hcju/proofwikiqa huggingface dataset:
Recall the definition of the cosine function: :$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$ Thus $\cos x$ is expressible in the form of a power series. From Cosine Function is Absolutely Convergent, we have that the interval of convergence of $\cos x$ is the whole of $\R$.  From Power Series is Differentiable on Interval of Convergence, it follows that $\cos x$ is continuous on the whole of $\R$. {{qed}}   Category:Cosine Function cihwj3iuzzbt0cyzcf6yfmunqgwqvjh
Recall the definition of the cosine function: :$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n}}{\left({2 n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$ For: :$\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n} } {\left({2 n}\right)!}$  to be absolutely convergent, we want: :$\displaystyle \sum_{n \mathop = 0}^\infty \left|{\left({-1}\right)^n \frac {x^{2 n} } {\left({2 n}\right)!}}\right| = \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^{2 n} } {\left({2 n}\right)!}$  to be convergent. But  :$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^{2 n} } {\left({2 n}\right)!}$  is just the terms of  :$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^n} {n!}$ for even $n$. Thus: :$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^{2n}}{\left({2n}\right)!} < \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^n}{n!}$ But: :$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left|{x}\right|^n}{n!} = \exp \left|{x}\right|$  from the Taylor Series Expansion for Exponential Function of $\left|{x}\right|$, which converges for all $x \in \R$. Also, the sequence of partial sums: :$\displaystyle \sum_{n \mathop = 0}^k \frac {\left|{x}\right|^{2n}}{\left({2n}\right)!}$ is increasing. The result follows from an application of the Monotone Convergence Theorem (Real Analysis). {{qed}}
Recall the definition of the cosine function: :$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$ Thus: :$\displaystyle \cos 0 = 1 - \frac {0^2} {2!} + \frac {0^4} {4!} - \cdots = 1$ {{qed}}
Recall the definition of the cosine function: :$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$ From Cosine of Zero is One, we have that: :$\cos 0 = 1$ This takes care of the case $n = 0$. From Sine and Cosine are Periodic on Reals: : $\cos \paren {x + 2 \pi}= \cos x$ and thus: :$\forall m \in \Z: \cos \paren {x + 2 m \pi} = \cos x$. The above observations combine to establish that: :$\cos 2 m \pi = 1$ Again from Sine and Cosine are Periodic on Reals: :$\cos \paren {x + \pi} = -\cos x$ Combining this with the above reasoning, it follows that: :$\forall m \in \Z: \cos \paren {2 m + 1} \pi = -1$ Note that $\forall n \in \Z$: :If $n$ is even, it is of the form $n = 2m$ for some $m \in \Z$, and so $\cos n \pi = 1$ :If $n$ is odd, it is of the form $n = 2 m + 1$ for some $m \in \Z$, and so $\cos n \pi = -1$ This leads to the desired conclusion: :$\cos n \pi = \paren {-1}^n$ {{qed}}
