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23773
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\section{Non-Zero Natural Numbers under Multiplication form Commutative Monoid}
Tags: Natural Numbers, Examples of Monoids, Monoid Examples, Examples of Commutative Monoids
\begin{theorem}
Let $\N_{>0}$ be the set of natural numbers without zero, i.e. $\N_{>0} = \N \setminus \set 0$.
The structure $\struct{\N_{>0}, \times}$ forms a commutative monoid.
\end{theorem}
\begin{proof}
From Non-Zero Natural Numbers under Multiplication form Commutative Semigroup, $\struct {\N_{>0}, \times}$ forms a commutative semigroup.
From Identity Element of Natural Number Multiplication is One, $\struct {\N_{>0}, \times}$ has an identity element which is $1$.
Hence the result, by definition of commutative monoid.
{{qed}}
\end{proof}
|
23774
|
\section{Non-Zero Rational Numbers Closed under Multiplication}
Tags: Algebraic Closure, Rational Multiplication, Rational Numbers
\begin{theorem}
The set of non-zero rational numbers is closed under multiplication.
\end{theorem}
\begin{proof}
Recall that Rational Numbers form Field under the operations of addition and multiplication.
By definition of a field, the algebraic structure $\struct {\Q_{\ne 0}, \times}$ is a group.
Thus, by definition, $\times$ is closed in $\struct {\Q_{\ne 0}, \times}$.
{{qed}}
\end{proof}
|
23775
|
\section{Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group}
Tags: Abelian Groups, Examples of Abelian Groups, Group Examples, Abelian Groups: Examples, Infinite Groups: Examples, Rational Multiplication, Examples of Infinite Groups, Abelian Group Examples, Rational Numbers
\begin{theorem}
Let $\Q_{\ne 0}$ be the set of non-zero rational numbers:
:$\Q_{\ne 0} = \Q \setminus \set 0$
The structure $\struct {\Q_{\ne 0}, \times}$ is a countably infinite abelian group.
\end{theorem}
\begin{proof}
From the definition of rational numbers, the structure $\struct {\Q, + \times}$ is constructed as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
Hence from Multiplicative Group of Field is Abelian Group, $\struct {\Q_{\ne 0}, \times}$ is an abelian group.
From Rational Numbers are Countably Infinite, we have that $\struct {\Q_{\ne 0}, \times}$ is a countably infinite group.
{{qed}}
\end{proof}
|
23776
|
\section{Non-Zero Real Numbers Closed under Multiplication}
Tags: Real Numbers, Algebraic Closure, Non-Zero Real Numbers Closed under Multiplication, Real Multiplication
\begin{theorem}
The set of non-zero real numbers is closed under multiplication:
:$\forall x, y \in \R_{\ne 0}: x \times y \in \R_{\ne 0}$
\end{theorem}
\begin{proof}
Recall that Real Numbers form Field under the operations of addition and multiplication.
By definition of a field, the algebraic structure $\left({\R_{\ne 0}, \times}\right)$ is a group.
Thus, by definition, $\times$ is closed in $\left({\R_{\ne 0}, \times}\right)$.
{{qed}}
\end{proof}
|
23777
|
\section{Non-Zero Real Numbers under Multiplication form Abelian Group}
Tags: Non-Zero Real Numbers under Multiplication form Abelian Group, Abelian Groups, Examples of Abelian Groups, Group Examples, Abelian Groups: Examples, Real Numbers, Infinite Groups: Examples, Examples of Infinite Groups, Abelian Group Examples, Real Multiplication
\begin{theorem}
Let $\R_{\ne 0}$ be the set of real numbers without zero:
:$\R_{\ne 0} = \R \setminus \set 0$
The structure $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group.
\end{theorem}
\begin{proof}
Taking the group axioms in turn:
\end{proof}
|
23778
|
\section{Non-Zero Value of Continuous Real-Valued Function has Neighborhood not including Zero}
Tags: Continuous Mappings on Metric Spaces
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $f: M \to \R$ be a continuous real-valued function.
Let $\map f a > 0$ for some $a \in M$.
Then there exists $\delta \in \R_{>0}$ such that:
:$\forall x \in \map {B_\delta} a$
where $\map {B_\delta} a$ denotes the open $\delta$-ball of $a$ in $M$.
\end{theorem}
\begin{proof}
Let $\epsilon \in \R_{>0}$ such that $\epsilon < \map f a$.
Let $\map {N_\epsilon} {\map f a} = \openint {\map f a - \epsilon} {\map f a + \epsilon}$ be the $\epsilon$-neighborhood of $\map f a$.
Then $\map {N_\epsilon} {\map f a}$ is an open interval of $\map f a$ such that:
:$\forall y \in \map {N_\epsilon} {\map f a}: y > 0$
We are given that $f$ is continuous.
So, by definition:
:$\exists U \subseteq A: \forall x \in U: \map f x \in \map {N_\epsilon} {\map f a}$
where $U$ is an open set of $M$.
As $a \map {N_\epsilon} {\map f a}$, it follows that:
:$a \in U$
By definition of open set:
:$\exists \delta \in \R_{>0}: \map {B_\delta} a \subseteq U$
Hence:
:$\forall x \in \map {B_\delta} a: \map f x \in \map {N_\epsilon} {\map f a}$
As we have already noted that:
:$\forall y \in \map {N_\epsilon} {\map f a}: y > 0$
the result follows.
{{qed}}
\end{proof}
|
23779
|
\section{Nonconstant Periodic Function with no Period is Discontinuous Everywhere}
Tags: Periodic Functions
\begin{theorem}
Let $f$ be a real periodic function that does not have a period.
Then $f$ is either constant or discontinuous everywhere.
\end{theorem}
\begin{proof}
Let $f$ be a real periodic function that does not have a period.
Let $x \in \R$.
If there is no $y \in \R$ such that $\map f x \ne \map f y$, then the result is verified by Constant Function has no Period.
If $f$ is non-constant, then let $y$ be such a value.
Let $\LL_{f > 0}$ be the set of all positive periodic elements of $f$.
This set is non-empty by Absolute Value of Periodic Element is Peroidic.
It is seen that $<$ forms a right-total relation on $\LL_{f > 0}$, for if not then $f$ would have a period.
By the Axiom of Dependent Choice there must exist a strictly-decreasing sequence $\sequence {L_n}$ in $\LL_{f > 0}$.
Since this sequence is bounded below by zero, it follows via Monotone Convergence Theorem that this sequence converges.
Also, by Convergent Sequence is Cauchy Sequence the sequence is Cauchy.
Since $\sequence {L_n}$ is Cauchy, the sequence $\sequence {d_n}_{n \mathop \ge 1}$ formed by taking $d_n = L_n - L_{n - 1}$ is null.
From Combination of Periodic Elements it follows that $\sequence {d_n}_{n \mathop \ge 1}$ is contained in $\LL_{f > 0}$, for every $d_n$ is a periodic element of $f$.
Consider the sequence $\sequence {\paren {x - y} \bmod d_n}_{n \mathop \ge 1}$.
It is seen by Limit of Modulo Operation that this sequence is also null.
Let $\epsilon \in \R_{> 0}$ such that $\epsilon < \size {\map f x - \map f y}$.
For any $\delta \in \R_{> 0}$, there is a $n \in \N_{> 0}$ such that:
{{begin-eqn}}
{{eqn | l = \paren {x - y} \bmod d_n
| r = \paren {x - y} - d_n \floor {\frac {x - y} {d_n} }
| c = {{Defof|Modulo Operation}}
}}
{{eqn | r = x - \paren {y + d_n \floor {\frac {x - y} {d_n} } }
}}
{{eqn | o = <
| r = \delta
}}
{{end-eqn}}
But:
{{begin-eqn}}
{{eqn | l = \size {\map f x - \map f {y + d_n \floor {\frac {x - y} {d_n} } } }
| r = \size {\map f x - \map f y}
| c = General Periodicity Property
}}
{{eqn | o = >
| r = \epsilon
}}
{{end-eqn}}
And so $f$ is not continuous at $x$.
But our choice of $x$ was completely arbitrary.
Hence the result.
{{qed}}
{{ADC}}
Category:Periodic Functions
\end{proof}
|
23780
|
\section{Nonempty Grothendieck Universe contains Von Neumann Natural Numbers}
Tags: Set Theory, Grothendieck Universes
\begin{theorem}
Let $\mathbb U$ be a non-empty Grothendieck universe.
Let $\N$ denote the set of von Neumann natural numbers.
Then $\N$ is a subset of $\mathbb U$.
\end{theorem}
\begin{proof}
We prove the claim by induction.
\end{proof}
|
23781
|
\section{Nonlimit Ordinal Cofinal to One}
Tags: Ordinals
\begin{theorem}
Let $x$ be a nonlimit non-empty ordinal.
Let $\operatorname{cof}$ denote the cofinal relation.
Let $1$ denote the ordinal one.
Then:
:$\operatorname{cof} \left({x, 1}\right)$
\end{theorem}
\begin{proof}
Since $1 = 0^+$, $1$ is not a limit ordinal.
Let $\le$ denote the subset relation.
It follows that $0 < 1$ by Ordinal is Less than Successor.
Moreover, $1 \le x$ follows by the fact that $0 < x$ and Successor of Element of Ordinal is Subset.
Thus we have $0 < 1 \le x$ and so by Condition for Cofinal Nonlimit Ordinals:
:$\operatorname{cof} \left({x, 1}\right)$
{{qed}}
\end{proof}
|
23782
|
\section{Nonnegative Quadratic Functional implies no Interior Conjugate Points}
Tags: Calculus of Variations
\begin{theorem}
If the quadratic functional
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x$
where:
:$\forall x \in \closedint a b: \map P x > 0$
is nonnegative for all $\map h x$:
:$\map h a = \map h b = 0$
then the closed interval $\closedint a b$ contains no inside points conjugate to $a$.
In other words, the open interval $\openint a b$ contains no points conjugate to $a$.
{{explain|Rewrite the above so it makes better sense. For example, should the "nonnegative" comment be above the condition on $\map P x$?}}
\end{theorem}
\begin{proof}
Consider the functional:
:$\forall t \in \closedint 0 1: \ds \int_a^b \paren {t \paren {P h^2 + Q h'^2} + \paren {1 - t} h'^2} \rd x$
By assumption:
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x \ge 0$
For $t = 1$, Euler's Equation reads:
:$\map {h''} x = 0$
which, along with condition $\map h a = 0$, is solved by:
:$\map h x = x - a$
for which there are no conjugate points in $\closedint a b$.
In other words:
:$\forall x \in \openint a b: \map h x > 0$
Hence:
:$\forall t \in \closedint 0 1: \ds \int_a^b \paren {t \paren {P h'^2 + Q h^2} + \paren {1 - t} h'^2} \rd x \ge 0$
The corresponding Euler's Equation is:
:$2 Q h t - \map {\dfrac \d {\d x} } {2 t P h' + 2 h' \paren {1 - t} } = 0$
which is equivalent to:
:$-\map {\dfrac \d {\d x} } {\paren {t P + \paren {1 - t} } h'} + t Q h = 0$
Let $\map h {x, t}$ be a solution to this such that:
:$\forall t \in \closedint 0 1: \map h {a, t} = 0$
:$\map {h_x} {a, t} = 1$
{{explain|What is $h_x$?}}
Suppose that for $\map h {x, t}$ there exists a conjugate point $\tilde a$ to $a$ in $\closedint a b$.
In other words:
:$\exists \tilde a \in \closedint a b: \map h {\tilde a, 1} = 0$
By definition, $a \ne \tilde a$.
Suppose $\tilde a = b$.
Then by lemma 1 of Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite:
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x = 0$
This agrees with the assumption.
Therefore, it is allowed that $\tilde a = b$.
For $t = 1$, any other conjugate point of $\map h {x, t}$ may reside only in $\openint a b$.
Consider the following set of all points $\tuple {x, t}$:
:$\set {\tuple {x, t}: \paren {\forall x \in \closedint a b} \paren {\forall t \in \closedint 0 1} \paren {\map h {x, t} = 0} }$
If it is non-empty, it represents a curve in $x - t$ plane, such that $h_x \left({x, t}\right) \ne 0$.
{{explain|Rather than using "it", give it a name and reference that name.}}
By the Implicit Function Theorem, $\map x t$ is continuous.
By hypothesis, $\tuple {\tilde a, 1}$ lies on this curve.
Suppose that the curve starts at this point.
The curve can terminate either inside the rectangle or its boundary.
If it terminates inside the rectangle $\closedint a b \times \closedint 0 1$, it implies that there is a discontinuous jump in the value of $h$.
{{explain|Again, rather than using "it", refer to the object in question directly, so it is clear what is being referred to.}}
:Therefore, it contradicts the continuity of $\map h {x, t}$ in the interval $t \in \closedint 0 1$.
{{explain|Specify what contradicts what, by invoking the AimForCont construct}}
If it intersects the line segment $x = b, 0 \le t \le 1$, then by lemma 2 of Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite it vanishes.
{{explain|"it" again}}
:This contradicts positive-definiteness of the functional for all $t$.
{{explain|See above note on the above contradiction proof}}
If it intersects the line segment $a \le x \le b, t = 1$, then $\exists t_0: \paren {\map h {x, t_0} = 0} \land \paren {\map {h_x} {x, t_0} = 0}$.
{{explain|$h_x$ again}}
If it intersects $a \le x \le b, t = 0$, then Euler's equation reduces to $h'' = 0$ with solution $h = x - a$, which vanishes only for $x = a$.
