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22773
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\section{Transplanting Theorem/Corollary}
Tags: Transplanting Theorem
\begin{theorem}
Let $\struct {S, \circ}$ be an algebraic structure.
Let $f: S \to S$ be an automorphism on $\struct {S, \circ}$.
Then the transplant of $\circ$ under $f$ is $\circ$ itself.
\end{theorem}
\begin{proof}
From the Transplanting Theorem there exists one and only one operation $\circ$ such that $f: \struct {S, \circ} \to \struct {S, \circ}$ is an automorphism.
{{qed}}
\end{proof}
|
22774
|
\section{Transpose of Linear Transformation is a Linear Transformation}
Tags: Linear Transformations
\begin{theorem}
Let $R$ be a commutative ring.
Let $G$ and $H$ be $R$-modules.
Let $G^*$ and $H^*$ be the algebraic duals of $G$ and $H$ respectively.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $u \in \map {\LL_R} {G, H}$.
Let $u^t: H^* \to G^*$ be the transpose of $u$.
Then $u^t: H^* \to G^*$ is itself a linear transformation.
\end{theorem}
\begin{proof}
By definition of evaluation linear transformation:
:$\forall x \in G: y \in H^*: \innerprod x {\map {u^t} y} = \innerprod {\map u x} y$
Since we have:
{{begin-eqn}}
{{eqn | l = \innerprod x {\map {u^t} {y + z} }
| r = \innerprod {\map u x} {y + z}
| c =
}}
{{eqn | r = \innerprod {\map u x} y + \innerprod {\map u x} z
| c =
}}
{{eqn | r = \innerprod x {\map {u^t} y} + \innerprod x {\map {u^t} z}
| c =
}}
{{eqn | r = \innerprod x {\map {u^t} y + \map {u^t} z}
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \innerprod x {\map {u^t} {\lambda y} }
| r = \innerprod {\map u x} {\lambda y}
| c =
}}
{{eqn | r = \lambda \innerprod {\map u x} y
| c =
}}
{{eqn | r = \lambda \innerprod x {\map {u^t} y}
| c =
}}
{{eqn | r = \innerprod x {\lambda \map {u^t} y}
| c =
}}
{{end-eqn}}
it follows that $u^t: H^* \to G^*$ is a linear transformation.
{{qed}}
\end{proof}
|
22775
|
\section{Transpose of Matrix Product}
Tags: Transpose Matrices, Matrix Algebra, Linear Algebra, Transposes of Matrices, Conventional Matrix Multiplication
\begin{theorem}
Let $\mathbf A$ and $\mathbf B$ be matrices over a commutative ring such that $\mathbf A \mathbf B$ is defined.
Then $\mathbf B^\intercal \mathbf A^\intercal$ is defined, and:
:$\paren {\mathbf A \mathbf B}^\intercal = \mathbf B^\intercal \mathbf A^\intercal$
where $\mathbf X^\intercal$ is the transpose of $\mathbf X$.
\end{theorem}
\begin{proof}
Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{n p}$
Let $\mathbf A \mathbf B = \sqbrk c_{m p}$.
Then from the definition of matrix product:
:$\ds \forall i \in \closedint 1 m, j \in \closedint 1 p: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
So, let $\paren {\mathbf A \mathbf B}^\intercal = \sqbrk r_{p m}$.
The dimensions are correct, because $\mathbf A \mathbf B$ is an $m \times p$ matrix, thus making $\paren {\mathbf A \mathbf B}^\intercal$ a $p \times m$ matrix.
Thus:
:$\ds \forall j \in \closedint 1 p, i \in \closedint 1 m: r_{j i} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
Now, let $\mathbf B^\intercal \mathbf A^\intercal = \sqbrk s_{p m}$
Again, the dimensions are correct because $\mathbf B^\intercal$ is a $p \times n$ matrix and $\mathbf A^\intercal$ is an $n \times m$ matrix.
Thus:
:$\ds \forall j \in \closedint 1 p, i \in \closedint 1 m: s_{j i} = \sum_{k \mathop = 1}^n b_{k j} \circ a_{i k}$
As the underlying structure of $\mathbf A$ and $\mathbf B$ is a commutative ring, then $a_{i k} \circ b_{k j} = b_{k j} \circ a_{i k}$.
Note the order of the indices in the term in the summation sign on the {{RHS}} of the above.
They are reverse what they would normally be because we are multiplying the transposes together.
Thus it can be seen that $r_{j i} = s_{j i}$ and the result follows.
{{qed}}
\end{proof}
|
22776
|
\section{Transpose of Row Matrix is Column Matrix}
Tags: Matrix Algebra, Transposes of Matrices
\begin{theorem}
Let $\mathbf x = \sqbrk x_{1 n} = \begin {bmatrix} x_1 & x_2 & \cdots & x_n \end {bmatrix}$ be a row matrix.
Then $\mathbf x^\intercal$, the transpose of $\mathbf x$, is a column matrix:
:$\begin {bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}^\intercal = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$
\end{theorem}
\begin{proof}
Self-evident.
{{Qed}}
\end{proof}
|
22777
|
\section{Transpose of Transpose of Matrix}
Tags: Matrix Algebra, Transposes of Matrices
\begin{theorem}
Let $\mathbf A$ be a matrix.
Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$.
Then:
:$\paren {\mathbf A^\intercal}^\intercal = \mathbf A$
\end{theorem}
\begin{proof}
Follows directly from the definition of the transpose of a matrix.
{{Qed}}
\end{proof}
|
22778
|
\section{Transpose of Upper Triangular Matrix is Lower Triangular}
Tags: Matrix Algebra, Triangular Matrices, Transposes of Matrices
\begin{theorem}
The transpose of an upper triangular matrix is a lower triangular matrix.
\end{theorem}
\begin{proof}
Let $\mathbf U = \sqbrk a_{m n}$ be an upper triangular matrix.
By definition:
:$\forall a_{i j} \in \mathbf U: i > j \implies a_{i j} = 0$
Let $\mathbf U^\intercal = \sqbrk b_{n m}$ be the transpose of $\mathbf U$.
That is:
:$\mathbf U^\intercal = \sqbrk b_{n m}: \forall i \in \closedint 1 n, j \in \closedint 1 n: b_{i j} = a_{j i}$
Thus:
:$\forall b_{j i} \in \mathbf U^\intercal: i > j \implies b_{j i} = 0$
By exchanging $i$ and $j$ in the notation of the above:
:$\forall b_{i j} \in \mathbf U^\intercal: i < j \implies b_{i j} = 0$
Thus by definition it is seen that $\mathbf U^\intercal$ is a lower triangular matrix.
{{qed}}
Category:Triangular Matrices
Category:Transposes of Matrices
\end{proof}
|
22779
|
\section{Transposition is Self-Inverse}
Tags: Symmetric Group, Symmetric Groups
\begin{theorem}
All transpositions are self-inverse.
\end{theorem}
\begin{proof}
Let $\pi = \begin{bmatrix} k_1 & k_2 \end{bmatrix}$ be a transposition.
Writing $\pi \pi$ in cycle notation gives:
:$\begin{bmatrix} k_1 & k_2 \end{bmatrix} \begin{bmatrix} k_1 & k_2 \end{bmatrix}$
from which we see that $k_1 \to k_2 \to k_1$ and $k_2 \to k_1 \to k_2$.
The result follows from the definition of self-inverse.
{{qed}}
Category:Symmetric Groups
\end{proof}
|
22780
|
\section{Transposition is of Odd Parity}
Tags: Symmetric Group, Symmetric Groups
\begin{theorem}
Let $S_n$ denote the set of permutations on $n$ letters.
Let $\pi \in S_n$ be a transposition.
Then $\pi$ is of odd parity.
\end{theorem}
\begin{proof}
Let $\pi = \begin{pmatrix} 1 & 2 \end{pmatrix}$ be a transposition.
Let $\Delta_n$ be defined as Product of Differences.
Then $\forall n \in \N_{>0}: \pi \cdot \Delta_n$ produces only one sign change in $\Delta_n$, that is, the one occurring in the factor $\paren {x_1 - x_2}$.
Thus:
:$\begin{pmatrix} 1 & 2 \end{pmatrix} \cdot \Delta_n = - \Delta_n$
and thus $\begin{pmatrix} 1 & 2 \end{pmatrix}$ is odd.
From Conjugates of Transpositions:
:$\begin{pmatrix} 1 & k \end{pmatrix} = \begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}$
Thus as $\begin{pmatrix} 2 & k \end{pmatrix}$ is self-inverse:
:$\begin{pmatrix} 1 & k \end{pmatrix} = \begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}^{-1}$
But from Parity of Conjugate of Permutation:
:$\map \sgn {\begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}^{-1} } = \sgn {\begin{pmatrix} 1 & 2 \end{pmatrix} }$
Thus:
:$\sgn {\begin{pmatrix} 1 & k \end{pmatrix} } = \sgn {\begin{pmatrix} 1 & 2 \end{pmatrix} }$
and thus $\begin{pmatrix} 1 & k \end{pmatrix}$ is odd.
Finally:
{{begin-eqn}}
{{eqn | l = \begin{pmatrix} h & k \end{pmatrix}
| r = \begin{pmatrix} 1 & h \end{pmatrix} \begin{pmatrix} 1 & k \end{pmatrix} \begin{pmatrix} 1 & h \end{pmatrix}
| c = Conjugates of Transpositions
}}
{{eqn | r = \begin{pmatrix} 1 & h \end{pmatrix} \begin{pmatrix} 1 & k \end{pmatrix} \begin{pmatrix} 1 & h \end{pmatrix}^{-1}
| c = Transposition is Self-Inverse
}}
{{end-eqn}}
But from Parity of Conjugate of Permutation:
:$\map \sgn {\begin{pmatrix} 1 & h \end{pmatrix} \begin{pmatrix} 1 & k \end{pmatrix} \begin{pmatrix} 1 & h \end{pmatrix}^{-1} } = \sgn {\begin{pmatrix} 1 & k \end{pmatrix} }$
Thus:
:$\sgn {\begin{pmatrix} h & k \end{pmatrix} } = \sgn {\begin{pmatrix} 1 & k \end{pmatrix} }$
and thus $\begin{pmatrix} h & k \end{pmatrix}$ is odd.
{{qed}}
\end{proof}
|
22781
|
\section{Tree has Center or Bicenter}
Tags: Tree Theory, Graph Theory, Trees
\begin{theorem}
Every tree has either:
: $(1): \quad$ Exactly one center
or:
: $(2): \quad$ Exactly one bicenter,
but never both.
That is, every tree is either central or bicentral.
\end{theorem}
\begin{proof}
A tree whose order is $1$ or $2$ is already trivially central or bicentral.
Let $T$ be a tree of order at least $3$.
First we establish that the construction of a center or bicenter actually works.
From Finite Tree has Leaf Nodes, there are always at least two nodes of degree $1$ to be removed.0
By the Handshake Lemma, it is clear that $T$ must also have at least one node whose degree is greater than $1$:
:$\ds \sum_{i \mathop = 1}^n \map {\deg_G} {v_i} = 2 q$
where $q$ is the number of edges in $T$.
But $q = n-1$ from Size of Tree is One Less than Order.
So if each node has degree $1$, then $n = 2 \paren {n - 1}$ and so $n = 2$.
Therefore if $n > 2$ there must be at least one node in $T$ of degree is greater than $1$.
Next, from Connected Subgraph of Tree is Tree, after having removed those nodes, what is left is still a tree.
Therefore the construction is valid.
We need to show the following:
:$(1): \quad T$ has only one center or bicenter
:$(2): \quad $T$ has either a center or a bicenter.
Suppose $T$ has more than one center or bicenter.
It would follow that at least one of the iterations constructing the center or bicenter disconnects $T$ into more than one component.
That could only happen if we were to remove an edge between two nodes of degree greater than $1$.
Hence $T$ has at most one center or bicenter.
Now to show that $T$ has at least one center or bicenter.
The proof works by the Principle of Complete Induction.
We know that a tree whose order is $1$ or $2$ is already trivially central or bicentral. This is our base case.
Suppose that all tree whose order is $n$ have at most one center or bicenter. This is our induction hypothesis.
Take a tree $T$ whose order is $n+1$ where $n > 2$.
Let $T$ have $k$ nodes of degree $1$.
We remove all these $k$ nodes.
This leaves us with a tree with $n + 1 - k$ nodes.
As we have seen that $T$ has at least one node whose degree is greater than $1$, $n + 1 - k \ge 1$.
As there are always at least two nodes of degree $1$, $n+1-k \le n-1$.
So after the first iteration, we are left with a tree whose order is between $1$ and $n-1$ inclusive.
By the induction hypothesis, this tree has either a center or bicenter.
The result follows by the Principle of Complete Induction.
{{qed}}
Category:Tree Theory
\end{proof}
|
22782
|
\section{Trefoil Knot is Homeomorphic to Circle}
Tags: Examples of Homeomorphisms
\begin{theorem}
The trefoil knot is homeomorphic to the circle.
\end{theorem}
\begin{proof}
Despite the fact that you cannot actually rearrange a trefoil knot actually into a circle in the usual $\R^3$ space, you can set up a mappping from one to the other.
{{finish}}
\end{proof}
|
22783
|
\section{Triangle Angle-Side-Angle Equality}
Tags: Triangles, Proofs by Contradiction
\begin{theorem}
If two triangles have:
:two angles equal to two angles, respectively
:the sides between the two angles equal
then the remaining angles are equal, and the remaining sides equal the respective sides.
That is to say, if two pairs of angles and the included sides are equal, then the triangles are congruent.
{{:Euclid:Proposition/I/26}}
\end{theorem}
\begin{proof}
:400px
Let $\angle ABC = \angle DEF$, $\angle BCA = \angle EFD$, and $BC = EF$.
{{AimForCont}} that $AB \ne DE$.
If this is the case, one of the two must be greater.
{{WLOG}}, we let $AB > DE$.
We construct a point $G$ on $AB$ such that $BG = ED$.
Using Euclid's first postulate, we construct the segment $CG$.
Now, since we have:
:$BG = ED$
:$\angle GBC = \angle DEF$
:$BC = EF$
it follows from Triangle Side-Angle-Side Equality that:
:$\angle GCB = \angle DFE$
But from Euclid's fifth common notion:
:$\angle DFE = \angle ACB > \angle GCB$
which is a contradiction.
Therefore, $AB = DE$.
So from Triangle Side-Angle-Side Equality:
:$\triangle ABC = \triangle DEF$
{{qed}}
{{Euclid Note|26|I|It appears to have originally been created by {{AuthorRef|Thales of Miletus}}.|part=first}}
\end{proof}
|
22784
|
\section{Triangle Conjugacy is Mutual}
Tags: Definitions: Polars of Points, Conjugate Triangles
\begin{theorem}
Let $\CC$ be a circle.
Let $\triangle PQR$ be a triangle.
Let $\triangle P'Q'R'$ be such that:
:$P'$ is the pole of $QR$
:$Q'$ is the pole of $PR$
:$R'$ is the pole of $PQ$
with respect to $\CC$.
Then:
:$P$ is the pole of $Q'R'$
:$Q$ is the pole of $P'R'$
:$R$ is the pole of $P'Q'$
with respect to $\CC$.
That is, $\triangle PQR$ and $\triangle P'Q'R'$ are '''conjugate triangles with respect to $\CC$'''.
</onlyinclude>
\end{theorem}
\begin{proof}
We have that:
:the polar of $P'$ is $QR$
:the polar of $Q'$ is $PR$
and so both polars pass through $R$.
Therefore:
:the polar of $R$ is $P'Q'$.
Similarly:
:the polar of $P$ is $Q'R'$
and:
:the polar of $Q$ is $P'R'$.
{{qed}}
\end{proof}
|
22785
|
\section{Triangle Inequality/Complex Numbers}
Tags: Complex Analysis, Triangle Inequality, Complex Modulus, Named Theorems
\begin{theorem}
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ denote the modulus of $z$.
Then:
:$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$
\end{theorem}
\begin{proof}
Let $z_1 = a_1 + i a_2, z_2 = b_1 + i b_2$.
Then from the definition of the modulus, the above equation translates into:
:$\paren {\paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2}^{\frac 1 2} \le \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} + \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}$
This is a special case of Minkowski's Inequality, with $n = 2$.
{{qed}}
\end{proof}
|
22786
|
\section{Triangle Inequality/Complex Numbers/Corollary 1}
Tags: Complex Modulus
\begin{theorem}
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\cmod z$ be the modulus of $z$.
Then:
: $\cmod {z_1 + z_2} \ge \cmod {z_1} - \cmod {z_2}$
\end{theorem}
\begin{proof}
Let $z_3 := z_1 + z_2$.
