 \, 302 \, 758 \, 401 - 1
| r = 776 \, 277 \, 962 \, 605 \, 516 \, 801
| c = which is prime
}}
{{eqn | n = 11
| l = 2 \times 776 \, 277 \, 962 \, 605 \, 516 \, 801 - 1
| r = 1 \, 552 \, 555 \, 925 \, 211 \, 033 \, 601
| c = which is prime
}}
{{eqn | n = 12
| l = 2 \times 1 \, 552 \, 555 \, 925 \, 211 \, 033 \, 601 - 1
| r = 3 \, 105 \, 111 \, 850 \, 422 \, 067 \, 201
| c = which is prime
}}
{{eqn | n = 13
| l = 2 \times 3 \, 105 \, 111 \, 850 \, 422 \, 067 \, 201 - 1
| r = 6 \, 210 \, 223 \, 700 \, 844 \, 134 \, 401
| c = which is $41 \times 691 \times 739 \times 4957 \times 59 \, 838 \, 677$ and so not prime
}}
{{end-eqn}}
Establishing that this is indeed the smallest such Cunningham chain of the second kind of length $13$ can be done by a computer search.
{{qed}}
\end{proof}
|
21516
|
\section{Smallest Element WRT Restricted Ordering}
Tags: Smallest Elements, Order Theory
\begin{theorem}
Let $S$ be a set or class.
Let $\preceq$ be an ordering on $S$.
Let $T$ be a subset or subclass of $S$.
Let $\preceq'$ be the restriction of $\preceq$ to $T$.
Let $m \in T$.
Then $m$ is the $\preceq$-smallest element of $T$ {{iff}} $m$ is the $\preceq'$-smallest element of $T$.
\end{theorem}
\begin{proof}
{{proof wanted|The same sort of utterly trivial thing as at Minimal WRT Restriction}}
Category:Smallest Elements
\end{proof}
|
21517
|
\section{Smallest Element is Infimum}
Tags: Order Theory, Smallest Elements, Infima
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T$ have a smallest element $m$.
Then $m$ is the infimum of $T$ in $S$.
\end{theorem}
\begin{proof}
Let $M$ be the smallest element of $T$.
Then by definition:
:$\forall x \in T: m \preceq x$
By definition of infimum, it is necessary to show that:
:$(1): \quad m$ is a lower bound of $T$ in $S$
:$(2): \quad L \preceq m$ for all lower bounds $L$ of $T$ in $S$.
By Smallest Element is Lower Bound, $m$ is a lower bound of $T$ in $S$.
It remains to be shown that:
:$L \preceq m$ for all lower bounds $L$ of $T$ in $S$.
Let $L \in S$ be a lower bound of $T$ in $S$.
By definition of lower bound:
:$\forall t \in T: L \preceq t$
We have that $m \in T$.
Therefore:
:$L \preceq m$
Hence the result.
{{qed}}
\end{proof}
|
21518
|
\section{Smallest Element is Initial Object}
Tags: Poset Categories, Order Categories, Preorder Categories
\begin{theorem}
Let $\mathbf P$ be an order category.
Suppose the objects $\mathbf P_0$ of $\mathbf P$, considered as an ordered set, have a smallest element $p$.
Then $p$ is an initial object of $\mathbf P$.
\end{theorem}
\begin{proof}
Since $p$ is the smallest element of $\mathbf P_0$, we have:
:$\forall q \in \mathbf P_0: p \le q$
that is, for every object $q$ of $\mathbf P$ there is a unique morphism $p \to q$.
That is, $p$ is initial.
{{qed}}
\end{proof}
|
21519
|
\section{Smallest Element is Lower Bound}
Tags: Smallest Elements, Boundedness, Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T$ have a smallest element $m \in T$.
Then $m$ is a lower bound of $T$.
It follows by definition that $T$ is bounded below.
\end{theorem}
\begin{proof}
Let $m \in T$ be a smallest element of $T$.
By definition:
:$\forall y \in T: m \preceq y$
But as $T \subseteq S$, it follows that $m \in S$.
Hence:
:$\exists m \in S: \forall y \in T: m \preceq y$
Thus $T$ is bounded below by the lower bound $m$.
{{qed}}
\end{proof}
|
21520
|
\section{Smallest Element is Minimal}
Tags: Minimal Elements, Smallest Elements, Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set which has a smallest element.
Let $m$ be the smallest element of $\struct {S, \preceq}$.
Then $m$ is a minimal element.
\end{theorem}
\begin{proof}
By definition of smallest element:
:$\forall y \in S: m \preceq y$
Suppose $y \preceq m$.
As $\preceq$ is an ordering, $\preceq$ is by definition antisymmetric.
Thus it follows by definition of antisymmetry that $m = y$.
That is:
:$y \preceq m \implies m = y$
which is precisely the definition of a minimal element.
{{qed}}
\end{proof}
|
21521
|
\section{Smallest Element is Unique}
Tags: Smallest Elements, Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
If $S$ has a smallest element, then it can have only one.
That is, if $a$ and $b$ are both smallest elements of $S$, then $a = b$.
\end{theorem}
\begin{proof}
Let $a$ and $b$ both be smallest elements of $S$.
Then by definition:
:$\forall y \in S: a \preceq y$
:$\forall y \in S: b \preceq y$
Thus it follows that:
:$a \preceq b$
:$b \preceq a$
But as $\preceq$ is an ordering, it is antisymmetric.
Hence by definition of antisymmetric, $a = b$.
{{qed}}
\end{proof}
|
21522
|
\section{Smallest Even Integer whose Euler Phi Value is not the Euler Phi Value of an Odd Integer}
Tags: 33,817,088, Euler Phi Function
\begin{theorem}
The smallest even integer whose Euler $\phi$ value is shared by no odd integer is $33 \, 817 \, 088$.
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = \map \phi {33 \, 817 \, 088}
| r = 16 \, 842 \, 752
| c = {{EulerPhiLink|33,817,088|33 \, 817 \, 088}}
}}
{{eqn | r = 2^{16} \times 257
| c =
}}
{{end-eqn}}
Consider the equation:
:$(1): \quad \map \phi x = 2^{16} \times 257$
Let $p$ be an odd prime factor of $x$.
Then as Euler Phi Function is Multiplicative, either:
:$p - 1 = 2^j$ for some $j \in \Z$ such that $1 \le j \le 16$
which corresponds to the Fermat primes, or:
:$p - 1 = 2^j \times 257$ for some $j \in \Z$ such that $1 \le j \le 16$
But $2^j \times 257 + 1$ is composite for $1 \le j \le 16$:
{{begin-eqn}}
{{eqn | q = j = 1
| l = 2^1 \times 257 + 1
| r = 515
| rr= = 5 \times 103
}}
{{eqn | q = j = 2
| l = 2^2 \times 257 + 1
| r = 1029
| rr= = 3 \times 7^3
}}
{{eqn | q = j = 3
| l = 2^3 \times 257 + 1
| r = 2057
| rr= = 11^2 \times 17
}}
{{eqn | q = j = 4
| l = 2^4 \times 257 + 1
| r = 4113
| rr= = 3^2 \times 457
}}
{{eqn | q = j = 5
| l = 2^5 \times 257 + 1
| r = 8225
| rr= = 5^2 \times 7 \times 47
}}
{{eqn | q = j = 6
| l = 2^6 \times 257 + 1
| r = 16 \, 449
| rr= = 3 \times 5483
}}
{{eqn | q = j = 7
| l = 2^7 \times 257 + 1
| r = 32 \, 897
| rr= = 67 \times 491
}}
{{eqn | q = j = 8
| l = 2^8 \times 257 + 1
| r = 65 \, 793
| rr= = 3 \times 7 \times 13 \times 241
}}
{{eqn | q = j = 9
| l = 2^9 \times 257 + 1
| r = 131 \, 585
| rr= = 5 \times 26 \, 317
}}
{{eqn | q = j = 10
| l = 2^{10} \times 257 + 1
| r = 263 \, 169
| rr= = 3^6 \times 19^2
}}
{{eqn | q = j = 11
| l = 2^{11} \times 257 + 1
| r = 526 \, 337
| rr= = 7 \times 17 \times 4423
}}
{{eqn | q = j = 12
| l = 2^{12} \times 257 + 1
| r = 1 \, 052 \, 673
| rr= = 3 \times 350 \, 891
}}
{{eqn | q = j = 13
| l = 2^{13} \times 257 + 1
| r = 2 \, 105 \, 345
| rr= = 5 \times 11 \times 101 \times 379
}}
{{eqn | q = j = 14
| l = 2^{14} \times 257 + 1
| r = 4 \, 210 \, 689
| rr= = 3 \times 7 \times 43 \times 4663
}}
{{eqn | q = j = 15
| l = 2^{15} \times 257 + 1
| r = 8 \, 421 \, 377
| rr= = 1123 \times 7499
}}
{{eqn | q = j = 16
| l = 2^{16} \times 257 + 1
| r = 16 \, 842 \, 753
| rr= = 3^2 \times 1 871417
}}
{{end-eqn}}
So the only possible odd prime factors of $x$ are the Fermat primes not exceeding $2^{16} + 1$:
:$3, 5, 17, 257, 65 \, 537$
such that:
:$257$ occurs with multiplicity $2$, since we need $257$ to be a factor of $\map \phi x$ but not $257^2$;
:all other prime factors occurs with multiplicity $1$, since we do not want these primes to be factors of $\map \phi x$.
As we have:
{{begin-eqn}}
{{eqn | l = \map \phi {257^2 \times 65 \, 537}
| r = 257 \times 256 \times 2^{16}
}}
{{eqn | o = >
| r = 2^{16} \times 257
}}
{{end-eqn}}
it follows that $257$ and $65 \, 537$ cannot appear together.
We can now write:
:$x = 2^y \times 3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2$
where each $\epsilon_i = 0$ or $1$.
{{AimForCont}} $x$ is odd, that is, $y = 0$.
We have that:
{{begin-eqn}}
{{eqn | l = \map \phi x
| r = \map \phi {3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2}
}}
{{eqn | r = 2^{\epsilon_0} \times 4^{\epsilon_1} \times 16^{\epsilon_2} \times 256 \times 257
| c = as $\epsilon_i = 0$ or $1$
}}
{{eqn | r = 2^{\epsilon_0 + 2 \epsilon_1 + 4 \epsilon_2 + 8} \times 257
| c =
}}
{{eqn | o = \le
| r = 2^{1 + 2 + 4 + 8} \times 257
| c = as $\epsilon_i = 0$ or $1$
}}
{{eqn | o = <
| r = 2^{16} \times 257
| c =
}}
{{eqn | r = \map \phi x
}}
{{end-eqn}}
which is a contradiction.
Thus there can be no odd integer $x$ satisfying $(1)$, that is, $y \ge 1$.