If it intersects $x = a, 0 \le t \le 1$, then $\exists t_0: \map {h_x} {a, t_0} = 0$
{{explain|"it" again}}
{{Help|explain cases $t = 1$ and $x = a$}}
By Proof by Cases, no such curve exists.
Thus, the point $\tuple {\tilde a, 1}$ does not exist, since it belongs to this curve.
Hence there are no conjugate points of $\map h {x, 1} = \map h x$ in the interval $\openint a b$.
{{qed}}
\end{proof}
|
23783
|
\section{Nontrivial Zeroes of Riemann Zeta Function are Symmetrical with respect to Critical Line}
Tags: Riemann Zeta Function
\begin{theorem}
The nontrivial zeroes of the Riemann $\zeta$ function are distributed symmetrically {{WRT}} the critical line.
That is, suppose $s_1 = \sigma_1 + i t$ is a nontrivial zero of $\map \zeta s$.
Then there exists another nontrivial zero $s_2$ of $\map \zeta s$ such that:
:$s_2 = 1 - s_1$
\end{theorem}
\begin{proof}
From Functional Equation for Riemann Zeta Function, we have:
:$\ds \pi^{-s/2} \map \Gamma {\dfrac s 2} \map \zeta s = \pi^{\paren {s/2 - 1/2 } } \map \Gamma {\dfrac {1 - s} 2} \map \zeta {1 - s}$
We suppose $s_1 = \sigma_1 + i t$ is a nontrivial zero of $\map \zeta s$.
Then we have:
{{begin-eqn}}
{{eqn | l = \map \zeta {s_1}
| r = 0
| c =
}}
{{eqn | l = \pi^{-s_1/2 } \map \Gamma {\dfrac {s_1} 2} \map \zeta {s_1}
| r = 0
| c = {{LHS}}
}}
{{eqn | l = \pi^{\paren {s_1/2 - 1/2 } } \map \Gamma {\dfrac {1 - s_1} 2} \map \zeta {1 - s_1}
| r = 0
| c = {{RHS}}
}}
{{end-eqn}}
It remains to be shown that of the three terms on the {{RHS}}, $\map \zeta {1 - s_1}$ MUST equal zero.
We can rewrite the first term in terms of the exponential function
:$\ds \pi^{\paren {s_1 / 2 - 1 / 2} } = \map \exp {\map \ln {\pi^{\paren {s_1 / 2 - 1 / 2} } } }$
From the definition of the exponential function, we know $\map \exp {\map \ln {\pi^{\paren { {s_1}/2 - 1/2 } } }}$ never equals zero.
From Zeroes of Gamma Function, we know $\map \Gamma {\dfrac {1 - s_1} 2}$ never equals zero.
Therefore:
:$\map \zeta {1 - s_1} = 0$
{{qed}}
\end{proof}
|
23784
|
\section{Nonzero Ideal of Polynomial Ring over Field has Unique Monic Generator}
Tags: Polynomial Rings
\begin{theorem}
Let $K$ be a field.
Let $K \sqbrk x$ be the polynomial ring in one variable over $K$.
Let $I \subseteq K \sqbrk x$ be a nonzero ideal.
Then $I$ is generated by a unique monic polynomial.
\end{theorem}
\begin{proof}
{{proof wanted}}
Category:Polynomial Rings
\end{proof}
|
23785
|
\section{Nonzero natural number is another natural number successor}
Tags: Natural Numbers, Proofs by Induction
\begin{theorem}
Let $\N$ be the 0-based natural numbers:
:$\N = \left\{{0, 1, 2, \ldots}\right\}$
Let $s: \N \to \N: \map s n = n + 1$ be the successor function.
Then:
:$\forall n \in \N \setminus \set 0 \paren {\exists m \in \N: \map s m = n }$
\end{theorem}
\begin{proof}
The proof will proceed by the Principle of Finite Induction on $\N \setminus \set 0$
\end{proof}
|
23786
|
\section{Norm Sequence of Cauchy Sequence has Limit}
Tags: Cauchy Sequences, Normed Division Rings
\begin{theorem}
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a Cauchy sequence in $R$.
Then $\sequence {\norm {x_n} }$ has a limit in $\R$.
That is,
:$\exists l \in \R: \ds \lim_{n \mathop \to \infty} \norm {x_n} = l$
\end{theorem}
\begin{proof}
It is first shown that $\sequence {\norm {x_n} }$ is a real Cauchy sequence in $\R$.
Let $\epsilon \in \R_{>0}$ be given.
By the definition of Cauchy sequence then:
:$\exists N \in \N: \forall n, m > N, \norm {x_n - x_m} < \epsilon$
By the reverse triangle inequality, then:
:$\forall n, m > N: \cmod {\norm {x_n} - \norm {x_m} } \le \norm {x_n - x_m} < \epsilon$
From the definition of a real Cauchy sequence it follows that $\sequence {\norm {x_n} }$ is a real Cauchy sequence in $\R$.
By Cauchy's Convergence Criterion on Real Numbers $\sequence {\norm {x_n} }$ has a limit in $\R$.
{{qed}}
\end{proof}
|
23787
|
\section{Norm is Complete Iff Equivalent Norm is Complete}
Tags: Normed Division Rings, Complete Metric Spaces
\begin{theorem}
Let $R$ be a division ring.
Let $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ be equivalent norms on $R$.
Then:
:$\struct {R,\norm {\,\cdot\,}_1}$ is complete {{iff}} $\struct {R,\norm {\,\cdot\,}_2}$ is complete.
\end{theorem}
\begin{proof}
By Cauchy Sequence Equivalence, for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_1$ {{iff}} $\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_2$.
By Convergent Equivalence, for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ converges in $\norm{\,\cdot\,}_1$ {{iff}} $\sequence {x_n}$ converges in $\norm{\,\cdot\,}_2$.
Hence:
:every Cauchy sequence in $\norm{\,\cdot\,}_1$ converges in $\norm{\,\cdot\,}_1$ {{iff}} every Cauchy sequence in $\norm{\,\cdot\,}_2$ converges in $\norm{\,\cdot\,}_2$.
The result follows.
{{qed}}
Category:Normed Division Rings
Category:Complete Metric Spaces
\end{proof}
|
23788
|
\section{Norm is Continuous}
Tags: Continuity, Norm Theory, Continuous Mappings
\begin{theorem}
Let $\struct {V, \norm {\,\cdot\,} }$ be a normed vector space.
Then the mapping $x \mapsto \norm x$ is continuous.
Here, the metric used is the metric $d$ induced by $\norm {\,\cdot\,}$.
\end{theorem}
\begin{proof}
Since $\norm x = \map d {x, \mathbf 0}$, the result follows directly from Distance Function of Metric Space is Continuous.
{{qed}}
Category:Norm Theory
Category:Continuous Mappings
\end{proof}
|
23789
|
\section{Norm of Adjoint}
Tags: Adjoints
\begin{theorem}
Let $H, K$ be Hilbert spaces.
Let $A \in \map B {H, K}$ be a bounded linear transformation.
Then the norm of $A$ satisfies:
:$\norm A^2 = \norm {A^*}^2 = \norm {A^* A}$
where $A^*$ denotes the adjoint of $A$.
\end{theorem}
\begin{proof}
Let $h \in H$ such that $\norm h_H \le 1$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {A h}_K^2
| r = \innerprod {A h} {A h}_K
| c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \innerprod {A^* A h} h_H
| c = {{Defof|Adjoint Linear Transformation}}
}}
{{eqn | o = \le
| r = \norm {A^*A h}_H \norm h_H
| c = Cauchy-Bunyakovsky-Schwarz Inequality
}}
{{eqn | o = \le
| r = \norm {A^* A} \norm h_H^2
| c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{eqn | o = \le
| r = \norm {A^* A}
| c = Assumption on $h$
}}
{{eqn | o = \le
| r = \norm {A^*} \norm A
| c = Norm on Bounded Linear Transformation is Submultiplicative
}}
{{end-eqn}}
By definition $(1)$ for $\norm A$, it follows that:
:$\norm A^2 \le \norm {A^* A} \le \norm {A^*} \norm A$
That is:
:$\norm A \le \norm {A^*}$.
By substituting $A^*$ for $A$, and using $A^{**} = A$ from Adjoint is Involutive, the reverse inequality is obtained.
Hence $\norm A^2 = \norm {A^* A} = \norm {A^*}^2$.
{{qed}}
\end{proof}
|
23790
|
\section{Norm of Eisenstein Integer}
Tags: Number Theory, Complex Analysis, Algebraic Number Theory
\begin{theorem}
Let $\alpha$ be an Eisenstein integer.
That is, $\alpha = a + b \omega$ for some $a, b \in \Z$, where $\omega = e^{2\pi i /3}$.
Then:
:$\cmod \alpha^2 = a^2 - a b + b^2$
where $\cmod {\, \cdot \,}$ denotes the modulus of a complex number.
\end{theorem}
\begin{proof}
We find that:
{{begin-eqn}}
{{eqn | l = \cmod \alpha^2
| r = \alpha \overline \alpha
| c = Modulus in Terms of Conjugate
}}
{{eqn | r = \paren {a + b \omega} \paren {\overline {a + b \omega} }
| c = Modulus in Terms of Conjugate
}}
{{eqn | r = \paren {a + b \omega} \paren {\overline a + \overline b \overline \omega}
| c = Sum of Complex Conjugates and Product of Complex Conjugates
}}
{{eqn | r = \paren {a + b \omega} \paren {a + b \overline \omega}
| c = Complex Number equals Conjugate iff Wholly Real
}}
{{eqn | r = a^2 + \paren {\omega + \overline \omega} a b + \omega \overline \omega b^2
| c =
}}
{{end-eqn}}
By the definition of the polar form of a complex number:
:$\omega = \exp \paren {\dfrac {2 \pi i} 3} = \map \cos {\dfrac {2 \pi} 3} + i \, \map \sin {\dfrac {2 \pi} 3} = -\dfrac 1 2 + i \dfrac {\sqrt 3} 2$
Thus by Sum of Complex Number with Conjugate:
:$\omega + \overline \omega = 2 \cdot \paren {-\dfrac 1 2} = -1$
Also:
{{begin-eqn}}
{{eqn | l = \omega \overline \omega
| r = \map \exp {\dfrac {2 \pi i} 3} \, \overline {\map \exp {\dfrac {2 \pi i} 3} }
| c =
}}
{{eqn | r = \map \exp {\dfrac {2 \pi i} 3} \, \map \exp {-\dfrac {2 \pi i} 3}
| c = Polar Form of Complex Conjugate
}}
{{eqn | r = \map \exp {\dfrac {2 \pi i} 3 - \dfrac {2 \pi i} 3}
| c = Exponential of Sum
}}
{{eqn | r = \map \exp 0
| c =
}}
{{eqn | r = 1
| c = Exponential of Zero
}}
{{end-eqn}}
Therefore:
:$\cmod \alpha^2 = a^2 + \paren {\omega + \overline \omega} a b + \omega \overline \omega b^2 = a^2 - a b + b^2$
as required.
{{qed}}
Category:Algebraic Number Theory
\end{proof}
|
23791
|
\section{Norm of Hermitian Operator}
Tags: Adjoints, Definitions: Adjoints
\begin{theorem}
Let $\mathbb F \in \set {\R, \C}$.
Let $\HH$ be a Hilbert space over $\mathbb F$.
Let $A : \HH \to \HH$ be a bounded Hermitian operator.
Let $\innerprod \cdot \cdot_\HH$ denote the inner product on $\HH$.
Let $\norm \cdot_\HH$ be the inner product norm on $\HH$.
Then the norm of $A$ satisfies:
:$\norm A = \sup \set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1}$
\end{theorem}
\begin{proof}
Let:
:$M = \sup \set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1}$
To show that $M = \norm A$ we first show that:
:$M \le \norm A$
We will then show that:
:$\norm A \le M$
Let $x \in \HH$ be such that:
:$\norm x_\HH = 1$.
Then we have:
{{begin-eqn}}
{{eqn | l = \size {\innerprod {A x} x_\HH}
| o = \le
| r = \norm {A x}_\HH \norm x_\HH
| c = Cauchy-Bunyakovsky-Schwarz Inequality: Inner Product Spaces
}}
{{eqn | o = \le
| r = \norm A \norm x_\HH^2
| c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{eqn | r = \norm A
| c = since $\norm x_\HH = 1$
}}
{{end-eqn}}
So taking the supremum over:
:$\set {x \in \HH : \norm x_\HH = 1}$
we have:
:$M \le \norm A$
We will now show that:
:$\norm A \le M$
Let $x, y \in \HH$.