Then:
{{begin-eqn}}
{{eqn | l = \cmod {z_3} + \cmod {\paren {-z_2} }
| o = \ge
| r = \cmod {z_3 + \paren {-z_2} }
| c = Triangle Inequality for Complex Numbers
}}
{{eqn | ll= \leadsto
| l = \cmod {z_3} + \cmod {z_2}
| o = \ge
| r = \cmod {z_3 - z_2}
| c = Complex Modulus of Additive Inverse
}}
{{eqn | ll= \leadsto
| l = \cmod {z_1 + z_2} + \cmod {z_2}
| o = \ge
| r = \cmod {z_1}
| c = substituting $z_3 = z_1 + z_2$
}}
{{eqn | ll= \leadsto
| l = \cmod {z_1 + z_2}
| o = \ge
| r = \cmod {z_1} - \cmod {z_2}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22787
|
\section{Triangle Inequality/Complex Numbers/General Result}
Tags: Triangle Inequality, Complex Modulus, Complex Analysis
\begin{theorem}
Let $z_1, z_2, \dotsc, z_n \in \C$ be complex numbers.
Let $\cmod z$ be the modulus of $z$.
Then:
:$\cmod {z_1 + z_2 + \dotsb + z_n} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_n}$
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\cmod {z_1 + z_2 + \dotsb + z_n} \le \cmod {z_1} + \cmod {z_2} + \dotsb + \cmod {z_n}$
$\map P 1$ is true by definition of the usual ordering on real numbers:
:$\cmod {z_1} \le \cmod {z_1}$
\end{proof}
|
22788
|
\section{Triangle Inequality/Examples/4 Points}
Tags: Examples of Triangle Inequality
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $x, y, z, t \in A$.
Then:
:$\map d {x, z} + \map d {y, t} \ge \size {\map d {x, y} - \map d {z, t} }$
\end{theorem}
\begin{proof}
We have that $\map d {x, z}$, $\map d {y, t}$, $\map d {x, y}$, $\map d {z, t}$ are themselves all real numbers.
Hence the Euclidean metric on the real number line can be applied:
{{begin-eqn}}
{{eqn | l = \size {\map d {x, y} - \map d {z, t} }
| o = \le
| r = \size {\map d {x, y} - \map d {y, z} } + \size {\map d {y, z} - \map d {z, t} }
| c =
}}
{{eqn | o = \le
| r = \map d {x, z} + \map d {y, t}
| c = Reverse Triangle Inequality twice
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22789
|
\section{Triangle Inequality/Vectors in Euclidean Space}
Tags: Triangle Inequality, Named Theorems, Linear Algebra
\begin{theorem}
Let $\mathbf x, \mathbf y$ be vectors in $\R^n$.
Let $\norm {\, \cdot \,}$ denote vector length.
Then:
:$\norm {\mathbf x + \mathbf y} \le \norm {\mathbf x} + \norm {\mathbf y}$
If the two vectors are scalar multiples where said scalar is non-negative, an equality holds:
:$\exists \lambda \in \R, \lambda \ge 0: \mathbf x = \lambda \mathbf y \iff \norm {\mathbf x + \mathbf y} = \norm {\mathbf x} + \norm {\mathbf y}$
\end{theorem}
\begin{proof}
Let $\mathbf x, \mathbf y \in \R^n$.
We have:
{{begin-eqn}}
{{eqn | l = \norm {\mathbf x + \mathbf y}^2
| r = \paren {\mathbf x + \mathbf y} \cdot \paren {\mathbf x + \mathbf y}
| c = Dot Product of Vector with Itself
}}
{{eqn | r = \mathbf x \cdot \mathbf x + \mathbf x \cdot \mathbf y + \mathbf y \cdot \mathbf x + \mathbf y \cdot \mathbf y
| c = Dot Product Distributes over Addition
}}
{{eqn | r = \mathbf x \cdot \mathbf x + 2 \paren {\mathbf x \cdot \mathbf y} + \mathbf y \cdot \mathbf y
| c = Dot Product Operator is Commutative
}}
{{eqn | r = \norm {\mathbf x}^2 + 2 \paren {\mathbf x \cdot \mathbf y} + \norm {\mathbf y}^2
| c = Dot Product of Vector with Itself
}}
{{end-eqn}}
From the Cauchy-Bunyakovsky-Schwarz Inequality:
{{begin-eqn}}
{{eqn | l = \size {\mathbf x \cdot \mathbf y}
| o = \le
| r = \norm {\mathbf x} \norm {\mathbf y}
}}
{{eqn | ll= \leadsto
| l = \mathbf x \cdot \mathbf y
| o = \le
| r = \norm {\mathbf x} \norm {\mathbf y}
| c = Negative of Absolute Value
}}
{{eqn | l = \norm {\mathbf x}^2 + 2 \paren {\mathbf x \cdot \mathbf y} + \norm {\mathbf y}^2
| o = \le
| r = \norm {\mathbf x}^2 + 2 \paren {\norm {\mathbf x} \norm {\mathbf y} } + \norm {\mathbf y}^2
}}
{{eqn | r = \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2
}}
{{eqn | ll= \leadsto
| l = \norm {\mathbf x + \mathbf y}^2
| o = \le
| r = \paren {\norm {\mathbf x} + \norm {\mathbf y} }^2
}}
{{eqn | ll= \leadsto
| l = \norm {\mathbf x + \mathbf y}
| o = \le
| r = \norm {\mathbf x} + \norm {\mathbf y}
| c = taking the square root of both sides
}}
{{end-eqn}}
{{qed}}
To prove that the equality holds if the vectors are scalar multiples of each other, assume:
:$\exists \lambda \in \R, \lambda \ge 0: \mathbf v = \lambda \mathbf w$
\end{proof}
|
22790
|
\section{Triangle Inequality for Contour Integrals}
Tags: Complex Analysis, Named Theorems, Contour Integration, Triangle Inequality
\begin{theorem}
Let $C$ be a contour.
Let $f: \Img C \to \C$ be a continuous complex function, where $\Img C$ denotes the image of $C$.
Then:
:$\ds \size {\int_C \map f z \rd z} \le \max_{z \mathop \in \Img C} \size {\map f z} \map L C$
where $\map L C$ denotes the length of $C$.
\end{theorem}
\begin{proof}
By definition of contour, $C$ is a concatenation of a finite sequence $C_1, \ldots, C_n$ of directed smooth curves.
Let $C_i$ be parameterized by the smooth path $\gamma_i: \closedint {a_i} {b_i} \to \C$ for all $i \in \set {1, \ldots, n}$.
Then:
{{begin-eqn}}
{{eqn | l = \size {\int_C \map f z \rd z}
| r = \size {\sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t}
| c = {{Defof|Complex Contour Integral}}
}}
{{eqn | o = \le
| r = \sum_{i \mathop = 1}^n \size {\int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t}
| c = Triangle Inequality for Complex Numbers
}}
{{eqn | o = \le
| r = \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \size {\map f {\map {\gamma_i} t} } \size {\map {\gamma_i'} t} \rd t
| c = Modulus of Complex Integral
}}
{{eqn | o = \le
| r = \sum_{i \mathop = 1}^n \max_{t \mathop \in \closedint {a_i} {b_i} } \size {\map f {\map {\gamma_i} t} } \int_{a_i}^{b_i} \size {\map {\gamma_i'} t} \rd t
| c = Linear Combination of Integrals
}}
{{eqn | o = \le
| r = \sum_{i \mathop = 1}^n \max_{z \mathop \in \Img C} \size {\map f z} \int_{a_i}^{b_i} \size {\map {\gamma_i'} t} \rd t
| c = as $\map {\gamma_i} t \in \Img C$
}}
{{eqn | r = \max_{z \mathop \in \Img C} \size {\map f z} \map L C
| c = {{Defof|Length of Contour (Complex Plane)}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22791
|
\section{Triangle Inequality for Generalized Sums}
Tags: Triangle Inequality, Banach Spaces, Generalized Sums
\begin{theorem}
Let $V$ be a Banach space.
Let $\norm {\,\cdot\,}$ denote the norm on $V$.
Let $\family {v_i}_{i \mathop \in I}$ be an indexed subset of $V$.
Let the generalized sum $\ds \sum \set {v_i: i \in I}$ converge absolutely.
Then:
:$(1): \quad \ds \norm {\sum \set {v_i: i \in I} } \le \sum \set {\norm {v_i}: i \in I}$
\end{theorem}
\begin{proof}
First of all, note that Absolutely Convergent Generalized Sum Converges assures us that the {{LHS}} in $(1)$ is defined.
{{AimForCont}} there exists an $\epsilon > 0$ such that:
:$\ds \norm {\sum \set {v_i: i \in I} } > \sum \set {\norm {v_i}: i \in I} + \epsilon$
This supposition is seen to be equivalent to:
:$\ds \norm {\sum \set {v_i: i \in I} } > \sum \set {\norm {v_i}: i \in I}$
{{explain|Expand below sentence to be more clear}}
Then, by definition of a generalized sum, there necessarily exists a finite subset $F$ of $I$ with:
:$\ds \norm {\sum_{i \mathop \in F} v_i} > \norm {\sum \set {v_i: i \in I} } - \epsilon > \sum \set {\norm {v_i}: i \in I}$
However, using the standard triangle equality on this ''finite'' sum (that is, {{NormAxiomMult|3}}, repetitively), we also have:
:$\ds \norm {\sum_{i \mathop \in F} v_i} \le \sum_{i \mathop \in F} \norm {v_i} \le \sum \set {\norm {v_i}: i \in I}$
Here the second inequality follows from Generalized Sum is Monotone.
These two estimates constitute a contradiction, and therefore such an $\epsilon$ cannot exist.
Hence:
:$\ds \norm {\sum \set {v_i: i \in I} } \le \sum \set {\norm {v_i}: i \in I}$
{{qed}}
Category:Generalized Sums
Category:Banach Spaces
Category:Triangle Inequality
\end{proof}
|
22792
|
\section{Triangle Inequality for Indexed Summations}
Tags: Triangle Inequality, Triangle Inequalities, Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a,b$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Let $f : \closedint a b \to \mathbb A$ be a mapping.
Let $\size {\, \cdot \,}$ denote the standard absolute value.
Let $\vert f \vert$ be the absolute value of $f$.
Then we have the inequality of indexed summations:
:$\ds \size {\sum_{i \mathop = a}^b \map f i} \le \sum_{i \mathop = a}^b \size {\map f i}$
\end{theorem}
\begin{proof}
The proof goes by induction on $b$.
\end{proof}
|
22793
|
\section{Triangle Inequality for Integrals}
Tags: Measure Theory, Triangle Inequality, Triangle Inequality for Integrals
\begin{theorem}
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\mu$-integrable function.
Then:
:$\ds \size {\int_X f \rd \mu} \le \int_X \size f \rd \mu$
\end{theorem}
\begin{proof}
Let $\ds z = \int_X f \rd \mu \in \C$.
By Complex Multiplication as Geometrical Transformation, there is a complex number $\alpha$ with $\cmod \alpha = 1$ such that:
:$\alpha z = \cmod z \in \R$
Let $u = \map \Re {\alpha f}$, where $\Re$ denotes the real part of a complex number.
By Modulus Larger than Real Part, we have that:
:$u \le \cmod {\alpha f} = \cmod f$
Thus we get the inequality:
{{begin-eqn}}
{{eqn | l = \cmod {\int_X f \rd \mu}
| r = \alpha \int_X f \rd \mu
| c =
}}
{{eqn | r = \int_X \alpha f \rd \mu
| c = Integral of Integrable Function is Homogeneous
}}
{{eqn | r = \int_X u \rd \mu
| c =
}}
{{eqn | o = \le
| r = \int_X \cmod f \rd \mu
| c = Integral of Integrable Function is Monotone
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22794
|
\section{Triangle Inequality for Summation over Finite Set}
Tags: Triangle Inequality, Summations
\begin{theorem}
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f : S \to \mathbb A$ be a mapping.
Let $\size {\, \cdot\,}$ denote the standard absolute value.
Let $\size f$ be the absoute value of $f$.
Then we have the inequality of summations on finite sets:
:$\ds \size {\sum_{s \mathop \in S} \map f s} \le \sum_{s \mathop \in S} \size {\map f s}$
\end{theorem}
\begin{proof}
Let $n$ be the cardinality of $S$.
Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{<n}$ is an initial segment of the natural numbers.
By definition of summation, we have to prove the following inequality of indexed summations:
:$\ds \size {\sum_{i \mathop = 0}^{n - 1} \map f {\map \sigma i} } \le \sum_{i \mathop = 0}^{n - 1} \map {\paren {\size f \circ \sigma} } i$
By Absolute Value of Mapping Composed with Mapping:
:$\size f \circ \sigma = \size {f \circ \sigma}$
The above equality now follows from Triangle Inequality for Indexed Summations.
{{qed}}
\end{proof}
|
22795
|
\section{Triangle Inequality on Distance from Point to Subset}
Tags: Triangle Inequality, Distance Function
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$.
Then:
:$\forall x, y \in A: \map d {x, H} \le \map d {x, y} + \map d {y, H}$
where $\map d {x, H}$ denotes the distance between $x$ and $H$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | q = \forall z \in H
| l = \map d {y, z}
| o = \ge
| r = \map d {x, z} - \map d {x, y}
| c = {{Metric-space-axiom|2}}
}}
{{eqn | ll= \leadsto
| q = \forall z \in H
| l = \map d {y, z}
| o = \ge
| r = \map d {x, H} - \map d {x, y}
| c = {{Defof|Distance between Element and Subset of Metric Space}}
}}
{{eqn | ll= \leadsto
| l = \map d {y, H}
| o = \ge
| r = \map d {x, H} - \map d {x, y}
| c = {{Defof|Distance between Element and Subset of Metric Space}}
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22796
|
\section{Triangle Right-Angle-Hypotenuse-Side Equality}
Tags: Triangles, Triangle, Right Triangles
\begin{theorem}
If two triangles have:
: one right angle each
: the sides opposite to the right angle equal
: another two respective sides equal
they will also have:
: their third sides equal
: the remaining two angles equal to their respective remaining angles.
\end{theorem}
\begin{proof}
Let $\triangle ABC$ and $\triangle DEF$ be two triangles having sides $AB = DE$ and $AC = DF$, and with $\angle ABC = \angle DEF = 90^\circ$.
By Pythagoras' Theorem:
: $BC = \sqrt {AB^2 + AC^2}$
and:
: $EF = \sqrt {DE^2 + DF^2}$
:$\therefore BC = \sqrt {AB^2 + AC^2} = \sqrt {DE^2 + DF^2} = EF$
The part that the remaining two angles are equal to their respective remaining angles follows from Triangle Side-Side-Side Equality.
{{qed}}
\end{proof}
|
22797
|
\section{Triangle Side-Angle-Angle Equality}
Tags: Triangles, Proofs by Contradiction
\begin{theorem}
If two triangles have:
:two angles equal to two angles, respectively
:the sides opposite one pair of equal angles equal
then the remaining angles are equal, and the remaining sides equal the respective sides.
That is to say, if two pairs of angles and a pair of opposite sides are equal, then the triangles are congruent.
{{:Euclid:Proposition/I/26}}
\end{theorem}
\begin{proof}
:360px
Let:
:$\angle ABC = \angle DEF$
:$\angle BCA = \angle EFD$
:$AB = DE$
{{AimForCont}} that $BC \ne EF$.
If this is the case, one of the two must be greater.
{{WLOG}}, let $BC > EF$.
We construct a point $H$ on $BC$ such that $BH = EF$, and then we construct the segment $AH$.
Now, since we have:
:$BH = EF$
:$\angle ABH = \angle DEF$
:$AB = DE$
from Triangle Side-Angle-Side Equality we have:
:$\angle BHA = \angle EFD$
But from External Angle of Triangle Greater than Internal Opposite, we have:
:$\angle BHA > \angle HCA = \angle EFD$
which is a contradiction.
Therefore $BC = EF$.
So from Triangle Side-Angle-Side Equality:
:$\triangle ABC = \triangle DEF$
{{qed}}
{{Euclid Note|26|I|part = second}}
\end{proof}
|
22798
|
\section{Triangle Side-Angle-Side Equality}
Tags: Triangles, Euclid Book I
\begin{theorem}
If $2$ triangles have:
: $2$ sides equal to $2$ sides respectively
: the angles contained by the equal straight lines equal
they will also have:
: their third sides equal
: the remaining two angles equal to their respective remaining angles, namely, those which the equal sides subtend.