Note that:
{{begin-eqn}}
{{eqn | l = \map \phi {2^y \times 3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2}
| r = 2^{y - 1} \times 2^{\epsilon_0} \times 4^{\epsilon_1} \times 16^{\epsilon_2} \times 256 \times 257
| c = as $\epsilon_i = 0$ or $1$
}}
{{eqn | r = 2^{y - 1 + \epsilon_0 + 2 \epsilon_1 + 4 \epsilon_2 + 8} \times 257
| c =
}}
{{eqn | r = 2^{16} \times 257
| c =
}}
{{eqn | ll= \leadsto
| l = y - 1 + \epsilon_0 + 2 \epsilon_1 + 4 \epsilon_2 + 8
| r = 16
}}
{{eqn | ll= \leadsto
| l = y
| r = 9 - \epsilon_0 - 2 \epsilon_1 - 4 \epsilon_2
}}
{{end-eqn}}
Therefore even solutions to $(1)$ are of the form:
:$x = 2^{9 - \epsilon_0 - 2 \epsilon_1 - 4 \epsilon_2} \times 3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2$
Since each of:
:$2^{-1} \times 3^1$
:$2^{-2} \times 5^1$
:$2^{-4} \times 17^1$
are greater than $1$, taking $\epsilon_i = 0$ for $i = 0, 1, 2$ would result in the smallest even integer satisfying $(1)$, that is:
:$x = 2^9 \times 257^2 = 33 \, 817 \, 088$
It can be established by computer that $33 \, 817 \, 088$ is the smallest even integer whose Euler $\phi$ value is shared by no odd integer.
{{qed}}
\end{proof}
|
21523
|
\section{Smallest Field is Field}
Tags: Fields, Field Theory
\begin{theorem}
The ring $\struct {\set {0_R, 1_R}, +, \circ}$ is the smallest algebraic structure which is a field.
\end{theorem}
\begin{proof}
From Field Contains at least 2 Elements, a field must contain at least two elements.
Hence the null ring, which contains one element, is ''not'' a field.
For $\struct {\set {0_R, 1_R}, +, \circ}$ to be a field:
:$\struct {\set {0_R, 1_R}, +}$ must be an abelian group.
This is fulfilled as this is the parity group.
:$\struct {\set {0_R, 1_R}, \circ}$ must be a commutative division ring.
This is fulfilled, as $\struct {\set {0_R, 1_R}^*, \circ} = \struct {\set {1_R}, \circ}$ is the trivial group.
:$\circ$ needs to distribute over $+$. This follows directly from Ring Product with Zero and the behaviour of the identity element in a group.
{{qed}}
\end{proof}
|
21524
|
\section{Smallest Fifth Power which is Sum of 6 Fifth Powers}
Tags: Fifth Powers
\begin{theorem}
The smallest fifth power which is the sum of $6$ fifth powers is $12^5 = 248 \, 832$:
:$12^5 = 4^5 + 5^5 + 6^5 + 7^5 + 9^5 + 11^5$
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = 12^5
| r = 248 \, 832
| c =
}}
{{eqn | r = 1024 + 3125 + 7776 + 16 \, 807 + 59 \, 049 + 161 \, 051
| c =
}}
{{eqn | r = 4^5 + 5^5 + 6^5 + 7^5 + 9^5 + 11^5
| c =
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that this is the smallest.}}
\end{proof}
|
21525
|
\section{Smallest Fourth Power as Sum of 5 Distinct Fourth Powers}
Tags: 50,625, 15, Fourth Powers
\begin{theorem}
The smallest $4$th power which can be expressed as the sum of $5$ distinct $4$th powers is:
:$15^4 = 4^4 + 6^4 + 8^4 + 9^4 + 14^4$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 15^4
| r = 50 \, 625
| c =
}}
{{eqn | r = 256 + 1296 + 4096 + 6561 + 38 \, 416
| c =
}}
{{eqn | r = 4^4 + 6^4 + 8^4 + 9^4 + 14^4
| c =
}}
{{end-eqn}}
By Fermat's Little Theorem, for $5 \nmid a$:
:$a^4 \equiv 1 \pmod 5$
For $5 \divides a$:
:$a^4 \equiv 0 \pmod 5$
Therefore in order for the equality to hold, one of the following must be true:
The $4$th power on the {{RHS}} is not divisible by $5$, while exactly $1$ of the $5$ $4$th powers on the {{LHS}} is not divisible by $5$; or
The $4$th power on the {{RHS}} is divisible by $5$, while all of the $5$ $4$th powers on the {{LHS}} is divisible by $5$; or
The $4$th power on the {{RHS}} is divisible by $5$, while none of the $5$ $4$th powers on the {{LHS}} is divisible by $5$.
In the first two cases, there must be at least $4$ distinct $4$th powers divisible by $5$.
The smallest multiples of $5$ are $0, 5, 10, 15$, so the $4$th powers on the {{LHS}} must exceed $15^4$, and thus these cases cannot produce a smaller example.
This leaves us with the third case, where {{LHS}} is divisible by $5$.
Since $5^4 \ne 4^4 + 3^4 + 2^4 + 1^4 + 0^4$, we only need to check that $10^4$ is not a sum of $5$ distinct $4$th powers.
As $8^4 + 7^4 + 6^4 + 5^4 + 4^4 = 8674 < 10^4$, we must have $9^4$ on the {{RHS}}.
As $9^4 + 8^4 = 10657 > 10^4$, we cannot have $8^4$ on the {{RHS}}.
As $9^4 + 6^4 + 5^4 + 4^4 + 3^4 = 8819 < 10^4$, we must have $7^4$ on the {{RHS}}.
As $9^4 + 7^4 + 6^4 = 10258 > 10^4$, we cannot have $6^4$ on the {{RHS}}.
But $9^4 + 7^4 + 5^4 + 4^4 + 3^4 = 9924 < 10^4$.
Therefore $10^4$ is not a sum of $5$ distinct $4$th powers.
This proves that $15^4$ is the smallest $4$th power which can be expressed as the sum of $5$ distinct $4$th powers.
{{qed}}
\end{proof}
|
21526
|
\section{Smallest Fourth Power which is Sum of 5 Fourth Powers}
Tags: 625, Fourth Powers
\begin{theorem}
$625$ is the smallest fourth power which is the sum of $5$ fourth powers:
:$625 = 5^4 = 2^4 + 2^4 + 3^4 + 4^4 + 4^4$
\end{theorem}
\begin{proof}
We check that for $n = 2, 3, 4$, $n^4$ is not a sum of $5$ smaller fourth powers.
We have:
:$5 \times 1^4 = 5 < 16 = 2^4$
:$5 \times 2^4 = 80 < 81 = 3^4$
so $2^4, 3^4$ are not sums of $5$ fourth powers.
For $n = 4$:
:$\dfrac {4^4} {3^4} < 4$
so such a sum can include at most $3$ $3^4$'s.
However:
:$3 \times 3^4 + 2^4 + 1^4 = 260 > 256 = 4^4$
:$3 \times 3^4 + 2 \ \ \times 1^4 = 245 < 256 = 4^4$
:$2 \times 3^4 + 3 \ \ \times 2^4 = 220 < 256 = 4^4$
therefore $4^4$ is not a sum of $5$ smaller fourth powers.
This shows that $5^4$ is the smallest fourth power which is the sum of $5$ fourth powers.
{{qed}}
\end{proof}
|
21527
|
\section{Smallest Integer Divisible by All Numbers from 1 to 100}
Tags: Divisors, Divisibility, 69,720,375,229,712,477,164,533,808,935,312,303,556,800
\begin{theorem}
The smallest positive integer which is divisible by each of the integers from $1$ to $100$ is:
:$69 \, 720 \, 375 \, 229 \, 712 \, 477 \, 164 \, 533 \, 808 \, 935 \, 312 \, 303 \, 556 \, 800$
\end{theorem}
\begin{proof}
Let $N$ be divisible by each of the integers from $1$ to $100$.
Each prime number between $2$ and $97$ must be a divisor of $N$.
Also:
:$2^6 = 64 \divides N$
:$3^4 = 81 \divides N$
:$5^2 = 25 \divides N$
:$7^2 = 49 \divides N$
Every other integer between $1$ and $100$ is the product of a subset of all of these.
Hence by Euclid's Lemma:
{{begin-eqn}}
{{eqn | l = N
| o = \ge
| r = 2^6 \times 3^4 \times 5^2 \times 7^2 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47 \times 53 \times 59 \times 61 \times 67 \times 71 \times 73 \times 79 \times 83 \times 89 \times 97
| c =
}}
{{eqn | r = 69 \, 720 \, 375 \, 229 \, 712 \, 477 \, 164 \, 533 \, 808 \, 935 \, 312 \, 303 \, 556 \, 800
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21528
|
\section{Smallest Integer not Sum of Two Ulam Numbers}
Tags: 23, Ulam Numbers
\begin{theorem}
The smallest integer greater than $1$ which is not the sum of two Ulam numbers is $23$.
\end{theorem}
\begin{proof}
Recall the Ulam numbers:
{{:Definition:Ulam Number/Sequence}}
We have:
{{begin-eqn}}
{{eqn | l = 2
| r = 1 + 1
}}
{{eqn | l = 3
| r = 2 + 1
}}
{{eqn | l = 4
| r = 3 + 1
}}
{{eqn | r = 2 + 2
}}
{{eqn | l = 5
| r = 4 + 1
}}
{{eqn | r = 3 + 2
}}
{{eqn | l = 6
| r = 4 + 2
}}
{{eqn | r = 3 + 3
}}
{{eqn | l = 7
| r = 6 + 1
}}
{{eqn | r = 4 + 3
}}
{{eqn | l = 8
| r = 6 + 2
}}
{{eqn | r = 4 + 4
}}
{{eqn | l = 9
| r = 8 + 1
}}
{{eqn | r = 6 + 3
}}
{{eqn | l = 10
| r = 8 + 2
}}
{{eqn | r = 6 + 4
}}
{{eqn | l = 11
| r = 8 + 3
}}
{{eqn | l = 12
| r = 11 + 1
}}
{{eqn | r = 8 + 4
}}
{{eqn | r = 6 + 6
}}
{{eqn | l = 13
| r = 11 + 2
}}
{{eqn | l = 14
| r = 13 + 1
}}
{{eqn | r = 11 + 3
}}
{{eqn | r = 8 + 6
}}
{{eqn | l = 15
| r = 13 + 2
}}
{{eqn | r = 11 + 4
}}
{{eqn | l = 16
| r = 13 + 3
}}
{{eqn | r = 8 + 8
}}
{{eqn | l = 17
| r = 16 + 1
}}
{{eqn | r = 13 + 4
}}
{{eqn | r = 11 + 6
}}
{{eqn | l = 18
| r = 16 + 2
}}
{{eqn | l = 19
| r = 18 + 1
}}
{{eqn | r = 16 + 3
}}
{{eqn | r = 13 + 6
}}
{{eqn | r = 11 + 8
}}
{{eqn | l = 20
| r = 18 + 2
}}
{{eqn | r = 16 + 4
}}
{{eqn | l = 21
| r = 18 + 3
}}
{{eqn | r = 13 + 8
}}
{{eqn | l = 22
| r = 18 + 4
}}
{{eqn | r = 16 + 6
}}
{{eqn | r = 11 + 11
}}
{{end-eqn}}
Now consider the the difference between $23$ and successive Ulam numbers:
{{begin-eqn}}
{{eqn | l = 23 - 18
| r = 5
| c = not a Ulam number
}}
{{eqn | l = 23 - 16
| r = 7
| c = not a Ulam number
}}
{{eqn | l = 23 - 13
| r = 10
| c = not a Ulam number
}}
{{eqn | l = 23 - 11
| r = 12
| c = not a Ulam number
}}
{{end-eqn}}
and it is not necessary to go further back.