Since $A$ is linear operator, we have:
:$\innerprod {\map A {u + v} } {u + v}_\HH = \innerprod {A u + A v} {u + v}_\HH$
and:
:$\innerprod {\map A {u - v} } {u - v}_\HH = \innerprod {A u - A v} {u - v}_\HH$
Using Inner Product is Sesquilinear, we have:
:$\innerprod {\map A {u + v} } {u + v}_\HH = \innerprod {A u} u_\HH + \innerprod {A u} v_\HH + \innerprod {A v} u_\HH + \innerprod {A v} v_\HH$
and:
:$\innerprod {\map A {u - v} } {u - v}_\HH = \innerprod {A u} u_\HH - \innerprod {A u} v_\HH - \innerprod {A v} u_\HH + \innerprod {A v} v_\HH$
We therefore obtain:
:$\innerprod {\map A {u + v} } {u + v}_\HH - \innerprod {\map A {u - v} } {u - v}_\HH = 2 \innerprod {A u} v_\HH + 2 \innerprod {A v} u_\HH$
We have:
{{begin-eqn}}
{{eqn | l = 2 \innerprod {A u} v_\HH + 2 \innerprod {A v} u_\HH
| r = 2 \paren {\innerprod {A u} v_\HH + \innerprod {A v} u_\HH}
}}
{{eqn | r = 2 \paren {\innerprod {A u} v_\HH + \innerprod v {A u}_\HH }
| c = {{Defof|Hermitian Operator}}
}}
{{eqn | r = 2 \paren {\innerprod {A u} v_\HH + \overline {\innerprod {A u} v_\HH} }
| c = using conjugate symmetry of the inner product
}}
{{eqn | r = 4 \map \Re {\innerprod {A u} v_\HH}
| c = Sum of Complex Number with Conjugate
}}
{{end-eqn}}
so we have:
:$4 \map \Re {\innerprod {A u} v_\HH} = \innerprod {\map A {u + v} } {u + v}_\HH - \innerprod {\map A {u - v} } {u - v}_\HH$
Recall that for each $h \in \HH$ with $\norm h_\HH = 1$, we have:
:$\size {\innerprod {A h} h_\HH} \le M$
by the definition of supremum.
From Operator is Hermitian iff Inner Product is Real, we have:
:$\innerprod {A h} h_\HH$ is a real number.
So, from the definition of the absolute value, we have:
:$\innerprod {A h} h_\HH \le M$
We can therefore see that:
:$\ds \innerprod {A \frac h {\norm h_\HH} } {\frac h {\norm h_\HH} }_\HH \le M$
for each $h \in \HH \setminus \set 0$.
So from Inner Product is Sesquilinear, we obtain:
:$\innerprod {A h} h_\HH \le M \norm h_\HH^2$
Note that if $h = 0$, from Inner Product with Zero Vector we have:
:$\innerprod {A h} h_\HH = 0$
and:
:$\norm h_\HH = 0$
so the inequality also holds for $h = 0$, and we obtain:
:$\innerprod {A h} h_\HH \le M \norm h_\HH^2$
for all $h \in \HH$.
So, we obtain:
{{begin-eqn}}
{{eqn | l = 4 \map \Re {\innerprod {A u} v_\HH}
| r = \innerprod {\map A {u + v} } {u + v}_\HH - \innerprod {\map A {u - v} } {u - v}_\HH
}}
{{eqn | o = \le
| r = M \norm {u + v}_\HH^2 - M \norm {u - v}_\HH^2
| c = applying the above inequality with $h = u + v$ and $h = u - v$
}}
{{eqn | o = \le
| r = M \paren {\norm {u + v}_\HH^2 + \norm {u - v}_\HH^2}
| c = {{Defof|Norm on Vector Space}}
}}
{{eqn | r = 2 M \paren {\norm u_\HH^2 + \norm v_\HH^2}
| c = Parallelogram Law (Inner Product Space)
}}
{{end-eqn}}
Now, take $u \in \HH$ with $A u \ne 0$.
Let:
:$v = \dfrac {\norm u_\HH} {\norm {A u}_\HH} A u$
Then, we have:
{{begin-eqn}}
{{eqn | l = \norm v_\HH
| r = \norm {\frac {\norm u_\HH} {\norm {A u}_\HH} A u}_\HH
}}
{{eqn | r = \frac {\norm u_\HH} {\norm {A u}_\HH} \norm {A u}_\HH
| c = using positive homogeneity of the norm
}}
{{eqn | r = \norm u_\HH
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \innerprod {A u} v_\HH
| r = \innerprod {A u} {\frac {\norm u_\HH} {\norm {A u}_\HH} A u}_\HH
}}
{{eqn | r = \frac {\norm u_\HH} {\norm {A u}_\HH} \innerprod {A u} {A u}_\HH
| c = Inner Product is Sesquilinear
}}
{{eqn | r = \frac {\norm u_\HH} {\norm {A u}_\HH} \norm {A u}_\HH^2
| c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \norm u_\HH \norm {A u}_\HH
}}
{{end-eqn}}
Since from the definition of a norm:
:$\norm u_\HH \norm {A u}_\HH$ is a real number
we have that:
:$\innerprod {A u} v$ is a real number.
So we have:
:$4 \map \Re {\innerprod {A u} v} = 4 \norm u_\HH \norm {A u}_\HH$
giving:
:$4 \norm u_\HH \norm {A u}_\HH \le 2 M \paren {\norm u_\HH^2 + \norm v_\HH^2}$
Since $\norm u = \norm v$, we have:
:$4 \norm u_\HH \norm {A u}_\HH \le 4 M \norm u_\HH^2$
That is:
:$\norm u_\HH \norm {A u}_\HH \le M \norm u_\HH^2$
for all $u \in \HH$ with $A u \ne 0$.
Note that we have:
:$M \norm u_\HH^2 \ge 0$
so the inequality also holds for $u \in \HH$ with $A u = 0$.
So, for $u \in \HH \setminus \set 0$, we have:
:$\norm {A u}_\HH \le M \norm u_\HH$
Since $\map A 0 = 0$, this inequality also holds for $u = 0$.
So:
:$M \in \set {c > 0: \forall h \in \HH: \norm {A h}_\HH \le c \norm h_\HH}$
From definition 4 of the operator norm, we have:
:$\norm A = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\HH \le c \norm h_\HH}$
so, from the definition of infimum we have:
:$\norm A \le M$
Since:
:$M \le \norm A$
and:
:$\norm A \le M$
we have:
:$\norm A = M = \sup \set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1}$
{{qed}}
\end{proof}
|
23792
|
\section{Norm of Unit of Normed Division Algebra}
Tags: Norm Theory, Algebras
\begin{theorem}
Let $\struct {A_F, \oplus}$ be a normed division algebra.
Let the unit of $\struct {A_F, \oplus}$ be $1_A$.
Then:
:$\norm {1_A} = 1$
where $\norm {1_A}$ denotes the norm of $1_A$.
\end{theorem}
\begin{proof}
By definition:
:$\forall a, b \in A_F: \norm {a \oplus b} = \norm a \norm b$
So:
:$\norm {1_A} = \norm {1_A \oplus 1_A} = \norm {1_A} \norm {1_A}$
So $\norm {1_A} \in \R$ is idempotent under real multiplication.
From Idempotent Elements of Ring with No Proper Zero Divisors, only $0 \in \R$ and $1 \in \R$ fit that bill.
But $\norm {1_A}$ can not be $0$ as that would make:
:$\forall a \in A_F: \norm a = \norm {1_A \oplus a} = \norm {1_A} \norm a = 0 \norm a = 0$
which contradicts the definition of the norm.
Hence the result.
{{Qed}}
\end{proof}
|
23793
|
\section{Norm of Vector Cross Product}
Tags: Vector Cross Product, Vector Algebra
\begin{theorem}
Let $\mathbf a$ and $\mathbf b$ be vectors in the Euclidean space $\R^3$.
Let $\times$ denote the vector cross product.
Then:
:$(1): \quad$ $\left\Vert{ \mathbf a \times \mathbf b }\right\Vert^2 = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left({\mathbf a \cdot \mathbf b}\right)^2$
:$(2): \quad$ $\left\Vert{ \mathbf a \times \mathbf b }\right\Vert = \left\Vert{\mathbf a}\right\Vert \left\Vert{\mathbf b}\right\Vert \left\vert{\sin \theta}\right\vert$
where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$, or an arbitrary number if $\mathbf a$ or $\mathbf b$ is the zero vector.
\end{theorem}
\begin{proof}
Let $\mathbf a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$, and $\mathbf b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$.
Then:
{{begin-eqn}}
{{eqn | l = \left\Vert{ \mathbf a \times \mathbf b }\right\Vert^2
| r = \left({ \mathbf a \times \mathbf b }\right) \cdot \left({ \mathbf a \times \mathbf b }\right)
| c = {{Defof|Euclidean Norm}}
}}
{{eqn | r = \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{bmatrix} \cdot \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{bmatrix}
| c = {{Defof|Vector Cross Product}}
}}
{{eqn | r = a_2^2 b_3^2 + a_3^2 b_2^2 - 2a_2 a_3 b_2 b_3 + a_3^2 b_1^2 + a_1^2 b_3^2 - 2a_1 a_3 b_1 b_3 + a_1^2 b_2^2 + a_2^2 b_1^2 - 2a_1 a_2 b_1 b_2
| c = {{Defof|Dot Product}}
}}
{{eqn | r = \left({a_1^2 + a_2^2 + a_3^2}\right) \left({b_1^2 + b_2^2 + b_3^2}\right) - \left({a_1 b_1 + a_2 b_2 + a_3 b_3}\right)^2
| c = by algebraic manipulations
}}
{{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left({\mathbf a \cdot \mathbf b}\right)^2
}}
{{end-eqn}}
This proves $(1)$.
{{qed|lemma}}
If $\mathbf a$ or $\mathbf b$ is the zero vector, then $\left\Vert{\mathbf a}\right\Vert = 0$, or $\left\Vert{\mathbf b}\right\Vert = 0$ by the positive definiteness norm axiom.
By calculation, it follows that $\mathbf a \times \mathbf b$ is also the zero vector, so $\left\Vert{\mathbf a \times \mathbf b}\right\Vert = 0$.
Hence, equality $(2)$ holds.
If both $\mathbf a$ or $\mathbf b$ are non-zero vectors, we continue the calculations from the first section:
{{begin-eqn}}
{{eqn | l = \left\Vert{ \mathbf a \times \mathbf b }\right\Vert^2
| r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left({\mathbf a \cdot \mathbf b}\right)^2
}}
{{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 - \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 \cos^2 \theta
| c = Cosine Formula for Dot Product
}}
{{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 \left({1 - \cos^2 \theta}\right)
}}
{{eqn | r = \left\Vert{\mathbf a}\right\Vert^2 \left\Vert{\mathbf b}\right\Vert^2 \sin^2 \theta
| c = Sum of Squares of Sine and Cosine
}}
{{end-eqn}}
Equality $(2)$ now follows after taking the square root of both sides of the equality.
This is possible as Square of Real Number is Non-Negative.
{{qed}}
\end{proof}
|
23794
|
\section{Norm on Bounded Linear Functional is Finite}
Tags: Hilbert Spaces
\begin{theorem}
Let $H$ be a Hilbert space.
Let $L$ be a bounded linear functional on $H$.
Let $\norm L$ denote the norm on $L$ defined as:
:$\norm L = \inf \set {c > 0: \forall h \in H: \size {L h} \le c \norm h_H}$
Then:
:$\norm L < \infty$
\end{theorem}
\begin{proof}
By definition of a bounded linear functional:
:$\exists c \in \R_{> 0}: \forall h \in H: \size{L h} \le c \norm h_H$
Hence:
:$\set {\lambda > 0: \forall h \in H: \size {L h} \le \lambda \norm h_H} \ne \O$
By definition:
$\set {\lambda > 0: \forall h \in H: \size {L h} \le \lambda \norm h_H}$ is bounded below by $0$.
From Corollary to Continuum Property:
:$\norm L = \inf \set {\lambda > 0: \forall h \in H: \size {L h} \le \lambda \norm h_H}$ exists.
We have:
{{begin-eqn}}
{{eqn | l = \norm L
| o = \le
| r = c
| c = {{Defof|Infimum}}
}}
{{eqn | o = <
| r = \infty
| c = because $c \in \R_{>0}$
}}
{{end-eqn}}
The result follows.
{{qed}}
Category:Hilbert Spaces
\end{proof}
|
23795
|
\section{Norm on Bounded Linear Transformation is Finite}
Tags: Hilbert Spaces
\begin{theorem}
Let $H, K$ be Hilbert spaces.
Let $A: H \to K$ be a bounded linear transformation.
Let $\norm A$ denote the norm of $A$ defined by:
:$\norm A = \inf \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$
Then:
:$\norm A < \infty$
\end{theorem}
\begin{proof}
By definition of a bounded linear transformation:
:$\exists c \in \R_{> 0}: \forall h \in H: \norm{A h}_K \le c \norm h_H$
Hence:
:$\set {\lambda > 0: \forall h \in H: \norm {A h}_K \le \lambda \norm h_H} \ne \O$
By definition:
:$\set {\lambda > 0: \forall h \in H: \norm {A h}_K \le \lambda \norm h_H}$ is bounded below.
From Corollary to Continuum Property:
:$\norm A = \inf \set {\lambda > 0: \forall h \in H: \norm {A h}_K \le \lambda \norm h_H}$ exists.
We have:
{{begin-eqn}}
{{eqn | l = \norm A
| o = \le
| r = c
| c = {{Defof|Infimum}}
}}
{{eqn | o = <
| r = \infty
| c = As $c \in \R_{> 0}$
}}
{{end-eqn}}
The result follows.
{{qed}}
Category:Hilbert Spaces
\end{proof}
|
23796
|
\section{Norm on Bounded Linear Transformation is Submultiplicative}
Tags: Linear Transformations on Hilbert Spaces, Bounded Linear Transformations
\begin{theorem}
Let $\struct {X, \norm \cdot_X}$, $\struct {Y, \norm \cdot_Y}$ and $\struct {Z, \norm \cdot_Z}$ be normed vector spaces.
Let $A : X \to Y$ and $B : Y \to Z$ be bounded linear transformations.
Let $\norm \cdot$ be the norm on the space of bounded linear transformations.