{{:Euclid:Proposition/I/4}}
\end{theorem}
\begin{proof}
500px
Let $\triangle ABC$ and $\triangle DEF$ be $2$ triangles having sides $AB = DE$ and $AC = DF$, and with $\angle BAC = \angle EDF$.
If $\triangle ABC$ is placed on $\triangle DEF$ such that:
: the point $A$ is placed on point $D$, and
: the line $AB$ is placed on line $DE$
then the point $B$ will also coincide with point $E$ because $AB = DE$.
So, with $AB$ coinciding with $DE$, the line $AC$ will coincide with the line $DF$ because $\angle BAC = \angle EDF$.
Hence the point $C$ will also coincide with the point $F$, because $AC = DF$.
But $B$ also coincided with $E$.
Hence the line $BC$ will coincide with line $EF$.
(Otherwise, when $B$ coincides with $E$ and $C$ with $F$, the line $BC$ will not coincide with line $EF$ and two straight lines will enclose a region which is impossible.)
Therefore $BC$ will coincide with $EF$ and be equal to it.
Thus the whole $\triangle ABC$ will coincide with the whole $\triangle DEF$ and thus $\triangle ABC = \triangle DEF$.
The remaining angles on $\triangle ABC$ will coincide with the remaining angles on $\triangle DEF$ and be equal to them.
{{qed}}
{{Euclid Note|4|I}}
\end{proof}
|
22799
|
\section{Triangle Side-Side-Side Equality}
Tags: Triangles, Euclid Book I
\begin{theorem}
Let two triangles have all $3$ sides equal.
Then they also have all $3$ angles equal.
Thus two triangles whose sides are all equal are themselves congruent.
{{:Euclid:Proposition/I/8}}
\end{theorem}
\begin{proof}
:500px
Let $\triangle ABC$ and $\triangle DEF$ be two triangles such that:
: $AB = DE$
: $AC = DF$
: $BC = EF$
Suppose $\triangle ABC$ were superimposed over $\triangle DEF$ so that point $B$ is placed on point $E$ and the side $BC$ on $EF$.
Then $C$ will coincide with $F$, as $BC = EF$ and so $BC$ coincides with $EF$.
{{AimForCont}} $BA$ does not coincide with $ED$ and $AC$ does not coincide with $DF$.
Then they will fall as, for example, $EG$ and $GF$.
Thus there will be two pairs of straight line segments constructed on the same line segment, on the same side as it, meeting at different points.
This contradicts the theorem Two Lines Meet at Unique Point.
Therefore $BA$ coincides with $ED$ and $AC$ coincides with $DF$.
Therefore $\angle BAC$ coincides with $\angle EDF$ and is equal to it.
The same argument can be applied to the other two sides, and thus we show that all corresponding angles are equal.
{{qed}}
{{Euclid Note|8|I}}
\end{proof}
|
22800
|
\section{Triangle is Convex Set}
Tags: Vector Spaces
\begin{theorem}
The interior of a triangle embedded in $\R^2$ is a convex set.
\end{theorem}
\begin{proof}
Denote the triangle as $\triangle$, and the interior of the boundary of $\triangle$ as $\Int \triangle$.
From Boundary of Polygon is Jordan Curve, it follows that the boundary of $\triangle$ is equal to the image of a Jordan curve, so $\Int \triangle$ is well-defined.
Denote the vertices of $\triangle$ as $A_1, A_2, A_3$.
For $i \in \set {1, 2, 3}$, put $j = i \bmod 3 + 1$, $k = \paren {i + 1} \bmod 3 + 1$, and:
:$U_i = \set {A_i + s t \paren {A_j - A_i} + \paren {1 - s} t \paren {A_k - A_i} : s \in \openint 0 1, t \in \R_{>0} }$
Suppose that the angle $\angle A_i$ between is $A_j - A_i$ and $A_k - A_i$ is non-convex.
As $\angle A_i$ is an internal angle in $\triangle$, it follows from definition of polygon that $\angle A_i$ cannot be zero or straight.
Then $\angle A_i$ is larger than a straight angle, which is impossible by Sum of Angles of Triangle Equals Two Right Angles.
It follows that $\angle A_i$ is convex.
From Characterization of Interior of Triangle, it follows that:
:$\ds \Int \triangle = \bigcap_{i \mathop = 1}^3 U_i$
From Interior of Convex Angle is Convex Set, it follows for $i \in \set {1, 2, 3}$ that $U_i$ is a convex set.
The result now follows from Intersection of Convex Sets is Convex Set (Vector Spaces).
{{qed}}
Category:Vector Spaces
\end{proof}
|
22801
|
\section{Triangle is Medial Triangle of Larger Triangle}
Tags: Triangles, Medial Triangles
\begin{theorem}
Let $\triangle ABC$ be a triangle.
$\triangle ABC$ is the medial triangle of a larger triangle.
\end{theorem}
\begin{proof}
By {{EuclidPostulateLink|Fifth}}, it is possible to construct exactly one straight line parallel to each of $AC$, $BC$ and $AC$.
:500px
From Parallelism implies Equal Corresponding Angles:
:$\angle ABC = \angle ECB = \angle DAB$
:$\angle ACB = \angle CBE = \angle CAF$
:$\angle CAB = \angle ABD = \angle ACF$
By Triangle Angle-Side-Angle Equality:
:$\triangle ABC = \triangle ABD$
:$\triangle ABC = \triangle CFA$
:$\triangle ABC = \triangle ECB$
Thus:
:$FC = CE$
:$DB = BE$
:$FA = AD$
The result follows by definition of medial triangle.
{{qed}}
\end{proof}
|
22802
|
\section{Triangle with Two Equal Angles is Isosceles}
Tags: Triangles, Isosceles Triangles, Triangle with Two Equal Angles is Isosceles, Euclid Book I
\begin{theorem}
If a triangle has two angles equal to each other, the sides which subtend the equal angles will also be equal to one another.
Hence, by definition, such a triangle will be isosceles.
{{:Euclid:Proposition/I/6}}
\end{theorem}
\begin{proof}
:200px
Let $\triangle ABC$ be a triangle in which $\angle ABC = \angle ACB$.
Suppose side $AB$ is not equal to side $AC$. Then one of them will be greater.
Suppose $AB > AC$.
We cut off from $AB$ a length $DB$ equal to $AC$.
We draw the line segment $CD$.
Since $DB = AC$, and $BC$ is common, the two sides $DB, BC$ are equal to $AC, CB$ respectively.
Also, $\angle DBC = \angle ACB$.
So by Triangle Side-Angle-Side Equality, $\triangle DBC = \triangle ACB$.
But $\triangle DBC$ is smaller than $\triangle ACB$, which is absurd.
Therefore, have $AB \le AC$.
A similar argument shows the converse, and hence $AB = AC$.
{{qed}}
{{Euclid Note|6|I|It is the converse of Proposition $5$: Isosceles Triangle has Two Equal Angles.}}
\end{proof}
|
22803
|
\section{Triangles with Integer Area and Integer Sides in Arithmetical Sequence}
Tags: Triangles, Areas of Triangles
\begin{theorem}
The triangles with the following sides in arithmetic sequence have integer areas:
:$3, 4, 5$
:$13, 14, 15$
:$15, 28, 41$
:$15, 26, 37$
Their areas are:
:$6, 84, 126, 156$
\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 $\triangle ABC$.
For $3, 4, 5$:
{{begin-eqn}}
{{eqn | l = s
| r = \frac {3 + 4 + 5} 2
| c =
}}
{{eqn | r = 6
| c =
}}
{{eqn | ll= \leadsto
| l = A
| r = \sqrt {6 \paren {6 - 3} \paren {6 - 4} \paren {6 - 5} }
| c =
}}
{{eqn | r = \sqrt {6 \times 3 \times 2 \times 1}
| c =
}}
{{eqn | r = \sqrt {6 \times 6}
| c =
}}
{{eqn | r = 6
| c =
}}
{{end-eqn}}
For $13, 14, 15$:
{{begin-eqn}}
{{eqn | l = s
| r = \frac {13 + 14 + 15} 2
| c =
}}
{{eqn | r = 21
| c =
}}
{{eqn | ll= \leadsto
| l = A
| r = \sqrt {21 \paren {21 - 13} \paren {21 - 14} \paren {21 - 15} }
| c =
}}
{{eqn | r = \sqrt {21 \times 8 \times 7 \times 6}
| c =
}}
{{eqn | r = \sqrt {\paren {3 \times 7} \times 2^3 \times 7 \times \paren {2 \times 3} }
| c =
}}
{{eqn | r = 3 \times 7 \times 2^2
| c =
}}
{{eqn | r = 84
| c =
}}
{{end-eqn}}
For $15, 28, 41$:
{{begin-eqn}}
{{eqn | l = s
| r = \frac {15 + 28 + 41} 2
| c =
}}
{{eqn | r = 42
| c =
}}
{{eqn | ll= \leadsto
| l = A
| r = \sqrt {42 \paren {42 - 15} \paren {42 - 28} \paren {42 - 41} }
| c =
}}
{{eqn | r = \sqrt {42 \times 27 \times 14 \times 1}
| c =
}}
{{eqn | r = \sqrt {\paren {2 \times 3 \times 7} \times 3^3 \times \paren {2 \times 7} }
| c =
}}
{{eqn | r = 2 \times 3^2 \times 7
| c =
}}
{{eqn | r = 126
| c =
}}
{{end-eqn}}
For $15, 26, 37$:
{{begin-eqn}}
{{eqn | l = s
| r = \frac {15 + 26 + 37} 2
| c =
}}
{{eqn | r = 39
| c =
}}
{{eqn | ll= \leadsto
| l = A
| r = \sqrt {39 \paren {39 - 15} \paren {39 - 26} \paren {39 - 37} }
| c =
}}
{{eqn | r = \sqrt {39 \times 24 \times 13 \times 2}
| c =
}}
{{eqn | r = \sqrt {\paren {3 \times 13} \times \paren {2^3 \times 3} \times 13 \times 2}
| c =
}}
{{eqn | r = 2^2 \times 3 \times 13
| c =
}}
{{eqn | r = 156
| c =
}}
{{end-eqn}}
{{expand|The citation below generalises this result, and this page could be turned into it.}}
\end{proof}
|
22804
|
\section{Triangles with One Equal Angle and Two Other Sides Proportional are Similar}
Tags: Triangles
\begin{theorem}
Let two triangles be such that one of the angles of one triangle equals one of the angles of the other.
Let two corresponding sides which are adjacent to one of the other angles, be proportional.
Let the third angle in both triangles be either both acute or both not acute.
Then all of the corresponding angles of these triangles are equal.
Thus, by definition, such triangles are similar.
{{:Euclid:Proposition/VI/7}}
\end{theorem}
\begin{proof}
Let $\triangle ABC, \triangle DEF$ be triangles such that:
: $\angle BAC = \angle EDF$
: two sides adjacent to $\angle ABC$ and $\angle DEF$ proportional, so that $AB : BC = DE : EF$
: $\angle ACB$ and $\angle DFE$ either both acute or both not acute.
We need to show that $\angle ABC = \angle DEF$ and $\angle BCA = \angle EFD$.
First assume that $\angle ACB$ and $\angle DFE$ are both acute.
:350px
Suppose $\angle ABC \ne \angle DEF$.
Then one of them is greater, WLOG $\angle ABC$.
On $AB$ and at point $B$, let $\angle ABG$ be constructed equal to $\angle DEF$.
Since $\angle BAG = \angle EDF$ and $\angle ABG = \angle DEF$, from Sum of Angles of Triangle Equals Two Right Angles:
: $\angle AGB = \angle DFE$
So $\triangle ABG$ is equiangular with $\triangle DEF$.
So from Equiangular Triangles are Similar:
: $AB : BG = DE : EF$
But by hypothesis:
: $DE : EF = AB : BC$
Thus from Equality of Ratios is Transitive:
: $AB : BG = AB : BC$
So from Magnitudes with Same Ratios are Equal:
: $BC = BG$
So from Isosceles Triangle has Two Equal Angles:
: $\angle BCG = \angle BGC$
By hypothesis $\angle BCG$ is less than a right angle.
So $\angle BGC$ is less than a right angle.
So from Two Angles on Straight Line make Two Right Angles $\angle AGB$ is greater than a right angle.
But this was proved equal to $\angle DFE$.
But by hypothesis $\angle DFE$ is less than a right angle.
From this contradiction we conclude that:
: $\angle ABC = \angle DEF$
We also have that $\angle BAC = \angle DEF$.
So from Sum of Angles of Triangle Equals Two Right Angles:
: $\angle ACB = \angle DFE$
So, as we wanted to show, $\triangle ABC$ is equiangular with $\triangle DEF$.
Now, suppose $\angle ACB$ and $\angle DFE$ are both not acute:
:350px
From the same construction we show similarly that $BC = BG$ and so from Isosceles Triangle has Two Equal Angles, $\angle BCG = \angle BGC$.
But $\angle BCG$ is not less than a right angle.
So neither is $\angle BGC$ less than a right angle.
So in $\triangle BGC$, $\angle BCG + \angle BGC$ are not less than two right angles.
From Two Angles of Triangle Less than Two Right Angles, this is impossible.
So from this contradiction we conclude that:
: $\angle ABC = \angle DEF$
Hence, similarly to above, $\triangle ABC$ is equiangular with $\triangle DEF$.
{{qed}}
{{Euclid Note|7|VI}}
\end{proof}
|
22805
|
\section{Triangles with One Equal Angle and Two Sides Proportional are Similar}
Tags: Triangles
\begin{theorem}
Let two triangles have two corresponding sides which are proportional.
Let the angles adjacent to both of these sides be equal.
Then all of their corresponding angles are equal.
Thus, by definition, such triangles are similar.
{{:Euclid:Proposition/VI/6}}
\end{theorem}
\begin{proof}
Let $\triangle ABC, \triangle DEF$ be triangles having $\angle BAC = \angle EDF$ and two sides proportional, so that $BA : AC = ED : DF$.
We need to show that $\triangle ABC$ is equiangular with $\triangle DEF$, such that $\angle ABC = \angle DEF$ and $\angle ACB = \angle DFE$.
:350px
Let $\angle FDG$ be constructed equal to either $\angle BAC$ or $\angle EDF$, and $\angle DFG = \angle ACB$.
From Sum of Angles of Triangle Equals Two Right Angles:
: $\angle ABC = \angle DGF$
Therefore $\triangle ABC$ is equiangular with $\triangle DGF$.
From Equiangular Triangles are Similar:
: $BA : AC = GD : DF$
But by hypothesis:
: $BA : AC = ED : DF$
So from Equality of Ratios is Transitive:
: $ED : DF = GD : DF$
From Magnitudes with Same Ratios are Equal:
: $ED = DG$
But $DF$ is common to $\triangle DEF$ and $\triangle DGF$, and $\angle EDF = \angle GDF$.
So from Triangle Side-Angle-Side Equality:
: $\triangle DEF = \triangle DGF$
But $\angle DFG = \angle ACB$ and so:
: $\angle ABC = \angle DFE$
But by hypothesis:
: $\angle BAC = \angle EDF$
Therefore by Sum of Angles of Triangle Equals Two Right Angles:
: $\angle ABC = \angle DEF$
Hence the result.
{{qed}}
{{Euclid Note|6|VI}}
\end{proof}
|
22806
|
\section{Triangles with Proportional Sides are Similar}
Tags: Triangles
\begin{theorem}
Let two triangles have corresponding sides which are proportional.
Then their corresponding angles are equal.
Thus, by definition, such triangles are similar.
{{:Euclid:Proposition/VI/5}}
\end{theorem}
\begin{proof}
Let $\triangle ABC, \triangle DEF$ be triangles whose sides are proportional, so that:
:$ AB : BC = DE : EF$
:$ BC : CA = EF : FD$
:$ BA : AC = ED : DF$
We need to show that
: $\angle ABC = \angle DEF$
: $\angle BCA = \angle EFD$
: $\angle BAC = \angle EDF$
:400px
On the straight line $EF$, and at the points $E, F$ on it, construct $\angle FEG = \angle ABC$ and $\angle EFG = \angle ACB$.
From Sum of Angles of Triangle Equals Two Right Angles, the remaining angle at $A$ equals the remaining angle at $G$.
Therefore $\triangle ABC$ is equiangular with $\triangle GEF$.
From Equiangular Triangles are Similar, the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.
So:
: $AB : BD = GE : EF$
But by hypothesis:
: $AB : BC = DE : EF$
So from Equality of Ratios is Transitive
: $DE : EF = GE : EF$
So each of $DE, GE$ has the same ratio to $EF$.
So from Magnitudes with Same Ratios are Equal:
: $DE = GE$
For the same reason:
: $DF = GF$
So we have that $DE = EG$, $EF$ is common and $DF = FG$.