Recall the definition of the cosine function: :$\displaystyle \cos z = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {z^{2 n} } {\left({2 n}\right)!} = 1 - \frac {z^2} {2!} + \frac {z^4} {4!} - \cdots$ From Even Power is Non-Negative: : $\forall n \in \N: z^{2 n} = \paren {-z}^{2 n}$ The result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \map \cos {-z}       | r = \frac {e^{i \paren {-z} } + e^{-i \paren {-z} } } 2       | c = Cosine Exponential Formulation }} {{eqn | r = \frac {e^{i z} + e^{-i z} } 2       | c = simplifying }} {{eqn | r = \cos z       | c = Cosine Exponential Formulation }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \map \cosh {-x} 	| r = \map \cos {-i x} 	| c = Hyperbolic Cosine in terms of Cosine }} {{eqn	| r = \map \cos {i x} 	| c = Cosine Function is Even }} {{eqn	| r = \cosh x 	| c = Hyperbolic Cosine in terms of Cosine }} {{end-eqn}} {{qed}}
Let $x_1, x_2 \in f' \sqbrk I: x_1 < x_2$. Let $\xi \in \openint {x_1} {x_2}$. We need to show that $\xi \in f' \sqbrk I$. Let $a, b \in I : \map {f'} a = x_1 \land \map {f'} b = x_2$. {{WLOG}}, assume $a < b$. The case $b < a$ is handled similarly. Let $\map g x = \map f x - \xi x$. Then: :$\map {g'} x = \map {f'} x - \xi$ By Differentiable Function is Continuous, $g$ is continuous. By Restriction of Continuous Mapping is Continuous, $g \restriction_{\closedint a b}$ is continuous. From Corollary 3 to Continuous Image of Compact Space is Compact: :$g \restriction_{\closedint a b}$ attains a minimum. By hypothesis, $\map {g'} a < 0$ and $\map {g'} b > 0$. Let $\map {N^*_\epsilon} x$ denote the deleted $\epsilon$-neighborhood  of $x$. Then from Behaviour of Function Near Limit: :$\exists \map {N^*_\epsilon} a: \forall x \in \map {N^*_\epsilon} a: \dfrac {\map g x - \map g a} {x - a} < 0$ and: :$\exists \map {N^*_\delta} b: \forall x \in \map {N^*_\delta} b: \dfrac {\map g x - \map g b} {x - b} > 0$ Thus: :$\exists x \in \map {N^*_\epsilon} a \cap \closedint a b: \map g x < \map g a$ and :$\exists x \in \map {N^*_\delta} b \cap \closedint a b: \map g x < \map g b$ Hence $\map g a$ and $\map g b$ are not minima of $g \restriction_{\closedint a b}$. So $g \restriction_{\closedint a b}$ must attain its minimum at some $m \in \openint a b$. By Derivative at Maximum or Minimum: :$\map {g'} m = 0$ Hence: :$\map {f'} m = \xi$ The result follows. {{qed}}
Since $\forall \left\{{a, b \in I: a < b}\right\}: \left({a \,.\,.\, b}\right) \subseteq I$, the result follows from Image of Interval by Derivative. {{qed}} Category:Real Analysis fxpe22z7ndq4g4jfjz3tqekkuk5i8b7
From the discussion of Sine and Cosine are Periodic on Reals, we have that: :$\map \sin {x + \dfrac \pi 2} = \cos x$ The result then follows directly from the Cosine of Multiple of Pi. {{qed}} Category:Sine Function 3vcd33anckhx2ex6a4galsl9evbwl9b
From Sine and Cosine are Periodic on Reals: Corollary: $\cos x$ is: :strictly positive on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ and: :strictly negative on the interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$ $\sin x$ is: :strictly positive on the interval $\openint 0 \pi$ and: :strictly negative on the interval $\openint \pi {2 \pi}$ The result follows directly from Sine and Cosine are Periodic on Reals. {{qed}} Category:Sine Function Category:Cosine Function 2nllk26o60f03t5j9of7k92y3kq77v0
This is established in Zeroes of Sine and Cosine. {{qed}} Category:Cosine Function rfuhggw0gmxdy0pt571mvoo0kbkotq4
Suppose that $\sqrt n = \dfrac a b$, with $a, b$ coprime integers and $b > 0$. Then we would have: : $n = \dfrac {a^2} {b^2}$ That is: :$n b^2 = a^2$ From Number divides Number iff Square divides Square: :$b^2 \divides a^2 \implies b \divides a$ However, since $a \perp b$ and $b \divides a$, this means that necessarily $b = 1$. That is, $\sqrt n = a$, an integer. Hence the result. {{qed}}
