hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Assume that $g$ exists. We need to prove that the limits $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist. $g$ is continuous at the end points $a$ and $b$ of its domain as $g$ is continuous on $\closedint a b$. $g$ is right-continuous at $a$ and left-continuous at $b$. This means that $\map g a = \displaystyle \lim_{x \mathop \to a^+} \map g x$ and $\map g b = \displaystyle \lim_{x \mathop \to b^-} \map g x$. In turn, this means that the expressions $\displaystyle \lim_{x \mathop \to a^+} \map g x$ and $\displaystyle \lim_{x \mathop \to b^-} \map g x$ exist. The two $x$ parameters as being part of the two limiting processes in these expressions can be considered confined to $\openint a b$. Therefore, $g$ in these two expressions can be replaced by $f$ as $g = f$ on $\openint a b$. We conclude that $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist. {{qed|lemma}} === Sufficient Condition === Let $\displaystyle \lim_{x \mathop \to a^+} \map f x$ and $\displaystyle \lim_{x \mathop \to b^-} \map f x$ exist. We need to prove that a function $g$ with the properties listed in the theorem exists. Define $g = f$ on $\openint a b$. Define: :$\map g a = \displaystyle \lim_{x \mathop \to a^+} \map f x$ :$\map g b = \displaystyle \lim_{x \mathop \to b^-} \map f x$ $g$ is defined at $a$ and $b$ as ensured by the existence of the limits on the right hand sides of these two equations. We know that $f$ is continuous on $\openint a b$. Therefore, $g$ too is continuous on $\openint a b$ as $g = f$ there. It remains to show that $g$ is continuous at $a$ and $b$. $\map g a = \displaystyle \lim_{x \mathop \to a^+} \map f x$ per definition. $x$ can be considered as being an element of $\openint a b$ as the limiting process at $a$ requires $x$ to approach $a$ from above. $\map g b = \displaystyle \lim_{x \mathop \to b^-} \map f x$ per definition. $x$ can be considered as being an element of $\openint a b$ as the limiting process at $b$ requires $x$ to approach $b$ from below. Therefore, the two equations above can be written: :$\map g a = \displaystyle \lim_{x \mathop \to a^+} \map g x$ :$\map g b = \displaystyle \lim_{x \mathop \to b^-} \map g x$ as $g = f$ on $\openint a b$. We conclude that $g$ is continuous at $a$ and $b$ as these two equations are exactly the definitions of continuity for $g$ at respectively $a$ and $b$. {{qed}} Category:Continuous Functions gwexoblpllrjrg5fmsbn6p203oqt2xg
We start by showing that $f$ is bounded. A real function $g$ defined on $\closedint a b$ exists by Extendability Theorem for Function Continuous on Open Interval such that: :$g$ equals $f$ on $\openint a b$ :$g$ is continuous on $\closedint a b$ Therefore: :$g$ is bounded on $\closedint a b$ by Continuous Function on Compact Subspace of Euclidean Space is Bounded and Closed Real Interval is Compact in Metric Space Accordingly, $g$ is bounded on $\openint a b$ as $\openint a b$ is a subset of $\closedint a b$. Therefore, $f$ is bounded on $\openint a b$ as $f$ equals $g$ on $\openint a b$. Also, $f$ is bounded on the set $\set {a, b}$ by the maximum $\map {\max} {\size {\map f a}, \size {\map f b}}$. Therefore: :$f$ is bounded on $\closedint a b$ We have: :$f$ is bounded on $\closedint a b$ :$f$ is continuous on $\openint a b$ Therefore: :$f$ is Darboux integrable on $\closedint a b$ by Bounded Function Continuous on Open Interval is Darboux Integrable. {{qed}} Category:Integral Calculus jlrf0rmt5onda0cxxy2wyqtomufrz4y