Then, we have:
:$\norm {B A} \le \norm B \norm A$
That is:
:$\norm \cdot$ is submultiplicative.
\end{theorem}
\begin{proof}
Let $x \in X$.
Then, we have:
{{begin-eqn}}
{{eqn | l = \norm {\paren {B A} x}_Z
| o = \le
| r = \norm B \norm {A x}_Y
| c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{eqn | o = \le
| r = \norm B \norm A \norm x_X
| c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{end-eqn}}
So, if:
:$\norm x_X = 1$
we have:
:$\norm {\paren {B A} x}_Z \le \norm B \norm A$
By the definition of supremum, we have:
:$\ds \sup_{\norm x_X = 1} \norm {\paren {B A} x}_Z \le \norm B \norm A$
So by the definition of the norm on the space of bounded linear transformations, we have:
:$\norm {B A} \le \norm B \norm A$
{{qed}}
Category:Bounded Linear Transformations
\end{proof}
|
23797
|
\section{Norm on Vector Space is Continuous Function}
Tags: Norm Theory
\begin{theorem}
Let $V$ be a vector space with norm $\norm {\, \cdot \,}$.
The function $\norm {\, \cdot \,}: V \to \R$ is continuous.
\end{theorem}
\begin{proof}
Let $x_n \to x$ in $V$.
We have:
:$x_n \to x \implies \norm {x_n - x} \to 0$
By the Reverse Triangle Inequality:
:$\size {\norm {x_n} - \norm x} \le \norm {x_n - x}$
Hence:
:$\size {\norm {x_n} - \norm x} \to 0$
{{Proofread|Check the steps up to here}}
Thus:
:$\norm {x_n} \to \norm x$
Hence the result from the definition of a continuous real function.
{{Proofread|Check whether the reason for continuity is correct}}
Category:Norm Theory
\end{proof}
|
23798
|
\section{Norm satisfying Parallelogram Law induced by Inner Product}
Tags: Normed Vector Spaces, Inner Product Spaces, Norm satisfying Parallelogram Law induced by Inner Product
\begin{theorem}
Let $V$ be a vector space over $\R$.
Let $\norm \cdot : V \to \R$ be a norm on $V$ such that:
:$\norm {x + y}^2 + \norm {x - y}^2 = 2 \paren {\norm x^2 + \norm y^2}$
for each $x, y \in V$.
Then the function $\innerprod \cdot \cdot : V \times V \to \R$ defined by:
:$\ds \innerprod x y = \frac {\norm {x + y}^2 - \norm {x - y}^2} 4$
for each $x, y \in V$, is an inner product on $V$.
Further, $\norm \cdot$ is the inner product norm of $\struct {V, \innerprod \cdot \cdot}$.
\end{theorem}
\begin{proof}
We first verify symmetry.
Let $x, y \in V$.
Then we have:
{{begin-eqn}}
{{eqn | l = \innerprod y x
| r = \frac {\norm {y + x}^2 - \norm {y - x}^2} 4
}}
{{eqn | r = \frac {\norm {x + y}^2 - \norm {-\paren {x - y} }^2} 4
}}
{{eqn | r = \frac {\norm {x + y}^2 - \norm {x - y}^2} 4
| c = from the positive homogeneity of the norm $\norm \cdot$
}}
{{eqn | r = \innerprod x y
}}
{{end-eqn}}
We now show non-negative definiteness and positiveness.
We have for each $x \in V$:
{{begin-eqn}}
{{eqn | l = \innerprod x x
| r = \frac {\norm {x + x}^2 - \norm {x - x}^2} 4
}}
{{eqn | r = \frac {\norm {2 x}^2 - \norm 0^2} 4
}}
{{eqn | r = \frac {4 \norm x^2} 4
| c = from the positive homogeneity of the norm $\norm \cdot$
}}
{{eqn | r = \norm x^2
}}
{{end-eqn}}
Since norms are positive definite, we then have:
:$\innerprod x x \ge 0$
for each $x \in V$, with:
:$\innerprod x x = 0$
{{iff}} $x = 0$.
We finally show linearity in the first argument.
We first show that for each $x, y, z \in V$ we have:
:$\innerprod {x + y} z = \innerprod x z + \innerprod y z$
We have:
{{begin-eqn}}
{{eqn | l = \innerprod x y + \innerprod y z
| r = \frac 1 4 \paren {\norm {x + y}^2 - \norm {x - y}^2 + \norm {y + z}^2 - \norm {y - z}^2}
}}
{{eqn | r = \frac 1 4 \paren {\norm {x + z}^2 + \norm y^2 - \norm {x - z}^2 - \norm y^2 + \norm x^2 + \norm {y + z}^2 - \norm x^2 - \norm {y - z}^2}
| c = adding $\dfrac 1 4 \paren {\norm y^2 - \norm y^2 + \norm x^2 - \norm x^2} = 0$
}}
{{end-eqn}}
Note that by hypothesis, we have:
:$\norm {f + g}^2 + \norm {f - g}^2 = 2 \paren {\norm f^2 + \norm g^2}$
for each $f, g \in V$.
By the hypothesis, setting $f = x + z$ and $g = y$, we have:
:$\norm {x + y + z}^2 + \norm {x - y + z}^2 = 2 \paren {\norm {x + z}^2 + \norm y^2}$
similarly, setting $f = x - z$ and $g = y$:
:$\norm {x + y - z}^2 + \norm {x - y - z}^2 = 2 \paren {\norm {x - z}^2 - \norm y^2}$
Setting $f = x$ and $g = y + z$ we also obtain:
:$\norm {x + y + z}^2 + \norm {x - y - z}^2 = 2 \paren {\norm x^2 + \norm {y + z}^2}$
Finally setting $f = x$ and $g = y - z$ we get:
:$\norm {x + y - z}^2 + \norm {x - y + z}^2 = 2 \paren {\norm x^2 + \norm {y - z}^2}$
Putting this all together, we have:
{{begin-eqn}}
{{eqn | l = \frac 1 4 \paren {\norm {x + z}^2 + \norm y^2 - \norm {x - z}^2 - \norm y^2 + \norm x^2 + \norm {y + z}^2 - \norm x^2 - \norm {y - z}^2}
| r = \frac 1 8 \paren {\norm {x + y + z}^2 + \norm {x - y + z}^2 - \norm {x + y - z}^2 - \norm {x - y - z}^2 + \norm {x + y + z}^2 + \norm {x - y - z}^2 - \norm {x + y - z}^2 - \norm {x - y + z}^2}
}}
{{eqn | r = \frac 2 8 \paren {\norm {x + y + z}^2 - \norm {x + y - z}^2}
}}
{{eqn | r = \frac 1 4 \paren {\norm {x + y + z}^2 - \norm {x + y - z}^2}
}}
{{eqn | r = \innerprod {x + y} z
}}
{{end-eqn}}
We now show that:
:$\innerprod {\alpha x} y = \alpha \innerprod x y$
for any $\alpha \in \R$ and $x, y \in V$.
We then have the lemma:
\end{proof}
|
23799
|
\section{Normal Space is Preserved under Homeomorphism}
Tags: Separation Axioms, Normal Spaces
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a normal space, then so is $T_B$.
\end{theorem}
\begin{proof}
We have that $\struct {S, \tau}$ is a normal space {{iff}}:
:$\struct {S, \tau}$ is a $T_4$ space
:$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.
From $T_4$ Space is Preserved under Homeomorphism:
:If $T_A$ is a $T_4$ space, then so is $T_B$.
From $T_1$ Space is Preserved under Homeomorphism:
:If $T_A$ is a $T_1$ (Fréchet) space, then so is $T_B$.
Hence the result.
{{qed}}
\end{proof}
|
23800
|
\section{Nicely Normed Cayley-Dickson Construction from Associative Algebra is Alternative}
Tags: Cayley-Dickson Construction
\begin{theorem}
Let $A = \left({A_F, \oplus}\right)$ be a $*$-algebra.
Let $A' = \left({A_F, \oplus'}\right)$ be constructed from $A$ using the Cayley-Dickson construction.
Then $A'$ is a nicely normed alternative algebra {{iff}} $A$ is a nicely normed associative algebra.
\end{theorem}
\begin{proof}
Let the conjugation operator on $A$ be $*$.
Let $\left({a, b}\right), \left({c, d}\right) \in A'$.
In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by product notation:
:$a \oplus b =: a b$
:$x \oplus' y =: x y$
The context will make it clear which is meant.
Let $A$ be a nicely normed associative algebra.
Then:
{{begin-eqn}}
{{eqn | l = \left({\left({a, b}\right) \left({a, b}\right)}\right) \left({c, d}\right)
| r = \left({a a - b b^*, a^* b + a b}\right) \left({c, d}\right)
| c =
}}
{{eqn | r = \left({\left({a a - b b^*}\right) c - d \left({a^* b + a b}\right)^*, \left({a a - b b^*}\right)^* d + c \left({a^* b + a b}\right)}\right)
| c =
}}
{{eqn | r = \left({\left({a a - b b^*}\right) c - d \left({b^* a + b^* a^*}\right), \left({a^* a^* - b b^*}\right) d + c \left({a^* b + a b}\right)}\right)
| c = from definition of conjugation
}}
{{eqn | r = \left({a a c - b b^* c - d b^* a - d b^* a^*, a^* a^* d - b b^* d + c a^* b + c a b}\right)
| c = $A$ is associative
}}
{{eqn | r = \left({a a c - b b^* c - d b^* \left({a + a^*}\right), a^* a^* d - b b^* d + c \left({a^* + a}\right) b}\right)
| c =
}}
{{eqn | r = \left({a a c - \left \Vert {b}\right \Vert^2 c - d b^* \left({2 \Re \left({a}\right)}\right), a^* a^* d - \left \Vert {b}\right \Vert^2 d + c \left({2 \Re \left({a}\right)}\right) b}\right)
| c = $A$ is nicely normed
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \left({a, b}\right) \left({\left({a, b}\right) \left({c, d}\right)}\right)
| r = \left({a, b}\right) \left({a c - d b^*, a^* d + c b}\right)
| c =
}}
{{eqn | r = \left({a \left({a c - d b^*}\right) - \left({a^* d + c b}\right) b^*, a^* \left({a^* d - c b}\right) + \left({a c + d b^*}\right) b}\right)
| c =
}}
{{eqn | r = \left({a a c - a d b^* - a^* d b^* - c b b^*, a^* a^* d + a^* c b + a c b - d b^* b}\right)
| c = $A$ is associative
}}
{{eqn | r = \left({a a c - \left({a + a^*}\right) d b^* - c \left \Vert {b}\right \Vert^2, a^* a^* d + \left({a^* + a}\right) c b - d b^* b}\right)
| c =
}}
{{eqn | r = \left({a a c - \left({2 \Re \left({a}\right)}\right) d b^* - c \left \Vert {b}\right \Vert^2, a^* a^* d + \left({2 \Re \left({a}\right)}\right) c b - d \left \Vert {b}\right \Vert^2}\right)
| c = $A$ is nicely normed
}}
{{end-eqn}}
Thus it can be seen that:
: $\left({\left({a, b}\right) \left({a, b}\right)}\right) \left({c, d}\right) = \left({a, b}\right) \left({\left({a, b}\right) \left({c, d}\right)}\right)$
Similarly it can be shown that:
: $\left({\left({c, d}\right) \left({a, b}\right)}\right) \left({a, b}\right) = \left({c, d}\right) \left({\left({a, b}\right) \left({a, b}\right)}\right)$
and so $A'$ is seen to be an alternative algebra.
It follows from reversing the chain of equalities that if $A'$ is a nicely normed and alternative algebra then $A$ has to be a nicely normed associative algebra.
{{qed|lemma}}
Then from Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed, we have that $A'$ is a nicely normed algebra {{iff}} $A$ is also a nicely normed algebra.
Hence the result.
{{qed}}
\end{proof}
|
23801
|
\section{Niemytzki Plane is Topology}
Tags: Niemytzki Plane
\begin{theorem}
Niemytzki plane is a topological space.
\end{theorem}
\begin{proof}
By definition $T = \struct {S, \tau}$ is the Niemytzki plane {{iff}}:
{{begin-eqn}}
{{eqn | n = 1
| l = S
| r = \set {\tuple {x, y}: y \ge 0}
}}
{{eqn | n = 2
| l = \map \BB {x, y}
| r = \set {\map {B_r} {x, y} \cap S: r > 0}
| c = if $x, y \in \R, y > 0$
}}
{{eqn | n = 3
| l = \map \BB {x, 0}
| r = \set {\map {B_r} {x, r} \cup \set {\tuple {x, 0} }: r > 0}
| c = if $x \in \R$
}}
{{eqn | n = 4
| l = \tau
| r = \set {\bigcup \GG: \GG \subseteq \bigcup_{\tuple {x, y} \in S} \map \BB {x, y} }
}}
{{end-eqn}}
where $\map {B_r} {x, y}$ denotes the open $r$-ball of $\tuple {x, y}$ in the real number plane $\R^2$ with the usual (Euclidean) metric.
According to Topology Defined by Neighborhood System it should be proved:
:$(\text N 0): \quad \forall \tuple {x, y} \in S: \map \BB {x, y}$ is non-empty set of subsets of $S$
:$(\text N 1): \quad \forall \tuple {x, y} \in S, U \in \map \BB {x, y}: \tuple {x, y} \in U$
:$(\text N 2): \quad \forall \tuple {x, z} \in S, U \in \map \BB {x, z}, \tuple {y, s} \in U: \exists V \in \map \BB {y, s}: V \subseteq U$
:$(\text N 3): \quad \forall \tuple {x, y} \in S, U_1, U_2 \in \map \BB {x, y}: \exists U \in \map \BB {x, y}: U \subseteq U_1 \cap U_2$
Ad $(\text N 0)$:
Let $\tuple {x, y} \in S$.
