hcju/proofwikiqa huggingface dataset:
Define $T$ as: :$T = \set {n \in \N : \forall k: 0 \le k \le n: k \in S}$ Since $n \le n$, it follows that $T \subseteq S$. Therefore, it will suffice to show that: :$\forall n \ge 0: n \in T$ Firstly, we have that $0 \in T$ {{iff}} the following condition holds: :$\forall k: 0 \le k \le 0 \implies k \in S$ Since $0 \in S$, it thus follows that $0 \in T$. Now suppose that $n \in T$; that is: :$\forall k: 0 \le k \le n \implies k \in S$ By $(2)$, this implies: :$n + 1 \in S$ Thus, we have: :$\forall k: 0 \le k \le n + 1 \implies k \in S$ {{MissingLinks|Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor for $\N$}} Therefore, $n + 1 \in T$. Hence, by the Principle of Finite Induction: :$\forall n \ge 0: n \in T$ That is: :$T = \N$ and as $S \subseteq \N$ it follows that: :$S = N$ {{Qed}} Category:Second Principle of Finite Induction 4mwpf2fajzm72tpav4jdn67u9l0guw1
Define $T$ as: :$T = \set {n \in \N_{>0}: \forall k: 1 \le k \le n: k \in S}$ Since $n \le n$, it follows that $T \subseteq S$. Therefore, it will suffice to show that: :$\forall n \ge 1: n \in T$ Firstly, we have that $1 \in T$ {{iff}} the following condition holds: :$\forall k: 1 \le k \le 1 \implies k \in S$ Since $1 \in S$, it thus follows that $1 \in T$. Now suppose that $n \in T$; that is: :$\forall k: 1 \le k \le n \implies k \in S$ By $(2)$, this implies: :$n + 1 \in S$ Thus, we have: :$\forall k: 1 \le k \le n + 1 \implies k \in S$ {{MissingLinks|Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor for $\N$}} Therefore, $n + 1 \in T$. Hence, by the Principle of Finite Induction: :$\forall n \ge 1: n \in T$ That is: :$T = \N_{>0}$ and as $S \subseteq \N_{>0}$ it follows that: :$S = N_{>0}$ {{Qed}}
For each $n \in \N$, let $\map {P'} n$ be defined as: :$\map {P'} n := \map P 0 \land \dots \land \map P n$ It suffices to show that $\map {P'} n$ is true for all $n \in \N$. It is immediate from the assumption $\map P 0$ that $\map {P'} 0$ is true. Now suppose that $\map {P'} n$ holds. By $(2)$, this implies that $\map P {n + 1}$ holds as well. Consequently, $\map {P'} n \land \map P {n + 1} = \map {P'} {n + 1}$ holds. Thus by the Principle of Mathematical Induction: :$\map {P'} n$ holds for all $n \in \N$ as desired. {{Qed}}
For each $n \in \N_{>0}$, let $\map {P'} n$ be defined as: :$\map {P'} n := \map P 1 \land \dots \land \map P n$ It suffices to show that $\map {P'} n$ is true for all $n \in \N_{>0}$. It is immediate from the assumption $\map P 1$ that $\map {P'} 1$ is true. Now suppose that $\map {P'} n$ holds. By $(2)$, this implies that $\map P {n + 1}$ holds as well. Consequently, $\map {P'} n \land \map P {n + 1} = \map {P'} {n + 1}$ holds. Thus by the Principle of Mathematical Induction: :$\map {P'} n$ holds for all $n \in \N_{>0}$ as desired. {{Qed}}
Let $S \subseteq \N_{>0}$ containing those $n \in \N_{>0}$ for which $\map P n$ does not hold. {{AimForCont}} $S \ne \O$. Then by the Well-Ordering Principle $S$ contains a minimal element $s$. We have that $s \ne 1$ because $\map P 1$ is true from $(1)$. Thus there must exist some $k \in \N_{>0}$ such that $s = k + 1$. As $k + 1$ is the minimal element of $S$ it follows that all $n$ such that $n < k + 1 = s$ are not in $S$ But this means that $\map P n$ is true for all $n < s$ But by $(2)$ it follows that $\map P s$ is true. That is, $s \notin S$. This contradicts our assertion that $s \in S$. Hence our assumption that $S \ne \O$ is false. It follows by Proof by Contradiction that $S = \O$. So $\map P n$ holds for all $n \in \N_{>0}$. {{Qed}}