Let $F$ be a primitive for $f$ on the interval $\closedint {-a} a$. Then, by Sum of Integrals on Adjacent Intervals for Integrable Functions, we have: {{begin-eqn}} {{eqn | l = \int_{-a}^a \map f x \rd x       | r = \int_{-a}^0 \map f x \rd x + \int_0^a \map f x \rd x       | c = }} {{end-eqn}} Therefore, it suffices to prove that: : $\displaystyle \int_{-a}^0 \map f x \rd x = \int_0^a \map f x \rd x$ To this end, let $\phi: \R \to \R$ be defined by $x \mapsto -x$.  From Derivative of Identity Function and Derivative of Constant Multiple, for all $x \in \R$, we have $\map {\phi'} x = -1$. Then, by means of Integration by Substitution, we compute: {{begin-eqn}} {{eqn | l = \int_{\map \phi a}^{\map \phi 0} \map f x \rd x       | r = \int_a^0 \map f {-u} \paren {-1} \rd u }} {{eqn | r = \int_0^a \map f {-u} \rd u       | c = {{Defof|Definite Integral}} }} {{eqn | r = \int_0^a \map f {-x} \rd x       | c = renaming integration variable }} {{eqn | r = \int_0^a \map f x \rd x       | c = $f$ is an even function }} {{end-eqn}} This concludes the proof. {{qed}}
Let $F$ be a primitive for $f$ on the interval $\closedint {-a} a$. Then, by Sum of Integrals on Adjacent Intervals for Integrable Functions, we have: {{begin-eqn}} {{eqn | l = \int_{-a}^a \map f x \rd x       | r = \int_{-a}^0 \map f x \rd x + \int_0^a \map f x \rd x       | c = }} {{end-eqn}} Therefore, it suffices to prove that: :$\displaystyle \int_{-a}^0 \map f x \rd x = -\int_0^a \map f x \rd x$ To this end, let $\phi: \R \to \R$ be defined by $x \mapsto -x$.  From Derivative of Identity Function and Derivative of Constant Multiple, for all $x \in \R$, we have $\map {\phi'} x = -1$. Then, by means of Integration by Substitution, we compute: {{begin-eqn}} {{eqn | l = \int_{\map \phi a}^{\map \phi 0} \map f x \rd x       | r = \int_a^0 \map f {-u} \paren {-1} \rd u }} {{eqn | r = \int_0^a \map f {-u} \rd u       | c = {{Defof|Definite Integral}} }} {{eqn | r = \int_0^a \map f {-x} \rd x       | c = Renaming integration variable }} {{eqn | r = -\int_0^a \map f x \rd x       | c = $f$ is an odd function }} {{end-eqn}} This concludes the proof. {{qed}}
Denote: : $\phi = \dfrac {1 + \sqrt 5} 2$, $\hat \phi = \dfrac {1 - \sqrt 5} 2$ and: : $\alpha = \dfrac {\phi} {\hat \phi} = - \dfrac {3 + \sqrt 5} 2$ From the Euler-Binet Formula: :$f_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$ Let $n \ge 1$. It immediately follows that: {{begin-eqn}} {{eqn | l = \frac {f_{n + 1} } {f_n}        | r = \dfrac {\phi^{n + 1} - \hat \phi^{n + 1} } {\phi^n - \hat \phi^n} }} {{eqn | r = \dfrac {\left(\phi^{n + 1} - \phi \hat \phi^n\right) + \left(\phi \hat \phi^n - \hat \phi^{n + 1}\right)} {\phi^n - \hat \phi^n} }} {{eqn | r = \phi + \dfrac {\hat \phi^n\left(\phi - \hat \phi\right)} {\phi^n - \hat \phi^n} }} {{eqn | r = \phi + \dfrac {\sqrt 5} {\alpha^n - 1} }} {{end-eqn}} From the definition of $\alpha$: :$|\alpha| > 1$ Therefore: :$\displaystyle \lim_{n \mathop \to \infty} \frac {f_{n + 1}} {f_n} = \lim_{n \mathop \to \infty}\ \phi + \dfrac {\sqrt 5} {\alpha^n - 1} = \phi$ {{MissingLinks|A link to calculation of used limit.}} {{qed}}