Hence the result.
{{qed}}
\end{proof}
|
21529
|
\section{Smallest Integer using Three Words in English Description}
Tags: Recreational Mathematics
\begin{theorem}
The smallest integer which uses exactly $3$ words in its standard (British) English description is:
:$21 \, 000$: '''twenty-one thousand'''
counting hyphenations as separate words.
\end{theorem}
\begin{proof}
All integers up to $100$ ('''one hundred''') use either $1$ or $2$ words:
:'''one'''
:'''sixty'''
:'''seventeen'''
:'''ninety-eight'''
All integers of the form $100 n$ for $n = 1, 2, \ldots 9$ use exactly $2$ words:
:'''one hundred'''
:'''seven hundred'''
:'''nine hundred'''
In British English, the technique for describing integers from $101$ to $199$, and $201$ to $299$ and so on, is to use '''and''' between the number of hundreds and the rest:
:'''one hundred and one'''
:'''three hundred and thirteen'''
:'''four hundred and twenty-six'''
:'''seven hundred and seventy'''
thus using either $4$ or $5$ words.
All integers of the form $1000 n$ for $n = 1, 2, \ldots 10$ use exactly $2$ words:
:'''two thousand'''
:'''five thousand'''
:'''eight thousand'''
:'''twelve thousand'''
:'''nineteen thousand'''
:'''twenty thousand'''
Similarly with hundreds, the technique for describing integers of the form $1000 m + n$ for $1 \le n \le 99$ is to use '''and''' between the number of thousands and the rest:
:'''five thousand and eighteen'''
:'''sixteen thousand and forty-eight'''
:'''thirty-seven thousand and sixty'''
thus using either $4$ or $5$ words.
All other numbers between $1100$ and $20 \, 999$ trivially use more than $3$ words:
:'''four thousand, eight hundred'''
:'''sixteen thousand, one hundred and seventy-seven'''
:'''twenty thousand, nine hundred and ninety-nine'''
and so on.
The smallest integer to use exactly $2$ words is $21$:
:'''twenty-one'''
Hence:
:'''twenty-one thousand'''
{{qed}}
\end{proof}
|
21530
|
\section{Smallest Integer which is Product of 4 Triples all with Same Sum}
Tags: Recreational Mathematics, 25,200
\begin{theorem}
The smallest integer which can be expressed as the product of $4$ different triplets of integers each of which has the same sum is:
{{begin-eqn}}
{{eqn | l = 25 \, 200
| r = 6 \times 56 \times 75
}}
{{eqn | r = 7 \times 40 \times 90
}}
{{eqn | r = 9 \times 28 \times 100
}}
{{eqn | r = 12 \times 20 \times 105
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = 6 \times 56 \times 75
| r = \paren {2 \times 3} \times \paren {2^3 \times 7} \times \paren {3 \times 5^2}
| c =
}}
{{eqn | r = 2^4 \times 3^2 \times 5^2 \times 7
| c =
}}
{{eqn | l = 6 + 56 + 75
| r = 137
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 7 \times 40 \times 90
| r = 7 \times \paren {2^3 \times 5} \times \paren {2 \times 3^2 \times 5^2}
| c =
}}
{{eqn | r = 2^4 \times 3^2 \times 5^2 \times 7
| c =
}}
{{eqn | l = 7 + 40 + 90
| r = 137
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 9 \times 28 \times 100
| r = 3^2 \times \paren {2^2 \times 7} \times \paren {2^2 \times 5^2}
| c =
}}
{{eqn | r = 2^4 \times 3^2 \times 5^2 \times 7
| c =
}}
{{eqn | l = 9 + 28 + 100
| r = 137
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 12 \times 20 \times 105
| r = \paren {2^2 \times 3} \times \paren {2^2 \times 5} \times \paren {3 \times 5 \times 7}
| c =
}}
{{eqn | r = 2^4 \times 3^2 \times 5^2 \times 7
| c =
}}
{{eqn | l = 12 + 20 + 105
| r = 137
| c =
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that this is the smallest such number with this property.}}
\end{proof}
|
21531
|
\section{Smallest Integer which is Sum of 3 Sixth Powers in 2 Ways}
Tags: 160,426,514, Sixth Powers
\begin{theorem}
The smallest positive integer which can be expressed as the sum of $3$ sixth powers in $2$ different ways is:
{{begin-eqn}}
{{eqn | l = 160 \, 426 \, 514
| r = 3^6 + 19^6 + 22^6
| c =
}}
{{eqn | r = 10^6 + 15^6 + 23^6
| c =
}}
{{end-eqn}}
Also note that:
{{begin-eqn}}
{{eqn | l = 854
| r = 3^2 + 19^2 + 22^2
| c =
}}
{{eqn | r = 10^2 + 15^2 + 23^2
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
We have that:
{{begin-eqn}}
{{eqn | l = 160 \, 426 \, 514
| r = 729 + 47 \, 045 \, 881 + 113 \, 379 \, 904
| c =
}}
{{eqn | r = 3^6 + 19^6 + 22^6
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 160 \, 426 \, 514
| r = 1 \, 000 \, 000 + 11 \, 390 \, 625 + 148 \, 035 \, 889
| c =
}}
{{eqn | r = 10^6 + 15^6 + 23^6
| c =
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown there are no smaller}}
\end{proof}
|
21532
|
\section{Smallest Magic Constant of Order 3 Multiplicative Magic Square}
Tags: Multiplicative Magic Squares
\begin{theorem}
The magic constant of the smallest possible multiplicative magic square with the smallest magic constant is as follows.
{{:Multiplicative Magic Square/Examples/Order 3/Smallest}}
</onlyinclude>
\end{theorem}
\begin{proof}
From Smallest Multiplicative Magic Square is of Order 3, the smallest possible multiplicative magic square is of order $3$.
Let $C$ be a magic constant for an order $3$ multiplicative magic square.
We prove that $C \ge 216$ is smallest in $2$ steps:
:Step $1$: If $C$ is a prime power $p^k$, then $k \ge 8$.
:Step $2$: For each prime $p$ dividing $C$, we must have $p^3$ dividing $C$ as well.
Then $216 = 2^3 \cdot 3^3$ is the smallest such constant.
\end{proof}
|
21533
|
\section{Smallest Magic Cube is of Order 3}
Tags: Magic Cubes
\begin{theorem}
Apart from the trivial order $1$ magic cube:
{{:Magic Cube/Examples/Order 1}}
the smallest magic cube is the order $3$ magic cube:
{{:Magic Cube/Examples/Order 3}}
\end{theorem}
\begin{proof}
Suppose there were an order $2$ magic cube.
Take one row of this magic cube.
From Magic Constant of Magic Cube, the row and column total is $9$.
Any row or column with a $1$ in it must therefore also have an $8$ in it.
But there are:
:one row
:one column
both of which have a $1$ in them.
Therefore the $8$ would need to go in $2$ distinct cells.
But $8$ appears in a magic cube exactly once.
Hence there can be no order $2$ magic cube.
{{qed}}
Category:Magic Cubes
\end{proof}
|
21534
|
\section{Smallest Magic Square is of Order 3}
Tags: Magic Squares
\begin{theorem}
Apart from the trivial order $1$ magic square:
{{:Magic Square/Examples/Order 1}}
the smallest magic square is the order $3$ magic square:
{{:Magic Square/Examples/Order 3}}
\end{theorem}
\begin{proof}
Suppose there were an order $2$ magic square.
The row and column total is $\dfrac {1 + 2 + 3 + 4} 2 = 5$.
Any row or column with a $1$ in it must therefore also have a $4$ in it.
But there are:
:one row
:one column
:one diagonal
all of which have a $1$ in them.
Therefore the $4$ would need to go in all $3$ cells.
But $4$ appears in a magic square exactly once.
Hence there can be no order $2$ magic square.
{{qed}}
\end{proof}
|
21535
|
\section{Smallest Multiple of 9 with all Digits Even}
Tags: 288, Classic Problems, Classic Puzzles, Recreational Mathematics, 9, Smallest Multiple of 9 with all Digits Even
\begin{theorem}
$288$ is the smallest integer multiple of $9$ all of whose digits are even.
\end{theorem}
\begin{proof}
By the brute force technique:
{{begin-eqn}}
{{eqn | l = 1 \times 9
| r = 9
}}
{{eqn | l = 2 \times 9
| r = 18
}}
{{eqn | l = 3 \times 9
| r = 27
}}
{{eqn | l = 4 \times 9
| r = 36
}}
{{eqn | l = 5 \times 9
| r = 45
}}
{{eqn | l = 6 \times 9
| r = 54
}}
{{eqn | l = 7 \times 9
| r = 63
}}
{{eqn | l = 8 \times 9
| r = 72
}}
{{eqn | l = 9 \times 9
| r = 81
}}
{{eqn | l = 10 \times 9
| r = 90
}}
{{eqn | l = 11 \times 9
| r = 99
}}
{{eqn | l = 12 \times 9
| r = 108
}}
{{end-eqn}}
All integer multiples of $9$ $k \times 9$ for $k = 12$ through to $k = 22$ begin with $1$, and so have at least one odd digit.
Then:
{{begin-eqn}}
{{eqn | l = 22 \times 9
| r = 198
}}
{{eqn | l = 23 \times 9
| r = 207
}}
{{eqn | l = 24 \times 9
| r = 216
}}
{{eqn | l = 25 \times 9
| r = 225
}}
{{eqn | l = 26 \times 9
| r = 234
}}
{{eqn | l = 27 \times 9
| r = 243
}}
{{eqn | l = 28 \times 9
| r = 252
}}
{{eqn | l = 39 \times 9
| r = 261
}}
{{eqn | l = 30 \times 9
| r = 270
}}
{{eqn | l = 31 \times 9
| r = 279
}}
{{eqn | l = 32 \times 9
| r = 288
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
21536
|
\section{Smallest Multiplicative Magic Square is of Order 3}
Tags: Multiplicative Magic Squares
\begin{theorem}
The order of the smallest multiplicative magic square is $3$, for example:
{{:Multiplicative Magic Square/Examples/Order 3/Smallest}}
\end{theorem}
\begin{proof}
Suppose there were an order $2$ multiplicative magic square $M$.
Let $a$ be the element of row $1$ and column $1$.
Let $a b$ be the magic constant of $M$.
Then $b$ is:
:the element of row $1$ and column $2$, to make the product of row $1$ equal to $a b$
:the element of row $2$ and column $1$, to make the product of column $1$ equal to $a b$
and so on.
But it is a principle of multiplicative magic squares that the elements are all distinct.
Hence no order $2$ multiplicative magic square can exist.
{{qed}}
Category:Multiplicative Magic Squares
\end{proof}
|
21537
|
\section{Smallest Multiply Perfect Number of Order 5}
Tags: Multiply Perfect Numbers, 14,182,439,040
\begin{theorem}
The number $14 \, 182 \, 439 \, 040$ is multiply perfect of order $5$:
:$\map {\sigma_1} {14 \, 182 \, 439 \, 040} = 70 \, 912 \, 195 \, 200 = 5 \times 14 \, 182 \, 439 \, 040$
It is the smallest positive integer to be so.