Let $S$ be the set defined as: :$S := \set {n \in \N_{>0}: \map P n \text { is false} }$ {{AimForCont}} $S \ne \O$. From the Well-Ordering Principle it follows that $S$ has a minimal element $m$. From $(1)$ we have that $\map P 1$ holds. Hence $1 \notin S$. Therefore $m \ne 1$. Therefore $m - 1 \in \N_{>0}$. But $m$ is the minimal element of $S$. So $m - 1 \notin S$. Therefore $\map P {m - 1}$ is true. Hence by $(2)$ it follows that $\map P m$. But then $m \notin S$. This contradicts our supposition that $m \in S$. Hence there can be no such $m \in S$. So $S = \O$ and the result follows. {{qed}}
Let $M$ be the set of all $n \in \N_{>0}$ for which $\map P n$ holds. By $(1)$ we have that $1 \in M$. By $(2)$ we have that if $k \in M$ then $k + 1 \in M$. From the Axiomatization of $1$-Based Natural Numbers, Axiom $(F)$, it follows that $M = \N_{>0}$. {{qed}}
We have that Natural Numbers are Non-Negative Integers. Then we have that Integers form Well-Ordered Integral Domain. The result follows from Induction on Well-Ordered Integral Domain. {{qed}}
By Null Sequences form Maximal Left and Right Ideal then $\NN$ is a maximal left ideal of $\CC$.  A field is by definition a commutative ring. In a commutative ring, a maximal left ideal is by definition a maximal ideal. {{qed}}
Let $S$ be the set of all divisors of $n$. Then from Absolute Value of Integer is not less than Divisors: :$\forall m \in S: -n \le m \le n$ Thus $S$ is finite. {{qed}} Category:Number Theory Category:Divisors dhwii31ikzevtdctqmctmo6ixevhs9s
Proof by Counterexample: We have from GCD of Integer and Divisor: :$\gcd \set {n, n} = n$ and so, for example: :$\gcd \set {2, 2} = 2$ and so: :$2 \not \perp 2$ Hence $\perp$ is not reflexive. But we also note that: :$\gcd \set {1, 1} = 1$ and so: :$1 \perp 1$ demonstrating that $\perp$ is not antireflexive either. The result follows by definition of non-reflexive relation. {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \perp       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = \gcd \set {x, y}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \gcd \set {y, x}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = y       | o = \perp       | r = x       | c =  }} {{end-eqn}} Hence the result by definition of symmetric relation. {{qed}}
Proof by Counterexample: We have: :$\gcd \set {3, 5} = 1 = \gcd \set {5, 3}$ and so: :$3 \perp 5$ and $5 \perp 3$ However, it is not the case that $3 = 5$. The result follows by definition of antisymmetric relation. {{qed}} Category:Coprime Integers 44ni75ofq0kkckmeabv42zb9lt25kkl
Proof by Counterexample: We have: {{begin-eqn}} {{eqn | l = \gcd \set {2, 3}       | r = 1       | c =  }} {{eqn | l = \gcd \set {3, 4}       | r = 1       | c =  }} {{eqn | l = \gcd \set {2, 4}       | r = 2       | c =  }} {{end-eqn}} Hence we have: :$2 \perp 3$ and $3 \perp 4$ However, it is not the case that $2 \perp 4$. Thus $\perp$ is not transitive. Then we have: {{begin-eqn}} {{eqn | l = \gcd \set {2, 3}       | r = 1       | c =  }} {{eqn | l = \gcd \set {3, 5}       | r = 1       | c =  }} {{eqn | l = \gcd \set {2, 5}       | r = 1       | c =  }} {{end-eqn}} :$2 \perp 3$ and $3 \perp 5$ and also: :$2 \perp 5$ Thus $\perp$ is not antitransitive either. The result follows by definition of non-transitive relation. {{qed}}