From Continued Fraction Expansion of Golden Mean: Successive Convergents, the $n$th convergent of the continued fraction expansion of $\phi$ is: :$C_n = \dfrac {f_{n + 1} } {f_n}$ The result follows from Continued Fraction Expansion of Irrational Number Converges to Number Itself. {{qed}}
Proof by induction: For all $n \in \N_{>0}$, let $P_n$ be the proposition: :$\card {f \paren A} = \card A$ when $\card A = n$ Suppose $\card A = 0$. {{begin-eqn}} {{eqn | l = \card A       | r = 0       | c = }} {{eqn | ll= \leadsto       | l = A       | r = \O       | c = Cardinality of Empty Set }} {{eqn | ll= \leadsto       | l = f \paren A       | r = \O       | c = Image of Empty Set is Empty Set }} {{eqn | ll= \leadsto       | l = \card {f \paren A}       | r = 0       | c = Cardinality of Empty Set }} {{end-eqn}} So $P_0$ holds. === Basis for the Induction === Suppose $\card A = 1$. Then let $A = \set a$. So: :$f \paren A = \set {f \paren a}$ and so: :$\card {f \paren A} = 1$ So $P_1$ is true. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P_k$ is true, where $k \ge 1$, then it logically follows that $P_{k + 1}$ is true. So this is our induction hypothesis: :$\card {f \paren A} = \card A$ when $\card A = k$ Then we need to show: :$\card {f \paren A} = \card A$ when $\card A = k + 1$ === Induction Step === This is our induction step: Consider $A$ where $\card A = k+1$. Let $s \in A$ and consider $A' = A \setminus \set s$. Let $f {\restriction_{A'} }$ be the restriction of $f$ to $A'$. By Restriction of Injection is Injection we have that $f {\restriction_{A'} }$ is also an injection. Then by the induction hypothesis: :$\card {f {\restriction_{A'} } \paren {A'} } = \card {A'}$ because $\card {A'} = k$. Now consider $f \paren s$. Suppose $f \paren s \in f {\restriction_{A'} } \paren {A'}$. Then: :$\exists t \in A': f \paren t = f \paren s$ and so $f$ would not be an injection. So: :$f \paren s \notin f {\restriction_{A'} } \paren {A'}$ and so: {{begin-eqn}} {{eqn | l = f \paren A       | r = f \paren {A' \cup \set s}       | c = }} {{eqn | r = f \paren {A'} \cup f \paren {\set s}       | c = Image of Union under Mapping }} {{eqn | ll= \leadsto       | l = \card {f \paren A}       | r = \card {f \paren {A'} } + \card {f \paren {\set s} }       | c = Cardinality of Set Union, as $f \paren {A'}$ and $f \paren {\set s}$ are disjoint }} {{eqn | r = k + 1       | c=  }} {{end-eqn}} So $P_k \implies P_{k + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed}}
;Proof by Counterexample: Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings. First note that unless $S_1 = S_3$ then $f_2 \circ f_1$ is not even defined. So in that case $f_2 \circ f_1$ is definitely not the same thing as $f_1 \circ f_2$. So, let us suppose $S_1 = S_3$ and so we define $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_1$. If $S_1 \ne S_2$ then: :$f_2 \circ f_1: S_1 \to S_1$ :$f_1 \circ f_2: S_2 \to S_2$ and so by Equality of Mappings they are unequal because their domains and codomains are different. Finally, suppose $S_1 = S_2$, and consider the following. :$S_1 = S_2 = \set {a, b}$ :$f_1 = \set {\tuple {a, a}, \tuple {b, a} }$ :$f_2 = \set {\tuple {a, b}, \tuple {b, b} }$ It is straightforward to check that $f_1$ and $f_2$ are mappings, and that: :$f_1 \circ f_2 = \set {\tuple {a, b}, \tuple {b, b} }$ :$f_2 \circ f_1 = \set {\tuple {a, a}, \tuple {b, a} }$ Thus, even in this limitingly simple case, we see that: :$f_2 \circ f_1 \ne f_1 \circ f_2$ {{qed}}