\end{theorem}
\begin{proof}
From {{DSFLink|14,182,439,040|14 \, 182 \, 439 \, 040}}:
:$\map {\sigma_1} {14 \, 182 \, 439 \, 040} = 70 \, 912 \, 195 \, 200$
{{ProofWanted|That it is the smallest one remains to be proved.}}
\end{proof}
|
21538
|
\section{Smallest Multiply Perfect Number of Order 6}
Tags: Multiply Perfect Numbers, 154,345,556,085,770,649,600
\begin{theorem}
The number $154 \, 345 \, 556 \, 085 \, 770 \, 649 \, 600$ is multiply perfect of order $6$:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {154 \, 345 \, 556 \, 085 \, 770 \, 649 \, 600}
| r = 926 \, 073 \, 336 \, 514 \, 623 \, 897 \, 600
| c =
}}
{{eqn | r = 6 \times 154 \, 345 \, 556 \, 085 \, 770 \, 649 \, 600
| c =
}}
{{end-eqn}}
It is the smallest positive integer to be so.
\end{theorem}
\begin{proof}
From {{DSFLink|154,345,556,085,770,649,600|154 \, 345 \, 556 \, 085 \, 770 \, 649 \, 600}}:
:$\map {\sigma_1} {154 \, 345 \, 556 \, 085 \, 770 \, 649 \, 600} = 926 \, 073 \, 336 \, 514 \, 623 \, 897 \, 600$
{{ProofWanted|That it is the smallest one remains to be proved.}}
\end{proof}
|
21539
|
\section{Smallest Non-Palindromic Number with Palindromic Square}
Tags: 26, Palindromic Numbers, Square Numbers
\begin{theorem}
$26$ is the smallest non-palindromic integer whose square is palindromic.
\end{theorem}
\begin{proof}
Checking the squares of all non-palindromic integers in turn from $10$ upwards, until a palindromic integer is reached:
{{begin-eqn}}
{{eqn | l = 10^2
| r = 100
}}
{{eqn | l = 12^2
| r = 144
}}
{{eqn | l = 13^2
| r = 169
}}
{{eqn | l = 14^2
| r = 196
}}
{{eqn | l = 15^2
| r = 225
}}
{{eqn | l = 16^2
| r = 256
}}
{{eqn | l = 17^2
| r = 289
}}
{{eqn | l = 18^2
| r = 324
}}
{{eqn | l = 19^2
| r = 361
}}
{{eqn | l = 20^2
| r = 400
}}
{{eqn | l = 21^2
| r = 441
}}
{{eqn | l = 23^2
| r = 529
}}
{{eqn | l = 24^2
| r = 576
}}
{{eqn | l = 25^2
| r = 625
}}
{{eqn | l = 26^2
| r = 676
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21540
|
\section{Smallest Normal Subgroup containing Set}
Tags: Normal Subgroups
\begin{theorem}
Let $S \subseteq G$ where $G$ is a group.
Then there exists a unique smallest normal subgroup of $G$ which contains $S$.
\end{theorem}
\begin{proof}
Let $\Bbb S$ be the set of all normal subgroups of $G$ that contain $S$.
Since $S \subseteq G \lhd G$, $\Bbb S \ne \O$.
Let $N = \bigcap \Bbb S$, that is, the intersection of all elements of $\Bbb S$.
By Intersection of Normal Subgroups is Normal, $N \lhd G$.
By the definition of intersection, $S \subseteq N$.
By the method of construction, $N$ is the smallest such subgroup.
By the definition of "smallest", $N$ is unique.
{{qed}}
Category:Normal Subgroups
\end{proof}
|
21541
|
\section{Smallest Number Expressible as Sum of at most Three Triangular Numbers in 4 ways}
Tags: 21, Triangular Numbers
\begin{theorem}
$21$ is the smallest number which can be expressed as the sum of at most $3$ triangular numbers in $4$ ways.
\end{theorem}
\begin{proof}
By inspection:
{{begin-eqn}}
{{eqn | l = 1
| r = T_1
| c = $1$ way
}}
{{eqn | l = 2
| r = T_1 + T_1
| c = $1$ way
}}
{{eqn | l = 3
| r = T_1 + T_1 + T_1
| c =
}}
{{eqn | r = T_2
| c = $2$ ways
}}
{{eqn | l = 4
| r = T_2 + T_1
| c = $1$ way
}}
{{eqn | l = 5
| r = T_2 + T_1 + T_1
| c = $1$ way
}}
{{eqn | l = 6
| r = T_2 + T_2
}}
{{eqn | r = T_3
| c = $2$ ways
}}
{{eqn | l = 7
| r = T_2 + T_2 + T_1
}}
{{eqn | r = T_3 + T_1
| c = $2$ ways
}}
{{eqn | l = 8
| r = T_3 + T_1 + T_1
| c = $1$ way
}}
{{eqn | l = 9
| r = T_3 + T_2
}}
{{eqn | r = T_2 + T_2 + T_2
| c = $2$ ways
}}
{{eqn | l = 10
| r = T_4
}}
{{eqn | r = T_3 + T_2 + T_1
| c = $2$ ways
}}
{{eqn | l = 11
| r = T_4 + T_1
| c = $1$ way
}}
{{eqn | l = 12
| r = T_4 + T_1 + T_1
}}
{{eqn | r = T_3 + T_3
| c =
}}
{{eqn | r = T_3 + T_2 + T_2
| c = $3$ ways
}}
{{eqn | l = 13
| r = T_4 + T_2
| c =
}}
{{eqn | r = T_3 + T_3 + T_1
| c = $2$ ways
}}
{{eqn | l = 14
| r = T_4 + T_2 + T_1
| c = $1$ way
}}
{{eqn | l = 15
| r = T_5
}}
{{eqn | r = T_3 + T_3 + T_2
| c = $2$ ways
}}
{{eqn | l = 16
| r = T_5 + T_1
}}
{{eqn | r = T_4 + T_3
| c =
}}
{{eqn | r = T_4 + T_2 + T_2
| c = $3$ ways
}}
{{eqn | l = 17
| r = T_5 + T_1 + T_1
}}
{{eqn | r = T_4 + T_3 + T_1
| c = $2$ ways
}}
{{eqn | l = 18
| r = T_5 + T_2
}}
{{eqn | r = T_3 + T_3 + T_3
| c = $2$ ways
}}
{{eqn | l = 19
| r = T_5 + T_2 + T_1
}}
{{eqn | r = T_4 + T_3 + T_2
| c = $2$ ways
}}
{{eqn | l = 20
| r = T_4 + T_4
| c = $1$ way
}}
{{eqn | l = 21
| r = T_6
}}
{{eqn | r = T_5 + T_3
| c =
}}
{{eqn | r = T_5 + T_2 + T_2
| c =
}}
{{eqn | r = T_4 + T_4 + T_1
| c = $4$ ways
}}
{{end-eqn}}
{{OEIS|A002636}}{{qed}}
\end{proof}
|
21542
|
\section{Smallest Number not Expressible as Sum of Fewer than 19 Fourth Powers}
Tags: Fourth Powers, 79, Hilbert-Waring Theorem
\begin{theorem}
The smallest positive integer which cannot be expressed as the sum of fewer than $19$ fourth powers is $79$:
:$79 = 15 \times 1^4 + 4 \times 2^4$
\end{theorem}
\begin{proof}
We have $1^4 = 1, 2^4 = 16, 3^4 = 81 > 79$.
Hence for each $n < 79$, we can only use $1^4$ and $2^4$ in our sum.
Write $n = 2^4 a + 1^4 b$.
We can use the greedy algorithm to generate these expressions, since replacing $2^4$ with $16 \times 1^4$ increases the number of fourth powers required.
Suppose $n < 64$.
By Division Theorem, there is a unique way to write $n = 16 q + r$, with $q \in \Z$, $0 \le r < 16$.
{{begin-eqn}}
{{eqn | l = 16 q + r
| r = n
}}
{{eqn | ll = \leadsto
| l = 16 q + r
| o = <
| r = 64
}}
{{eqn | ll = \leadsto
| l = 16 q
| o = <
| r = 64
}}
{{eqn | ll = \leadsto
| l = q
| o = <
| r = 4
}}
{{end-eqn}}
Thus $q + r \le 3 + 15 = 18$.
It follows that each positive integer less than $64$ can be expressed in not more than $18$ fourth powers.
Suppose $64 \le n \le 78$.
We cannot use more than $4$ $2^4$, since $5 \times 2^4 = 80 > n$.
Thus we write $n = 4 \times 2^4 + \paren {n - 64} \times 1^4$.
Since $n - 64 \le 78 - 64 = 14$, we can use not more than $18$ fourth powers to express $n$.
For $n = 79$, we still cannot use more than $4$ $2^4$, since $5 \times 2^4 = 80 > n$.
Thus $n = 4 \times 2^4 + 15 \times 1^4$ uses the least number of fourth powers.
Hence $79$ is the smallest positive integer which cannot be expressed as the sum of fewer than $19$ fourth powers.
{{qed}}
\end{proof}
|
21543
|
\section{Smallest Number to appear 6 Times in Pascal's Triangle}
Tags: 120, Pascal's Triangle
\begin{theorem}
The smallest positive integer greater than $1$ to appear $6$ times in Pascal's Triangle is $120$.
\end{theorem}
\begin{proof}
We have:
:$\dbinom {120} 1 = \dbinom {16} 2 = \dbinom {10} 3 = \dbinom {10} 7 = \dbinom {16} {14} = \dbinom {120} {119} = 120$
To verify that this is the smallest, we look at binomial coefficients that are no more than $120$.
Observe that for $n > 120$, $1 \le k \le n - 1$:
:$\dbinom n k \ge \dbinom n 1 = n > 120$
For $n > 16$, $2 \le k \le n - 2$:
:$\dbinom n k \ge \dbinom n 2 = \dfrac {n \paren {n - 1}} {2!} > \dfrac {16 \paren {15}} {2!} = 120$
For $n > 10$, $3 \le k \le n - 3$:
:$\dbinom n k \ge \dbinom n 3 = \dfrac {n \paren {n - 1} \paren {n - 2}} {3!} > \dfrac {10 \paren 9 \paren 8} {3!} = 120$
Any number, except $2$, appears twice in the form $\dbinom n 1$ and $\dbinom n {n - 1}$.
Therefore we scout for numbers appearing four times, not as those forms, by listing $\dbinom n k$ for:
:$n \le 10$ and $2 \le k \le n - 2$;
:$n \le 16$ and $k = 2$ or $n - 2$
They are:
:$6$
:$10, 10$
:$15, 20, 15$
:$21, 35, 35, 21$
:$28, 56, 70, 56, 28$
:$36, 84, 126, 126, 84, 36$
:$45, 120, *, *, *, 120, 45$
:$55, 55$
:$66, 66$
:$78, 78$
:$91, 91$
:$105, 105$
:$120, 120$
and we see that the only number appearing four times is $120$.