So from Triangle Side-Side-Side Equality:
: $\triangle DEF = \triangle GEF$
That is:
: $\angle DEF = \angle GEF, \angle DFE = \angle GFE, \angle EDF = \angle EGF$
As $\angle GEF = \angle ABC$ it follows that:
: $\angle ABC = \angle DEF$
For the same reason $\angle ACB = \angle DFE$ and $\angle BAC = \angle EDF$.
Hence the result.
{{Qed}}
{{Euclid Note|5|VI}}
\end{proof}
|
22807
|
\section{Triangles with Two Equal Angles are Similar}
Tags: Triangles, Euclidean Geometry
\begin{theorem}
Two triangles which have two corresponding angles which are equal are similar.
\end{theorem}
\begin{proof}
Let $\triangle ABC$ and $\triangle DEF$ be triangles such that $\angle ABC = \angle DEF$ and $\angle BAC = \angle EDF$.
Then from Sum of Angles of Triangle Equals Two Right Angles $\angle ACB$ is equal to two right angles minus $\angle ABC + \angle BAC$.
Also from Sum of Angles of Triangle Equals Two Right Angles $\angle DFE$ is equal to two right angles minus $\angle DEF + \angle EDF$.
That is, $\angle DFE$ is equal to two right angles minus $\angle ABC + \angle BAC$.
So $\angle DFE = \angle ACB$ and so all three corresponding angles of $\triangle ABC$ and $\triangle DEF$ are equal.
The result follows from Equiangular Triangles are Similar.
{{qed}}
\end{proof}
|
22808
|
\section{Triangular Fermat Number}
Tags: Fermat Numbers, Triangular Numbers
\begin{theorem}
The only one Fermat number which is triangular is $3$.
\end{theorem}
\begin{proof}
Let $F_n$ be a Fermat number which is triangular.
We use the weaker form of Divisor of Fermat Number:
:Every divisor of $F_n$ is of the form $k \times 2^{n + 1} + 1$
as this formulation is valid for all $n \in \N$.
From Closed Form for Triangular Numbers:
:$F_n = \dfrac {m \paren {m + 1} } 2$
Either $m$ or $m + 1$ is even.
Suppose $m$ is even.
Then:
{{begin-eqn}}
{{eqn | q = \exists k \in \N
| l = \frac m 2
| r = k \times 2^{n + 1} + 1
| c = Divisor of Fermat Number
}}
{{eqn | ll= \leadsto
| l = m + 1
| r = 2 \paren {k \times 2^{n + 1} + 1} + 1
}}
{{eqn | r = k \times 2^{n + 2} + 3
}}
{{eqn | q = \exists j \in \N
| l = m + 1
| r = j \times 2^{n + 1} + 1
| c = Divisor of Fermat Number
}}
{{eqn | ll= \leadsto
| l = k \times 2^{n + 2} + 2
| r = j \times 2^{n + 1}
}}
{{eqn | ll= \leadsto
| l = k \times 2^{n + 1} + 1
| r = j \times 2^n
}}
{{end-eqn}}
As the {{LHS}} is always odd, we must have $2^n = 1$.
This forces $n = 0$, with $F_n = 3$.
We check that $3$ is indeed a triangular number.
Suppose $m + 1$ is even.
Then:
{{begin-eqn}}
{{eqn | q = \exists k \in \N
| l = \frac {m + 1} 2
| r = k \times 2^{n + 1} + 1
| c = Divisor of Fermat Number
}}
{{eqn | ll= \leadsto
| l = m
| r = 2 \paren {k \times 2^{n + 1} + 1} - 1
}}
{{eqn | r = k \times 2^{n + 2} + 1
}}
{{eqn | l = F_n
| r = \frac {m \paren {m + 1} } 2
}}
{{eqn | ll= \leadsto
| l = 2^{2^n} + 1
| r = \paren {k \times 2^{n + 2} + 1} \paren {k \times 2^{n + 1} + 1}
}}
{{eqn | r = k^2 \times 2^{2 n + 3} + k \paren {2^{n + 2} + 2^{n + 1} } + 1
}}
{{eqn | ll= \leadsto
| l = 2^{2^n}
| r = k^2 \times 2^{2 n + 3} + k \paren {2^{n + 2} + 2^{n + 1} }
}}
{{eqn | r = k \times 2^{n + 1} \times \paren {k \times 2^{n + 2} + 3 }
}}
{{end-eqn}}
As $k \times 2^{n + 2} + 3 > 1$ and is odd, $k \times 2^{n + 2} + 3$ cannot be a power of $2$.
So $m + 1$ cannot be even, and we have exhausted all possibilities.
{{qed}}
\end{proof}
|
22809
|
\section{Triangular Fibonacci Numbers}
Tags: Fibonacci Numbers, Triangular Numbers
\begin{theorem}
The only Fibonacci numbers which are also triangular are:
:$0, 1, 3, 21, 55$
{{OEIS|A039595}}
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 0
| r = \dfrac {0 \times 1} 2
}}
{{eqn | l = 1
| r = \dfrac {1 \times 2} 2
}}
{{eqn | l = 3
| r = \dfrac {2 \times 3} 2
| rr= = 1 + 2
}}
{{eqn | l = 21
| r = \dfrac {6 \times 7} 2
| rr= = 8 + 13
}}
{{eqn | l = 55
| r = \dfrac {10 \times 11} 2
| rr= = 21 + 34
}}
{{end-eqn}}
It remains to be shown that these are the only ones.
Let $F_n$ be the $n$th Fibonacci number.
From Odd Square is Eight Triangles Plus One, $F_n$ is triangular {{iff}} $8 F_n + 1$ is square.
It remains to be demonstrated that $8 F_n + 1$ is square {{iff}}:
:$n \in \set{\pm 1, 0, 2, 4, 8, 10}$
So, let $8 F_n + 1$ be square.
Then:
:$n \equiv \begin{cases} \pm 1 \pmod {2^5 \times 5} & : n \text { odd} \\ 0, 2, 4, 8, 10 \pmod {2^5 \times 5^2 \times 11} & : n \text { even} \end{cases}$
{{ProofWanted|The proof needs to be studied.}}
\end{proof}
|
22810
|
\section{Triangular Matrices forms Subring of Square Matrices}
Tags: Rings of Square Matrices, Matrix Algebra, Triangular Matrices, Square Matrices, Rings, Subrings
\begin{theorem}
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map {\MM_R} n$ be the order $n$ square matrix space over a ring $R$.
Let $\struct {\map {\MM_R} n, +, \times}$ denote the ring of square matrices of order $n$ over $R$.
Let $\map {U_R} n$ be the set of upper triangular matrices of order $n$ over $R$.
Then $\map {U_R} n$ forms a subring of $\struct {\map {\MM_R} n, +, \times}$.
Similarly, let $\map {L_R} n$ be the set of lower triangular matrices of order $n$ over $R$.
Then $\map {L_R} n$ forms a subring of $\struct {\map {\MM_R} n, +, \times}$.
\end{theorem}
\begin{proof}
From Negative of Triangular Matrix, if $\mathbf B \in \map {U_R} n$ then $-\mathbf B \in \map {U_R} n$.
Then from Sum of Triangular Matrices, if $\mathbf A, -\mathbf B \in \map {U_R} n$ then $\mathbf A + \paren {-\mathbf B} \in \map {U_R} n$.
From Product of Triangular Matrices, if $\mathbf A, \mathbf B \in \map {U_R} n$ then $\mathbf A \mathbf B \in \map {U_R} n$.
The result follows from the Subring Test.
The same argument can be applied to matrices in $\map {L_R} n$.
{{qed}}
\end{proof}
|
22811
|
\section{Triangular Number Modulo 3 and 9}
Tags: Polygonal Numbers, Triangular Numbers
\begin{theorem}
Let $n$ be a triangular number.
Then one of the following two conditions applies:
:$n \equiv 0 \pmod 3$
:$n \equiv 1 \pmod 9$
\end{theorem}
\begin{proof}
Let $n = T_r$.
Then from Closed Form for Triangular Numbers:
:$n = \dfrac {r \paren {r + 1} } 2$
It suffices from Euclid's Lemma to investigate the nature of $r \paren {r + 1}$ modulo $3$.
There are three cases to consider:
:$r \equiv 0 \pmod 3$
:$r \equiv 1 \pmod 3$
:$r \equiv 2 \pmod 3$
Let $r \equiv 0 \pmod 3$.
Then:
:$r \paren {r + 1} \equiv 0 \pmod 3$
and so $T_r \equiv 0 \pmod 3$.
{{qed|lemma}}
Let $r \equiv 2 \pmod 3$.
Then:
:$r + 1 \equiv 3 \equiv 0 \pmod 3$
So:
:$r \paren {r + 1} \equiv 0 \pmod 3$
and so $T_r \equiv 0 \pmod 3$.
{{qed|lemma}}
Let $r \equiv 1 \pmod 3$.
Then $\exists k \in \Z: r = 3 k + 1$.
So $r \paren {r + 1} = \paren {3 k + 1} \paren {3 k + 2}$
{{begin-eqn}}
{{eqn | l = r \paren {r + 1}
| r = \paren {3 k + 1} \paren {3 k + 2}
| c =
}}
{{eqn | r = 9 k^2 + 9 k + 2
| c =
}}
{{eqn | r = 9 k \paren {k + 1} + 2
| c =
}}
{{end-eqn}}
So:
:$T_r = 9 \dfrac {k \paren {k + 1} } 2 + 1$
Thus $T_r \equiv 1 \pmod 9$.
{{qed}}
Category:Triangular Numbers
\end{proof}
|
22812
|
\section{Triangular Number Pairs with Triangular Sum and Difference}
Tags: Triangular Number Pairs with Triangular Sum and Differences, Triangular Number Pairs with Triangular Sum and Difference, Triangular Numbers
\begin{theorem}
The sequence of pairs of triangular numbers whose sum and difference are also both triangular begins:
:$\tuple {15, 21}, \tuple {105, 171}, \tuple {378, 703}, \tuple {780, 990}, \tuple {1485, 4186}, \tuple {2145, 3741}, \tuple {5460, 6786}, \tuple {7875, 8778}$
{{OEIS|A185129|order = first}}
{{OEIS|A185128|order = second}}
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = T_5
| r = \frac {5 \times 6} 2
| c =
}}
{{eqn | r = 15
| c =
}}
{{eqn | l = T_6
| r = \frac {6 \times 7} 2
| c =
}}
{{eqn | r = 21
| c =
}}
{{eqn | ll= \leadsto
| l = T_6 - T_5
| r = 21 - 15
| c =
}}
{{eqn | r = 6
| c =
}}
{{eqn | r = \dfrac {3 \times 4} 2
| c =
}}
{{eqn | r = T_3
| c =
}}
{{eqn | l = T_6 + T_5
| r = 21 + 15
| c =
}}
{{eqn | r = 36
| c =
}}
{{eqn | r = \dfrac {8 \times 9} 2
| c =
}}
{{eqn | r = T_8
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = T_{39}
| r = \frac {39 \times 40} 2
| c =
}}
{{eqn | r = 780
| c =
}}
{{eqn | l = T_{44}
| r = \frac {44 \times 45} 2
| c =
}}
{{eqn | r = 990
| c =
}}
{{eqn | ll= \leadsto
| l = T_{44} - T_{39}
| r = 990 - 780
| c =
}}
{{eqn | r = 210
| c =
}}
{{eqn | r = \dfrac {20 \times 21} 2
| c =
}}
{{eqn | r = T_{20}
| c =
}}
{{eqn | l = T_{44} + T_{39}
| r = 990 + 780
| c =
}}
{{eqn | r = 1770
| c =
}}
{{eqn | r = \dfrac {59 \times 60} 2
| c =
}}
{{eqn | r = T_{59}
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = T_{1869}
| r = \frac {1869 \times 1870} 2
| c =
}}
{{eqn | r = 1 \, 747 \, 515
| c =
}}
{{eqn | l = T_{2090}
| r = \frac {2090 \times 2091} 2
| c =
}}
{{eqn | r = 2 \, 185 \, 095
| c =
}}
{{eqn | ll= \leadsto
| l = T_{2090} - T_{1869}
| r = 2 \, 185 \, 095 - 1 \, 747 \, 515
| c =
}}
{{eqn | r = 437 \, 580
| c =
}}
{{eqn | r = \dfrac {935 \times 936} 2
| c =
}}
{{eqn | r = T_{935}
| c =
}}
{{eqn | l = T_{2090} + T_{1869}
| r = 2 \, 185 \, 095 - 1 \, 747 \, 515
| c =
}}
{{eqn | r = 3 \, 932 \, 610
| c =
}}
{{eqn | r = \dfrac {2804 \times 2805} 2
| c =
}}
{{eqn | r = T_{2804}
| c =
}}
{{end-eqn}}
That these are the smallest such pairs can be ascertained by trial by exhaustion.
{{qed}}
\end{proof}
|
22813
|
\section{Triangular Number as Alternating Sum and Difference of Squares}
Tags: Triangular Numbers, Square Numbers
\begin{theorem}
{{begin-eqn}}
{{eqn | q = \forall n \in \N
| l = \frac {n \paren {n + 1} } 2
| r = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j \paren {n - j}^2
| c =
}}
{{eqn | r = n^2 - \paren {n - 1}^2 + \paren {n - 2}^2 - \cdots + \paren {-1}^{n - 1}
| c =
}}
{{end-eqn}}
Thus the $n$th triangular number can be expressed as the alternating sum and difference of squares:
So:
{{begin-eqn}}
{{eqn | l = 1
| r = 1^2
| c =
}}
{{eqn | l = 3
| r = 2^2 - 1^2
| c =
}}
{{eqn | l = 6
| r = 3^2 - 2^2 + 1^2
| c =
}}
{{eqn | l = 10
| r = 4^2 - 3^2 + 2^2 - 1^2
| c =
}}
{{end-eqn}}
and so on.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \frac {n \paren {n + 1} } 2 = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j \paren {n - j}^2$
\end{proof}
|
22814
|
\section{Triangular Number cannot be Cube}
Tags: Cube Numbers, Triangular Numbers
\begin{theorem}
Let $T_n$ be the $n$th triangular number such that $n > 1$.
Then $T_n$ cannot be a cube.
\end{theorem}
\begin{proof}
Suppose $T_n = x^3$ for some $x \in \Z$.
Then by Odd Square is Eight Triangles Plus One:
:$\exists y \in \Z: 8 T_n + 1 = \paren {2 x}^3 + 1 = y^2$
By Cube which is One Less than a Square:
:$2 x = 2$, $y = 3$
giving the unique solution:
:$T_n = 1^3 = 1$
{{qed}}
\end{proof}
|
22815
|
\section{Triangular Number whose Square is Triangular}
Tags: Triangular Numbers, 6
\begin{theorem}
The only triangular number with less than $660$ digits, whose square is also triangular, is $6$.
\end{theorem}
\begin{proof}
We have that:
::${T_3}^2 = 6^2 = 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8$
To establish that it is the only one yet known can be established by an exhaustive search.
{{qed}}
\end{proof}
|
22816
|
\section{Triangular Numbers in Geometric Sequence}
Tags: Geometric Progressions, Geometric Sequences, Triangular Numbers
\begin{theorem}
The numbers:
:$1, 6, 36$
are the smallest triangular numbers in geometric sequence.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 6 \div 1
| r = 6
| c =
}}
{{eqn | l = 36 \div 6
| r = 6
| c =
}}
{{end-eqn}}
Hence the common ratio is $6$.
{{qed}}
\end{proof}
|
22817
|
\section{Triangular Numbers which are Sum of Two Cubes}
Tags: 28, Cube Numbers, Triangular Numbers
\begin{theorem}
The sequence of triangular numbers which are the sum of $2$ cubes begins:
:$28, 91, 351, 2926, 8001, 46971, 58653, 93528, 97461, \dots$
{{OEIS|A113958}}
\end{theorem}
\begin{proof}
Can be demonstrated by brute force.