From Intersection with Relative Complement is Empty: : $A \cap \relcomp S A = \O$ From Image of Intersection under Injection: :$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$ {{iff}} $f$ is an injection. Hence the result. {{qed}}
We have: :Restriction of Injection is Injection :Restriction of Mapping to Image is Surjection Thus we have that: :$f {\restriction_{S \times X}}: S \to X$ is an injection and :$f {\restriction_{S \times X}}: S \to X$ is a surjection Hence the result by definition of bijection. {{qed}}
From Cardinality of Image of Injection we have that $\card {f \sqbrk S} = \card S$. From the nature of an injection, for all $s \in S$, the $\mathcal R_f$-class of $s$ is a singleton. Hence the result. {{qed}}
From the definition of a surjection: :$\forall t \in T: \exists s \in S: \map f s = t$ Thus there are as many $\mathcal R_f$-classes of $f$ as there are elements of $T$. Hence the result. {{qed}}
As $e^n = e \in G^n$, the set $G^n$ is not empty. Let $x^n, y^n \in G^n$. From Power of Product in Abelian Group, we have: :$x^n \paren {y^n}^{-1} = x^n \paren {y^{-1} }^n = \paren {x y^{-1} }^n$ It follows that $x^n \paren {y^n}^{-1} \in G^n$. From the One-Step Subgroup Test, we conclude that $G^n$ is a subgroup of $G$. Now assume $G$ is finite. Let $n \ne 1$ such that $n \divides \order G$. For $G^n$ to be a proper subgroup, it is enough to show that the map $x \mapsto x^n$ is not injective. Now as $n \ne 1$, from the Fundamental Theorem of Arithmetic it is possible to find a prime $p$ such that $p \divides n$. By Cauchy's Lemma (Group Theory), there is an element $x$ with order $p$. It then follows that $x^n = e$, but $x \ne e$. Therefore, the map $x \mapsto x^n$ is not injective. Hence the result. {{qed}} Category:Abelian Groups 5rybzdcslzin548decnogsxbwp6aj2g
Let $a \in S$. Because $\struct {S, \circ}$ is a semigroup, $\circ$ is associative on $S$. The proof proceeds by the Principle of Mathematical Induction. Let $P \paren m$ be the proposition: :$\forall n \in \N_{>0}: a^{n + m} = a^n \circ a^m$ that is: :$\forall n \in \N_{>0}: \circ^{n + m} a = \paren {\circ^n a} \circ \paren {\circ^m a}$ === Basis for the Induction === {{begin-eqn}} {{eqn | lo= \forall n \in \N_{>0}:       | l = \circ^{n + 1} a       | r = \paren {\circ^n a} \circ a       | c = Definition of $\circ^n a: \N_{>0} \to S$ }} {{eqn | r = \paren {\circ^n a} \circ \paren {\circ^1 a}       | c = Definition of $\circ^n a$ for $n = 1$ }} {{end-eqn}} So $P \paren 1$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \paren k$ is true, where $k \ge 1$, then it logically follows that $P \paren {k + 1}$ is true. So this is our induction hypothesis: :$\circ^{n + k} a = \paren {\circ^n a} \circ \paren {\circ^k a}$ It is then to be shown that: :$\circ^{n + \paren {k + 1} } a = \paren {\circ^n a} \circ \paren {\circ^{k + 1} a}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \circ^{n + \paren {k + 1} } a       | r = \circ^{\paren {n + k} + 1} a       | c = Natural Number Addition is Associative }} {{eqn | r = \paren {\circ^{n + k} a} \circ \paren {\circ^1 a}       | c = Basis for the Induction }} {{eqn | r = \paren {\paren {\circ^n