{{qed}}
\end{proof}
|
21544
|
\section{Smallest Number which is Multiplied by 99 by Appending 1 to Each End}
Tags: 112,359,550,561,797,752,809, Recreational Mathematics, 112,359,550,561,797,732,809
\begin{theorem}
The smallest positive integer which is multiplied by $99$ when $1$ is appended to each end is:
:$112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809$
\end{theorem}
\begin{proof}
We have that:
:$112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809 = 101 \times 1 \, 052 \, 788 \, 969 \times 1 \, 056 \, 689 \, 261$
while:
{{begin-eqn}}
{{eqn | l = 11 \, 123 \, 595 \, 505 \, 617 \, 977 \, 528 \, 091
| r = 3^2 \times 11 \times 101 \times 1 \, 052 \, 788 \, 969 \times 1 \, 056 \, 689 \, 261
| c =
}}
{{eqn | r = 3^2 \times 11 \times 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809
| c =
}}
{{end-eqn}}
Let $N$ be the smallest integer satisfying $99 N = \sqbrk {1N1}$ when expressed in decimal notation.
Suppose $N$ is $k$ digits long.
Then:
:$\sqbrk {1 N 1} = 10^{k + 1} + 10 N + 1$
Subtracting $10 N$ from $99 N$ gives:
:$89 N = 10^{k + 1} + 1$
One can show, by trial and error, that the smallest $k$ where $10^{k + 1} + 1$ is divisible by $89$ is $21$.
Then $N = \dfrac {10^{22} + 1} {89} = 112 \, 359 \, 550 \, 561 \, 797 \, 752 \, 809$.
{{qed}}
\end{proof}
|
21545
|
\section{Smallest Number with 16 Divisors}
Tags: Divisor Counting Function, Tau Function, 120
\begin{theorem}
The smallest positive integer with $16$ divisors is $120$.
\end{theorem}
\begin{proof}
From {{DCFLink|120}}:
:$\map {\sigma_0} {120} = 16$
The result is a specific instance of Smallest Number with $2^n$ Divisors:
:$120 = 2 \times 3 \times 4 \times 5$
{{qed}}
\end{proof}
|
21546
|
\section{Smallest Numbers with 240 Divisors}
Tags: Divisor Counting Function, Tau Function
\begin{theorem}
The smallest integers with $240$ divisors are:
:$720 \, 720, 831 \, 600, 942 \, 480, 982 \, 800, 997 \, 920, \ldots$
\end{theorem}
\begin{proof}
In the below, $\map {\sigma_0} n$ denotes the divisor counting function of $n$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\sigma_0} {720 \, 720}
| r = 240
| c = {{DCFLink|720,720|720 \, 720}}
}}
{{eqn | l = \map {\sigma_0} {831 \, 600}
| r = 240
| c = {{DCFLink|831,600|831 \, 600}}
}}
{{eqn | l = \map {\sigma_0} {942 \, 480}
| r = 240
| c = {{DCFLink|942,480|942 \, 480}}
}}
{{eqn | l = \map {\sigma_0} {982 \, 800}
| r = 240
| c = {{DCFLink|982,800|982 \, 800}}
}}
{{eqn | l = \map {\sigma_0} {997 \, 920}
| r = 240
| c = {{DCFLink|997,920|997 \, 920}}
}}
{{end-eqn}}
{{ProofWanted|Prove that these are indeed the smallest}}
\end{proof}
|
21547
|
\section{Smallest Odd Abundant Number not Divisible by 3}
Tags: Abundant Numbers, 5,391,411,025
\begin{theorem}
The smallest odd abundant number not divisible by $3$ is $5 \, 391 \, 411 \, 025$.
\end{theorem}
\begin{proof}
We have:
:$5 \, 391 \, 411 \, 025 = 5^2 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29$
showing it is not divisible by $3$.
Then from {{DSFLink|5,391,411,025|5 \, 391 \, 411 \, 025}} we have:
:$\map {\sigma_1} {5 \, 391 \, 411 \, 025} = 10 \, 799 \, 308 \, 800 = 2 \times 5 \, 391 \, 411 \, 025 + 16 \, 486 \, 750$
demonstrating that $5 \, 391 \, 411 \, 025$ is abundant.
{{ProofWanted|It is still to be shown it is the smallest such.}}
\end{proof}
|
21548
|
\section{Smallest Odd Number not of form 2 a squared plus p}
Tags: Prime Numbers, Goldbach's Lesser Conjecture, Polynomial Expressions for Primes, 17
\begin{theorem}
$17$ is the smallest odd number $n$ greater than $3$ which cannot be expressed in the form:
:$n = 2 a^2 + p$
where:
:$p$ is prime
:$a \in \Z_{>0}$ is a (strictly) positive integer.
\end{theorem}
\begin{proof}
First note that $3$ is not so expressible:
{{begin-eqn}}
{{eqn | l = 3 - 2 \times 1^2
| r = 1
| c = which is not prime
}}
{{end-eqn}}
and so $3$ cannot be written in such a form.
Then we have:
{{begin-eqn}}
{{eqn | l = 5
| r = 2 \times 1^2 + 3
| c =
}}
{{eqn | l = 7
| r = 2 \times 1^2 + 5
| c =
}}
{{eqn | l = 9
| r = 2 \times 1^2 + 7
| c =
}}
{{eqn | l = 11
| r = 2 \times 2^2 + 3
| c =
}}
{{eqn | l = 13
| r = 2 \times 1^2 + 11
| c =
}}
{{eqn | r = 2 \times 2^2 + 5
| c =
}}
{{eqn | l = 15
| r = 2 \times 1^2 + 13
| c =
}}
{{eqn | r = 2 \times 2^2 + 7
| c =
}}
{{end-eqn}}
However, we have:
{{begin-eqn}}
{{eqn | l = 17 - 2 \times 1^2
| r = 15
| c = which is not prime
}}
{{eqn | l = 17 - 2 \times 2^2
| r = 9
| c = which is not prime
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21549
|
\section{Smallest Pair of Quasiamicable Numbers}
Tags: Quasi-Amicable Numbers, Quasiamicable Numbers, 48, 75
\begin{theorem}
The smallest pair of quasiamicable numbers is $48$ and $75$.
\end{theorem}
\begin{proof}
From Quasiamicable Numbers: $48$ and $75$ we have that $48$ and $75$ are quasiamicable numbers.
It remains to be demonstrated that these are indeed the smallest such.
Let $n$ be the smaller number of the quasiamicable pair.
Then we must have $\map {\sigma_1} n - n - 1 > n$.
Since $\map {\sigma_1} n > 2 n$, $n$ is abundant.
The abundant numbers for $n < 48$ are:
:$12, 18, 20, 24, 30, 36, 40, 42$
And we have:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {12} - 13
| r = \paren {1 + 2 + 4} \paren {1 + 3} - 13
}}
{{eqn | r = 15
}}
{{eqn | l = \map {\sigma_1} {15} - 16
| r = \paren {1 + 3} \paren {1 + 5} - 16
}}
{{eqn | r = 8
| rr = \ne 12
}}
{{eqn | l = \map {\sigma_1} {18} - 19
| r = \paren {1 + 2} \paren {1 + 3 + 9} - 19
}}
{{eqn | r = 20
}}
{{eqn | l = \map {\sigma_1} {20} - 21
| r = \paren {1 + 2 + 4} \paren {1 + 5} - 21
}}
{{eqn | r = 21
| rr = \ne 18
}}
{{eqn | l = \map {\sigma_1} {21} - 22
| r = \paren {1 + 3} \paren {1 + 7} - 22
}}
{{eqn | r = 10
| rr = \ne 20
}}
{{eqn | l = \map {\sigma_1} {24} - 25
| r = \paren {1 + 2 + 4 + 8} \paren {1 + 3} - 25
}}
{{eqn | r = 35
}}
{{eqn | l = \map {\sigma_1} {35} - 36
| r = \paren {1 + 5} \paren {1 + 7} - 36
}}
{{eqn | r = 12
| rr = \ne 24
}}
{{eqn | l = \map {\sigma_1} {30} - 31
| r = \paren {1 + 2} \paren {1 + 3} \paren {1 + 5} - 31
}}
{{eqn | r = 41
}}
{{eqn | l = \map {\sigma_1} {41} - 42
| r = 0
| rr = \ne 30, 36
}}
{{eqn | l = \map {\sigma_1} {36} - 37
| r = \paren {1 + 2 + 4} \paren {1 + 3 + 9} - 37
}}
{{eqn | r = 41
}}
{{eqn | l = \map {\sigma_1} {40} - 41
| r = \paren {1 + 2 + 4 + 8} \paren {1 + 5} - 41
}}
{{eqn | r = 49
}}
{{eqn | l = \map {\sigma_1} {49} - 50
| r = \paren {1 + 7 + 49} - 50
}}
{{eqn | r = 7
| rr = \ne 40
}}
{{eqn | l = \map {\sigma_1} {42} - 43
| r = \paren {1 + 2} \paren {1 + 3} \paren {1 + 7} - 43
}}
{{eqn | r = 53
}}
{{eqn | l = \map {\sigma_1} {53} - 54
| r = 0
| rr = \ne 42
}}
{{end-eqn}}
hence $n$ is at least $48$, proving the result.
{{qed}}
\end{proof}
|
21550
|
\section{Smallest Pandigital Square}
Tags: 1,026,753,849, Pandigital Integers, Square Numbers
\begin{theorem}
The smallest pandigital square is $1 \, 026 \, 753 \, 849$:
:$1 \, 026 \, 753 \, 849 = 32 \, 043^2$
\end{theorem}
\begin{proof}
We check all the squares of numbers from $\ceiling {\sqrt {1 \, 023 \, 456 \, 789} } = 31 \, 992$ up to $32 \, 042$, with the following constraints:
Since all these squares has $10$ as its two leftmost digits, the number cannot end with $0$, $1$ or $9$.
A pandigital number is divisible by $9$, so our number must be divisible by $3$.
These constraints leaves us with the following $12$ candidates:
{{begin-eqn}}
{{eqn | l = 31 \, 992^2
| r = 1 \, 023 \, 488 \, 064
}}
{{eqn | l = 31 \, 995^2
| r = 1 \, 023 \, 680 \, 025
}}
{{eqn | l = 31 \, 998^2
| r = 1 \, 023 \, 872 \, 004
}}
{{eqn | l = 32 \, 004^2
| r = 1 \, 024 \, 256 \, 016
}}
{{eqn | l = 32 \, 007^2
| r = 1 \, 024 \, 448 \, 049
}}
{{eqn | l = 32 \, 013^2
| r = 1 \, 024 \, 832 \, 169
}}
{{eqn | l = 32 \, 016^2
| r = 1 \, 025 \, 024 \, 256
}}
{{eqn | l = 32 \, 022^2
| r = 1 \, 025 \, 408 \, 484
}}
{{eqn | l = 32 \, 025^2
| r = 1 \, 025 \, 600 \, 625
}}
{{eqn | l = 32 \, 028^2
| r = 1 \, 025 \, 792 \, 784
}}
{{eqn | l = 32 \, 034^2
| r = 1 \, 026 \, 177 \, 156
}}
{{eqn | l = 32 \, 037^2
| r = 1 \, 026 \, 369 \, 369
}}
{{end-eqn}}
By inspection, none of these numbers are pandigital.