For example:
{{begin-eqn}}
{{eqn | l = 28
| r = 1 + 27
| c =
}}
{{eqn | r = 1^3 + 3^3
| c =
}}
{{eqn | r = \dfrac {7 \paren {7 + 1} } 2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 91
| r = 27 + 64
| c =
}}
{{eqn | r = 3^3 + 4^3
| c =
}}
{{eqn | r = \dfrac {13 \paren {13 + 1} } 2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 351
| r = 125 + 216
| c =
}}
{{eqn | r = 5^3 + 6^3
| c =
}}
{{eqn | r = \dfrac {26 \paren {26 + 1} } 2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 2976
| r = 729 + 2197
| c =
}}
{{eqn | r = 5^3 + 6^3
| c =
}}
{{eqn | r = \dfrac {76 \paren {76 + 1} } 2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 8001
| r = 1 + 8000
| c =
}}
{{eqn | r = 1^3 + 20^3
| c =
}}
{{eqn | r = \dfrac {126 \paren {126 + 1} } 2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 46 \, 971
| r = 4096 + 42 \, 875
| c =
}}
{{eqn | r = 16^3 + 35^3
| c =
}}
{{eqn | r = \dfrac {306 \paren {306 + 1} } 2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 58 \, 653
| r = 8000 + 50 \, 653
| c =
}}
{{eqn | r = 20^3 + 37^3
| c =
}}
{{eqn | r = \dfrac {342 \paren {343 + 1} } 2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 93 \, 528
| r = 42 \, 875 + 50 \, 653
| c =
}}
{{eqn | r = 35^3 + 37^3
| c =
}}
{{eqn | r = \dfrac {432 \paren {433 + 1} } 2
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 97 \, 461
| r = 125 + 97 \, 336
| c =
}}
{{eqn | r = 5^3 + 46^3
| c =
}}
{{eqn | r = \dfrac {441 \paren {442 + 1} } 2
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22818
|
\section{Triangular Numbers which are also Square}
Tags: Triangular Numbers, Square Numbers
\begin{theorem}
Let $A_n$ be the $n$th non-negative integer whose square is also a triangular number.
Then:
:$A_n = \begin{cases} 0 & : n = 0 \\
1 & : n = 1 \\
6 A_{n - 1} - A_{n - 2} & : n > 1 \end{cases}$
\end{theorem}
\begin{proof}
Let $n \in \Z_{>0}$ be such that $n^2$ is a triangular number.
Then we have:
{{begin-eqn}}
{{eqn | q = \exists m \in \Z_{>0}
| l = n^2
| r = \dfrac {m \paren {m + 1} } 2
| c = Closed Form for Triangular Numbers
}}
{{eqn | ll= \leadstoandfrom
| l = 2 n^2
| r = m^2 + m
| c =
}}
{{eqn | r = \dfrac {\paren {2 m + 1}^2 - 1} 4
| c = Completing the Square
}}
{{eqn | ll= \leadstoandfrom
| l = 8 n^2 + 1
| r = \paren {2 m + 1}^2
| c = simplifying
}}
{{eqn | ll= \leadstoandfrom
| l = x^2 - 8 y^2
| r = 1
| c = setting $x = 2 m + 1$ and $y = n$
}}
{{end-eqn}}
This is a Pellian Equation.
From Pell's Equation: $x^2 - 8 y^2 = 1$, the smallest positive integral solution is:
:$\tuple {x, y} = \tuple {3, 1}$
Thus we have that the sequence of $n$ such that $n^2$ is triangular is the sequence of numerators of the Continued Fraction Expansion of $\sqrt 8$ whose indices are even.
Let $\tuple {x, y} = \tuple {p_k, q_k}$ be a solution to $x^2 - 8 y^2 = 1$.
Then from Continued Fraction Expansion of $\sqrt 8$ we have:
{{begin-eqn}}
{{eqn | l = p_{k + 1}
| r = a_{k + 1} p_k + p_{k - 1}
| c =
}}
{{eqn | r = 4 p_k + p_{k - 1}
| c = as $a_{k + 1} = 4$
}}
{{eqn | l = p_{k + 2}
| r = a_{k + 2} p_{k + 1} + p_k
| c =
}}
{{eqn | r = p_{k + 1} + p_k
| c = as $a_{k + 2} = 1$
}}
{{eqn | r = 4 p_k + p_{k - 1} + p_k
| c = from above
}}
{{eqn | r = 5 p_k + p_{k - 1}
| c = simplfying
}}
{{eqn | l = p_{k + 3}
| r = a_{k + 3} p_{k + 2} + p_{k + 1}
| c =
}}
{{eqn | r = 4 p_{k + 2} + p_{k + 1}
| c = as $a_{k + 3} = 4$
}}
{{eqn | l = p_{k + 4}
| r = a_{k + 4} p_{k + 3} + p_{k + 2}
| c =
}}
{{eqn | r = p_{k + 3} + p_{k + 2}
| c = as $a_{k + 4} = 1$
}}
{{eqn | r = 4 p_{k + 2} + p_{k + 1} + p_{k + 2}
| c = from above
}}
{{eqn | r = 5 p_{k + 2} + p_{k + 1}
| c = simplifying
}}
{{eqn | r = 5 p_{k + 2} + \paren {p_{k + 2} - p_k}
| c = substituting for $p_{k + 1}$
}}
{{eqn | r = 6 p_{k + 2} + p_k
| c = simplifying
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22819
|
\section{Trichotomy Law for Real Numbers}
Tags: Trichotomy Law for Real Numbers, Trichotomy Law, Inequalities, Analysis, Real Numbers
\begin{theorem}
The real numbers obey the Trichotomy Law.
That is, $\forall a, b \in \R$, exactly one of the following holds:
{{begin-axiom}}
{{axiom | n = 1
| lc= $a$ is greater than $b$:
| m = a > b
}}
{{axiom | n = 2
| lc= $a$ is equal to $b$:
| m = a = b
}}
{{axiom | n = 3
| lc= $a$ is less than $b$:
| m = a < b
}}
{{end-axiom}}
\end{theorem}
\begin{proof}
This follows directly from the fact that the real numbers form a totally ordered field.
{{Qed}}
\end{proof}
|
22820
|
\section{Trichotomy is Antireflexive}
Tags: Reflexive Relations, Trichotomies
\begin{theorem}
Let $\RR$ be a trichotomy.
Then $\RR$ is an antireflexive relation.
\end{theorem}
\begin{proof}
Let $\RR$ be a trichotomy on a set $S$.
Let $x \in S$.
By definition of a trichotomy, for all $a, b \in S$, either:
:$a \mathrel \RR b$
:$a = b$
:$b \mathrel \RR a$
As $x = x$ it follows directly that $x \not < x$.
Hence the result by definition of antireflexive relation.
{{qed}}
Category:Reflexive Relations
Category:Trichotomies
\end{proof}
|
22821
|
\section{Trimorphic Number is not necessarily Automorphic}
Tags: Trimorphic Numbers, Automorphic Numbers
\begin{theorem}
Let $n \in \Z_{>0}$ be a trimorphic number.
Then it is not necessarily the case that $n$ is also an automorphic number.
\end{theorem}
\begin{proof}
Take as an example $n = 49$.
We have that:
:$49^3 = 117 \, 6 \mathbf{49}$
demonstrating that $49$ is trimorphic.
However, we also have that:
:$49^2 = 2401$
demonstrating that $49$ is not automorphic.
{{qed}}
\end{proof}
|
22822
|
\section{Triple Angle Formulas/Cosine/2 cos 3 theta + 1}
Tags: Cosine Function, Triple Angle Formula for Cosine
\begin{theorem}
:$2 \cos 3 \theta + 1 = \paren {\cos \theta - \cos \dfrac {2 \pi} 9} \paren {\cos \theta - \cos \dfrac {4 \pi} 9} \paren {\cos \theta - \cos \dfrac {8 \pi} 9}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = z^6 + z^3 + 1
| r = \paren {z^2 - 2 z \cos \dfrac {2 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {4 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {8 \pi} 9 + 1}
| c = Complex Algebra Examples: $z^6 + z^3 + 1$
}}
{{eqn | ll= \leadsto
| l = z^3 + z^0 + z^{-3}
| r = \paren {z - 2 \cos \dfrac {2 \pi} 9 + z^{-1} } \paren {z - 2 \cos \dfrac {4 \pi} 9 + z^{-1} } \paren {z - 2 \cos \dfrac {8 \pi} 9 + z^{-1} }
| c =
}}
{{end-eqn}}
Setting $z = e^{i \theta}$:
{{begin-eqn}}
{{eqn | l = e^{3 i \theta} + 1 + e^{-3 i \theta}
| r = \paren {e^{i \theta} - 2 \cos \dfrac {2 \pi} 9 + e^{-i \theta} } \paren {e^{i \theta} - 2 \cos \dfrac {4 \pi} 9 + e^{-i \theta} } \paren {e^{i \theta} - 2 \cos \dfrac {8 \pi} 9 + e^{-i \theta} }
| c =
}}
{{eqn | ll= \leadsto
| l = 2 \dfrac {e^{3 i \theta} + e^{-3 i \theta} } 2 + 1
| r = \paren {2 \dfrac {e^{i \theta} + e^{-i \theta} } 2 - 2 \cos \dfrac {2 \pi} 9} \paren {2 \dfrac {e^{i \theta} + e^{-i \theta} } 2 - 2 \cos \dfrac {4 \pi} 9} \paren {2 \dfrac {e^{i \theta} + e^{-i \theta} } 2 - 2 \cos \dfrac {8 \pi} 9}
| c = rearranging
}}
{{eqn | ll= \leadsto
| l = 2 \cos 3 \theta + 1
| r = \paren {2 \cos \theta - 2 \cos \dfrac {2 \pi} 9} \paren {2 \cos \theta - 2 \cos \dfrac {4 \pi} 9} \paren {2 \cos \theta - 2 \cos \dfrac {8 \pi} 9}
| c = Cosine Exponential Formulation
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22823
|
\section{Triple of Triangular Numbers whose Pairwise Sums are Triangular}
Tags: Triangular Numbers
\begin{theorem}
The following triplet of triangular numbers has the property that the sum of each pair of them, and their total, are all triangular numbers:
:$66, 105, 105$
\end{theorem}
\begin{proof}
Throughout we use Closed Form for Triangular Numbers, which gives that the $n$th triangular number can be expressed as:
:$T_n = \dfrac {11 \times 12} 2$
We have:
{{begin-eqn}}
{{eqn | l = 66
| r = \frac {11 \times 12} 2
| c = and so is triangular
}}
{{eqn | l = 105
| r = \frac {14 \times 15} 2
| c = and so is triangular
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = 66 + 105
| r = 171
| c =
}}
{{eqn | r = \frac {18 \times 19} 2
| c = and so is triangular
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 105 + 105
| r = 210
| c =
}}
{{eqn | r = \frac {20 \times 21} 2
| c = and so is triangular
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 66 + 105 + 105
| r = 210
| c =
}}
{{eqn | r = \frac {23 \times 24} 2
| c = and so is triangular
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22824
|
\section{Triple with Product Quadruple the Sum}
Tags:
\begin{theorem}
Let $a, b, c \in \N$ such that $a \le b \le c$.
Then the solutions to:
:$a b c = 4 \paren {a + b + c}$
are:
:$\tuple {0, 0, 0}, \tuple {1, 5, 24}, \tuple {1, 6, 14}, \tuple {1, 8, 9}, \tuple {2, 3, 10}, \tuple {2, 4, 6}$
\end{theorem}
\begin{proof}
Suppose $a \ge 4$.
Then:
{{begin-eqn}}
{{eqn | l = a b c
| o = \ge
| r = 16 c
| c = as $4 \le a \le b$
}}
{{eqn | o = \ge
| r = 4 \paren {a + b + c + c}
| c = as $a \le b \le c$
}}
{{eqn | o = >
| r = 4 \paren {a + b + c}
| c = as $c > 0$
}}
{{end-eqn}}
hence $0 \le a \le 3$.
For $a = 0$, we have $4 \paren {b + c} = 0$.
This forces $b = c = 0$, giving the trivial solution:
:$\tuple {0, 0, 0}$
{{qed|lemma}}
For $a \ge 1$, note that:
{{begin-eqn}}
{{eqn | l = 4 \paren {a + b + c}
| r = a b c
}}
{{eqn | ll= \leadstoandfrom
| l = 4 \paren {a + b}
| r = \paren {a b - 4} c
}}
{{eqn | ll= \leadstoandfrom
| l = c
| r = \frac {4 \paren {a + b} } {a b - 4}
}}
{{end-eqn}}
In the second step, {{LHS}} is strictly positive.
Hence $a b > 4$ and the final step is justified.
For $a = 1$, we have $b c = 4 \paren {1 + b + c}$.
Since $a b > 4$, $b > 4$.
Suppose $b \ge 9$.
Then:
{{begin-eqn}}
{{eqn | l = b c
| o = \ge
| r = 9 c
}}
{{eqn | o = \ge
| r = 4 \paren {b + c} + c
| c = as $b \le c$
}}
{{eqn | o = >
| r = 4 \paren {1 + b + c}
| c = as $c \ge b > 4$
}}
{{end-eqn}}
hence $b \le 8$.
We find the value of $c$ by substituting $a$ and $b$:
:$b = 5: c = \dfrac {4 \paren {1 + 5} } {5 - 4} = 24$
:$b = 6: c = \dfrac {4 \paren {1 + 6} } {6 - 4} = 14$
:$b = 7: c = \dfrac {4 \paren {1 + 7} } {7 - 4} = \dfrac {32} 3$
:$b = 8: c = \dfrac {4 \paren {1 + 8} } {8 - 4} = 9$
and we see that $\tuple {1, 5, 24}, \tuple {1, 6, 14}, \tuple {1, 8, 9}$ are valid solutions.
{{qed|lemma}}
For $a = 2$, we have $2 b c = 4 \paren {2 + b + c}$.
Since $a b > 4$, $b > 2$.
Suppose $b \ge 5$.
Then:
{{begin-eqn}}
{{eqn | l = 2 b c
| o = \ge
| r = 10 c
}}
{{eqn | o = \ge
| r = 4 \paren {b + c} + 2 c
| c = as $b \le c$
}}
{{eqn | o = >
| r = 4 \paren {2 + b + c}
| c = as $c \ge b > 4$
}}
{{end-eqn}}
hence $b \le 4$.
We find the value of $c$ by substituting $a$ and $b$:
:$b = 3: c = \dfrac {4 \paren {2 + 3} } {2 \times 3 - 4} = 10$
:$b = 4: c = \dfrac {4 \paren {2 + 4} } {2 \times 4 - 4} = 6$
and we see that $\tuple {2, 3, 10}, \tuple {2, 4, 6}$ are valid solutions.
{{qed|lemma}}
For $a = 3$, we have $3 b c = 4 \paren {3 + b + c}$.
Suppose $b \ge 4$.
Then:
{{begin-eqn}}
{{eqn | l = 3 b c
| o = \ge
| r = 12 c
}}
{{eqn | o = \ge
| r = 4 \paren {b + c} + 4 c
| c = as $b \le c$
}}
{{eqn | o = >
| r = 4 \paren {3 + b + c}
| c = as $c \ge b > 1$
}}
{{end-eqn}}
hence $b \le 3$.
Since $3 = a \le b$, this forces $b = 3$
We find the value of $c$ by substituting $a$ and $b$:
:$c = \dfrac {4 \paren {3 + 3} } {3 \times 3 - 4} = \dfrac {24} 5$
and we see that it is not a valid solution.
{{qed|lemma}}
We have considered all possible values of $a$.
Hence the result.
{{qed}}
\end{proof}
|
22825
|
\section{Triple with Sum and Product Equal}
Tags: 6
\begin{theorem}
For $a, b, c \in \Z$, $a \le b \le c$, the solutions to the equation:
:$a + b + c = a b c$
are:
:$\tuple {1, 2, 3}$
:$\tuple {-3, -2, -1}$
and the trivial solution set:
:$\set {\tuple {-z, 0, z}: z \in \N}$
\end{theorem}
\begin{proof}
Suppose one of $a, b, c$ is zero.
Then $a b c = 0 = a + b + c$.
The remaining two numbers sum to $0$, giving the solution set:
:$\set {\tuple {-z, 0, z}: z \in \N}$
{{qed|lemma}}
Suppose $a < 0$ and $0 < b \le c$.
Then $a b c \le a < a + b + c$.
Hence equality never happens.
Similarly, for $a \le b < 0$ and $c > 0$:
:$a b c \ge c > a + b + c$
Hence equality never happens.
{{qed|lemma}}
Now it remains the case $0 < a \le b \le c$.
Suppose $a \ge 2$.
Then:
{{begin-eqn}}
{{eqn | l = a b c
| o = \ge
| r = 4 c
}}
{{eqn | o = \ge
| r = a + b + c + c
}}
{{eqn | o = >
| r = a + b + c
}}
{{end-eqn}}
hence $a = 1$.
Suppose $b \ge 3$.
Then:
{{begin-eqn}}
{{eqn | l = b c
| o = \ge
| r = c + c + c
}}
{{eqn | o = \ge
| r = b + c + c
}}
{{eqn | o = >
| r = b + c
}}
{{end-eqn}}
hence $b = 1$ or $b = 2$.