a} \circ \paren {\circ^k a} } \circ \paren {\circ^1 a}       | c = Induction Hypothesis }} {{eqn | r = \paren {\circ^n a} \circ \paren {\paren {\circ^k a} \circ \paren {\circ^1 a} }       | c = $\circ$ is associative }} {{eqn | r = \paren {\circ^n a} \circ \paren {\circ^{k + 1} a}       | c = Basis for the Induction }} {{end-eqn}} So $P \paren {k + 1}$ is true. So by the Principle of Mathematical Induction, this result is true for all $m, n \in \N_{>0}$: :$\forall m, n \in \N_{>0}: \circ^{n + m} a = \paren {\circ^n a} \circ \paren {\circ^m a}$ or: :$\forall m, n \in \N_{>0}: a^{n + m} = a^n \circ a^m$ {{qed}}
Let $b = \circ^m a$. Let $h: \N_{>0} \to S$ be the mapping defined as: :$\forall n \in \N_{>0}: h \paren n = \circ^{n m} a$ Let the mapping $f_b: \N_{>0} \to S$ be recursively defined as: :$\forall n \in \N_{>0}: f_b \paren n = \circ^n b$ From the Principle of Recursive Definition: :$f_b$ is the unique mapping which satisfies: :$\forall n \in \N_{>0}: f_b \paren n = \begin{cases} b & : n = 1 \\ f_b \paren r \circ b & : n = r \circ 1 \end{cases}$ But $h \paren 1 = \circ^{1 \times m} a = \circ^m a = b$. So: {{begin-eqn}} {{eqn | lo= \forall n \in \N_{>0}:       | l = h \paren {n + 1}       | r = \circ^{\paren {n + 1} m} a       | c =  }} {{eqn | r = \circ^{\paren {n m} + m} a       | c = Natural Number Multiplication Distributes over Addition }} {{eqn | r = \paren {\circ ^{n \ast m} } \circ \paren {\circ^m a}       | c = Index Laws for Semigroup: Sum of Indices }} {{eqn | r = h \paren n) \circ b       | c =  }} {{end-eqn}} Thus $h = f_b$, and so: :$\forall n, m \in \N_{>0}: \circ^{n m} = \circ^n \paren {\circ^m a}$ From Natural Number Multiplication is Commutative: :$\forall n, m \in \N_{>0}: \circ^m \paren {\circ^n a} = \circ^{m n} = \circ^{n m}$ {{qed}}
For all sets $S$, $S \subseteq S$, that is, $S$ is a subset of itself. Thus $\struct {S, \circ}$ is a semigroup which is a subset of $\struct {S, \circ}$, and therefore a subsemigroup of $\struct {S, \circ}$. {{Qed}}
As $\struct {S, \circ}$ is an algebraic structure, the result Identity is Unique can be applied directly. {{qed}}
By the definition of a group, $\struct {G, \circ}$ is also a monoid. The result follows by applying the result Identity of Monoid is Unique. {{qed}}
Let $e$ and $f$ both be identity elements of a group $\struct {G, \circ}$. Then: {{begin-eqn}} {{eqn | l = e       | r = e \circ f       | c = $f$ is an identity }} {{eqn | r = f       | c = $e$ is an identity }} {{end-eqn}} So $e = f$ and there is only one identity after all. {{qed}}
From Group has Latin Square Property, there exists a unique $x \in G$ such that: :$a x = b$ and there exists a unique $y \in G$ such that: :$y a = b$ Setting $b = a$, this becomes: There exists a unique $x \in G$ such that: :$a x = a$ and there exists a unique $y \in G$ such that: :$y a = a$ These $x$ and $y$ are both $e$, by definition of identity element. {{qed}}
From the definition of an integral domain, $\struct {D, +, \times}$ is a commutative ring such that $\struct {D^*, \circ}$ is a monoid. The result follows from Identity of Monoid is Unique. {{qed}}
By definition, an integral domain is a ring. The result the follows from Ring Zero is Unique. {{qed}}
From the definition of an integral domain, $\struct {D, +, \times}$ is a ring. The result follows from Ring Negative is Unique. {{qed}}