{{qed}}
\end{proof}
|
21551
|
\section{Smallest Penholodigital Square}
Tags: Pandigital Integers, Penholodigital Integers, 11,826, 139,854,276, Square Numbers
\begin{theorem}
The smallest penholodigital square number is:
:$11 \, 826^2 = 139 \, 854 \, 276$
\end{theorem}
\begin{proof}
Let $n$ be the smallest positive integer whose square is penholodigital.
First it is noted that the smallest penholodigital number is $123 \, 456 \, 789$.
Hence any square penholodigital number must be at least as large as that.
Thus we can can say that:
:$n \ge \ceiling {\sqrt {123 \, 456 \, 789} } = 11 \, 112$
where $\ceiling {\, \cdot \,}$ denotes the ceiling function.
It remains to be demonstrated that no positive integer between $11 \, 112$ and $11 \, 826$ has a penholodigital square.
{{ProofWanted|This is over $700$ numbers. The task can be filtered by, for example, disregarding all $n$ ending in $1$ and $9$ because their squares will both begin and end in $1$ at this low range of the $10000$s, and of course all $n$ ending in $0$ because their squares will end in $0$.}}
\end{proof}
|
21552
|
\section{Smallest Perfect Square Dissection}
Tags: Number Theory, 112, Squares, 21, Dissections, Recreational Mathematics
\begin{theorem}
The smallest perfect square dissection is of an integer square of side $112$ into $21$ parts.
\end{theorem}
\begin{proof}
:700px
{{ProofWanted|That this is the smallest still needs to be proved.}}
\end{proof}
|
21553
|
\section{Smallest Polyomino with Hole}
Tags: Heptominoes, Polyominoes
\begin{theorem}
The smallest polyomino with a hole is the heptomino in the form of a $3 \times 3$
square with the center $1 \times 1$ square and a corner $1 \times 1$ square missing:
:200px
\end{theorem}
\begin{proof}
From 35 Hexominoes and by inspection, none of the $35$ hexominoes has a hole.
From Number of Heptominoes and by inspection, exactly one of the $108$ heptominoes has a hole.
This is the smallest polyomino with a hole.
{{qed}}
\end{proof}
|
21554
|
\section{Smallest Positive Integer Combination is Greatest Common Divisor}
Tags: Integer Combinations, Greatest Common Divisor, Smallest Positive Integer Combination is Greatest Common Divisor
\begin{theorem}
Let $a, b \in \Z_{>0}$ be (strictly) positive integers.
Let $d \in \Z_{>0}$ be the smallest positive integer such that:
: $d = a s + b t$
where $s, t \in \Z$.
Then:
:$(1): \quad d \divides a \land d \divides b$
:$(2): \quad c \divides a \land c \divides b \implies c \divides d$
where $\divides$ denotes divisibility.
That is, $d$ is the greatest common divisor of $a$ and $b$.
\end{theorem}
\begin{proof}
Let $D$ be the subset of $\Z_{>0}$ defined as:
:$D = \left\{{a s + b t: s, t \in \Z, a s + b t > 0}\right\}$
Setting $s = 1$ and $t = 0$ it is clear that $a = \left({a \times 1 + b \times 0}\right) \in D$.
So $D \ne \varnothing$.
By Integers Bounded Below has Smallest Element, $D$ has a smallest element $d$, say.
Thus $d = a s + b t$ for some $s, t \in \Z$.
\end{proof}
|
21555
|
\section{Smallest Positive Integer which is Sum of 2 Fourth Powers in 2 Ways}
Tags: Fourth Powers, 635,318,657
\begin{theorem}
The smallest positive integer which can be expressed as the sum of $2$ fourth powers in $2$ different ways is:
{{begin-eqn}}
{{eqn | l = 635 \, 318 \, 657
| r = 59^4 + 158^4
| c =
}}
{{eqn | r = 133^4 + 134^4
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
The fact that these are the smallest can be demonstrated by calculation.
{{qed}}
\end{proof}
|
21556
|
\section{Smallest Positive Integer which is Sum of 2 Odd Primes in 6 Ways}
Tags: Prime Numbers
\begin{theorem}
The smallest positive integer which is the sum of $2$ odd primes in $6$ different ways is $60$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 60
| r = 7 + 53
| c = $6$ ways
}}
{{eqn | r = 13 + 47
| c =
}}
{{eqn | r = 17 + 43
| c =
}}
{{eqn | r = 19 + 41
| c =
}}
{{eqn | r = 23 + 37
| c =
}}
{{eqn | r = 29 + 31
| c =
}}
{{end-eqn}}
It is determined that there are no smaller numbers with this property by inspection:
{{begin-eqn}}
{{eqn | l = 58
| r = 53 + 5
| c = $3$ ways
}}
{{eqn | r = 47 + 11
| c =
}}
{{eqn | r = 41 + 17
| c =
}}
{{eqn | r = 29 + 29
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 56
| r = 53 + 3
| c = $3$ ways
}}
{{eqn | r = 43 + 13
| c =
}}
{{eqn | r = 37 + 19
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 54
| r = 47 + 7
| c = $5$ ways
}}
{{eqn | r = 43 + 11
| c =
}}
{{eqn | r = 41 + 13
| c =
}}
{{eqn | r = 37 + 17
| c =
}}
{{eqn | r = 31 + 23
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 52
| r = 47 + 5
| c = $3$ ways
}}
{{eqn | r = 41 + 11
| c =
}}
{{eqn | r = 29 + 23
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 50
| r = 47 + 3
| c = $4$ ways
}}
{{eqn | r = 43 + 7
| c =
}}
{{eqn | r = 37 + 13
| c =
}}
{{eqn | r = 31 + 19
| c =
}}
{{end-eqn}}
From the workings of Smallest Positive Integer which is Sum of 2 Odd Primes in n Ways, $48$ is the smallest positive integer which is the sum of $2$ odd primes in $5$ different ways.
Thus there is no need to explore smaller numbers.
{{qed}}
\end{proof}
|
21557
|
\section{Smallest Positive Integer which is Sum of 2 Odd Primes in n Ways}
Tags: Prime Numbers
\begin{theorem}
The sequence of positive integers $n$ which are the smallest such that they are the sum of $2$ odd primes in $k$ different ways begins as follows:
:{| border="1"
|-
! align="right" style = "padding: 2px 10px" | $k$
! align="right" style = "padding: 2px 10px" | $n$
|-
| align="right" style = "padding: 2px 10px" | $1$
| align="right" style = "padding: 2px 10px" | $6$
|-
| align="right" style = "padding: 2px 10px" | $2$
| align="right" style = "padding: 2px 10px" | $10$
|-
| align="right" style = "padding: 2px 10px" | $3$
| align="right" style = "padding: 2px 10px" | $22$
|-
| align="right" style = "padding: 2px 10px" | $4$
| align="right" style = "padding: 2px 10px" | $34$
|-
| align="right" style = "padding: 2px 10px" | $5$
| align="right" style = "padding: 2px 10px" | $48$
|-
| align="right" style = "padding: 2px 10px" | $6$
| align="right" style = "padding: 2px 10px" | $60$
|}
{{OEIS|A001172}}
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 6
| r = 3 + 3
| c = $1$ way
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 8
| r = 5 + 3
| c = $1$ way
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 10
| r = 7 + 3
| c = $2$ ways
}}
{{eqn | r = 5 + 5
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 12
| r = 7 + 5
| c = $1$ way
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 14
| r = 11 + 3
| c = $2$ ways
}}
{{eqn | r = 7 + 7
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 16
| r = 13 + 3
| c = $2$ ways
}}
{{eqn | r = 11 + 5
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 18
| r = 13 + 5
| c = $2$ ways
}}
{{eqn | r = 11 + 7
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 20
| r = 17 + 3
| c = $2$ ways
}}
{{eqn | r = 13 + 7
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 22
| r = 19 + 3
| c = $3$ ways
}}
{{eqn | r = 17 + 5
| c =
}}
{{eqn | r = 11 + 11
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 24
| r = 19 + 5
| c = $2$ ways
}}
{{eqn | r = 13 + 11
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 26
| r = 23 + 3
| c = $3$ ways
}}
{{eqn | r = 19 + 7
| c =
}}
{{eqn | r = 13 + 13
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 28
| r = 23 + 5
| c = $2$ ways
}}
{{eqn | r = 17 + 11
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 30
| r = 23 + 7
| c = $3$ ways
}}
{{eqn | r = 19 + 11
| c =
}}
{{eqn | r = 17 + 13
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 32
| r = 29 + 3
| c = $2$ ways
}}
{{eqn | r = 19 + 13
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 34
| r = 31 + 3
| c = $4$ ways
}}
{{eqn | r = 29 + 5
| c =
}}
{{eqn | r = 23 + 11
| c =
}}
{{eqn | r = 17 + 17
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 36
| r = 31 + 5
| c = $4$ ways
}}
{{eqn | r = 29 + 7
| c =
}}
{{eqn | r = 23 + 13
| c =
}}
{{eqn | r = 19 + 17
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 38
| r = 31 + 7
| c = $2$ ways
}}
{{eqn | r = 19 + 19
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 40
| r = 37 + 3
| c = $3$ ways
}}
{{eqn | r = 29 + 11
| c =
}}
{{eqn | r = 23 + 17
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 42
| r = 37 + 5
| c = $4$ ways
}}
{{eqn | r = 31 + 11
| c =
}}
{{eqn | r = 29 + 13
| c =
}}
{{eqn | r = 23 + 19
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 44
| r = 41 + 3
| c = $3$ ways
}}
{{eqn | r = 37 + 7
| c =
}}
{{eqn | r = 31 + 13
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 46
| r = 43 + 3
| c = $4$ ways
}}
{{eqn | r = 41 + 5
| c =
}}
{{eqn | r = 29 + 17
| c =
}}
{{eqn | r = 23 + 23
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 48
| r = 43 + 5
| c = $5$ ways
}}
{{eqn | r = 41 + 7
| c =
}}
{{eqn | r = 37 + 11
| c =
}}
{{eqn | r = 31 + 17
| c =
}}
{{eqn | r = 29 + 19
| c =
}}
{{end-eqn}}
From Smallest Positive Integer which is Sum of 2 Odd Primes in 6 Ways, the smallest positive integer which is the sum of $2$ odd primes in $6$ different ways is $60$.
{{qed}}
\end{proof}
|
21558
|
\section{Smallest Prime Number not Difference between Power of 2 and Power of 3}
Tags: Powers, 41, Powers of 2, Powers of 3
\begin{theorem}
$41$ is the smallest prime number which is not the difference between a power of $2$ and a power of $3$.
\end{theorem}
\begin{proof}
First we have:
{{begin-eqn}}
{{eqn | l = 2
| r = 3^1 - 2^0
| c =
}}
{{eqn | l = 3
| r = 2^2 - 3^0
| c =
}}
{{eqn | l = 5
| r = 3^2 - 2^2
| c =
}}
{{eqn | l = 7
| r = 3^2 - 2^1
| c =
}}
{{eqn | l = 11
| r = 3^3 - 2^4
| c =
}}
{{eqn | l = 13
| r = 2^4 - 3^1
| c =
}}
{{eqn | l = 17
| r = 3^4 - 2^6
| c =
}}
{{eqn | l = 19
| r = 3^3 - 2^3
| c =
}}
{{eqn | l = 23
| r = 3^3 - 2^2
| c =
}}
{{eqn | l = 29
| r = 2^5 - 3^1
| c =
}}
{{eqn | l = 31
| r = 2^5 - 3^0
| c =
}}
{{eqn | l = 37
| r = 2^6 - 3^3
| c =
}}
{{end-eqn}}
{{AimForCont}} $41 = 2^n - 3^m$.