Consider the polynomial $x + 1$ in $\R \sqbrk X$. There exists no polynomial $\map f x$ such that: :$\paren {x + 1} \map f x = 1$ This is because the {{LHS}} has degree $1$, and the {{RHS}} has degree $0$. {{qed}}
From Integer Multiples form Commutative Ring, $\struct {2 \Z, +, \times}$ is a commutative ring. As $2 \ne 1$, we also have from Integer Multiples form Commutative Ring that $\struct {2 \Z, +, \times}$ has no unity. Hence by definition it is not an integral domain. {{qed}}
From Subrings of Integers are Sets of Integer Multiples, a ring of the form $\struct {n \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$ when $n \ge 1$. $\struct {2 \Z, +, \times}$ is such an example. {{qed}}
Proof by Counterexample: We have that: :$2 + 0 i \in \Z \sqbrk i$ However there is no $z \in \Z \sqbrk i$ such that: :$x \paren {2 + 0 i} = 1 + 0 i$ So, by definition, $\Z \sqbrk i$ is not a field. Thus $\Z \sqbrk i$ is not a subfield of $\C$. {{qed}}
{{MissingLinks}} {{proofread}} Let $J$ be an ideal of $\struct {\Z_m, +, \times}$. $\struct {J, +}$ is a subgroup of $\struct {\Z_m, +}$. Let $\struct {G, +}$ be a subgroup of $\struct {\Z_m, +}$. Then $\struct {G, +}$ is a cyclic subgroup generated by $\gen d$, where $d \divides m$. We know that for a finite cyclic group of order $k$, the order of every subgroup is a divisor of $k$. Also there is exactly one [Definition:Subgroup|subgroup]] for each divisor. It follows that all ideals of $\struct {\Z_m, +, \times}$ are of form $\gen d$, where $d$ is a positive divisor of $m$. {{qed}}
By Quotient Ring of Cauchy Sequences is Division Ring then $\CC \,\big / \NN$ is a division ring. By Corollary to Cauchy Sequences form Ring with Unity then $\CC$ is a commutative ring with unity. By Quotient Ring of Commutative Ring is Commutative then $\CC \,\big / \NN$ is a commutative division ring, that is, a field. {{qed}}
By Quotient Ring of Cauchy Sequences is Normed Division Ring then $\CC \,\big / \NN$ is a normed division ring. By Corollary to Quotient Ring of Cauchy Sequences is Normed Division Ring then $\CC \,\big / \NN$ is a field. The result follows. {{qed}}
By definition, $\sequence {x_n}$ converges to $a \in R$ {{iff}} $\displaystyle \lim_{n \to \infty} \norm {x_n - a} = 0$ Then: {{begin-eqn}} {{eqn | l = \displaystyle \lim_{n \mathop \to \infty} \norm {x_n - a} = 0       | o = \leadstoandfrom       | r = \sequence {x_n - a} \in \NN       | c = {{Defof|Null Sequence in Normed Division Ring}} }} {{eqn | r = \sequence {x_n} - \sequence {a, a, a, \dotsc} \in \NN       | o = \leadstoandfrom       | c = {{Defof|Ring of Cauchy Sequences|Addition in Ring of Cauchy Sequences}} }} {{eqn | r = \sequence {x_n} \in \sequence {a, a, a, \dotsc} + \NN       | o = \leadstoandfrom       | c = Element in Left Coset iff Product with Inverse in Subgroup }} {{end-eqn}} {{qed}} Category:Normed Division Rings kyl41kgiab0qw0p446n6xt5ds8zq1ne
{{begin-eqn}} {{eqn | l = \map \phi {a -_1 b}       | r = \map \phi {a +_1 \paren {-b} }       | c = {{Defof|Ring Subtraction}} }} {{eqn | r = \map \phi a +_2 \map \phi {-b}       | c = {{Defof|Ring Homomorphism}} }} {{eqn | r = \map \phi a +_2 \paren {-\map \phi b}       | c = Ring Homomorphism Preserves Negatives }} {{eqn | r = \map \phi a -_2 \map \phi b       | c = {{Defof|Ring Subtraction}} }} {{end-eqn}} {{qed}}