Let $a, b \in A_p$. Then $a = \dfrac {p n_1} {d_1}, b = \dfrac {p n_1} {d_1}$ where: :$n_1, n_2 \in \Z$ :$d_1, d_2 \in \Z_{>0}$ :$p n_1 \perp d_1, p n_2 \perp d_2$ Then: {{begin-eqn}} {{eqn | l = a \times b       | r = \frac {p n_1} {d_1} \times \frac {p n_2} {d_2}       | c =  }} {{eqn | r = \frac {p n_1 p n_2} {d_1 d_2}       | c = {{Defof|Rational Multiplication}} }} {{eqn | r = \frac {p^2 \left({n_1 n_2}\right)} {d_1 d_2}       | c =  }} {{end-eqn}} From Euclid's Lemma for Prime Divisors, if $p \mathop \backslash d_1 d_2$ then either $p \mathop \backslash d_1$ or $p \mathop \backslash d_2$. But neither of these is the case, so $p \nmid d_1 d_2$. Hence by Prime not Divisor implies Coprime: :$p \perp d_1 d_2$ where $\perp$ denotes coprimeness. So when $\dfrac {p^2 \left({n_1 n_2}\right)} {d_1 d_2}$ is expressed in canonical form, $p$ will still be a divisor of the numerator. Hence the result. {{qed}}
From Sum of Geometric Sequence: :$\displaystyle \frac {q^n - 1} {q - 1} = \sum_{k \mathop = 0}^{n - 1} q^k$ That is: :$q^n - 1 = r \paren {q - 1}$ where $r = 1 + q + q^2 + \cdots + q^{n - 1}$. As Integer Addition is Closed and Integer Multiplication is Closed, it follows that $r \in \Z$. Hence the result by definition of divisor of integer. {{qed}}
Let $n$ be odd. Then from Sum of Odd Positive Powers: :$\displaystyle q^n + 1 = \paren {q + 1} \sum_{k \mathop = 1}^n \paren {-1}^k q^{k - 1}$ Let $n$ be even. Consider the equation: :$q^n + 1 = 0$ We have: :$\paren {-1}^n + 1 = 2$ So $-1$ is not a root of $q^n + 1 = 0$. By Polynomial Factor Theorem, $q + 1$ is not a divisor of $q^n + 1$ when $n$ is even. {{qed}}
Let $S_i$, $i \in \left\{{1, \ldots, n}\right\}$, $n \in \N_{>0}$, denote sets within the domain of $f$. Define $S = \displaystyle \bigcup_{i \mathop = 1}^{n} S_i$. Suppose first that $f$ is bounded on each of the sets $S_i$. We need to prove that $f$ is bounded on $S$. Let $b_i$ be a bound for $f$ on $S_i$, $i \in \left\{{1, \ldots, n}\right\}$. Let $b = \max \left({b_1, \ldots, b_n}\right)$ where $\max$ is the maximum operation. We note that $b \ge b_i$ for every $i$ in $\left\{{1, \ldots, n}\right\}$. Now, by definition of bound, a number greater than or equal to a bound is also a bound. Therefore, since $b_i$ is a bound for $f$ on $S_i$ and $b \ge b_i$, $b$ is a bound for $f$ on $S_i$ for every $i$. Since $b$ is a bound for $f$ on every set $S_i$, $b$ is a bound for $f$ on the union of the sets $S_i$. In other words, $f$ is bounded on $S$. This finishes the "if" part of the proof. Now, suppose that $f$ is bounded on $S$. We need to prove that $f$ is bounded on each of the sets $S_i$, $i \in \left\{{1, \ldots, n}\right\}$. Since $f$ is bounded on $S$, there is a bound $K$ that satisfies $K \ge \vert f\left( {s} \right) \vert$ for every element $s$ in $S$. Pick a set $S_i$. Since $K \ge \vert f\left( {s} \right) \vert$ is true of every element $s$ of $S$, this is also true of every element of $S_i$  as $S_i$ is a subset of $S$. Consequently, $K$ is a bound for $f$ on $S_i$. Since $i$ is arbitrary, we conclude that $f$ is bounded on $S_i$ for every $i$ in $\left\{{1, \ldots, n}\right\}$. {{qed}} Category:Real Analysis Category:Boundedness tpyk1kucdrcysb9i50xore6nw3psmsa
Proof by induction: For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition: :For all $p \in \Z$, $p$ is even {{iff}} $p^n$ is even. First it is worth confirming that $P \left({0}\right)$ does not hold: :$\forall p \in \Z: p^0 = 1$ which is not even whatever the parity of $p$. $P \left({1}\right)$ is true, as this just says: : $p$ is even {{iff}} $p$ is even which is a tautology. === Basis for the Induction === $P \left({2}\right)$ is the case: : $p$ is even {{iff}} $p^2$ is even which is demonstrated in Parity of Integer equals Parity of its Square. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: : $p$ is even {{iff}} $p^k$ is even Then we need to show: : $p$ is even {{iff}} $p^{k+1}$ is even === Induction Step === This is our induction step: Let $p$ be even. By the induction hypothesis, $p^k$ is also even. Then: {{begin-eqn}} {{eqn | l = p       | r = 2 r       | c = for some $r \in \Z$ }} {{eqn | l = p^k       | r = 2 s       | c = for some $s \in \Z$ }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = p^{k+1}       | r = p \cdot p^k       | c =  }} {{eqn | r = \left({2 r}\right) \left({2 s}\right)       | c =  }} {{eqn | r = 2 \left({2 r s}\right)       | c =  }} {{end-eqn}} So $p^{k+1}$ is even. Now suppose $p$ is not even (that is, odd). By the induction hypothesis, $p^k$ is also not even (that is, odd). Then: {{begin-eqn}} {{eqn | l = p       | r = 2 r + 1       | c = for some $r \in \Z$ }} {{eqn | l = p^k       | r = 2 s + 1       | c = for some $s \in \Z$ }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = p^{k+1}       | r = p \cdot p^k       | c =  }} {{eqn | r = \left({2 r + 1}\right) \left({2 s + 1}\right)       | c =  }} {{eqn | r = 4 r s + 2 \left({r + s}\right) + 1       | c =  }} {{eqn | r = 2 \left({2 r s + r + s}\right) + 1       | c =  }} {{end-eqn}} So $p^{k+1}$ is not even (that is, odd). So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :For all $n \in \Z_{>0}$, for all $p \in \Z$, $p$ is even {{iff}} $p^n$ is even. {{qed}} Category:Integers 4kflgjxhdq7h3hb59iyxn37ue7f0s11
By Parity of Integer equals Parity of Positive Power, $q^n$ is even {{iff}} $q$ is even. Thus it follows that $q^n + 1$ is even {{iff}} $q$ is odd. The result follows by definition of even integer. {{qed}}
Note that if $q = 1$ then $q^n + 1 = 2$ which ''is'' prime. Hence the condition on $q$ in the statement of the theorem. So by hypothesis $q > 1$. Let $q$ be odd. Then by Two divides Power Plus One iff Odd, $q^n + 1$ is not prime. Let $q$ be even. Let $n$ be expressed in the form $r 2^m$ where $r$ is odd. Then $q^n + 1$ can be expressed in the form: : $q^{r 2^m} + 1 = \left({q^{2^m}}\right)^r + 1$ By Number Plus One divides Power Plus One iff Odd, $q^{2^m} + 1$ is a divisor of $\left({q^{2^m}}\right)^r + 1$. So for all $r > 1$ it follows that $\left({q^{2^m}}\right)^r + 1$ is composite. Hence the result. {{qed}}
By the definition of Fermat number {{begin-eqn}} {{eqn | l = F_{n + k} - 1       | r = 2^{2^{n + k} }       | c = {{Defof|Fermat Number}} }} {{eqn | r = 2^{2^n 2^k}       | c =  }} {{eqn | r = \left({2^{2^n} }\right)^{2^k}       | c =  }} {{eqn | r = \left({F_n - 1}\right)^{2^k}       | c = {{Defof|Fermat Number}} }} {{end-eqn}} {{qed}}
A partial boolean interpretation for $S$ is a mapping from $S$ to the set of truth values $\set {T, F}$. By Cardinality of Set of All Mappings, the total number of mappings from $S$ to $T$ is: :$\card {T^S} = \card T^{\card S}$ The result follows directly. {{qed}}
Follows directly from Integers under Addition form Abelian Group. By definition, an abelian group is a group. Also by definition, a group is a monoid. Hence the result. {{qed}}
The cases where $n = 1$ and $n = 2$ are clear. Let $a = x \circ y \in P_n: x \in P_r, y \in P_s$. If $r > 1$ then we write $x = a_1 \circ z$ where $z = a_2 \circ a_3 \circ \ldots \circ a_r$ by induction. Then $x \circ y = \paren {a_1 \circ z} \circ y = a_1 \circ \paren {z \circ y} = a_1 \circ \paren {a_2 \circ a_3 \circ \ldots \circ a_n}$ (again by induction). If $r = 1$, then by induction $x \circ y = a_1 \circ y = a_1 \circ \paren {a_2 \circ a_3 \circ \ldots \circ a_n}$. Thus in either case, $x \circ y = a_1 \circ \paren {a_2 \circ a_3 \circ \ldots \circ a_n}$ which is a single element of $P_n$. Hence we see that $\map {P_n} {a_1, a_2, \ldots, a_n}$ consists of a single element. {{Qed}}