We have that $n > 3$.
Thus:
:$2^n \equiv 0 \pmod 8$
and as:
:$41 \equiv 1 \pmod 8$
we have:
:$1 \equiv -3^m \pmod 8$
which is not possible:
For any integer $k$,
{{begin-eqn}}
{{eqn | l = -3^{2 k}
| r = -9^k
}}
{{eqn | o = \equiv
| r = -1^k
| rr = \pmod 8
| c = Congruence of Powers
}}
{{eqn | o = \equiv
| r = -1
| rr = \pmod 8
}}
{{eqn | l = -3^{2 k + 1}
| r = 3 \paren {-3^{2 k} }
}}
{{eqn | o = \equiv
| r = 3 \paren {-1}
| rr = \pmod 8
| c = Modulo Multiplication is Well-Defined
}}
{{eqn | o = \equiv
| r = -3
| rr = \pmod 8
}}
{{end-eqn}}
So for any integer $m$, $1 \not \equiv -3^m \pmod 8$.
Now suppose $41 = 3^m - 2^n$.
We have that $m > 1$ and $n > 2$.
Taking $\mod 3$ on both sides:
{{begin-eqn}}
{{eqn | l = 41
| r = 3^m - 2^n
}}
{{eqn | ll = \leadsto
| l = -1
| o = \equiv
| r = - 2^n
| rr = \pmod 3
| c =
}}
{{eqn | ll = \leadsto
| l = 1
| o = \equiv
| r = 2^n
| rr = \pmod 3
| c =
}}
{{eqn | o = \equiv
| r = \paren {-1}^n
| rr = \pmod 3
| c = Congruence of Powers
}}
{{end-eqn}}
which shows that $n$ is even.
Taking $\mod 4$ on both sides:
{{begin-eqn}}
{{eqn | l = 41
| r = 3^m - 2^n
}}
{{eqn | ll = \leadsto
| l = 1
| o = \equiv
| r = 3^m
| rr = \pmod 4
| c =
}}
{{eqn | o = \equiv
| r = \paren {-1}^m
| rr = \pmod 4
| c = Congruence of Powers
}}
{{end-eqn}}
which shows that $m$ is even as well.
Now we take $\mod 5$ on both sides:
{{begin-eqn}}
{{eqn | l = 41
| r = 3^m - 2^n
}}
{{eqn | ll = \leadsto
| l = 1
| o = \equiv
| r = 3^m - 2^n
| rr = \pmod 5
| c =
}}
{{eqn | o = \equiv
| r = 9^{m / 2} - 4^{n / 2}
| rr = \pmod 5
| c =
}}
{{eqn | o = \equiv
| r = \paren {-1}^{m / 2} - \paren {-1}^{n / 2}
| rr = \pmod 5
| c = Congruence of Powers
}}
{{end-eqn}}
But $\paren {-1}^{m / 2} - \paren {-1}^{n / 2}$ can only be $0$ or $\pm 2$, not $1$, which is a contradiction.
Therefore $41$ is not the difference between a power of $2$ and a power of $3$.
{{qed}}
\end{proof}
|
21559
|
\section{Smallest Prime Number whose Period is of Maximum Length}
Tags: Fractions
\begin{theorem}
$7$ is the smallest prime number the period of whose reciprocal, when expressed in decimal notation, is maximum:
:$\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$
\end{theorem}
\begin{proof}
From Maximum Period of Reciprocal of Prime, the maximum period of $\dfrac 1 p$ is $p - 1$.
:$\dfrac 1 2 = 0 \cdotp 5$: not recurring.
:$\dfrac 1 3 = 0 \cdotp \dot 3$: recurring with period $1$.
:$\dfrac 1 5 = 0 \cdotp 2$: not recurring.
:$\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$: recurring with period $6$.
{{qed}}
\end{proof}
|
21560
|
\section{Smallest Pythagorean Quadrilateral with Integer Sides}
Tags: Pythagorean Triangles, Quadrilaterals
\begin{theorem}
The smallest Pythagorean quadrilateral in which the sides of the $4$ right triangles formed by its sides and perpendicular diagonals are all integers has an area of $21 \, 576$.
The sides of the right triangles in question are:
:$25, 60, 65$
:$91, 60, 109$
:$91, 312, 325$
:$25, 312, 313$
\end{theorem}
\begin{proof}
:800px
The $4$ right triangles are inspected:
{{begin-eqn}}
{{eqn | l = 25^2 + 60^2
| r = 625 + 3600
}}
{{eqn | r = 4225
| c =
}}
{{eqn | r = 65^2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 91^2 + 60^2
| r = 8281 + 3600
}}
{{eqn | r = 11 \, 881
| c =
}}
{{eqn | r = 109^2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 91^2 + 312^2
| r = 8281 + 97 \, 344
}}
{{eqn | r = 105 \, 625
| c =
}}
{{eqn | r = 325^2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 25^2 + 312^2
| r = 625 + 97 \, 344
}}
{{eqn | r = 97 \, 969
| c =
}}
{{eqn | r = 313^2
| c =
}}
{{end-eqn}}
The area of each right triangle is calculated:
{{begin-eqn}}
{{eqn | l = \dfrac {25 \times 60} 2
| r = 750
}}
{{eqn | l = \dfrac {91 \times 60} 2
| r = 2730
}}
{{eqn | l = \dfrac {91 \times 312} 2
| r = 14 \, 196
}}
{{eqn | l = \dfrac {25 \times 312} 2
| r = 3900
}}
{{end-eqn}}
Thus the total area is:
:$750 + 2730 + 14 \, 196 + 3900 = 21 \, 576$
{{ProofWanted|It needs to be proved that this is the smallest.}}
\end{proof}
|
21561
|
\section{Smallest Pythagorean Triangle is 3-4-5}
Tags: Pythagorean Triangles
\begin{theorem}
The smallest Pythagorean triangle has sides of length $3$, $4$ and $5$.
:300px
\end{theorem}
\begin{proof}
From Solutions of Pythagorean Equation, all Pythagorean triangles, the set of all primitive Pythagorean triples is generated by:
:$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$
where:
:$m, n \in \Z_{>0}$ are (strictly) positive integers
:$m \perp n$, that is, $m$ and $n$ are coprime
:$m$ and $n$ are of opposite parity
:$m > n$.
The smallest two (strictly) positive integers which satisfy the above criteria are:
:$n = 1$
:$m = 2$
Hence:
:$2 m n = 2 \times 2 \times 1 = 4$
:$m^2 - n^2 = 2^2 - 1^2 = 3$
:$m^2 + n^2 = 2^2 + 1^2 = 5$
and to confirm:
:$3^2 + 4^2 = 9 + 16 = 25 = 5^2$
{{qed}}
\end{proof}
|
21562
|
\section{Smallest Scalene Obtuse Triangle with Integer Sides and Area}
Tags: Scalne Triangles, Obtuse Triangles, Scalene Triangles
\begin{theorem}
The smallest scalene obtuse triangle with integer sides and area has sides of length $4, 13, 15$.
\end{theorem}
\begin{proof}
From Heron's Formula, the area $A$ of $\triangle ABC$ is given by:
:$A = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.
Here we have:
{{begin-eqn}}
{{eqn | l = s
| r = \dfrac {4 + 13 + 15} 2
| c =
}}
{{eqn | r = 16
| c =
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = A
| r = \sqrt {16 \paren {16 - 4} \paren {16 - 13} \paren {16 - 15} }
| c =
}}
{{eqn | r = \sqrt {16 \times 12 \times 3 \times 1}
| c =
}}
{{eqn | r = \sqrt {2^4 \times 2^2 \times 3 \times 3 \times 1}
| c =
}}
{{eqn | r = \sqrt {2^6 \times 3^2}
| c =
}}
{{eqn | r = 2^3 \times 3
| c =
}}
{{eqn | r = 24
| c =
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that it is the smallest. This can be done by exhaustion: the number of triples defining an obtuse scalene triangle are not that many.}}
\end{proof}
|
21563
|
\section{Smallest Sequence of Three Consecutive Semiprimes}
Tags: Semiprimes
\begin{theorem}
The smallest triple of consecutive semiprimes is:
:$33, 34, 35$
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = 33
| r = 3 \times 11
}}
{{eqn | l = 34
| r = 2 \times 17
}}
{{eqn | l = 35
| r = 5 \times 7
}}
{{end-eqn}}
It can be seen from the sequence of semiprimes that there exist no smaller such triples.
{{qed}}
\end{proof}
|
21564
|
\section{Smallest Set is Unique}
Tags: Set Theory, Smallest Elements, Order Theory
\begin{theorem}
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
Then the smallest set of $\TT$, if it exists, must be unique.
\end{theorem}
\begin{proof}
Let $A, B \in \TT$ both be smallest sets of $\TT$.
Since $A$ is the smallest set:
:$A \subseteq B$
Since $B$ is the smallest set:
:$B \subseteq A$
Hence, by definition of set equality:
:$A = B$
Therefore the smallest set of $\TT$ is unique.
{{qed}}
Category:Set Theory
Category:Smallest Elements
\end{proof}
|
21565
|
\section{Smallest Set may not Exist}
Tags: Set Theory, Smallest Elements, Order Theory
\begin{theorem}
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
The smallest set of $\TT$ may not exist.
\end{theorem}
\begin{proof}
Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$.
Then since $\set 0 \nsubseteq \set 1$:
:$\set 0$ is not the smallest set of $\TT$.
Similarly, since $\set 1 \nsubseteq \set 0$:
:$\set 1$ is not the smallest set of $\TT$.
Therefore $\TT$ has no smallest set.
{{qed}}
Category:Set Theory
Category:Smallest Elements
\end{proof}
|
21566
|
\section{Smallest Simple Graph with One Vertex Adjacent to All Others}
Tags: Graph Theory
\begin{theorem}
Let $G = \struct {V, E}$ be a simple graph of order $n$.
Let $G$ be the simple graph with the smallest size such that one vertex is adjacent to all other vertices of $G$.
Then $G$ is the star graph of order $n$ and is of size $n - 1$.
\end{theorem}
\begin{proof}
In order for the one vertex in question to be adjacent to all the others, it needs to be incident to $n - 1$ edges.
This is the smallest number of edges required.
Hence $G$ is a star graph.
From Size of Star Graph, the size of $G$ is thus $n - 1$.
{{qed}}
\end{proof}
|
21567
|
\section{Smallest Square Inscribed in Two Pythagorean Triangles}
Tags: Pythagorean Triangles
\begin{theorem}
The smallest square with integer sides that can be inscribed within two different Pythagorean triangles so that one side of the square lies on the hypotenuse has side length $780$.
The two Pythagorean triangles in question have side lengths $\tuple {1443, 1924, 2405}$ and $\tuple {1145, 2748, 2977}$.