For $b = 1$:
:$c = 1 + 1 + c$
which is a contradiction.
For $b = 2$:
:$2 c = 1 + 2 + c$
giving $c = 3$.
Therefore $\tuple {1, 2, 3}$ is the only solution for strictly positive integer values of $a, b, c$.
Similarly, for $a \le b \le c < 0$, we have $0 < -c \le -b \le -a$.
Also:
{{begin-eqn}}
{{eqn | l = \paren {-c} + \paren {-b} + \paren {-a}
| r = -\paren {a + b + c}
}}
{{eqn | r = -a b c
}}
{{eqn | r = \paren {-c} \paren {-b} \paren {-a}
}}
{{end-eqn}}
and we have already show that $-c = 1, -b = 2, -a = 3$.
Therefore we have $\tuple {a, b, c} = \tuple {-3, -2, -1}$.
{{qed|lemma}}
We have considered all possible signs of $a, b, c$.
Hence the result.
{{qed}}
Category:6
\end{proof}
|
22826
|
\section{Triples of Consecutive Sphenic Numbers}
Tags: Sphenic Numbers
\begin{theorem}
The sequence of triplets of consecutive sphenic numbers starts:
:$\tuple {1309, 1310, 1311}, \tuple {1885, 1886, 1887}, \tuple {2013, 2014, 2015}, \ldots$
{{OEIS|A066509|order = first}}
{{OEIS|A248202|order = middle}}
\end{theorem}
\begin{proof}
Note that there cannot be quadruplets of such numbers, since one of the quadruplets must be divisible by $4$, making it non-sphenic.
We have:
{{begin-eqn}}
{{eqn | l = 1309
| r = 7 \times 11 \times 17
}}
{{eqn | l = 1310
| r = 2 \times 5 \times 131
}}
{{eqn | l = 1311
| r = 3 \times 19 \times 23
}}
{{eqn | l = 1885
| r = 5 \times 13 \times 29
}}
{{eqn | l = 1886
| r = 2 \times 23 \times 41
}}
{{eqn | l = 1887
| r = 3 \times 17 \times 37
}}
{{eqn | l = 2013
| r = 3 \times 11 \times 61
}}
{{eqn | l = 2014
| r = 2 \times 19 \times 53
}}
{{eqn | l = 2015
| r = 5 \times 13 \times 31
}}
{{end-eqn}}
hence each number above is sphenic.
{{qed}}
\end{proof}
|
22827
|
\section{Triplet in Arithmetic Sequence with equal Divisor Sum}
Tags: Arithmetic Sequences, Sigma Function, 323, Arithmetic Progressions, 295, 267, Divisor Sum Function
\begin{theorem}
The smallest triple of integers in arithmetic sequence which have the same divisor sum is:
:$\map {\sigma_1} {267} = \map {\sigma_1} {295} = \map {\sigma_1} {323} = 360$
\end{theorem}
\begin{proof}
We have that:
{{begin-eqn}}
{{eqn | l = 295 - 267
| r = 28
| c =
}}
{{eqn | l = 323 - 295
| r = 28
| c =
}}
{{end-eqn}}
demonstrating that $267, 295, 323$ are in arithmetic sequence with common difference $28$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {267}
| r = 360
| c = {{DSFLink|267}}
}}
{{eqn | l = \map {\sigma_1} {295}
| r = 360
| c = {{DSFLink|295}}
}}
{{eqn | l = \map {\sigma_1} {323}
| r = 360
| c = {{DSFLink|323}}
}}
{{end-eqn}}
{{ProofWanted|It remains to be shown that these are the smallest.}}
\end{proof}
|
22828
|
\section{Triplets of Products of Two Distinct Primes}
Tags: Semiprimes
\begin{theorem}
The following triplets of consecutive positive integers are the smallest in which each number is the product of $2$ distinct prime numbers:
:$33, 34, 35$
:$85, 86, 87$
:$93, 94, 95$
:$141, 142, 143$
:$201, 202, 203$
:$213, 214, 215$
:$217, 218, 219$
\end{theorem}
\begin{proof}
Taking each triplet in turn:
{{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}}
{{begin-eqn}}
{{eqn | l = 85
| r = 5 \times 17
}}
{{eqn | l = 86
| r = 2 \times 43
}}
{{eqn | l = 87
| r = 3 \times 29
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 93
| r = 3 \times 31
}}
{{eqn | l = 94
| r = 2 \times 47
}}
{{eqn | l = 95
| r = 5 \times 19
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 141
| r = 3 \times 47
}}
{{eqn | l = 142
| r = 2 \times 71
}}
{{eqn | l = 143
| r = 11 \times 13
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 201
| r = 3 \times 67
}}
{{eqn | l = 202
| r = 2 \times 101
}}
{{eqn | l = 203
| r = 7 \times 29
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 213
| r = 3 \times 71
}}
{{eqn | l = 214
| r = 2 \times 107
}}
{{eqn | l = 215
| r = 5 \times 43
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 217
| r = 7 \times 31
}}
{{eqn | l = 218
| r = 2 \times 109
}}
{{eqn | l = 219
| r = 3 \times 73
}}
{{end-eqn}}
It is noted that the triplet:
:$121, 122, 123$
while consisting of semiprimes, not all of these are the product of $2$ distinct prime numbers, as $121 = 11^2$.
{{qed}}
\end{proof}
|
22829
|
\section{Trisecting the Angle/Archimedean Spiral}
Tags: Trisecting the Angle, Classic Problems, Plane Geometry
\begin{theorem}
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of an Archimedean spiral.
\end{theorem}
\begin{proof}
Let the equation of the Archimedean spiral be $r = a \theta$.
Then:
{{begin-eqn}}
{{eqn | l = \angle DOB
| r = \frac {OD} a
}}
{{eqn | r = \frac {OA/3} a
}}
{{eqn | r = \frac 1 3 \angle AOB
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22830
|
\section{Trisecting the Angle/Neusis Construction}
Tags: Definitions: Plane Geometry, Classic Problems, Plane Geometry, Trisecting the Angle
\begin{theorem}
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of a neusis construction.
\end{theorem}
\begin{proof}
We have that $\angle BCD + \angle ACB$ make a straight angle.
As $CD = AB$ by construction, $CD = BC$ by definition of radius of circle.
Thus $\triangle BCD$ is isosceles.
By Isosceles Triangle has Two Equal Angles:
:$\angle CBD = \angle CDB$
From Sum of Angles of Triangle equals Two Right Angles:
:$\angle BCD + 2 \angle CBD$ equals two right angles.
Thus:
:$2 \angle CBD = \angle ACB$
Similarly, by Isosceles Triangle has Two Equal Angles:
:$\angle ACB = \angle CAB$
and again from Sum of Angles of Triangle equals Two Right Angles:
:$\angle ABC + 2 \angle ACB$ equals two right angles.
and so:
:$\angle ABC + 4 \angle CBD$ equals two right angles.
But $\alpha + \angle ABC + \angle CBD$ make a straight angle.
Thus:
:$\alpha + \angle ABC + \angle CBD = \angle ABC + 4 \angle CBD$
and so:
:$\alpha = 3 \angle CBD$
{{qed}}
\end{proof}
|
22831
|
\section{Trisecting the Angle/Parabola}
Tags: Trisecting the Angle, Classic Problems, Plane Geometry
\begin{theorem}
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of a parabola.
However, the points on the parabola that are required for this construction cannot be found by using only a straightedge and compass.
\end{theorem}
\begin{proof}
First, notice that because $A$ lies on $\CC_1$:
:$A = \tuple {\cos \angle POQ, \sin \angle POQ}$
This means:
:$B = \tuple {0, \sin \angle POQ}$
Because $C$ is the midpoint of $AB$:
:$C = \tuple {\dfrac {\cos \angle POQ} 2, \sin \angle POQ}$
Because $D$ lies on $\CC_1$:
:$D = \tuple {0, 1}$
and so:
:$E = \tuple {\dfrac {\cos \angle POQ} 2, 1}$
From Equation of Circle, $C_2$ has the equation:
:$\paren {x - \dfrac {\cos \angle POQ} 2}^2 + \paren {y -1}^2 = \dfrac {\cos^2 \angle POQ} 4 + 1$
Because $F$ lies on both $\CC_2$ and $\PP$, we can solve for the $x$-coordinate of $F$:
{{begin-eqn}}
{{eqn | l = \paren {x - \dfrac {\cos \angle POQ} 2}^2 + \paren {2 x^2 - 1}^2
| r = \dfrac {\cos^2 \angle POQ} 4 + 1
}}
{{eqn | ll= \leadsto
| l = x^2 - x \cos \angle POQ + \dfrac {\cos^2 \angle POQ} 4 + 4 x^4 - 4 x^2 + 1
| r = \dfrac {\cos^2 \angle POQ} 4 + 1
}}
{{eqn | ll= \leadsto
| l = x^2 - x \cos \angle POQ + 4 x^4 - 4 x^2
| r = 0
}}
{{eqn | ll= \leadsto
| l = 4 x^4 - 3 x^2 - x \cos \angle POQ
| r = 0
}}
{{eqn | ll= \leadsto
| l = x \paren {4 x^3 - 3 x - \cos \angle POQ}
| r = 0
}}
{{end-eqn}}
Thus $x = 0$ or $4 x^3 - 3 x - \cos \angle POQ = 0$.
This confirms what is obvious by the construction, namely that $\CC_2$ and $\PP$ intersect at the origin.
We can also see that the solution we are after must lie in the first quadrant, where $x \ne 0$.
So there must be one positive $x$ such that:
:$4 x^3 - 3 x = \cos \angle POQ$
By the Triple Angle Formula for Cosine $\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$, it is clear that one solution that works is:
:$x = \map \cos {\dfrac {\angle POQ} 3}$
Because, by construction:
:there is a vertical line through $F$ and $H$
:$H$ lies on the unit circle $\CC_1$
it follows that:
:$H = \tuple {\cos \dfrac {\angle POQ} 3, \sin \dfrac {\angle POQ} 3}$
Therefore:
:$\angle POH = \dfrac {\angle POQ} 3$
{{qed}}
\end{proof}
|
22832
|
\section{Trisecting the Angle by Compass and Straightedge Construction is Impossible}
Tags: Trisecting the Angle, Classic Problems, Plane Geometry
\begin{theorem}
There is no compass and straightedge construction for the trisection of the general angle.
\end{theorem}
\begin{proof}
Let $OA$ and $OB$ intersect at $O$.
It will be shown that there is no general method using a compass and straightedge construction to construct $OC$ such that $\angle AOB = 3 \times \angle AOC$.
It is sufficient to demonstrate that this is impossible for one specific angle.
Hence we choose $\angle AOB = 60 \degrees$.
Let $A$ and $B$ be points on the unit circle whose center is at $\tuple {0, 0}$.
Let $A$ lie on the $x$-axis.
Thus:
:$O$ is the point $\tuple {0, 0}$
:$A$ is the point $\tuple {1, 0}$
:$B$ is the point $\tuple {\cos 60 \degrees, \sin 60 \degrees}$
These all belong to $\Q \sqbrk {\sqrt 3}$.
trisection of $AOB$ is equivalent to constructing the point $\tuple {\cos 20 \degrees, \sin 20 \degrees}$.
From Triple Angle Formula for Cosine:
:$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$
so:
:$8 \cos^3 20 \degrees - 6 \cos 20 \degrees = 2 \cos 60 \degrees = 1$
Thus $\cos 20 \degrees$ is a root of the polynomial:
:$8 x^3 = 6 x - 1$
which by Irreducible Polynomial: $8 x^3 - 6 x - 1$ in Rationals is irreducible over $\Q$.
Thus $\cos 20 \degrees$ is algebraic over $\Q$ with degree $3$.
Thus by Algebraic Element of Degree 3 is not Element of Field Extension of Degree Power of 2, $\cos 20 \degrees$ is not an element of any extension of $\Q$ of degree $2^m$.
The result follows from Point in Plane is Constructible iff Coordinates in Extension of Degree Power of 2.
{{qed}}
\end{proof}
|
22833
|
\section{Trivial Estimate for Cyclotomic Polynomials}
Tags: Cyclotomic Polynomials
\begin{theorem}
Let $n \ge 1$ be a natural number.
Let $\Phi_n$ be the $n$th cyclotomic polynomial.
Let $\phi$ be the Euler totient function.
Let $z \in \C$ be a complex number.
Then:
:$\size {\size z - 1}^{\map \phi n} \le \size {\map {\Phi_n} z} \le \paren {\size z + 1}^{\map \phi n}$
where:
:the first inequality becomes an equality only if:
:::$n = 1$ and $z \in \R_{\ge 0}$
::or:
:::$n = 2$ and $z \in \R_{\le 0}$
:the second inequality becomes an equality only if:
:::$n = 1$ and $z \in \R_{\le 0}$
::or:
:::$n = 2$ and $z \in \R_{\ge 0}$
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Cyclotomic Polynomials
\end{proof}
|
22834
|
\section{Trivial Field Extension is Galois}
Tags: Galois Theory
\begin{theorem}
Let $F$ be a field.
The trivial field extension $F/F$ is Galois.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Galois Theory
\end{proof}
|
22835
|
\section{Trivial Gradation is Gradation}
Tags: Ring Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {M, e, \cdot}$ be a monoid.
Let
:$\ds R = \bigoplus_{m \mathop \in M} R_m$
be the trivial $M$-gradation on $R$.
This is a gradation on $R$.
\end{theorem}
\begin{proof}
We are required to show that:
:$\forall x \in R_m, y \in R_n: x \circ y \in R_{m \cdot n}$
First suppose that $m = n = e$ are both the identity.
In this case, $R_m = R_n = R$.
Since by definition, $R$ is closed under $\circ$, it follows that
:$\forall x \in R, y \in R: x \circ y \in R$
as required.
Now suppose that either $m \ne e$ or $n \ne e$.
After possibly exchanging $m$ and $n$, we may as well assume that $n \ne e$.
In particular, $R_n = \mathbf 0$ is the zero ring.
So if $y \in R_n$, then $y = 0$.
Therefore, for every $x \in R_m$, by Ring Product with Zero, we must have
:$x \circ y = x \circ 0 = 0$
Since $R_{m \cdot n}$ is an abelian group it must by definition contain $0$.
Therefore $x \circ y \in R_{m \cdot n}$ as required.
{{Qed}}
Category:Ring Theory
\end{proof}
|
22836
|
\section{Trivial Group Action is Group Action}
Tags: Group Actions
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $S$ be a set.
Let $*: G \times S \to S$ be the trivial group action:
:$\forall \tuple {g, s} \in G \times S: g * s = s$
Then $*$ is indeed a group action.
\end{theorem}
\begin{proof}
The group action axioms are investigated in turn.
Let $g_1, g_2 \in G$ and $s \in S$.
Thus:
{{begin-eqn}}
{{eqn | l = g_1 * \paren {g_2 * s}
| r = g_1 * s
| c = Definition of $*$
}}
{{eqn | r = s
| c = Definition of $*$
}}
{{eqn | r = \paren {g_1 \circ g_2} * s
| c = Definition of $*$
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|1}} holds.
Then:
{{begin-eqn}}
{{eqn | l = e * s
| r = s
| c = Definition of $*$
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|2}} holds.
{{qed}}
Category:Group Actions
\end{proof}
|
22837
|
\section{Trivial Group is Abelian}
Tags: Abelian Groups, Trivial Group
\begin{theorem}
The trivial group is an abelian group.
\end{theorem}
\begin{proof}
From Trivial Group is Cyclic Group, it is shown that the algebraic structure $\struct {\set e, \circ}$ such that $e \circ e = e$ is a cyclic group.
The result follows from Cyclic Group is Abelian.
{{qed}}
Category:Trivial Group
Category:Abelian Groups
\end{proof}
|
22838
|
\section{Trivial Group is Cyclic Group}
Tags: Cyclic Groups, Trivial Group, Definitions: Group Examples, Group Examples
\begin{theorem}
The trivial group is a cyclic group.
\end{theorem}
\begin{proof}
In Trivial Group is Group it is shown that the algebraic structure $\struct {\set e, \circ}$ such that $e \circ e = e$ is in fact a group.
It remains to be shown that it is cyclic.
In order for $G$ to be a cyclic group, every element $x$ of $G$ has to be expressible in the form $x = g^n$ for some $g \in G$ and some $n \in \Z$.
In this case, for every integer $n$, every element of $G$ can be expressed in the form $e^n$.
Thus $G$ is trivially a cyclic group.