Proof by strong induction: For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le n$. $P \left({1}\right)$ is trivially true, as this just says $a_1 = a_1$. $P \left({2}\right)$ is the case: :$a_1 \circ a_2 = a_1 \circ a_2$ for which there is also nothing to prove. === Basis for the Induction === $P \left({3}\right)$ is the case: :$\left({a_1 \circ a_2}\right) \circ a_3 = a_1 \circ \left({a_2 \circ a_3}\right)$ which is the definition of associativity. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le k$. Then we need to show: :The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le k+1$. === Induction Step === This is our induction step: Consider the expressions: :$(1): \quad \left({a_1 \circ a_2 \circ \cdots \circ a_i}\right) \circ \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$ :$(2): \quad \left({a_1 \circ a_2 \circ \cdots \circ a_j}\right) \circ \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$ for some $i, j \in \Z: 1 \le i, j \le k$. We require to show that whatever the values of $i$ and $j$, these expressions are equal. {{WLOG}}, suppose that $i < j$. Then the above expressions can be written: :$\left({a_1 \circ a_2 \circ \cdots \circ a_i}\right) \circ \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_j}\right) \circ \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$ By the basis for the induction, the General Associativity Theorem holds for each of the parts in parenthesis. Let: : $a = \left({a_1 \circ a_2 \circ \cdots \circ a_i}\right)$ : $b = \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_j}\right)$ : $c = \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$ By definition of associative operation: :$\left({a \circ b}\right) \circ c = a \circ \left({b \circ c}\right)$ This demonstrates the equality of the expressions $(1)$ and $(2)$. So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Second Principle of Mathematical Induction. Therefore: :The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_n$ for $n \in \N$. {{qed}}
Follows directly from Rational Numbers under Addition form Abelian Group. By definition, an abelian group is a group. Also by definition, a group is a monoid. Hence the result. {{qed}}
From Real Numbers under Addition form Abelian Group, we have that $\struct {\R, +}$ forms an abelian group. From Non-Zero Real Numbers under Multiplication form Abelian Group, we have that $\struct {\R_{\ne 0}, \times}$ forms an abelian group. Next we have that Real Multiplication Distributes over Addition. Thus all the criteria are fulfilled, and $\struct {\R, +, \times}$ is a field. {{qed}}
Taking the monoid axioms in turn: === $\text S 0$: Closure === Real Addition is Closed. {{qed|lemma}} === $\text S 1$: Associativity === Real Addition is Associative. {{qed|lemma}} === $\text S 2$: Identity === From Real Addition Identity is Zero, we have that the identity element of $\struct {\R, +}$ is the real number $0$. {{qed|lemma}} Hence the result. {{qed}}
From Integers Modulo m under Addition form Cyclic Group, $\struct {\Z_m, +_m}$ is a cyclic group. The result follows from Cyclic Group is Abelian. {{Qed}}
Taking the monoid axioms in turn: === $\text S 0$: Closure === Complex Addition is Closed. {{qed|lemma}} === $\text S 1$: Associativity === Complex Addition is Associative. {{qed|lemma}} === $\text S 2$: Identity === From Complex Addition Identity is Zero, we have that the identity element of $\struct {\C, +}$ is the complex number $0 + 0 i$: :$\paren {x + i y} + \paren {0 + 0 i} = \paren {x + 0} + i \paren {y + 0} = x + i y$ and similarly for $\paren {0 + 0 i} + \paren {x + i y}$. {{qed|lemma}} Hence the result. {{qed}}