In both cases: $y > 0$ and $y = 0$, by $(2)$ and $(3)$:
:$\map \BB {x, y}$ is non-empty
Let $U \in \map \BB {x, y}$.
In a case when $y > 0$:
:$\exists r > 0: U = \map {B_r} {x, y} \cap S$
By Intersection is Subset:
:$U$ is a subset of $S$.
In a case when $y = 0$:
:$\exists r > 0: U = \map {B_r} {x, r} \cup \set {\tuple {x, 0} }$
By definition of subset:
:$\map {B_r} {x, r} \subseteq S$ and $\set {\tuple {x, 0} } \subseteq S$
By Union is Smallest Superset:
:$U$ is a subset of $S$.
Ad $(\text N 1)$:
Let $\tuple {x, y} \in S$.
Let $U \in \map \BB {x, y}$.
In a case when $y > 0$ by $(2)$:
:$\exists r > 0: U = \map {B_r} {x, y} \cap S$
By the metric space axioms:
:$\map d {\tuple {x, y}, \tuple {x, y} } = 0$
where $d$ denotes the distance function of $\R^2$ Euclidean space.
By definition of open ball:
:$\tuple {x, y} \in \map {B_r} {x, y}$
Thus by definition of intersection:
:$\tuple {x, y} \in U$
In a case when $y = 0$ by $(3)$:
:$\exists r > 0: U = \map {B_r} {x, r} \cup \set {\tuple {x, 0} }$
By definition of singleton:
:$\tuple {x, 0} \in \set {\tuple {x, 0} }$
Thus by definition of union:
:$\tuple {x, 0} \in U$
Thus in both cases:
:$\tuple {x, 0} \in U$
Ad $(\text N 2)$:
Let $\tuple {x, 0} \in S, U \in \map \BB {x, y}, \tuple {z, s} \in U$.
In a case when $y > 0$ by $(2)$:
:$\exists r > 0: U = \map {B_r} {x, y} \cap S$
By definition of intersection:
:$\tuple {z, s} \in \map {B_r} {x, y}$
By definition of open ball:
:$\map d {\tuple {x, y}, \tuple {z, s} } < r$
In a subcase when $s > 0$ define
:$r' := r - \map d {\tuple {x, y}, \tuple {z, s} }$
and
:$V := \map {B_{r'} } {z, s} \cap S$
By $(2)$:
:$V \in \map \BB {z, s}$
By proof of Open Ball of Point Inside Open Ball:
:$\map {B_{r'} } {z, s} \subseteq \map {B_r} {x, y}$
Thus by Set Intersection Preserves Subsets/Corollary:
:$V \subseteq U$
In a subcase when $s = 0$ define
:$r'' := \dfrac {r - \map d {\tuple {x, y}, \tuple {z, s} } } 2$
and
:$V := \map {B_{r''} } {z, r''} \cup \set {\tuple {z, 0} }$
By $(3)$:
:$V \in \map \BB {z, s}$
{{begin-eqn}}
{{eqn | l = \map d {\tuple {x, y}, \tuple {z, r''} }
| o = \le
| r = \map d {\tuple {x, y}, \tuple {z, 0} } + \map d {\tuple {z, 0}, \tuple {z, r''} }
| c = Metric Space Axioms
}}
{{eqn | r = \map d {\tuple {x, y}, \tuple {z, 0} } + r''
| c = {{Defof|Real Number Plane with Euclidean Metric}}
}}
{{eqn | r = r - r''}}
{{end-eqn}}
By Open Ball is Subset of Open Ball:
:$\map {B_{r''} } {z, r''} \subseteq \map {B_r} {x, y}$
and
:$\map {B_{r''} } {z, r''} \subseteq S$
By Intersection is Largest Subset:
:$\map {B_{r''} } {z, r''} \subseteq U$
By definition of subset:
:$\set {\tuple {z, 0} } \subseteq U$
Thus by Union is Smallest Superset:
:$V \subseteq U$
In a case when $y = 0$ by $(3)$:
:$\exists r > 0: U = \map {B_r} {x, r} \cup \set {\tuple {x, 0} }$
By definition of onion and singleton:
:$\tuple {z, s} \in \map {B_r} {x, r}$ or $\tuple {z, s} = \tuple {x, 0}$
In a subcase when $\tuple {z, s} = \tuple {x, 0}$ define:
:$V := U$
By $(3)$:
:$V \in \map \BB {z, s}$
Thus by Set is Subset of Itself:
:$V \subseteq U$
In a subcase when $\tuple {z, s} \in \map {B_r} {x, r}$, define:
:$r' := r - \map d {\tuple {x, r}, \tuple {z, s} }$
and
:$V := \map {B_{r'} } {z, s} \cap S$
By definition of open ball:
:$\map d {\tuple {x, r}, \tuple {z, s} } < r$
Then $r' > 0$
By $(2)$:
:$V \in \map \BB {z, s}$
By Open Ball of Point Inside Open Ball:
:$\map {B_{r'} } {z, s} \subseteq \map {B_{r'} } {x, r}$
By Intersection is Subset:
:$V = \map {B_{r'} } {z, s} \cap S \subseteq \map {B_{r'} } {z, s}$
By Set is Subset of Union:
:$\map {B_{r'} } {x, r} \subseteq U$
Thus by Subset Relation is Transitive:
$V \subseteq U$
Ad $(\text N 3)$:
Let $\tuple {x, y} \in S, U_1, U_2 \in \map \BB {x, y}$.
In a case when $y > 0$ by $(2)$:
:$\exists r_1 > 0: U_1 = \map {B_{r_1} } {x, y} \cap S$
and
:$\exists r_2 > 0: U_2 = \map {B_{r_2} } {x, y} \cap S$
By Trichotomy Law for Real Numbers:
:$r_1 \le r_2$ or $r_1 \ge r_2$
{{WLOG}}, suppose
:$r_1 \le r_2$
{{begin-eqn}}
{{eqn | l = r_2 - r_1
| o = \ge
| r = 0
}}
{{eqn | r = \map d {\tuple {x, y}, \tuple {x, y} }
| c = Metric Space Axioms
}}
{{end-eqn}}
Then by Open Ball is Subset of Open Ball:
:$\map {B_{r_1} } {x, y} \subseteq \map {B_{r_2} } {x, y}$
By Set Intersection Preserves Subsets/Corollary:
:$U_1 \subseteq U_2$
By Intersection with Subset is Subset:
:$U_1 = U_1 \cap U_2$
Thus by Set is Subset of Itself:
:$U_1 \subseteq U_1 \cap U_2$
In a case when $y = 0$ by $(3)$:
:$\exists r_1 > 0: U_1 = \map {B_{r_1} } {x, r_1} \cup \set {\tuple {x, 0} }$
and
:$\exists r_2 > 0: U_2 = \map {B_{r_2} } {x, r_2} \cup \set {\tuple {x, 0} }$
By Trichotomy Law for Real Numbers:
:$r_1 \le r_2$ or $r_1 \ge r_2$
{{WLOG}}, suppose
:$r_1 \le r_2$
{{begin-eqn}}
{{eqn | l = r_2 - r_1
| r = \map d {\tuple {x, r_2}, \tuple {x, r_1} }
| c = {{Defof|Real Number Plane with Euclidean Metric}}
}}
{{end-eqn}}
Then by Open Ball is Subset of Open Ball:
:$\map {B_{r_1} } {x, r_1} \subseteq \map {B_{r_2} } {x, r_2}$
By Set Union Preserves Subsets/Corollary:
:$U_1 \subseteq U_2$
By Intersection with Subset is Subset:
:$U_1 = U_1 \cap U_2$
Thus by Set is Subset of Itself:
:$U_1 \subseteq U_1 \cap U_2$
Thus by $(\text N 0)$-$(\text N 3)$, $(4)$ and Topology Defined by Neighborhood System:
:$T = \struct {S, \tau}$ is a topological space.
{{qed}}
\end{proof}
|
23802
|
\section{Nilpotent Element is Zero Divisor}
Tags: Nilpotent Ring Elements, Zero Divisors, Ring Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Suppose further that $R$ is not the null ring.
Let $x \in R$ be a nilpotent element of $R$.
Then $x$ is a zero divisor in $R$.
\end{theorem}
\begin{proof}
First note that when $R$ is the null ring the result is false.
This is because although $0_R$ is nilpotent element in the null ring, it is not actually a zero divisor.
Hence in this case $0_R$ is both nilpotent and a zero divisor.
So, let $R$ be a non-null ring.
By hypothesis, there exists $n \in \Z_{>0}$ such that $x^n = 0_R$.
If $n = 1$, then $x = 0_R$.
By hypothesis, $R$ is not the null ring, so we may choose $y \in R \setminus \set 0$.
By Ring Product with Zero:
:$y \circ x = y \circ 0_R = 0_R$
Therefore $x$ is a zero divisor in $R$.
If $n \ge 2$, define $y = x^{n - 1}$.
Then:
:$y \circ x = x^{n - 1} \circ x = x^n = 0_R$
so $x$ is a zero divisor in $R$.
{{Qed}}
Category:Nilpotent Ring Elements
Category:Zero Divisors
\end{proof}
|
23803
|
\section{Nilpotent Elements of Commutative Ring form Ideal}
Tags: Nilpotent Ring Elements, Ring Theory, Ideal Theory, Commutative Rings
\begin{theorem}
Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$ and whose unity is $1_R$.
The subset of nilpotent elements of $R$ form an ideal of $R$.
\end{theorem}
\begin{proof}
Let $N$ be the subset of nilpotent elements.
Because $0_R$ is nilpotent, $0_R \in N$ and so $N$ is non-empty.
Let $x \in N$ and $a \in R$.
We have:
{{begin-eqn}}
{{eqn | q = \exists n \in \Z_{>0}
| l = x^n
| r = 0_R
| c = {{Defof|Nilpotent Ring Element}}
}}
{{eqn | ll= \leadsto
| l = a^n \circ x^n
| r = 0_R
| c = {{Defof|Ring Zero}}
}}
{{eqn | ll= \leadsto
| l = \paren {a \circ x}^n
| r = 0_R
| c = Power of Product of Commutative Elements in Semigroup
}}
{{eqn | ll= \leadsto
| l = a x
| o = \in
| r = N
| c = Definition of $N$
}}
{{end-eqn}}
Let $x, y \in N$.
Let $x^n = 0$ and $y^m = 0$.
By the Binomial Theorem, $\paren {x - y}^p = 0$ for $p \ge n + m - 1$.
Thus $x - y \in N$.
Thus from Test for Ideal, $N$ is an ideal.
{{qed}}
\end{proof}
|
23804
|
\section{Nilpotent Ring Element plus Unity is Unit}
Tags: Nilpotence
\begin{theorem}
Let $A$ be a ring with unity.
Let $1 \in A$ be its unity.
Let $a \in A$ be nilpotent.
Then $1 + a$ is a unit of $A$.
\end{theorem}
\begin{proof}
Because $a$ is nilpotent, there exists a natural number $n > 0$ with $a^n = 0$.
By Sum of Geometric Sequence in Ring:
:$\paren {1 + a} \cdot \ds \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k = 1 + \paren {-a}^n$
:$\paren {\ds \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k} \cdot \paren {1 + a} = 1 + \paren {-a}^n$
where $\sum$ denotes summation.
By Negative of Nilpotent Ring Element:
:$\paren {-a}^n = 0$
Thus $1 + a$ is a unit.
{{qed}}
\end{proof}
|
23805
|
\section{Nine Regular Polyhedra}
Tags: Regular Polyhedra
\begin{theorem}
There exist $9$ regular polyhedra.
\end{theorem}
\begin{proof}
From Five Platonic Solids, there exist $5$ regular polyhedra which are convex:
:the regular tetrahedron
:the cube
:the regular octahedron
:the regular dodecahedron
:the regular icosahedron.
From Four Kepler-Poinsot Polyhedra:
{{:Four Kepler-Poinsot Polyhedra}}
All $4$ of the above are regular polyhedra which are non-convex.
making the total $9$.
\end{proof}
|
23806
|
\section{Niven's Theorem}
Tags: Trigonometry, Niven's Theorem
\begin{theorem}
Consider the angles $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$.
The only values of $\theta$ such that both $\dfrac \theta \pi$ and $\sin \theta$ are rational are:
:$\theta = 0: \sin \theta = 0$
:$\theta = \dfrac \pi 6: \sin \theta = \dfrac 1 2$
:$\theta = \dfrac \pi 2: \sin \theta = 1$
\end{theorem}
\begin{proof}
We will prove that if both $\dfrac \theta \pi$ and $\cos \theta$ are rational then:
:$\theta \in \set{ {0, \dfrac \pi 3, \dfrac \pi 2} }$
\end{proof}
|
23807
|
\section{Niven's Theorem/Lemma}
Tags: Proofs by Induction, Niven's Theorem
\begin{theorem}
For any integer $n \ge 1$, there exists a polynomial $\map {F_n} x$ such that:
:$\map {F_n} {2 \cos t} = 2 \cos n t$
In addition:
:$\deg F_n = n$
and $F_n$ is a monic polynomial with integer coefficients.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For $n = 1$, it is seen that:
:$\map {F_1} x = x$
fulfils the propositions.