By Range of Modulo Operation for Positive Modulus and Range of Modulo Operation for Negative Modulus we have: :$- \size y \lt x \bmod y \lt \size y$ The result follows from the Squeeze Theorem. {{qed}}
As $y \to \infty$: {{begin-eqn}} {{eqn | l = 0       | o = \le       | m = x       | mo= <       | r = y }} {{eqn | ll= \leadsto       | l = 0       | o = \le       | m = \frac x y       | mo= <       | r = 1 }} {{eqn | ll= \leadsto       | o =       | m = \floor {\frac x y}       | mo= =       | r = 0 }} {{end-eqn}} Therefore by the definition of modulo operation: {{begin-eqn}} {{eqn | l = \lim_{y \mathop \to \infty} x \bmod y       | r = \lim_{y \mathop \to \infty} x - y \floor {\dfrac x y} }} {{eqn | r = \lim_{y \mathop \to \infty} x - y \cdot 0 }} {{eqn | r = x }} {{end-eqn}} Hence the result. {{qed}}
=== Limit 1 === {{:Limit of Modulo Operation/Limit 1}} === Limit 2 === {{:Limit of Modulo Operation/Limit 2}} Category:Limits of Functions Category:Limit of Modulo Operation 44gsv3kmkf9ww8992bf3ug38ugwi974
By Cauchy Sequence Is Eventually Bounded Away From Non-Limit then: :$\exists N_1 \in \N$ and $C \in \R_{\gt 0}: \forall n \ge N_1: \norm {x_n} \gt C$ Since $\sequence {x_n}$ is a Cauchy sequence then: :$\exists N_2 \in \N: \forall n, m \ge N_2: \norm {x_n - x_m} < C$ Let $N = \max \set {N_1, N_2}$. Let $n, m \ge N$. Then: :$\norm {x_n - x_m} < C < \norm {x_n}$ By Corollary to Three Points in Ultrametric Space have Two Equal Distances then: :$\norm {x_n} = \norm {x_m}$ The result follows. {{qed}}
Let $\struct {R, +, \circ}$ be a ring with unity. === $(1)$ implies $(2)$ === Let $x \in R$ be a unit of $\struct {R, +, \circ}$ by definition 1. Then by definition: :$\exists y \in R: x \circ y = 1_R = y \circ x$ That is, by definition of divisor: :$x \divides 1_R$ Thus $x$ is a unit of $\struct {R, +, \circ}$ by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $x \in R$ be a unit of $\struct {R, +, \circ}$ by definition 2. Then by definition: :$x \divides 1_R$ By definition of divisor: :$\exists t \in R: 1_R = t \circ x$ Thus $x$ is a unit of $\struct {R, +, \circ}$ by definition 1. {{qed}} Category:Units of Rings l4gkmgz1c66ajs0aoygzxdbic0xbwud
{{Proofread}} We are given that $a, b \in D$ such that $a$ and $b$ are not both equal to $0$. {{WLOG}}, suppose specifically that $b \ne 0$. Let $S \subseteq D$ be the set defined as: :$S = \set {x \in D_{\ne 0}: x = m \times a + n \times b: m, n \in D}$ where $D_{\ne 0}$ denotes $D \setminus 0$. Setting $m = 0$ and $n = 1$, for example, it is noted that $b \in S$. Therefore $S \ne \O$. By definition, $\nu$ has the properties: :$(1): \quad \forall a, b \in D, b \ne 0: \exists q, r \in D$ such that $\map \nu r < \map \nu b$, or $r = 0$, such that: ::::$a = q \times b + r$ :$(2): \quad \forall a, b \in D, b \ne 0$: ::::$\map \nu a \le \map \nu {a \times b}$ Let $\nu \sqbrk S$ denote the image of $S$ under $\nu$. We have that: :$\nu \sqbrk S \subseteq \N$ Hence by the Well-Ordering Principle $\nu \sqbrk S$ has a smallest element. Let $d \in S$ be such that $\map \nu d$ is that smallest element of $\nu \sqbrk S$. By definition of $S$, we have that: :$d = u \times a + v \times b$ for some $u, v \in D$. Let $x \in S$. By $(2)$ above: :$x = q \times d + r$ such that either: :$\map \nu r < \map \nu d$ or: :$r = 0$ {{AimForCont}} $r \ne 0$. Then: {{begin-eqn}} {{eqn | lo= \exists m, n \in D:       | l = x       | r = m \times