Taking the monoid axioms in turn: === $\text S 0$: Closure === Rational Multiplication is Closed. {{qed|lemma}} === $\text S 1$: Associativity === Rational Multiplication is Associative. {{qed|lemma}} === $\text S 2$: Identity === Rational Multiplication Identity is $1$. {{qed|lemma}} Hence the result. {{qed}}
Taking the monoid axioms in turn: === $\text S 0$: Closure === Real Multiplication is Closed. {{qed|lemma}} === $\text S 1$: Associativity === Real Multiplication is Associative. {{qed|lemma}} === $\text S 2$: Identity === Real Multiplication Identity is $1$. {{qed|lemma}} Hence the result. {{qed}}
Taking the monoid axioms in turn: === $\text S 0$: Closure === Complex Multiplication is Closed. {{qed|lemma}} === $\text S 1$: Associativity === Complex Multiplication is Associative. {{qed|lemma}} === $\text S 2$: Identity === Complex Multiplication Identity is $1$. {{qed|lemma}} Hence the result. {{qed}}
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}}
From Natural Numbers under Addition form Commutative Monoid, $\left({\N, +}\right)$ forms a commutative monoid. From Natural Numbers Bounded Below under Addition form Commutative Semigroup, $\left({\N_{>0}, +}\right)$ forms a commutative semigroup. From Identity Element of Natural Number Addition is Zero, $0$ is the identity of $\left({\N, +}\right)$  From Natural Number Addition is Cancellable, all elements of $\left({\N, +}\right)$ are cancellable. Suppose $\left({\N_{>0}, +}\right)$ is a monoid. From Identity of Cancellable Monoid is Identity of Submonoid, if $\left({\N_{>0}, +}\right)$ has an identity then it is the same as the identity of $\left({\N, +}\right)$. But as $0 \notin \N_{>0}$, it follows that $\left({\N_{>0}, +}\right)$ has no identity. Hence $\left({\N_{>0}, +}\right)$ is not a monoid. {{qed}}
We have that: :Natural Number Addition is Associative :Natural Number Addition is Commutative From Restriction of Associative Operation is Associative, $+$ is associative on $\struct {M, +}$. From Restriction of Commutative Operation is Commutative, $+$ is commutative on $\struct {M, +}$. It remains to be shown that $+$ is closed on $\struct {M, +}$. Let $a, b \in M$. Then $\exists r, s \in \N: a = m + r, b = m + s$. Thus: {{begin-eqn}} {{eqn | l = a + b       | r = \paren {m + r} + \paren {m + s}       | c =  }} {{eqn | r = m + \paren {m + r + s}       | c =  }} {{eqn | o = \ge       | r = m       | c =  }} {{end-eqn}} So $a + b \in M$ and so $+$ is closed on $\struct {M, +}$. {{qed}} Category:Examples of Semigroups Category:Natural Numbers miqbplbsv6ao8s0vmpg3sajbedkaty7
From Real Numbers are Totally Ordered, $\R$ is a totally ordered set. From Max Operation on Toset is Semigroup, $\left({\R_{\ge 0}, \max}\right)$ is a semigroup. By definition of $\R_{\ge 0}$: :$\forall x \in \R_{\ge 0}: 0 \le x$ Thus by definition of the max operation: :$\forall x \in \R_{\ge 0}: \max \left({0, x}\right) = x = \max \left({x, 0}\right)$ So $0$ is the identity of $\left({\R_{\ge 0}, \max}\right)$ by definition. The result follows by definition of monoid. {{qed}}
By definition, the free commutative monoid $M$ on $\left\{{x}\right\}$ is: :$M = \left\{{e, x, x^2, x^3, \ldots}\right\}$ where $e$ denotes the null sequence of elements of $X$. Let $\phi$ denote the mapping from $M$ to $\N$ as: :$\forall a \in M: \phi \left({a}\right) = \begin{cases} 0 & : a = e \\ n & : a = x^n \end{cases}$ By definition of $\phi$: :$\phi$ is injective: $\phi \left({a}\right) = \phi \left({b}\right) \implies a = b$ :$\phi$ is surjective: $\forall a \in \N: \exists b \in M: \phi \left({b}\right) = a$ :$\phi$ is a monoid homomorphism: $\phi \left({a b}\right) = \phi \left({a + b}\right) = \phi \left({a}\right) + \phi \left({b}\right)$ Hence the result, by definition of isomorphism. {{qed}}