For $n = 2$:
:$\map {F_2} x = x^2 - 2$
For $n > 2$:
{{begin-eqn}}
{{eqn | l = 2 \map \cos {n - 1} t \cos t
| r = \cos n t + \map \cos {n - 2} t
| c =
}}
{{eqn | ll= \leadsto
| l = 2 \cos n t
| r = \paren {2 \map \cos {n - 1} t} \paren {2 \cos t} - 2 \map \cos {n - 2} t
| c =
}}
{{eqn | r = 2 \cos t \map {F_{n - 1} } {2 \cos t} - \map {F_{n - 2} } {2 \cos t}
| c =
}}
{{end-eqn}}
so:
: $\map {F_n} x = x \map {F_{n - 1}} x - \map {F_{n - 2}} x \in \Z \sqbrk x$
will fulfil:
: $\map {F_n} {2 \cos t} = 2 \cos n t$
Because $\deg F_{n - 1} = n - 1$ and $\deg F_{n - 2} = n - 2$, we can conclude that:
:$\deg F_n = \deg \paren {x \map {F_{n - 1}} x - \map {F_{n - 2}} x} = n$
In addition, the leading coefficient of $F_n$ is equal to the leading coefficient of $F_{n - 1}$, which is $1$.
Hence the result.
{{qed}}
Category:Proofs by Induction
Category:Niven's Theorem
\end{proof}
|
23808
|
\section{No 4 Fibonacci Numbers can be in Arithmetic Sequence}
Tags: Arithmetic Progressions, Arithmetic Sequences, Fibonacci Numbers
\begin{theorem}
Let $a, b, c, d$ be distinct Fibonacci numbers.
Then, except for the trivial case:
:$a = 0, b = 1, c = 2, d = 3$
it is not possible that $a, b, c, d$ are in arithmetic sequence.
\end{theorem}
\begin{proof}
Let:
:$a = F_i, b = F_j, c = F_k, d = F_l$
where $F_n$ denotes the $n$th Fibonacci number.
{{WLOG}}, further suppose that;
:$a < b < c < d$
or equivalently:
:$i < j < k < l$
Since $i, j, k, l$ are integers, the inequality could be written as:
:$i \le j - 1 \le k - 2 \le l - 3$
Now consider:
{{begin-eqn}}
{{eqn | l = d - c
| r = F_l - F_k
| c =
}}
{{eqn | o = \ge
| r = F_l - F_{l - 1}
| c = By assumption, $k - 2 \le l - 3$
}}
{{eqn | r = F_{l - 2}
| c = {{Defof|Fibonacci Number}}
}}
{{eqn | o = \ge
| r = F_j
| c = By assumption, $j - 1 \le l - 3$
}}
{{eqn | o = \ge
| r = F_j - F_i
| c =
}}
{{eqn | r = b - a
| c =
}}
{{end-eqn}}
For $a, b, c, d$ be in arithmetic sequence:
:$d - c = b - a$
This means that the all the inequalities above must be equalities:
:$F_l - F_k = F_l - F_{l - 1}$
:$F_{l - 2} = F_j$
:$F_j = F_j - F_i$
So:
:$F_i = 0$
and:
:$F_k = F_{l - 1}$
:$F_j = F_{l - 2}$
The only Fibonacci numbers having different index but have the same value is $F_1 = F_2 = 1$.
So one of the following is true:
:$F_k = F_{l - 1} = 1$
:$F_j = F_{l - 2} = 1$
:$j - 1 = k - 2 = l - 3$
Suppose the third statement is true.
Write $k = j + 1$, $l = j + 2$.
Then:
{{begin-eqn}}
{{eqn | l = F_{j + 2} - F_{j + 1}
| r = F_{j + 1} - F_j
| c = $F_j, F_{j + 1}, F_{j + 2}$ form an arithmetic sequence
}}
{{eqn | ll = \leadsto
| l = F_j
| r = F_{j - 1}
| c = {{Defof|Fibonacci Number}}
}}
{{eqn | ll = \leadsto
| l = F_j - F_{j - 1}
| r = 0
| c =
}}
{{eqn | ll = \leadsto
| l = F_{j - 2}
| r = 0
| c = {{Defof|Fibonacci Number}}
}}
{{end-eqn}}
The only zero term of the Fibonacci numbers is $F_0$.
This gives $j = 2$.
Therefore the only arithmetic sequence among Fibonacci numbers satisfying the condition above is:
:$F_0, F_2, F_3, F_4$
which corresponds to:
:$0, 1, 2, 3$
Now suppose $F_j = 1$.
Since $F_i, F_j, F_k, F_l$ form an arithmetic sequence:
:$F_k = F_j + \paren {F_j - F_i} = 2$
:$F_l = F_k + \paren {F_j - F_i} = 3$
Which again gives the arithmetic sequence $0, 1, 2, 3$.
Finally suppose $F_k = 1$.
Since $F_i, F_j, F_k$ form an arithmetic sequence:
:$F_j = \dfrac 1 2 \paren {F_i + F_k} = \dfrac 1 2$
which is not an integer.
So $F_k \ne 1$.
All cases have been accounted for, and the only arithmetic sequence that can be formed is $0, 1, 2, 3$.
{{qed}}
\end{proof}
|
23809
|
\section{No Bijection between Finite Set and Proper Subset}
Tags: Set Theory, No Bijection between Finite Set and Proper Subset, Subsets, Subset
\begin{theorem}
A finite set can not be in one-to-one correspondence with one of its proper subsets.
That is, a finite set is not Dedekind-infinite.
\end{theorem}
\begin{proof}
Follows directly from Same Cardinality Bijective Injective Surjective.
{{Stub}}
\end{proof}
|
23810
|
\section{No Bijection from Set to its Power Set}
Tags: Bijections, Power Set, Mappings
\begin{theorem}
Let $S$ be a set.
Let $\powerset S$ denote the power set of $S$.
There is no bijection $f: S \to \powerset S$.
\end{theorem}
\begin{proof}
A bijection is by its definition also a surjection.
By Cantor's Theorem there is no surjection from $S$ to $\powerset S$.
Hence the result.
{{Qed}}
\end{proof}
|
23811
|
\section{No Boolean Interpretation Models a WFF and its Negation}
Tags: Propositional Logic, Propositional Calculus
\begin{theorem}
Let $v$ be a boolean interpretation.
Let $\mathbf A$ be a WFF of propositional logic.
Then $v$ can not model both $\mathbf A$ and $\neg \mathbf A$.
\end{theorem}
\begin{proof}
Suppose that $v$ models $\mathbf A$:
:$v \models \mathbf A$
Then $v \left({\mathbf A}\right) = T$ by definition of models.
By definition of boolean interpretation, $v \left({\neg \mathbf A}\right) = F$.
In particular, $v (\neg \mathbf A) \ne T$, so that:
:$v \not\models \neg \mathbf A$
Hence the result.
{{qed}}
Category:Propositional Logic
\end{proof}
|
23812
|
\section{No Group has Two Order 2 Elements}
Tags: Group Theory, Order of Group Elements
\begin{theorem}
A group can not contain exactly two elements of order $2$.
\end{theorem}
\begin{proof}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Suppose:
: $s, t \in \struct {G, \circ}\: s \ne t, \order s = \order t = 2$
That is. they are self-inverse:
:$s^2 = e = t^2$
The identity is of order $1$.
Hence $s$ nor $t$ is the identity
Hence, as $s \ne t$, then $s \circ t \in G$ is distinct from both $s$ and $t$.
Also $s \circ t \ne e$ because $s \ne t^{-1}$.
Suppose $s$ and $t$ commute.
Then $\paren {s \circ t}^2 = e$ from Self-Inverse Elements Commute iff Product is Self-Inverse.
Thus there is a third element (at least) in $G$ which is of order $2$.
Now suppose $s$ and $t$ do ''not'' commute.
Then from Commutation Property in Group, $s \circ t \circ s^{-1}$ is another element of $G$ distinct from both $s$ and $t$.
But from Order of Conjugate Element equals Order of Element:
:$\order {s \circ t \circ s^{-1} } = \order t$
and thus $s \circ t \circ s^{-1}$ is another element of order $2$.
Thus there is a third element (at least) in $G$ which is of order $2$.
{{Qed}}
\end{proof}
|
23813
|
\section{No Infinitely Descending Membership Chains}
Tags: Set Theory
\begin{theorem}
Let $\omega$ denote the minimal infinite successor set.
Let $F$ be a function whose domain is $\omega$.
Then:
:$\exists n \in \omega: \map F {n^+} \notin \map F n$
\end{theorem}
\begin{proof}
Let $F$ be a function whose domain is $\omega$.
By the axiom of replacement, the range of $F$ is a set.
Let the range of $F$ be denoted $\map \WW F$.
Then:
{{begin-eqn}}
{{eqn | o =
| r = \exists x \in \map \WW F: \paren {\map \WW F \cap x} = \O
| c = Axiom of Foundation
}}
{{eqn | o = \leadsto
| r = \exists x: \paren {\paren {\map \WW F \cap x} = \O \land \exists n \in \omega: x = \map F n}
| c = as $x \in \map \WW F$, $x = \map F n$ for some $n \in \omega$
}}
{{eqn | o = \leadsto
| r = \exists n \in \omega: \paren {\map \WW F \cap \map F n} = \O
| c = Logical manipulation eliminating $x$
}}
{{end-eqn}}
But:
:$\map F {n^+} \in \map \WW F$
So:
:$\map F {n^+} \notin \map F n$
{{qed}}
\end{proof}
|
23814
|
\section{No Infinitely Descending Membership Chains/Corollary}
Tags: Axiom of Foundation
\begin{theorem}
There cannot exist a sequence $\sequence {x_n}$ whose domain is $\N_{\gt 0}$ such that:
:$\forall n \in \N_{\gt 0}: x_{n+1} \in x_n$
\end{theorem}
\begin{proof}
{{AimForCont}} there is a sequence like that.
From the definition of a sequence, let $f$ be the the mapping that is defined by $\sequence {x_n}$.
Let $\omega$ denote the minimal infinite successor set.
Let $g: \omega \to \N_{\gt 0}$ be defined as:
:$\map g \alpha = \alpha + 1$
Then the composition $f \circ g$ is a mapping whose domain is $\omega$ such that:
:$\forall n \in \omega: \map {\paren {f \circ g} } {n^+} \in \map {\paren {f \circ g} } n$
But this contradicts No Infinitely Descending Membership Chains.
Therefore by Proof by Contradiction there cannot exist such a sequence.
Hence the result.
{{qed}}
Category:Axiom of Foundation
\end{proof}
|
23815
|
\section{No Injection from Power Set to Set}
Tags: Power Set, No Injection from Power Set to Set, Injections
\begin{theorem}
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then there is no injection from $\powerset S$ into $S$.
\end{theorem}
\begin{proof}
The identity mapping $f: \mathcal P(S) \to \mathcal P(S)$ is a surjection by Identity Mapping is Surjection.
Thus by the lemma, there can be no injection from $\mathcal P(S)$ into $S$.
{{qed}}
\end{proof}
|
23816
|
\section{No Injection from Power Set to Set/Lemma}
Tags: Power Set, No Injection from Power Set to Set
\begin{theorem}
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then there does not exist a set $B$ such that there is an injection from $B$ into $S$ and a surjection from $B$ onto $\powerset S$.
\end{theorem}
\begin{proof}
{{AimForCont}} there exists such a $B$.
Let $i: B \to S$ be an injection.
Let $f: B \to \powerset S$ be a surjection.
Let $i^\gets: \powerset S \to \powerset B$ be the inverse image mapping of $i$.
By Mapping Induced by Inverse of Injection is Surjection, $i^\gets$ is a surjection.
Let $f^\to: \powerset B \to \powerset {\powerset S}$ be the direct image mapping of $f$.
By Direct Image Mapping of Surjection is Surjection, $f^\to$ is a surjection:
We have that $i^\gets: \powerset S \to \powerset B$ and $f^\to: \powerset B \to \powerset {\powerset S}$ are surjective.
By Composite of Surjections is Surjection, their composition $f^\to \circ i^\gets: \powerset S \to \powerset {\powerset S}$ is a surjection by Composite of Surjections is Surjection.
But this violates Cantor's Theorem, contradicting the assumption that such a $B$ exists.
{{qed}}
Category:No Injection from Power Set to Set
\end{proof}
|
23817
|
\section{No Isomorphism from Woset to Initial Segment}
Tags: Order Isomorphisms, Well-Orderings, Orderings, Order Morphisms
\begin{theorem}
Let $\struct {S, \preceq}$ be a woset.
Let $a \in S$, and let $S_a$ be the initial segment of $S$ determined by $a$.
Then there is no order isomorphism between $S$ and $S_a$.
\end{theorem}
\begin{proof}
{{AimForCont}} $f: S \to S_a$ is an order isomorphism.
By Order Isomorphism from Woset onto Subset, $\forall x \in S: x \preceq \map f x$.
In particular, then, $a \preceq \map f a$.
But the codomain of $f$ is $S_a$, so $\map f a \in S_a$.
Thus $\map f a \prec a$, which is a contradiction.
So there can be no such order isomorphism.
{{qed}}
\end{proof}
|
23818
|
\section{No Largest Ordinal}
Tags: Ordinals
\begin{theorem}
Let $a$ be a set of ordinals.
Then:
:$\forall x \in a: x \prec \paren {\bigcup a}^+$
\end{theorem}
\begin{proof}
For this proof, we shall use $\prec$, $\in$, and $\subset$ interchangeably.