a + n \times b       | c =  }} {{eqn | ll= \leadsto       | l = r       | r = x - q \times d       | c =  }} {{eqn | r = \paren {m \times a + n \times b} - q \paren {u \times a + v \times b}       | c =  }} {{eqn | r = \paren {m - q \times u} a + \paren {n - q \times v} b       | c =  }} {{eqn | ll= \leadsto       | o =       | r = \paren {r \in S} \land \paren {\map \nu r < \map \nu d}       | c =  }} {{end-eqn}} which contradicts the choice of $d$ as the element of $S$ such that $\map \nu d$ is the smallest element of $\nu \sqbrk S$. Therefore: :$\forall x \in S: x = q \times d$ for some $q \in D$. That is: :$\forall x \in S: d \divides x$ where $\divides$ denotes divisibility. In particular: :$d \divides a = 1 \times a + 0 \times b$ :$d \divides b = 0 \times a + 1 \times b$ Thus: :$d \divides a \land d \divides b \implies \map \nu 1 \le \map \nu d \le \map \nu {\gcd \set {a, b} }$ However, note that as $\gcd \set {a, b}$ also divides $a$ and $b$ (by definition), we have: {{begin-eqn}} {{eqn | l = \gcd \set {a, b}       | o = \divides       | r = \paren {u \times a + v \times b} = d       | c =  }} {{eqn | ll= \leadsto       | l = \gcd \set {a, b}       | o = \divides       | r = d       | c =  }} {{eqn | ll= \leadsto       | l = \map \nu {\gcd \set {a, b} }       | o = \le       | r = \map \nu d       | c =  }} {{end-eqn}} Since $d$ is the element of $S$ such that $\map \nu d$ is the smallest element of $\nu \sqbrk S$: :$\gcd \set {a, b} = d = u \times a + v \times b$ {{qed}} {{Namedfor|Étienne Bézout|cat = Bézout}}
By principal ideal domain, $D$ is a principal ideal domain Consider the principal ideal $U = \ideal b$ of $D$ generated by $b$. As $\nu$ is a valuation function, we have: :$\forall x \in D, x \ne 0: \map \nu b \le \map \nu {b \times x}$ As $U$ is a principal ideal: :$\forall a \in \ideal b: \exists x \in D: a = x \times b$ and so: :$\forall a \in U: \map \nu b \le \map \nu a$ Let $\map \nu b = \map \nu {b c}$. Then from above: :$\forall a \in U: \map \nu {b c} \le \map \nu a$ From the argument in Euclidean Domain is Principal Ideal Domain, we have that $U$ is also the ideal generated by $b c$. {{explain|The above statement is not very clear and needs to be expanded.}} Thus $a b$ is a divisor of $b$ itself, that is: :$\exists y \in D: b = b c y$ Thus: :$c y = 1$ and so $c$ is a unit of $D$. So if $c$ is not a unit, then it must be the case that $\map \nu b < \map \nu {b c}$. {{qed}}
For $a \in D$ we have that: :$\map \nu 1 \le \map \nu {1 a} = \map \nu a$ by definition of Euclidean valuation. Let $a$ be a unit of $D$. Then: :$\exists b \in D: a b = 1$ Then: :$\map \nu a \le \map \nu {a b} = \map \nu 1$ and so: :$\map \nu a = \map \nu 1$ {{qed|lemma}} Let $\map \nu a = \map \nu 1$. We can write this as: :$\map \nu {1 a} = \map \nu 1$ and it follows from Euclidean Valuation of Non-Unit is less than that of Product that $a$ is a unit of $D$. {{qed}}
Consider the coefficients of $f$ expressed as fractions. Let $k$ be any positive integer that is divisible by the denominators of all the coefficients of $f$. Such a number is bound to exist: just multiply all those denominators together, for example. Then $\map f X$ is a polynomial equal to $\dfrac 1 k$ multiplied by a polynomial with integral coefficients. Let $d$ be the GCD of all these integral coefficients. Then $\map f X$ is equal to $\dfrac h k$ multiplied by a primitive polynomial.