Group axioms $\text G 0$, $\text G 1$ and $\text G 2$ are satisfied by dint of $\struct {S, \circ}$ being a monoid. Recall the definition of right cancellable operation: :$\forall a, b, c \in S: a \circ c = b \circ c \implies a = b$ Let $\rho_c: S \to S$ be the right regular representation of $\struct {S, \circ}$ with respect to $c$. By Right Cancellable iff Right Regular Representation Injective, $\rho_c$ is an injection. By Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijection, $\rho_c$ is a bijection. Thus $a \circ b = e$ has a unique solution for all $a \in S$. That is, group axiom $\text G 3$ holds on $S$. {{qed}}
Group axioms $G0$, $G1$ and $G2$ are satisfied by dint of $\left({S, \circ}\right)$ being a monoid. Recall the definition of left cancellable operation: :$\forall a, b, c \in S: c \circ a = c \circ b \implies a = b$ Let $\lambda_c: S \to S$ be the left regular representation of $\left({S, \circ}\right)$ with respect to $c$. By Left Cancellable iff Left Regular Representation Injective, $\lambda_c$ is an injection. By Left Regular Representation wrt Left Cancellable Element on Finite Semigroup is Bijection, $\lambda_c$ is a bijection. Thus $a \circ b = e$ has a unique solution for all $a \in S$. That is, group axiom $G3$ holds on $S$. {{qed}}
From the field axioms: {{begin-axiom}} {{axiom | n = \text M 0         | lc= Closure under product         | q = \forall x, y \in F         | m = x \circ y \in F }} {{axiom | n = \text M 1         | lc= Associativity of product         | q = \forall x, y, z \in F         | m = \paren {x \circ y} \circ z = x \circ \paren {y \circ z} }} {{axiom | n = \text M 2         | lc= Commutativity of product         | q = \forall x, y \in F         | m = x \circ y = y \circ x }} {{axiom | n = \text M 3         | lc= Identity element for product         | q = \exists 1_F \in F, 1_F \ne 0_F: \forall x \in F         | m = x \circ 1_F = x = 1_F \circ x         | rc= $1_F$ is called the unity }} {{axiom | n = \text M 4         | lc= Inverse elements for product         | q = \forall x \in F^*: \exists x^{-1} \in F^*         | m = x \circ x^{-1} = 1_F = x^{-1} \circ x }} {{end-axiom}} Hence the result. {{qed}}
Recall that a field is a non-trivial commutative division ring. The result follows from Non-Zero Elements of Division Ring form Group. {{qed}}
By hypothesis, $S$ is not empty. As $0 \notin S$, it follows that $S \subseteq \Q_{\ne 0}$. Recall that $-1 \times -1 = 1$ and also $1 \times 1 = 1$. Thus: :$\forall x \in S: x \times y^{-1} \in S$ The result follows from the One-Step Subgroup Test. {{qed}}
Let $a, b, c \in S$. Let $\alpha$ be the transposition on $S$ which exchanges $a$ and $b$. Let $\beta$ be the transposition on $S$ which exchanges $b$ and $c$. Then: :$\alpha \circ \beta$ maps $\tuple {a, b, c}$ to $\tuple {c, a, b}$ while: :$\beta \circ \alpha$ maps $\tuple {a, b, c}$ to $\tuple {b, c, a}$ Thus $\alpha, \beta \in S_n$ such that $\alpha$ does not commute with $\beta$. Hence the result by definition of abelian group. {{qed}}
Let $\alpha \in S_n$ such that $\alpha$ is not the identity mapping. From Center of Symmetric Group is Trivial, $\alpha$ is not in the center $Z \paren {S_n}$ of $S_n$. Thus $S_n \ne Z \paren {S_n}$. The result follows by Group equals Center iff Abelian. {{qed}}
Let $\gcd \set {m, n} = k$ where $k > 1$. Then $m = k r$ and $n = k s$ for some $r, s \in \Z$. Thus $m n = k^2 r s$. From Integer is Expressible as Product of Primes there exists $p \in \Z$ such that $p$ is prime and $p \divides k$. That is: :$\exists t \in \Z: k = p t$ and so: :$m n = p^2 t^2 r s$ That is: :$p^2 \divides m $ Hence the result by definition of the Möbius function. {{qed}} Category:Multiplicative Functions Category:Möbius Function oxry1u2cq5rmz4yamh7ls14ie4hb64z