We are justified in doing this because of Ordering on Ordinal is Subset Relation and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = a
| c = {{Hypothesis}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \subseteq
| r = \bigcup A
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \preceq
| r = \bigcup A
| c =
}}
{{eqn | o = \prec
| r = \paren {\bigcup A}^+
| c = Ordinal is Less than Successor
}}
{{eqn | ll= \leadsto
| l = x
| o = \prec
| r = \paren {\bigcup A}^+
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23819
|
\section{No Membership Loops}
Tags: Set Theory, Class Theory, Axiom of Foundation, Axiom of Regularity
\begin{theorem}
For any proper classes or sets $A_1, A_2, \ldots, A_n$:
:$\neg \paren {A_1 \in A_2 \land A_2 \in A_3 \land \cdots \land A_n \in A_1}$
\end{theorem}
\begin{proof}
{{NotZFC}}
Either $A_1, A_2, \ldots, A_n$ are all sets, or there exists a proper class $A_m$ such that $1 \le m \le n$.
Suppose there exists a proper class $A_m$.
Then, by the definition of a proper class, $\neg A_m \in A_{m+1}$, since it is not a member of any class.
The result then follows directly.
Otherwise it follows that all $A_1, A_2, \ldots, A_n$ are sets.
Then, by the fact that Epsilon Relation is Strictly Well-Founded and a Strictly Well-Founded Relation has no Relational Loops, it follows that:
:$\neg \paren {A_1 \in A_2 \land A_2 \in A_3 \land \cdots \land A_n \in A_1}$
{{qed}}
\end{proof}
|
23820
|
\section{No Natural Number between Number and Successor}
Tags: Ordinals, No Ordinal Between Set and Successor, Ordering on Natural Numbers, No Natural Number between Number and Successor
\begin{theorem}
Let $\N$ be the natural numbers.
Let $n \in \N$.
Then no natural number $m$ exists strictly between $n$ and its successor:
:$\neg \exists m \in \N: \paren {n < m < n^+}$
That is:
:If $m \le n \le m^+$, then $m = n$ or $m = n^+$.
\end{theorem}
\begin{proof}
{{AimForCont}} such an ordinal $y$ exists.
Then, by Ordering on Ordinal is Subset Relation:
:$x \in y$
and from Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:
:$y \in x^+$
Applying the definition of a successor set, we have:
:$y \in x \lor y = x$
But this creates a membership loop, in that:
:$x \in y \in x \lor x \in x$
By No Membership Loops, we have created a contradiction.
The result follows from Proof by Contradiction.
{{qed}}
\end{proof}
|
23821
|
\section{No Non-Trivial Norm on Rational Numbers is Complete}
Tags: Normed Division Rings, Complete Metric Spaces
\begin{theorem}
No non-trivial norm on the set of the rational numbers is complete.
\end{theorem}
\begin{proof}
By P-adic Norm not Complete on Rational Numbers, no $p$-adic norm $\norm{\,\cdot\,}_p$ on the set of the rational numbers, for any prime $p$, is complete.
By Rational Number Space is not Complete Metric Space, the absolute value $\size{\,\cdot\,}$ on the set of the rational numbers is not complete.
By Norm is Complete Iff Equivalent Norm is Complete, no norm is complete if it is equivalent to either the absolute value $\size{\,\cdot\,}$ or the $p$-adic norm $\norm{\,\cdot\,}_p$ for some prime $p$.
By Ostrowski's Theorem, every non-trivial norm is equivalent to either the absolute value $\size{\,\cdot\,}$ or the $p$-adic norm $\norm{\,\cdot\,}_p$ for some prime $p$.
The result follows.
{{qed}}
\end{proof}
|
23822
|
\section{No Order Isomophism Between Distinct Initial Segments of Woset}
Tags: Well-Orderings
\begin{theorem}
Let $E$ be a well-ordered set.
Let $S_\alpha, S_\beta$ be initial segments of $E$ that are order isomorphic.
Then $S_\alpha = S_\beta$.
\end{theorem}
\begin{proof}
{{AimForCont}} $S_\alpha \ne S_\beta$.
Then $\alpha \ne \beta$.
By the trichotomy law, $\alpha \prec \beta$ or $\beta \prec \alpha$.
{{WLOG}} assume $\alpha \prec \beta$.
Then $S_\alpha \subsetneqq S_\beta$.
That is, $S_\alpha$ is an initial segment of $S_\beta$.
By hypothesis $S_\alpha$ and $S_\beta$ are order isomorphic.
Thus there is an order isomorphism between $S_\beta$ and an initial segment of $S_\beta$.
This contradicts No Isomorphism from Woset to Initial Segment.
Thus $S_\alpha = S_\beta$.
{{qed}}
\end{proof}
|
23823
|
\section{No Quadruple of Consecutive Sums of Squares Exists}
Tags: Sums of Squares
\begin{theorem}
It is not possible for a quadruple of consecutive positive integers each of which is the sum of two squares.
\end{theorem}
\begin{proof}
$4$ consecutive positive integers will be in the forms:
{{begin-eqn}}
{{eqn | l = n_0
| o = \equiv
| r = 0
| rr= \pmod 4
| c =
}}
{{eqn | l = n_1
| o = \equiv
| r = 1
| rr= \pmod 4
| c =
}}
{{eqn | l = n_2
| o = \equiv
| r = 2
| rr= \pmod 4
| c =
}}
{{eqn | l = n_3
| o = \equiv
| r = 3
| rr= \pmod 4
| c =
}}
{{end-eqn}}
in some order.
But from Sum of Two Squares not Congruent to 3 modulo 4, $n_3$ cannot be the sum of two squares.
The result follows.
{{qed}}
\end{proof}
|
23824
|
\section{No Simple Graph is Perfect}
Tags: Simple Graphs, Graph Theory, Perfect Graphs
\begin{theorem}
Let $G$ be a simple graph whose order is $2$ or greater.
Then $G$ is not perfect.
\end{theorem}
\begin{proof}
Recall that a perfect graph is one where each vertex is of different degree.
We note in passing that the simple graph consisting of one vertex trivially fulfils the condition for perfection.
{{AimForCont}} $G$ is a simple graph of order $n$ where $n \ge 2$ such that $G$ is perfect.
First, suppose that $G$ has no isolated vertices.
By the Pigeonhole Principle, for all vertices to have different degrees, one of them must be of degree at least $n$.
That means it must connect to at least $n$ other vertices.
But there are only $n - 1$ other vertices to connect to.
Therefore $G$ cannot be perfect.
Now suppose $G$ has an isolated vertex.
There can be only one, otherwise there would be two vertices of degree zero, and so $G$ would not be perfect.
Again by the Pigeonhole Principle, for all vertices to have different degrees, one of them must be of degree at least $n - 1$.
But of the remaining $n - 1$ vertices, one of them is of degree zero.
So it cannot be adjacent to any vertex.
So there are only $n - 2$ other vertices to connect to.
Therefore $G$ cannot be perfect.
{{qed}}
\end{proof}
|
23825
|
\section{No Valid Categorical Syllogism contains two Particular Premises}
Tags: Categorical Syllogisms
\begin{theorem}
Let $Q$ be a valid categorical syllogism.
Then at least one of the premises of $Q$ is universal.
\end{theorem}
\begin{proof}
Suppose both premises of $Q$ are particular.
Then the pattern of $Q$ is one of $\text{II}x$, $\text{IO}x$, $\text{OI}x$ or $\text{OO}x$, where $x$ is the conclusion.
$\text{I}$ is neither universal nor negative.
Thus the $\text{II}x$ pattern does not distribute the middle term of $Q$.
So $\text{II}x$ violates the rule Middle Term of Valid Categorical Syllogism is Distributed at least Once.
$\text{O}$ is negative.
Thus $\text{OO}x$ violates the rule No Valid Categorical Syllogism contains two Negative Premises.
It remains to investigate $\text{IO}x$ and $\text{OI}x$.
By Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative, the conclusion of $Q$ is negative, either $\text{E}$ or $\text{O}$.
By the definition of Distributed Predicate of Categorical Syllogism, the predicate of the conclusion of $Q$ is distributed.
Thus, by construction, the primary term $P$ of $Q$ is distributed.
By Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise, it follows that $P$ is distributed in the major premise of $Q$.
This eliminates $\text{IO}x$ as the particular affirmative $\text{I}$ distributes neither its subject nor its predicate.
The final pattern to be investigated is $\text{OI}x$.
By definition, $\text{O}$ distributes only its predicate.
As $\text{O}$ needs to distribute the primary term $P$, it cannot also distribute the middle term $M$ of $Q$.
But the particular affirmative $\text{I}$ (as has been seen above) also does not distribute $M$.
Thus $M$ remains undistributed, so violating the rule Middle Term of Valid Categorical Syllogism is Distributed at least Once.
This eliminates $\text{OI}x$.
{{qed}}
\end{proof}
|
23826
|
\section{No Valid Categorical Syllogism with Particular Premise has Universal Conclusion}
Tags: Categorical Syllogisms
\begin{theorem}
Let $Q$ be a valid categorical syllogism.
Let one of the premises of $Q$ be particular.
Then the conclusion of $Q$ is also particular.
\end{theorem}
\begin{proof}
Let the major premise of $Q$ be denoted $\text{Maj}$.
Let the minor premise of $Q$ be denoted $\text{Min}$.
Let the conclusion of $Q$ be denoted $\text{C}$.
From No Valid Categorical Syllogism contains two Particular Premises, either $\text{Maj}$ or $\text{Min}$ has to be universal.
Let the other premise of $Q$ be particular.
Suppose $\text{C}$ is the universal affirmative $\mathbf A \left({S, P}\right)$.
From Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative it follows that $\text{Maj}$ and $\text{Min}$ are also both affirmative.
The only patterns fitting the criteria are $\text{AI}x$ and $\text{IA}x$, both of which distribute only one term in $Q$.
We have that $\text{C}$ distributes $S$.
From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise, $S$ must also be distributed in $\text{Min}$.
From Middle Term of Valid Categorical Syllogism is Distributed at least Once, the middle term $M$ of $Q$ needs to be distributed.
Thus we require two distributed terms in the premises for $\text{C}$ to be the universal affirmative.
So if $\text{C}$ is affirmative, neither premise of $Q$ can be particular.
From this contradiction it follows that $\text{C}$ is not $\mathbf A \left({S, P}\right)$.
Suppose $\text{C}$ is the universal negative $\mathbf E \left({S, P}\right)$.
In this case, both $S$ and $P$ are distributed in $C$, as the universal negative distributes both $S$ and $P$.
By Middle Term of Valid Categorical Syllogism is Distributed at least Once, the middle term $M$ of $Q$ needs to be distributed.
Thus there are three terms needing to be distributed in two premises.
So at least one premises needs to be the universal negative.
The other premise is not $\mathbf O$ as this violates No Valid Categorical Syllogism contains two Negative Premises.
But it also is not $\mathbf I$, as $\mathbf I$ does not distribute any terms.
So if $\text{C}$ is the universal negative, neither premise can be particular.
From this contradiction it follows that $\text{C}$ is not $\mathbf E \left({S, P}\right)$.
{{qed}}
\end{proof}
|
23827
|
\section{Noether's Theorem (Calculus of Variations)}
Tags: Physics, Partial Differential Equations, Calculus of Variations
\begin{theorem}
Let $y_i$, $F$, $\Psi_i$, $\Phi$ be real functions.
Let $x, \epsilon \in \R$.
Let $\mathbf y = \sequence {y_i}_{1 \mathop \le i \mathop \le n}$ and $\mathbf \Psi = \sequence{\Psi_i}_{1 \mathop \le i \mathop \le n}$ be vectors.
Let
:$\Phi = \map \Phi {x, \mathbf y, \mathbf y'; \epsilon}, \quad \Psi_i = \map {\Psi_i} {x, \mathbf y, \mathbf y'; \epsilon}$
such that:
:$\map \Phi {x, \mathbf y, \mathbf y'; 0} = x, \quad \map {\Psi_i} {x, \mathbf y, \mathbf y'; 0} = y_i$
where $x, \mathbf y, \mathbf y', \epsilon$ are variables.
Let:
:$\ds J \sqbrk {\mathbf y} = \int_{x_0}^{x_1} \map F {x, \mathbf y, \mathbf y'} \rd x$
be a functional.
Let:
:$X = \map \Phi {x, \mathbf y, \mathbf y'; \epsilon}, \quad \mathbf Y = \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon}$
Suppose, $J \sqbrk {\mathbf y}$ is invariant under transformations $x \rightarrow X$ and $\mathbf y \rightarrow \mathbf Y$ for arbitrary $x_0$ and $x_1$.