{{qed}}
\end{proof}
|
22839
|
\section{Trivial Group is Group}
Tags: Cyclic Groups, Trivial Group
\begin{theorem}
The trivial group is a group.
\end{theorem}
\begin{proof}
Let $G = \struct {\set e, \circ}$ be an algebraic structure.
\end{proof}
|
22840
|
\section{Trivial Group is Initial Object}
Tags: Category of Groups, Trivial Group, Group Examples
\begin{theorem}
Let $\mathbf{Grp}$ be the category of groups.
Let $1 = \set e$ be the trivial group.
Then $1$ is an initial object of $\mathbf{Grp}$.
\end{theorem}
\begin{proof}
Let $\struct {G, \circ}$ be a group with identity $e_G$.
By Group Homomorphism Preserves Identity, any hypothetical group homomorphism $\phi: 1 \to G$ must satisfy:
:$\map \phi e = e_G$
Let us define the mapping $\phi$ in this way.
By Equality of Mappings, only one such mapping $1 \to G$ can exist, establishing uniqueness.
Now to verify that $\phi$ is actually a group homomorphism.
Since $1$ has only the element $e$, this is verified by:
{{begin-eqn}}
{{eqn | l = \map \phi e \circ \map \phi e
| r = e_G \circ e_G
| c = Definition of $\phi$
}}
{{eqn | r = e_G
| c = {{Defof|Identity Element}}
}}
{{eqn | r = \map \phi e
}}
{{eqn | r = \map \phi {e * e}
}}
{{end-eqn}}
where we used $*$ to denote the group operation on $1$.
Thus $\phi$ is a group homomorphism.
We have thus established that there is a unique morphism $1 \to G$ in $\mathbf{Grp}$ for all $G$.
That is, $1$ is an initial object in $\mathbf{Grp}$.
{{qed}}
\end{proof}
|
22841
|
\section{Trivial Group is Smallest Group}
Tags: Trivial Group
\begin{theorem}
Let $G = \struct {\set e, \circ}$ be a trivial group.
Then $G$ is the smallest group possible, in that there exists no set with lower cardinality which is the underlying set of a group.
\end{theorem}
\begin{proof}
From Trivial Group is Group, we have that there does exist a group of cardinality $1$.
From Group is not Empty, there can be no group of smaller order.
{{qed}}
\end{proof}
|
22842
|
\section{Trivial Group is Terminal Object of Category of Groups}
Tags: Category of Groups, Trivial Group, Group Examples
\begin{theorem}
Let $\mathbf {Grp}$ be the category of groups.
Let $\set e$ be the trivial group.
Then $\set e$ is a terminal object of $\mathbf {Grp}$.
\end{theorem}
\begin{proof}
Let $\struct {G, \circ}$ be any group.
By Singleton is Terminal Object of Category of Sets, there is precisely one mapping:
:$!: G \to \set e$
defined by:
:$\forall g \in G: ! (g) = e$
By definition, any group homomorphism is also a mapping.
Hence, there is at most one morphism $\struct {G, \circ} \to \set e$ in $\mathbf {Grp}$.
Now to verify that the mapping $!$ is a group homomorphism.
For any $g, h \in G$, we have (using $*$ for the group operation on $\set e$):
{{begin-eqn}}
{{eqn | l = ! (g) * ! (h)
| r = e * e
| c = Definition of $!$
}}
{{eqn | r = e
| c = Definition of $*$
}}
{{eqn | r = ! \paren {g \circ h}
| c = Definition of $!$
}}
{{end-eqn}}
That is, $!$ is a group homomorphism.
Thus for all groups $\struct {G, \circ}$, there is a unique group homomorphism $!: G \to \set e$.
That is, $\set e$ is a terminal object of $\mathbf {Grp}$.
{{qed}}
\end{proof}
|
22843
|
\section{Trivial Module is Module}
Tags: Module Theory
\begin{theorem}
Let $\struct {G, +_G}$ be an abelian group whose identity is $e_G$.
Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be the trivial $R$-module, such that:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$
Then $\struct {G, +_G, \circ}_R$ is a module.
\end{theorem}
\begin{proof}
Checking the module axioms in turn:
:{{Module-axiom|1}}: $\quad \lambda \circ \paren {x +_G y} = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$
:{{Module-axiom|2}}: $\quad \paren {\lambda +_R \mu} \circ x = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\mu \circ x}$
:{{Module-axiom|3}}: $\quad \paren {\lambda \times_R \mu} \circ x = e_G = \lambda \circ e_G = \lambda \circ \paren {\mu \circ x}$
Thus the trivial module is indeed a module.
{{qed}}
\end{proof}
|
22844
|
\section{Trivial Module is Not Unitary}
Tags: Module Theory
\begin{theorem}
Let $\struct {G, +_G}$ be an abelian group whose identity is $e_G$.
Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be the trivial $R$-module, such that:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$
Then unless $R$ is a ring with unity and $G$ contains only one element, this is ''not'' a unitary module.
\end{theorem}
\begin{proof}
By definition, for a trivial module to be unitary, $R$ needs to be a ring with unity.
For {{Module-axiom|4}} to apply, we require that:
:$\forall x \in G: 1_R \circ x = x$
But for the trivial module:
:$\forall x \in G: 1_R \circ x = e_G$
So {{Module-axiom|4}} can apply only when:
:$\forall x \in G: x = e_G$
Thus for the trivial module to be unitary, it is necessary that $G$ be the trivial group, and thus to contain one element.
{{qed}}
\end{proof}
|
22845
|
\section{Trivial Norm on Division Ring is Non-Archimedean}
Tags: Norm Examples, Trivial Norms, Norm Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a division ring whose ring zero is $0_R$.
Then the trivial norm $\norm {\, \cdot \,}: R \to \R_{\ge 0}$, which is given by:
:$\norm x = \begin{cases}
0 & : x = 0_R \\
1 & : \text{ otherwise}
\end{cases}$
is non-archimedean:
:$\norm {x + y} \le \max \set {\norm x, \norm y}$
\end{theorem}
\begin{proof}
Let $x, y = 0_R$.
Then:
:$\norm x, \norm y = 0$
Therefore:
:$\max \set {\norm x, \norm y} = 0$.
Hence:
{{begin-eqn}}
{{eqn | l = \norm {x + y}
| r = \norm {0_R + 0_R}
| c =
}}
{{eqn | r = \norm {0_R}
| c =
}}
{{eqn | r = 0
| c =
}}
{{eqn | r = \max \set {\norm x, \norm y}
| c =
}}
{{end-eqn}}
Let $x \ne 0_R$ or $y \ne 0_R$.
Then:
:$\norm x = 1$ or $\norm y = 1$
Therefore:
:$\max \set {\norm x, \norm y} = 1$.
Hence:
{{begin-eqn}}
{{eqn | l = \norm {x + y}
| o = \le
| r = 1
| c = {{Defof|Trivial Norm on Division Ring}}
}}
{{eqn | r = \max \set {\norm x, \norm y}
| c =
}}
{{end-eqn}}
{{qed}}
Category:Trivial Norms
\end{proof}
|
22846
|
\section{Trivial Norm on Division Ring is Norm}
Tags: Division Rings, Norm Examples, Trivial Norms, Norm Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a division ring, and denote its ring zero by $0_R$.
Then the trivial norm $\norm {\, \cdot \,}: R \to \R_{\ge 0}$, which is given by:
:$\norm x = \begin{cases}
0 & \text { if } x = 0_R\\
1 & \text { otherwise}
\end{cases}$
defines a norm on $R$.
\end{theorem}
\begin{proof}
Proving each of the norm axioms one by one:
\end{proof}
|
22847
|
\section{Trivial Ordering Compatibility in Boolean Ring}
Tags: Orderings, Huntington Algebras, Abstract Algebra, Boolean Rings, Order Theory
\begin{theorem}
Let $\struct {S, +, \circ}$ be a Boolean ring.
Then the trivial ordering is the only ordering on $S$ compatible with both its operations.
\end{theorem}
\begin{proof}
That the trivial ordering is compatible with $\circ$ and $*$ follows from Trivial Ordering is Universally Compatible.
Conversely, suppose that $\preceq$ is a ordering compatible with $\circ$ and $*$.
{{proof wanted}}
\end{proof}
|
22848
|
\section{Trivial Ordering is Universally Compatible}
Tags: Orderings, Order Theory
\begin{theorem}
Let $S$ be a set.
Let $\RR$ be the trivial ordering on $S$.
Then $\RR$ is universally compatible.
\end{theorem}
\begin{proof}
To prove that the trivial ordering is in fact an ordering, we need to checking each of the criteria for an ordering:
\end{proof}
|
22849
|
\section{Trivial Quotient Group is Quotient Group}
Tags: Quotient Groups, Examples of Quotient Groups, Normal Subgroups, Group Isomorphisms, Morphisms
\begin{theorem}
Let $G$ be a group.
Then the trivial quotient group:
:$G / \set {e_G} \cong G$
where:
:$\cong$ denotes group isomorphism
:$e_G$ denotes the identity element of $G$
is a quotient group.
\end{theorem}
\begin{proof}
From Trivial Subgroup is Normal:
:$\set {e_G} \lhd G$
Let $x \in G$.
Then:
:$x \set {e_G} = \set {x e_G} = \set x$
So each (left) coset of $G$ modulo $\set {e_G}$ has one element.
Now we set up the quotient epimorphism $\psi: G \to G / \set {e_G}$:
:$\forall x \in G: \map \phi x = x \set {e_G}$
which is of course a surjection.
We now need to establish that it is an injection.
Let $p, q \in G$.
{{begin-eqn}}
{{eqn | l = \map \phi p
| r = \map \phi q
| c =
}}
{{eqn | ll= \leadsto
| l = p \set {e_G}
| r = q \set {e_G}
| c = Definition of $\phi$
}}
{{eqn | ll= \leadsto
| l = \set p
| r = \set q
| c = from above
}}
{{eqn | ll= \leadsto
| l = p
| r = q
| c = {{Defof|Set Equality}}
}}
{{end-eqn}}
So $\psi$ is a group isomorphism and therefore:
:$G / \set {e_G} \cong G$
{{qed}}
\end{proof}
|
22850
|
\section{Trivial Quotient is a Bijection}
Tags: Bijections, Quotient Sets, Quotient Mappings
\begin{theorem}
Let $\Delta_S$ be the diagonal relation on a set $S$.
Let $q_{\Delta_S}: S \to S / \Delta_S$ be the trivial quotient of $S$.
Then $q_{\Delta_S}: S \to S / \Delta_S$ is a bijection.
\end{theorem}
\begin{proof}
The diagonal relation is defined as:
:$\Delta_S = \set {\tuple {x, x}: x \in S} \subseteq S \times S$
From the fact that $q_{\Delta_S}$ is a quotient mapping, we know that it is a surjection.
It relates each $x \in S$ to the singleton $\set x$.
Thus:
:$\set x = \eqclass x {\Delta_S} \subseteq S$
So $\map {q_{\Delta_S} } x = \set x$, and it follows that:
:$\eqclass x {\Delta_S} = \eqclass y {\Delta_S} \implies x = y$
Thus $q_{\Delta_S}$ is injective, and therefore by definition a bijection.
{{Qed}}
\end{proof}
|
22851
|
\section{Trivial Relation is Equivalence}
Tags: Examples of Equivalence Relations, Equivalence Relations, Relations, Trivial Relation
\begin{theorem}
The trivial relation on $S$:
:$\RR = S \times S$
is always an equivalence in $S$.
\end{theorem}
\begin{proof}
Let us verify the conditions for an equivalence in turn.
\end{proof}
|
22852
|
\section{Trivial Relation is Largest Equivalence Relation}
Tags: Equivalence Relations, Trivial Relation
\begin{theorem}
The trivial relation $\TT$ on $S$ is the largest equivalence in $S$, in the sense that:
:$\forall \EE \subseteq S \times S: \EE \subseteq \TT$
where $\EE$ denotes a general equivalence relation.
\end{theorem}
\begin{proof}
The trivial relation $\TT$ on $S$ is defined as:
:$\TT = S \times S$
It is confirmed from Trivial Relation is Equivalence that the trivial relation is in fact an equivalence relation.
Let $\EE$ be an arbitrary equivalence relation on $S$.
By definition of relation, $\EE \subseteq S \times S$ and so (trivially) $\EE \subseteq \TT$.
{{qed}}
\end{proof}
|
22853
|
\section{Trivial Relation is Universally Congruent}
Tags: Examples of Congruence Relations, Equivalence Relations, Trivial Relation, Abstract Algebra, Congruence Relations, Relations
\begin{theorem}
The trivial relation $\RR = S \times S$ on a set $S$ is universally congruent on $S$.
\end{theorem}
\begin{proof}
Let $\struct {S, \circ}$ be any algebraic structure which is closed for $\circ$.
By definition of trivial relation:
:$x \in S \land y \in S \implies x \mathrel \RR y$
So:
{{begin-eqn}}
{{eqn | l = x_1, x_2, y_1, y_2
| o = \in
| r = S
| c =
}}
{{eqn | ll= \leadsto
| l = x_1 \circ y_1, x_2 \circ y_2
| o = \in
| r = S
| c = {{Defof|Closed Algebraic Structure}}
}}
{{eqn | ll= \leadsto
| l = \paren {x_1 \circ y_1}
| o = \RR
| r = \paren {x_2 \circ y_2}
| c = {{Defof|Trivial Relation}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22854
|
\section{Trivial Ring from Abelian Group}
Tags: Abelian Groups, Rings, Trivial Rings, Ring Theory
\begin{theorem}
Any abelian group $\struct {G, +}$ may be turned into a trivial ring by defining the ring product to be:
:$\forall x, y \in G: x \circ y = e_G$
\end{theorem}
\begin{proof}
Follows directly from the definition of a trivial ring.
{{qed}}
\end{proof}
|
22855
|
\section{Trivial Ring is Commutative Ring}
Tags: Trivial Rings, Ring Theory, Commutative Rings
\begin{theorem}
Let $\struct {R, +, \circ}$ be a trivial ring.
Then $\struct {R, +, \circ}$ is a commutative ring.
\end{theorem}
\begin{proof}
First we need to show that a trivial ring is actually a ring in the first place.
Taking the ring axioms in turn:
\end{proof}
|
22856
|
\section{Trivial Solution of Homogeneous Linear 1st Order ODE}
Tags: Linear First Order ODEs
\begin{theorem}
The homogeneous linear first order ODE:
:$\dfrac {\d y} {\d x} + \map Q x y = 0$
has the particular solution:
:$\map y x = 0$
that is, the zero constant function.
This particular solution is referred to as the '''trivial solution'''.
\end{theorem}
\begin{proof}
We have:
:$\map {\dfrac {\d} {\d x} } 0 = 0$
from which:
:$\dfrac {\d y} {\d x} + \map Q x y = 0$
Hence the result.
{{Qed}}
Category:Linear First Order ODEs
\end{proof}
|
22857
|
\section{Trivial Solution of Homogeneous Linear 2nd Order ODE}
Tags: Linear Second Order ODEs, Homogeneous LSOODEs
\begin{theorem}
The homogeneous linear second order ODE:
:$\dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
has the particular solution:
:$\map y x = 0$
that is, the zero constant function.
This particular solution is referred to as the '''trivial solution'''.
\end{theorem}
\begin{proof}
We have:
:$\map {\dfrac {\d} {\d x} } 0 = 0$
and so:
:$\map {\dfrac {\d^2} {\d x^2} } 0 = 0$
from which:
:$\dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
Hence the result.
{{Qed}}
\end{proof}
|
22858
|
\section{Trivial Solution to System of Homogeneous Simultaneous Linear Equations is Solution}
Tags: Simultaneous Linear Equations
\begin{theorem}
Let $S$ be a '''system of homogeneous simultaneous linear equations''':
:$\ds \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = 0$
Consider the trivial solution to $A$:
:$\tuple {x_1, x_2, \ldots, x_n}$
such that:
:$\forall j \in \set {1, 2, \ldots, n}: x_j = 0$
Then the trivial solution is indeed a solution to $S$.
\end{theorem}
\begin{proof}
Let $i \in \set {1, 2, \ldots, m}$.
We have:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n \alpha_{i j} x_j
| r = \sum_{j \mathop = 1}^n \alpha_{i j} \times 0
| c =
}}
{{eqn | r = \sum_{j \mathop = 1}^n 0
| c =
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
This holds for all $i \in \set {1, 2, \ldots, m}$.
Hence:
:$\ds \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = 0$
and the result follows.
{{qed}}
Category:Simultaneous Linear Equations
\end{proof}
|
22859
|
\section{Trivial Subgroup is Normal}
Tags: Normal Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then the trivial subgroup $\struct {\set e, \circ}$ of $G$ is a normal subgroup in $G$.