By the definition of a group, $\struct {G, \circ}$ is a monoid each of whose elements has an inverse. The result follows directly from Inverse in Monoid is Unique. {{qed}}
Let $\struct {G, \circ}$ be a group whose identity element is $e$. By {{GroupAxiom|3}}, every element of $G$ has at least one inverse. Suppose that: :$\exists b, c \in G: a \circ b = e, a \circ c = e$ that is, that $b$ and $c$ are both inverse elements of $a$. Then: {{begin-eqn}} {{eqn | l = b       | r = b \circ e       | c = as $e$ is the identity element }} {{eqn | r = b \circ \paren {a \circ c}       | c = as $c$ is an inverse of $a$ }} {{eqn | r = \paren {b \circ a} \circ c       | c = {{GroupAxiom|1}} }} {{eqn | r = e \circ c       | c = as $b$ is an inverse of $a$ }} {{eqn | r = c       | c = as $e$ is the identity element }} {{end-eqn}} So $b = c$ and hence the result. {{qed}}
Let $x, y \in G$. We already have, from the definition of inverse element, that: :$\forall x \in G: \exists x^{-1} \in G: x \circ x^{-1} = e = x^{-1} \circ x$ By Group has Latin Square Property, there exists exactly one $a \in G$ such that $a \circ x = y$. Similarly, there exists exactly one $b \in G$ such that $x \circ b = y$. Substituting $e$ for $y$, it follows that $x^{-1}$ as defined above is unique. {{qed}}
By definition of ring with unity, the algebraic structure $\struct {R, \circ}$ is a monoid. The result follows from Inverse in Monoid is Unique. {{qed}}
By Units of Gaussian Integers form Group, $\struct {U_\C, \times}$ forms a group. Each of the elements of $U_\C$ is a complex number, and non-zero, and therefore $U_\C \subseteq \C \setminus \set 0$. The result follows by definition of subgroup. {{qed}}
Follows directly from the definition of subgroup: $H$ is a subset of $\map \Gamma S$, and $\struct {H, \circ}$ is a group. Hence the result. {{qed}}
By Complex Numbers form Field, $\C$ is a field. From Group of Units of Field it follows that: : $\C^\times = \C \setminus \set 0$ {{qed}} Category:Examples of Groups Category:Complex Numbers tss7spdnxtbd4mireozplygblsf6ij5
{{AimForCont}} $\struct {\Q_{>0}, \times}$ is cyclic. Then $\struct {\Q_{>0}, \times}$ has a generator $x$ such that $x > 1$. It would follow that: :$\Q = \set {\ldots, x^{-2}, x^{-1}, 1, x, x^2, \ldots}$ where the elements are arranged in ascending order. But then consider $y \in \Q: y = \dfrac {1 + x} 2$. So $1 < y < x$ and so $y \notin \Q$. From this contradiction it is concluded that there can be no such single generator of $\struct {\Q_{>0}, \times}$. Therefore, by definition, $\struct {\Q_{>0}, \times}$ is not a cyclic group. {{qed}}
From Determinant of Matrix Product: : $\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \, \map \det {\mathbf B}$ which is seen to be a group homomorphism by definition. {{qed}}
From Sum of Logarithms we have: :$\forall x, y \in \R_{>0}: \map {\log_b} {x y} = \map {\log_b} x + \map {\log_b} y$ That is $\log_b$ is a group homomorphism. From Change of Base of Logarithm, $\log_b$ is a constant multiplied by the natural logarithm function. Then we have that Logarithm is Strictly Increasing. From Strictly Monotone Real Function is Bijective, it follows that $\log_b$ is a bijection. So $\log_b$ is a bijective group homomorphism, and so a group isomorphism. {{qed}}