Then:
:$\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F} \phi = C$
where $C$ is a constant and:
:$\map {\boldsymbol \psi} {x, \mathbf y, \mathbf y'} = \dfrac {\partial \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg\rvert_{\epsilon \mathop = 0}$
:$\map \phi {x, \mathbf y, \mathbf y'} = \dfrac {\partial \map \Phi {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg\rvert_{\epsilon \mathop = 0}$
\end{theorem}
\begin{proof}
Apply Taylor's Theorem to the transformations $X$, $\mathbf Y$ at the point $\epsilon = 0$:
{{begin-eqn}}
{{eqn | l = X
| r = \map \Phi {x, \mathbf y, \mathbf y'; \epsilon}
}}
{{eqn | r = \map \Phi {x, \mathbf y, \mathbf y'; 0} + \epsilon \frac {\partial \map \Phi {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg\rvert_{\epsilon \mathop = 0} + \map \OO \epsilon
}}
{{eqn | r = x + \epsilon \phi + \map \OO \epsilon
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \mathbf Y
| r = \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon}
}}
{{eqn | r = \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; 0} + \epsilon \frac {\partial \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg \rvert_{\epsilon \mathop = 0} + \map \OO \epsilon
}}
{{eqn | r = y + \epsilon \boldsymbol \psi + \map \OO \epsilon
}}
{{end-eqn}}
Use the general variation formula, and suppose that the curve $\mathbf y = \map {\mathbf y} x$ is an extremal of $J \sqbrk {\mathbf y}$:
:$\delta x = \epsilon \phi, \quad \delta y = \epsilon\psi$
:$\delta J = \epsilon \bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F }\phi} {x \mathop = x_0} {x \mathop = x_1}$
Since $J \sqbrk {\mathbf y}$ is invariant under the transformation, the variation vanishes.
Then for any $\epsilon \ne 0$ we have:
:$\bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y'\cdot \nabla_{\mathbf y'} F} \phi} {x \mathop = x_0} {} = \bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F }\phi} {x \mathop = x_1} {}$
This has to hold for arbitrary $x_0$ and $x_1$.
Since only a constant mapping produces the same result for any input, the term in brackets has to be a constant.
{{qed}}
{{Namedfor|Emmy Noether|cat = Noether E}}
\end{proof}
|
23828
|
\section{Noether's Theorem (Hamiltonian Mechanics)}
Tags: Physics, Partial Differential Equations, Calculus of Variations, Hamiltonian Mechanics
\begin{theorem}
Let there be an infinitesimal transformation of generalised coordinates such that:
:$q_i \to \tilde q_i = q_i + q_i^\alpha \tuple{q, \dot q, t} \varepsilon_\alpha + \hbox {terms vanishing on shell}$
where $\varepsilon$ is not time-dependent.
Under this transformation, let the variation of the Lagrangian be:
:$L \tuple{q + \delta q, \dot q + \delta \dot q, t} - L \tuple{q, \dot q, t} = \dfrac {\rd}{\rd t} \LL^\alpha \tuple{q, \dot q, t} \varepsilon_\alpha$
Let $s$ be the number of degrees of freedom of the system.
Then the quantity:
:$\ds \JJ^\alpha = \sum_{i \mathop = 1}^s \frac {\partial L} {\partial \dot q_i} q_i^\alpha - \LL^\alpha$
is conserved.
\end{theorem}
\begin{proof}
{{proof wanted}}
{{Namedfor|Emmy Noether|cat = Noether E}}
Category:Partial Differential Equations
Category:Physics
Category:Hamiltonian Mechanics
Category:Calculus of Variations
\end{proof}
|
23829
|
\section{Noether Normalization Lemma}
Tags: Commutative Algebra, Named Theorems
\begin{theorem}
Let $k$ be a field.
Let $A$ be a non-trivial finitely generated $k$-algebra.
{{explain|the above link is for Definition:Non-Trivial Ring -- we need to define a Definition:Non-Trivial Algebra}}
Then there exists $n \in \N$ and a finite injective morphism of $k$-algebra:
:$k \sqbrk {x_1, \dotsc, x_n} \to A$
{{DefinitionWanted|Instead of using Definition:Finite Ring Homomorphism, another page is to be generated defining the module-theory version.}}
\end{theorem}
\begin{proof}
Since $A$ is finitely generated, we prove this by induction on the number $m$ of generators as a $k$-algebra.
\end{proof}
|
23830
|
\section{Noetherian Domain is Factorization Domain}
Tags: Factorization, Ring Theory
\begin{theorem}
Let $R$ be a noetherian integral domain.
Then $R$ is a factorization domain.
\end{theorem}
\begin{proof}
Let $\FF$ be the set of ideals of $R$ of the form $x R$, with $x$ not a unit and such that $x$ cannot be decomposed in the form:
:$x = u p_1 \dotsm p_r$
where $u$ is a unit and $p_1, \dotsc, p_r$ irreducible.
We show by contradiction that $\FF = \O$.
{{AimForCont}} $\FF \ne \O$.
Since $R$ is noetherian, we can choose a maximal element $a R \in \FF$.
By construction, $a$ is not irreducible, so we can write:
:$a = b c$
with $b, c$ non-units and not associates.
By the definition of associates in a commutative ring this means that $b R \subsetneq a R$ and $a R \subsetneq b R$.
Since $a R$ is assumed maximal, this means that $b R$ and $c R$ do not belong to $\FF$.
Therefore there exist units $u, v$ and irreducible elements $p_1, \dotsc, p_r, q_1, \dotsc, q_s$ such that:
:$b = u p_1 \dotsm p_r$
and:
:$c = v q_1 \dotsm q_s$
But this implies that:
:$a = b c = \paren {u v} p_1 \cdots p_r \cdot q_1 \dotsm q_s$
This is a contradiction, since {{hypothesis}} $a$ cannot be written in this form.
{{Qed}}
Category:Ring Theory
Category:Factorization
\end{proof}
|
23831
|
\section{Noetherian Space is Compact}
Tags: Noetherian Spaces
\begin{theorem}
Let $\struct {X, \tau}$ be a Noetherian topological space.
Then $\struct {X, \tau}$ is compact.
\end{theorem}
\begin{proof}
{{tidy}}
Let $\family {U_i}_{i \mathop \in I}$ be a cover of $X$.
That is, $\bigcup_{i \mathop \in I} U_i = X$.
Let $V$ be the collection of finite cover of $\family {U_i}_{i \mathop \in I}$.
Let $W = \set {\bigcup Y: Y \in V}$.
Then $W$ is a collection of open sets.
By Set of Open Sets in Noetherian Space has Maximal Element, $W$ has a maximal element.
Let $\ds U' = \bigcup_{j \mathop = 1}^n U_{i_j}$ be the maximal element.
{{AimForCont}} $U' \subsetneq X$.
Let $x \in X \setminus U'$.
Let $U_{i_{n + 1} }$ be a neighborhood of $x$, where $i_{n + 1} \in I$.
Then $U' \cup U_{i_{n + 1} }$ is larger than $U'$ and contradicts maximality condition.
Hence $U'$ is a finite subcover.
This shows that $\struct {X, \tau}$ is compact.
{{qed}}
Category:Noetherian Spaces
\end{proof}
|
23832
|
\section{Non-Abelian Group of Order p Cubed has Exactly One Normal Subgroup of Order p}
Tags: P-Groups
\begin{theorem}
Let $p$ be a prime number.
Let $G$ be a non-abelian group of order $p^3$.
Then $G$ contains exactly one normal subgroup of order $p$.
\end{theorem}
\begin{proof}
From Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is not the trivial subgroup.
From Quotient of Group by Center Cyclic implies Abelian, $G / \map G Z$ cannot be cyclic and non-trivial.
Thus $\order {G / \map G Z}$ cannot be $p$ and so must be $p^2$.
Thus $\order {\map G Z} = p$.
Let $N$ be a normal subgroup of $G$ of order $p$.
Then from Normal Subgroup of p-Group of Order p is Subset of Center:
:$N \subseteq \map G Z$
Thus there is no normal subgroup of order $p$ different from $\map G Z$.
Hence the result.
{{qed}}
\end{proof}
|
23833
|
\section{Non-Abelian Order 10 Group has Order 5 Element}
Tags: Non-Abelian Order 10 Group has Order 5 Element, Groups of Order 10, Order of Group Elements
\begin{theorem}
Let $G$ be a non-abelian group of order $10$.
Then $G$ has at least one element of order $5$.
\end{theorem}
\begin{proof}
By Lagrange's Theorem, all the elements of $G$ have orders $1$, $2$, $5$ or $10$.
From Identity is Only Group Element of Order 1, $9$ elements of $G$ have orders greater than $1$.
From Cyclic Group is Abelian, $G$ is not cyclic.
If $g \in G$ was of order $10$ then $g$ would generate the cyclic group $C_{10}$.
Thus $G$ has no element of order $10$.
So all elements of $G$ except the identity have orders $2$ or $5$.
{{AimForCont}} $G$ has no element of order $5$.
Then all elements are of order $2$.
Then $G$ is a Boolean group.
Then by Boolean Group is Abelian, $G$ is abelian.
This contradicts the assertion that $G$ is non-abelian.
Hence the result by Proof by Contradiction.
{{qed}}
\end{proof}
|
23834
|
\section{Non-Abelian Order 2p Group has Order p Element}
Tags: Order of Group Elements, Groups of Order 2 p
\begin{theorem}
Let $p$ be an odd prime.
Let $G$ be a non-abelian group of order $2 p$.
Then $G$ has at least one element of order $p$.
\end{theorem}
\begin{proof}
By Lagrange's Theorem, all the elements of $G$ have orders $1$, $2$, $p$ or $2p$.
From Identity is Only Group Element of Order 1, $2 p - 1$ elements of $G$ have orders greater than $1$.
From Cyclic Group is Abelian, $G$ is not the cyclic group $2 p$.
If $g \in G$ was of order $2 p$ then $g$ would generate the cyclic group $C_{2 p}$.
Thus $G$ has no element of order $2 p$.
So all elements of $G$ except the identity have orders $2$ or $p$.
{{AimForCont}} $G$ has no element of order $p$.
Then all elements are of order $2$.
Then $G$ is a Boolean group.
Then by Boolean Group is Abelian, $G$ is abelian.
This contradicts the assertion that $G$ is non-abelian.
Hence the result by Proof by Contradiction.
{{qed}}
\end{proof}
|
23835
|
\section{Non-Abelian Order 8 Group has Order 4 Element}
Tags: Groups of Order 8
\begin{theorem}
Let $G$ be a non-abelian group of order $8$.
Then $G$ has at least one element of order $4$.
\end{theorem}
\begin{proof}
Let $e \in G$ be the identity of $G$.
Let $g \in G$ be an arbitrary element of $G$ such that $g \ne e$.
From Identity is Only Group Element of Order 1, only $e$ has order $1$.
Thus from Order of Element Divides Order of Finite Group:
:$\order g \in \set {2, 4, 8}$
Suppose $\order g = 8$.
Then $G$ is cyclic.
So by Cyclic Group is Abelian, $G$ would be abelian.
So $g$ cannot be of order $8$.
Suppose all elements of $G \setminus \set e$ are of order $2$.
Then by definition $G$ is a Boolean group.
By Boolean Group is Abelian, $G$ would be abelian.
So the only option left is for at least one element of $G$ to be of order $4$.
{{qed}}
\end{proof}
|
23836
|
\section{Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group}
Tags: Quaternion Group, Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group, Groups of Order 8
\begin{theorem}
Let $G$ be a group with the following properties:
:$(1): \quad G$ is non-abelian.
:$(2): \quad G$ is of order $8$.
:$(3): \quad G$ has precisely one element of order $2$.
Then $G$ is isomorphic to the quaternion group $Q$.
\end{theorem}
\begin{proof}
From Order of Element Divides Order of Finite Group all the elements in $G$ have order $1,2,4$ or $8$.
From Cyclic Group is Abelian, $\left({1}\right)$ and $\left({2}\right)$, no elements in $G$ have order $8$, i.e. they all have order $1,2$ or $4$.
Let the identity element be $1$ and the one with order $2$ be $-1$.
Also denote $1 \circ a$ as $+a$, $-1 \circ a$ as $-a$, $\left\{ {\pm a}\right\} = \left\{ {a, -a}\right\}$ for simplicity.
\end{proof}
|
23837
|
\section{Non-Abelian Simple Finite Groups are Infinitely Many}
Tags: Simple Groups
\begin{theorem}
There exist infinitely many types of group which are non-abelian and finite.
\end{theorem}
\begin{proof}
We have that Alternating Group is Simple except on 4 Letters.
So for all $n \in \N$ such that $n \ne 4$, the alternating group $A_n$ is a simple group.
We also have that $A_n$ is non-abelian for all $n > 3$.
Hence the result.
{{qed}}
\end{proof}
|
23838
|
\section{Non-Archimedean Division Ring Iff Non-Archimedean Completion}
Tags: Complete Metric Spaces, Definitions: Norm Theory, Division Rings, Normed Division Rings, Definitions: Division Rings, Norm Theory, Definitions: Complete Metric Spaces, Non-Archimedean Norms
\begin{theorem}
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\struct {R', \norm {\, \cdot \,}' }$ be a normed division ring completion of $\struct {R, \norm {\, \cdot \,} }$
Then:
:$\norm {\, \cdot \,}$ is non-archimedean {{iff}} $\norm {\, \cdot \,}'$ is non-archimedean.
\end{theorem}
\begin{proof}
By the definition of a normed division ring completion then:
:$(1): \quad$ there exists a distance-preserving ring monomorphism $\phi: R \to R'$.
:$(2): \quad \struct {R', \norm {\, \cdot \,}' }$ is a complete metric space.
:$(3): \quad \phi \sqbrk R$ is a dense subspace in $\struct {R', \norm {\, \cdot \,}' }$.
By Normed Division Ring is Dense Subring of Completion, $\phi \sqbrk R$ is a dense normed division subring of $R'$ and $\phi: R \to \map \phi R$ is an isometric isomorphism.
By Isometrically Isomorphic Non-Archimedean Division Rings, $\struct {R, \norm {\, \cdot \,} }$ is non-archimedean {{iff}} $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is.
So it remains to show that $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ non-archimedean {{iff}} $\struct {R', \norm {\, \cdot \,}' }$ is.
\end{proof}
|
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