\end{theorem}
\begin{proof}
By Inscribed Squares in Right-Angled Triangle/Side Lengths/Side Lies on Hypotenuse:
:For a Pythagorean triangle with side lengths $a, b, c$, the required inscribed square has side length given by:
::$\dfrac {abc}{ab + c^2}$
For primitive Pythagorean triples, $a, b, c$ are pairwise coprime, so the above fraction is in canonical form.
In other words, if the required side length is an integer, the triangle cannot be primitive, and this side length would be equal to some multiple of the product $abc$ of its primitive version.
Therefore in order to find the smallest such square, we would need to compare and find two sets of primitive Pythagorean triples such that their product would have a sufficiently small lowest common multiple.
The $\tuple {3, 4, 5}$ triple has a product of $60$.
The $\tuple {5, 12, 13}$ triple has a product of $780$.
These two products have a lowest common multiple of $780$.
Since $\sqrt [3] {780} < 10$, there is no need to search any further for triangles with smallest side length greater than $10$.
The ones remaining are:
:$7, 24, 25$
:$8, 15, 17$
:$9, 40, 41$
which all have products greater than $780$.
Therefore the solution must be generated with the two smallest triangles.
The inscribed square side length for the $\tuple {3, 4, 5}$ triangle is:
:$\dfrac {60} {3 \times 4 + 5^2} = \dfrac {60} {37}$
so it must be enlarged $481$ times to have a side length of $780$.
The inscribed square side length for the $\tuple {5, 12, 13}$ triangle is:
:$\dfrac {780} {5 \times 12 + 13^2} = \dfrac {780} {229}$
so it must be enlarged $229$ times to have a side length of $780$.
After enlargement, we get the triples $\tuple {1443, 1924, 2405}$ and $\tuple {1145, 2748, 2977}$, which is the result we have.
{{qed}}
\end{proof}
|
21568
|
\section{Smallest Square which is Sum of 3 Fourth Powers}
Tags: Fourth Powers, Square Numbers
\begin{theorem}
The smallest positive integer whose square is the sum of $3$ fourth powers is $481$:
:$481^2 = 12^4 + 15^4 + 20^4$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 12^4 + 15^4 + 20^4
| r = 20 \, 736 + 50 \, 625 + 160 \, 000
| c =
}}
{{eqn | r = 231 \, 361
| c =
}}
{{eqn | r = 481^2
| c =
}}
{{end-eqn}}
The smallest solution must be fourth powers of coprime integers, otherwise dividing by their GCD would yield a smaller solution.
By Fermat's Right Triangle Theorem, $x^4 + y^4 = z^2$ has no solutions.
Thus none of the fourth powers is $0$.
Consider the equation $x^4 + y^4 + z^4 = n^2$.
By Square Modulo 4, since fourth powers are squares, at most one of $x, y, z$ is odd.
If none of them are odd, the numbers are not coprime.
Therefore there is exactly one odd number.
{{qed|lemma}}
By Fermat's Little Theorem, for $3 \nmid a$:
:$a^2 \equiv 1 \pmod 3$
Therefore by Congruence of Powers:
:$a^4 \equiv 1^2 \equiv 1 \pmod 3$
For $3 \divides a$:
:$a^2 \equiv a^4 \equiv 0 \pmod 3$
Therefore we cannot have two of $x, y, z$ not divisible by $3$.
By our coprimality condition we cannot have $3 \divides x, y, z$.
Hence either $3 \nmid x, y, z$ or exactly one of $x, y, z$ is not divisible by $3$.
{{qed|lemma}}
Similarly, by Fermat's Little Theorem, for $5 \nmid a$:
:$a^4 \equiv 1 \pmod 5$
For $5 \divides a$:
:$a^4 \equiv 0 \pmod 5$
By Square Modulo 5:
:$a^2 \equiv 0, 1, 4 \pmod 5$
Therefore there is exactly one number not divisible by $5$.
{{qed|lemma}}
Since $\sqrt {481} < 22$, we only need to check fourth powers of numbers less than $22$, satisfying the criteria above.
Since there are exactly two multiples of $2$ and $5$ among $x, y, z$, at least one of them must be a multiple of $10$.
Suppose $5 \nmid x$, $5 \divides y$ and $10 \divides z$.
\end{proof}
|
21569
|
\section{Smallest Strictly Positive Rational Number does not Exist}
Tags: Rational Numbers
\begin{theorem}
There exists no smallest element of the set of strictly positive rational numbers.
\end{theorem}
\begin{proof}
{{AimForCont}} $x = \dfrac p q$ is the smallest strictly positive rational number.
By definition of strictly positive:
:$0 < \dfrac p q$
Let us calculate the mediant of $0$ and $\dfrac p q$:
:$\dfrac 0 1 < \dfrac {0 + p} {1 + q} < \dfrac p q$
The inequalities follow from Mediant is Between.
Thus $\dfrac p {1 + q}$ is a strictly positive rational number which is smaller than $\dfrac p q$.
Thus $\dfrac p q$ cannot be the smallest strictly positive rational number.
The result follows by Proof by Contradiction.
{{qed}}
\end{proof}
|
21570
|
\section{Smallest Strong Fibonacci Pseudoprime of Type I}
Tags: 443,372,888,629,441, Strong Fibonacci Pseudoprimes
\begin{theorem}
The smallest strong Fibonacci pseudoprime of type I is $443 \, 372 \, 888 \, 629 \, 441$.
\end{theorem}
\begin{proof}
Let $N := 443 \, 372 \, 888 \, 629 \, 441$, to save writing it out in full each time.
From the definition of a strong Fibonacci pseudoprime of type I:
{{:Definition:Strong Fibonacci Pseudoprime of Type I}}
We have that:
:$N = 17 \times 31 \times 41 \times 43 \times 89 \times 97 \times 167 \times 331$
Of these, we see that:
{{begin-eqn}}
{{eqn | l = 17
| r = 4 \times 4 + 1
| c = which is a prime number of the form $4 n + 1$
}}
{{eqn | l = 31
| r = 4 \times 8 - 1
| c = which is a prime number of the form $4 n - 1$
}}
{{eqn | l = 41
| r = 4 \times 10 + 1
| c = which is a prime number of the form $4 n + 1$
}}
{{eqn | l = 43
| r = 4 \times 11 - 1
| c = which is a prime number of the form $4 n - 1$
}}
{{eqn | l = 89
| r = 4 \times 22 + 1
| c = which is a prime number of the form $4 n + 1$
}}
{{eqn | l = 97
| r = 4 \times 24 + 1
| c = which is a prime number of the form $4 n + 1$
}}
{{eqn | l = 167
| r = 4 \times 42 - 1
| c = which is a prime number of the form $4 n - 1$
}}
{{eqn | l = 331
| r = 4 \times 83 - 1
| c = which is a prime number of the form $4 n - 1$
}}
{{end-eqn}}
Thus there are $4$ (an even number) of prime factors of $N$ of the form $4 n - 1$.
We note that:
{{begin-eqn}}
{{eqn | l = N - 1
| r = 288 \times 1 \, 539 \, 489 \, 196 \, 630
| c = and so $\paren {17^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 960 \times 461 \, 846 \, 758 \, 989
| c = and so $\paren {31^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 1680 \times 263 \, 912 \, 433 \, 708
| c = and so $\paren {41^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 1848 \times 239 \, 920 \, 394 \, 280
| c = and so $\paren {43^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 7920 \times 55 \, 981 \, 425 \, 332
| c = and so $\paren {89^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 9408 \times 47 \, 127 \, 220 \, 305
| c = and so $\paren {97^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 27 \, 888 \times 15 \, 898 \, 339 \, 380
| c = and so $\paren {167^2 - 1} \divides \paren {N - 1}$
}}
{{eqn | r = 109 \, 560 \times 4 \, 046 \, 850 \, 024
| c = and so $\paren {331^2 - 1} \divides \paren {N - 1}$
}}
{{end-eqn}}
Thus for all $p \divides N$, we have that $\paren {p^2 - 1} \divides \paren {N - 1}$.
From Difference of Two Squares we have that:
:$p^2 - 1 = \paren {p + 1} \paren {p - 1}$
and so for all $p \divides N$:
:$\paren {p - 1} \divides \paren {N - 1}$
We also have that $N$ is square-free.
Thus by Korselt's Theorem, $N$ is a Carmichael number.
Now we also have that:
{{begin-eqn}}
{{eqn | l = \paren {p^2 - 1}
| o = \divides
| r = \paren {N - 1}
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {p + 1} \paren {p - 1}
| o = \divides
| r = \paren {N - 1}
| c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
| l = 2 \paren {p + 1} \paren {\frac {p - 1} 2}
| o = \divides
| r = \paren {N - 1}
| c = as $p - 1$ is even
}}
{{eqn | ll= \leadsto
| l = 2 \paren {p + 1}
| o = \divides
| r = \paren {N - 1}
| c =
}}
{{end-eqn}}
Thus we have that:
:$\forall p \divides N: 2 \paren {p + 1} \divides \paren {N - 1}$
This is actually stronger than the conditions for which $N$ is a strong Fibonacci pseudoprime of type I:
{{:Definition:Strong Fibonacci Pseudoprime of Type I}}
It can be established by an exhaustive search that there are no smaller Carmichael numbers with this property.
Hence the result.
{{qed}}
\end{proof}
|
21571
|
\section{Smallest Titanic Palindromic Prime}
Tags: Prime Numbers
\begin{theorem}
The smallest titanic prime that is also palindromic is:
:$10^{1000} + 81 \, 918 \times 10^{498} + 1$
which can be written as:
:$1 \underbrace {000 \ldots 000}_{497} 81918 \underbrace {000 \ldots 000}_{497} 1$
\end{theorem}
\begin{proof}
{{Alpertron-factorizer|date = $6$th March $2022$|time = $1.7$ seconds}}
It remains to be demonstrated that it is the smallest such palindromic prime with $1000$ digits or more.
By 11 is Only Palindromic Prime with Even Number of Digits, there are no palindromic primes with exactly $1000$ digits.
Hence such a prime must be greater than $10^{1000}$.
We need to check all numbers of the form:
:$1 \underbrace {000 \ldots 000}_{497} abcba \underbrace {000 \ldots 000}_{497} 1$
with $\sqbrk {abc} < 819$.
Using the [https://www.alpertron.com.ar/ECM.HTM Alpertron integer factorization calculator] and the argument:
x=0;x=x+1;x<820;10^1000+x*10^500+RevDigits(x/10+10^499,10)
it is verified that there are no primes in the range $\sqbrk {abc} < 819$.
Therefore the number above is the smallest titanic palindromic prime.
{{qed}}
\end{proof}
|
21572
|
\section{Smallest Triple of Consecutive Sums of Squares}
Tags: Sums of Squares
\begin{theorem}
The smallest triple of consecutive positive integers each of which is the sum of two squares is:
:$\tuple {232, 233, 234}$
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = 232
| r = 14^2 + 6^2
| c =
}}
{{eqn | l = 233
| r = 13^2 + 8^2
| c =
}}
{{eqn | l = 234
| r = 15^2 + 3^2
| c =
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown this is the smallest such triple.}}
\end{proof}
|
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