Suppose that $a \cdot \map f X = b \cdot \map g X$ where $a, b \in \Q$ and $f, g$ are primitive. Then: :$\map g X = \dfrac a b \map f X$ where $\dfrac a b$ is some rational number which can be expressed as $\dfrac m n$ where $m$ and $n$ are coprime. Then: :$\map g X = \dfrac m n \map f X$ that is: :$m \cdot \map f X = n \cdot \map g X$ Suppose $m > 1$. Then from Euclid's Lemma $m$ has a divisor $p$ which does not divide $n$ (as $m \perp n$). So $m$ must divide every coefficient of $g$. But this can not be so, as $g$ is primitive, so $m = 1$. In a similar way, $n = 1$. So $f = g$ and $a = b$, so demonstrating uniqueness. {{qed}}
Let $\cont h$ denote the content of $\map h X$. From Polynomial has Integer Coefficients iff Content is Integer: :$\cont h \in \Z$ Let $\map h X = \map f X \, \map g X$ as suggested. Then from Rational Polynomial is Content Times Primitive Polynomial: {{begin-eqn}} {{eqn | l = \map h X       | r = \cont f \cont g \cdot \map {f'} X \, \map {g'} X       | c = Rational Polynomial is Content Times Primitive Polynomial }} {{eqn | r = \cont h \cdot \map {f'} X \, \map {g'} X       | c = Content of Rational Polynomial is Multiplicative }} {{end-eqn}} From the above, $\map {f'} X$ and $\map {g'} X$ are primitive. Hence by definition: :$\cont {f'} = \cont {g'} = 1$ From Polynomial has Integer Coefficients iff Content is Integer, both $\map {f'} X$ and $\map {g'} X$ have coefficients which are all integers. We also have by definition of content that $\cont f$ and $\cont g$ are rational numbers. The result follows. {{qed}}
{{AimForCont}} $\map f X = \map g X \, \map h X$ for some $\map g X, \map h X \in \Q \sqbrk X$. By hypothesis: :$\map f X \in \Z \sqbrk X$ and so by definition has coefficients all of which are integers. But from Factors of Polynomial with Integer Coefficients have Integer Coefficients it follows that $\map f X$ can be expressed as: :$\map f X = \map {g'} X \, \map {h'} X$ where both $\map {g'} X$ and $\map {h'} X$ are elements of $\Q \sqbrk X$ which have coefficients all of which are integers. That is: :$\map {g'} X \in \Z \sqbrk X$ and: :$\map {h'} X \in \Z \sqbrk X$ This contradicts the statement that $\map f X$ is irreducible in $\Z \sqbrk X$. Hence the result by Proof by Contradiction. {{qed}}
Let $x \in \R$. Let $L \in \Q$. If $x \in \Q$, then: {{begin-eqn}} {{eqn | l = \map D {x + L}       | r = c       | c = Rational Addition is Closed }} {{eqn | r = \map D x }} {{end-eqn}} If $x \notin \Q$, then: {{begin-eqn}} {{eqn | l = \map D {x + L}       | r = d       | c = Rational Number plus Irrational Number is Irrational }} {{eqn | r = \map D x }} {{end-eqn}} Combining the above two shows that: :$\forall x \in \R: \map D x = \map D {x + L}$ Hence the result. {{qed}} Category:Periodic Functions tl67hs3t48g1041uc6cl3wfo3tmenr4
Let $D: \R \to \R$ be a Dirichlet function. In proving that the Dirichlet Function is Periodic, it was shown that every non-zero rational number is a periodic element of $D$. Therefore, the period of $D$ must be the smallest element of $\Q_{> 0}$. But from Rational Numbers are not Well-Ordered under Conventional Ordering there cannot exist such an element. Hence the result. {{qed}} Category:Periodic Functions gbw385yz4t49ue86fmt3m04luu36wc7
By Dirichlet Function is Periodic and Dirichlet Function has no Period, it is seen that the Dirichlet functions are such an example. {{qed}} Category:Periodic Functions h8bvusygz1495o4a7vflouyy4m8j6jr
From Module of All Mappings is Module, we have that $\struct {G^S, +_G', \circ}_R$ is an $R$-module. To show that $\struct {G^S, +_G', \circ}_R$ is a unitary $R$-module, we verify the following: :$\forall f \in G^S: 1_R \circ f = f$ Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module. Then: :$\forall x \in G: 1_R \circ x = x$ Thus: {{begin-eqn}} {{eqn | l = \map {\paren {1_R \circ f} } x       | r = 1_R \circ \paren {\map f x}       | c =  }} {{eqn | r = \map f x       | c =  }} {{end-eqn}} Hence the result. {{qed}}
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 $\mathcal L_{f > 0}$ 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 $\mathcal L_{f > 0}$, 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 $\mathcal L_{f > 0}$. 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 $\mathcal L_{f > 0}$, 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 t87r6owlmp5wcgtu8s0a2c4rqgdqan3
This is a special case of the Vector Space of All Mappings, where $S$ is the set $\closedint 1 n \subset \N^*$. {{qed}}
This is a special case of a direct product of vector spaces where each of the $G_k$ is the $K$-vector space $K$. {{qed}}
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}}