\end{theorem}
\begin{proof}
First, by Trivial Subgroup is Subgroup, $\struct {\set e, \circ}$ is a subgroup of $G$.
To show $\struct {\set e, \circ}$ is normal in $G$:
:$\forall a \in G: a \circ e \circ a^{-1} = a \circ a^{-1} = e$
Hence the result.
{{qed}}
\end{proof}
|
22860
|
\section{Trivial Subgroup is Subgroup}
Tags: Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then the trivial subgroup $\struct {\set e, \circ}$ is indeed a subgroup of $\struct {G, \circ}$.
\end{theorem}
\begin{proof}
Using the One-Step Subgroup Test:
: $(1): \quad e \in \set e \leadsto \set e \ne \O$
: $(2): \quad e \in \set e \leadsto e \circ e^{-1} = e \in \set e$
{{qed}}
\end{proof}
|
22861
|
\section{Trivial Topological Space is Indiscrete}
Tags: Trivial Topological Space
\begin{theorem}
Let $X$ be a trivial topological space.
Then $X$ is indiscrete.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Trivial Topological Space
\end{proof}
|
22862
|
\section{Trivial Topological Space is Irreducible}
Tags: Trivial Topological Space, Irreducible Spaces
\begin{theorem}
Let $X$ be a trivial topological space.
Then $X$ is irreducible.
\end{theorem}
\begin{proof}
Follows from:
:Trivial Topological Space is Indiscrete
:Indiscrete Space is Irreducible
{{qed}}
Category:Irreducible Spaces
Category:Trivial Topological Space
\end{proof}
|
22863
|
\section{Trivial Topological Space is Non-Meager}
Tags: Trivial Topological Space, Second Category Spaces, Non-Meager Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a trivial topological space.
Then $T$ is non-meager.
\end{theorem}
\begin{proof}
As $T$ is a trivial topological space, by definition $S$ is a singleton: $S = \set s$, say.
Then $\set s$ is an open set.
That is, $s$ is an open point.
The result follows from Space with Open Point is Non-Meager.
{{qed}}
\end{proof}
|
22864
|
\section{Trivial Zeroes of Riemann Zeta Function are Even Negative Integers}
Tags: Zeta Function, Riemann Zeta Function, Analytic_Number_Theory, Analytic Number Theory
\begin{theorem}
Let $\rho = \sigma + i t$ be a zero of the Riemann zeta function not contained in the critical strip:
:$0 \le \map \Re s \le 1$
Then:
:$s \in \set {-2, -4, -6, \ldots}$
These are called the '''trivial zeros''' of $\zeta$.
\end{theorem}
\begin{proof}
{{finish|First it needs to be established that those points are in fact zeroes. This follows directly and trivially from Riemann Zeta Function at Non-Positive Integers, but needs to be made explicit here.}}
First we note that by Zeroes of Gamma Function, $\Gamma$ has no zeroes on $\C$.
Therefore, the completed Riemann zeta function:
:$\map \xi s = \dfrac 1 2 s \paren {s - 1} \pi^{-s/2} \map \Gamma {\dfrac s 2} \, \map \zeta s$
has the same zeroes as $\zeta$.
Additionally by Functional Equation for Riemann Zeta Function, we have:
:$\map \xi s = \map \xi {1 - s}$
for all $s \in \C$.
Therefore if:
:$\map \zeta s \ne 0$
for all $s$ with $\map \Re s > 1$, then also:
:$\map \zeta s \ne 0$
for all $s$ with $\map \Re s < 0$.
Let us consider $\map \Re s > 1$.
We have:
:$\ds \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
where here and in the following $p$ ranges over the primes.
Therefore, we have:
:$\ds \map \zeta s \prod_p \paren {1 - p^{-s} } = 1$
All of the factors of this infinite product can be found in the product:
:$\ds \prod_{n \mathop = 2}^\infty \paren {1 - n^{-s} }$
which converges absolutely.
This follows because from Convergence of P-Series:
:$\ds \sum_{k \mathop = 1}^\infty k^{-s}$ converges absolutely.
Hence:
:$\ds \prod_p \paren {1 - p^{-s} }$ converges absolutely.
So by the fact that:
:$\ds \map \zeta s \prod_p \paren {1 - p^{-s} } = 1$
we know $\map \zeta s$ cannot possibly be zero for any point in the region in question.
{{qed}}
\end{proof}
|
22865
|
\section{True Weight from False Balance/Unequal Arms}
Tags: False Balances
\begin{theorem}
Let $B$ be a body whose weight is $W$.
Let $B$ be weighed in a false balance whose falseness is because its arms are of different lengths.
Let the readings of the weight of $B$ be $a$ and $b$ when placed in opposite pans.
Then:
:$W = \sqrt {a b}$
\end{theorem}
\begin{proof}
Let the lengths of the arms of the false balance be $x$ and $y$.
{{WLOG}}, placing $B$ in the pan at the end of $x$ gives:
:$W x = a y$
and placing $B$ in the pan at the end of $y$ gives:
:$W y = b x$
{{TheoremWanted|We need to invoke the physics of couples to justify the above statements, but we haven't done the work yet to cover it.}}
Then:
{{begin-eqn}}
{{eqn | l = W x
| r = a y
| c =
}}
{{eqn | l = W y
| r = a x
| c =
}}
{{eqn | ll= \leadsto
| l = \dfrac x y
| r = \dfrac W a
| c =
}}
{{eqn | r = \dfrac b W
| c =
}}
{{eqn | ll= \leadsto
| l = W^2
| r = a b
| c = multiplying both sides by $W a$
}}
{{eqn | ll= \leadsto
| l = W
| r = \sqrt {a b}
| c =
}}
{{end-eqn}}
{{qed}}
Category:False Balances
\end{proof}
|
22866
|
\section{Tschirnhaus Transformation yields Depressed Polynomial}
Tags: Named Theorems, Algebra, Polynomial Theory
\begin{theorem}
Let $\map f x$ be a polynomial of order $n$:
:$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 $
Then the Tschirnhaus transformation:
$y = x + \dfrac {a_{n-1}} {n a_n}$
converts $f$ into a depressed polynomial:
:$b_n y^n + b_{n-1} y^{n-1} + \cdots + b_1 y + b_0$
where $b_{n-1} = 0$.
\end{theorem}
\begin{proof}
Substituting $y = x + \dfrac {a_{n-1}} {n a_n}$ gives us $x = y - \dfrac {a_{n-1}} {n a_n}$.
By the Binomial Theorem:
:$a_n x^n = a_n \paren{ y^n - \dfrac {a_{n-1}} {a_n} y^{n-1} + \map { f'_{n-2} } y }$
where $\map { f'_{n-2} } y$ is a polynomial in $y$ of order $n-2$.
Now we note that:
:$a_{n-1} x^{n-1} = a_{n-1} y^{n-1} - \map { f''_{n-2} } y$
where $\map { f''_{n-2} } y$ is another polynomial in $y$ of order $n-2$.
The terms in $y^{n-1}$ cancel out.
Hence the result.
{{qed}}
{{namedfor|Ehrenfried Walther von Tschirnhaus|cat=Tschirnhaus}}
Category:Polynomial Theory
\end{proof}
|
22867
|
\section{Tukey-Teichmüller Lemma}
Tags: Set Theory
\begin{theorem}
Let $\FF$ be a non-empty set of finite character.
Then $\FF$ has an element which is maximal with respect to the set inclusion relation.
\end{theorem}
\begin{proof}
Let $\NN \subseteq \FF$ be a nest.
We will show that $\bigcup \NN \in \FF$.
Let $F$ be a finite subset of $\bigcup \NN$.
By the definitions of subset and of union, each element of $F$ is an element of at least one element of $\NN$.
By the Principle of Finite Choice, there is a mapping $c: F \to \NN$ such that:
:$\forall x \in F: x \in \map c x$
Then $c \sqbrk F$ is a finite subset of $\NN$.
From Finite Totally Ordered Set is Well-Ordered, $f \sqbrk F$ has a greatest element $P \in \NN \subseteq \FF$.
Then:
:$F$ is a finite subset of $P$
and:
:$P \in \FF$
Since $\FF$ has finite character:
:$F \in \FF$
We have thus shown that every finite subset of $\bigcup \NN$ is in $\FF$.
Since $\FF$ is of finite character:
:$\bigcup \NN \in \FF$
Thus by Zorn's Lemma, $\FF$ has a maximal element.
{{qed}}
{{AoC|Zorn's Lemma}}
{{Namedfor|John Wilder Tukey|name2 = Oswald Teichmüller|cat = Tukey|cat2 = Teichmüller}}
\end{proof}
|
22868
|
\section{Tuning Fork Delta Sequence}
Tags: Examples of Delta Sequences, Dirac Delta Distribution
\begin{theorem}
thumb300pxThe graph of the tuning fork delta sequence. As $n$ grows, the rectangles becomes thinner and longer. The area of each shape is equal to $1$, where the area under the axis contributes negatively. Note that at $x = 0$ the graph is always negative, and its value here approaches $-\infty$. Only taking the entire graph we can ensure the proper behaviour for $n$ approaching $\infty$,
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:$\map {\delta_n} x := \begin{cases}
-n & : \size x < \frac 1 {2n} \\
2n & : \frac 1 {2n} \le \size x \le \frac 1 n \\
0 & : \size x > \frac 1 n \end{cases}$
Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.
That is, in the distributional sense it holds that:
:$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$
or
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$
where $\phi \in \map \DD \R$ is a test function, $\delta$ is the Dirac delta distribution, and $\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.
\end{theorem}
\begin{proof}
Let $\phi \in \map \DD \R$ be a test function.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x
| r = \lim_{n \mathop \to \infty} \paren {\int_{- \infty}^{-\frac 1 n } \map \phi x \map {\delta_n} x \rd x + \int_{- \frac 1 n }^{- \frac 1 {2n} } \map \phi x \map {\delta_n} x \rd x + \int_{- \frac 1 {2n} }^{\frac 1 {2n} } \map \phi x \map {\delta_n} x \rd x + \int_{\frac 1 {2n} }^{\frac 1 n} \map \phi x \map {\delta_n} x \rd x + \int_{\frac 1 n}^{\infty} \map \phi x \map {\delta_n} x \rd x}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {2n \int_{- \frac 1 n }^{- \frac 1 {2n} } \map \phi x \rd x - n \int_{- \frac 1 {2n} }^{\frac 1 {2n} } \map \phi x \rd x + 2n \int_{\frac 1 {2n} }^{\frac 1 n} \map \phi x \rd x}
}}
{{eqn | r = \lim_{n \mathop \to \infty} n \paren {2 \map \phi {\xi_1} \paren {- \frac 1 {2n} - \paren {- \frac 1 n} } - \map \phi {\xi_2} \paren {\frac 1 {2n} - \paren {- \frac 1 {2n} } } + 2 \map \phi {\xi_3} \paren {\frac 1 n - \frac 1 {2n} } }
| c = Mean Value Theorem for Integrals, $\xi_1 \in \closedint {-\frac 1 n } {-\frac 1 {2n} }$, $\xi_2 \in \closedint {-\frac 1 {2n} } {\frac 1 {2n} }$, $\xi_3 \in \closedint {\frac 1 {2n} } {\frac 1 n }$
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren { \map \phi {\xi_1} - \map \phi {\xi_2} + \map \phi {\xi_3} }
}}
{{eqn | r = \map \phi {\lim_{n \mathop \to \infty} \xi_1} - \map \phi {\lim_{n \mathop \to \infty} \xi_2} + \map \phi {\lim_{n \mathop \to \infty} \xi_3}
| c = Limit of Image of Sequence on Real Numbers
}}
{{eqn | r = \map \phi 0 - \map \phi 0 + \map \phi 0
| c = $\ds \lim_{n \mathop \to \infty} \frac 1 n = 0$, Squeeze Theorem for Real Sequences
}}
{{eqn | r = \map \delta \phi
| c = {{Defof|Dirac Delta Distribution}}
}}
{{end-eqn}}
Furthermore:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty \map {\delta_n} x \rd x
| r = 2n \int_{- \frac 1 n }^{- \frac 1 {2n} } \rd x - n \int_{- \frac 1 {2n} }^{\frac 1 {2n} } \rd x + 2n \int_{\frac 1 {2n} }^{\frac 1 n} \rd x
}}
{{eqn | r = n \paren {2 \paren {- \frac 1 {2n} - \paren {- \frac 1 n} } - \paren {\frac 1 {2n} - \paren {- \frac 1 {2n} } } + 2 \paren {\frac 1 n - \frac 1 {2n} } }
}}
{{eqn | r = n \paren {\frac 2 {2n} - \frac 2 {2n} + \frac 2 {2n} }
}}
{{eqn | r = 1
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22869
|
\section{Tusi Couple is Diameter of Stator}
Tags: Hypocycloids, Tusi Couples
\begin{theorem}
A Tusi couple is a degenerate case of the hypocycloid whose form is a straight line that forms a diameter of the stator.
\end{theorem}
\begin{proof}
Let $C_1$ be a circle of radius $b$ rolling without slipping around the inside of a circle $C_2$ of radius $a$.
Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.
Let $P$ be a point on the circumference of $C_1$.
Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.
Let $H$ be the hypocycloid formed by the locus of $P$.
From Number of Cusps of Hypocycloid from Integral Ratio of Circle Radii we have that $H$ will have $2$ cusps {{iff}}:
:$a = 2 b$
By Equation of Hypocycloid a hypocycloid can be expressed in parametric form as:
:$x = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta}$
:$y = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta}$
Hence:
{{begin-eqn}}
{{eqn | l = x
| r = \paren {2 b - b} \cos \theta + b \map \cos {\paren {\dfrac {2 b - b} b} \theta}
| c =
}}
{{eqn | r = b \cos \theta + b \cos \theta
| c =
}}
{{eqn | r = 2 b \cos \theta
| c =
}}
{{end-eqn}}
Thus the $x$ coordinate of the $2$ cusp hypocycloid has a range $\closedint {-b} b$.
Similarly:
{{begin-eqn}}
{{eqn | l = y
| r = \paren {2 b - b} \sin \theta - b \map \sin {\paren {\dfrac {2 b - b} b} \theta}
| c =
}}
{{eqn | r = b \sin \theta - b \sin \theta
| c =
}}
{{eqn | r = 0
| c =
}}
{{end-eqn}}
Thus the $y$ coordinate of the $2$ cusp hypocycloid is fixed at $y = 0$.
Thus the $2$ cusp hypocycloid consists of the line segment:
:$x \in \closedint {-b} b, y = 0$.
which is a diameter of the containing circle.
{{qed}}
\end{proof}
|
22870
|
\section{Twelve Factorial plus One is divisible by 13 Squared}
Tags: Factorials, 12, 13
\begin{theorem}
:$12! + 1$ is divisible by $13^2$.
\end{theorem}
\begin{proof}
By calculuation:
{{begin-eqn}}
{{eqn | l = 12! + 1
| r = 479 \, 001 \, 601
| c =
}}
{{eqn | r = 2 \, 834 \, 329 \times 13 \times 13
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22871
|
\section{Twice Differentiable Real Function with Negative Second Derivative is Strictly Concave}
Tags: Differential Calculus, Concave Real Functions
\begin{theorem}
Let $f$ be a real function which is twice differentiable on the open interval $\openint a b$ such that:
:$\map {f''} x < 0$ for each $x \in \openint a b$.
Then $f$ is strictly concave on $\openint a b$ {{iff}} its second derivative $f'' < 0$ on $\openint a b$.
\end{theorem}
\begin{proof}
From Real Function is Strictly Concave iff Derivative is Strictly Decreasing, $f$ is strictly concave {{iff}} $f'$ is strictly decreasing.
Since $f'' < 0$, we have that $f'$ is strictly decreasing from Real Function with Strictly Negative Derivative is Strictly Decreasing.
{{qed}}
\end{proof}
|
22872
|
\section{Twice Differentiable Real Function with Positive Second Derivative is Strictly Convex}
Tags: Differential Calculus, Convex Real Functions
\begin{theorem}
Let $f$ be a real function which is twice differentiable on the open interval $\openint a b$ such that:
:$\map {f''} x > 0$ for each $x \in \openint a b$.
Then $f$ is strictly convex on $\openint a b$.
\end{theorem}
\begin{proof}
From Real Function is Strictly Convex iff Derivative is Strictly Increasing, $f$ is strictly convex {{iff}} $f'$ is strictly increasing.
Since $f'' > 0$, we have that $f'$ is strictly increasing from Real Function with Strictly Positive Derivative is Strictly Increasing.
{{qed}}
\end{proof}
|
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