hcju/proofwikiqa huggingface dataset:
We assume the truth of '''WOP'''. Let $S \subseteq \N$ which satisfy: :$(D): \quad 0 \in S$ :$(E): \quad n \in S \implies n+1 \in S$. We want to show that $S = \N$, that is, the '''PFI''' is true. {{AimForCont}} that: :$S \ne \N$ Consider $S' = \N \setminus S$, where $\setminus$ denotes set difference. From Set Difference is Subset, $S' \subseteq \N$. So from '''WOP''', $S'$ has a minimal element. A lower bound of $\N$ is $0$. By Lower Bound for Subset, $0$ is also a lower bound for $S'$. By hypothesis, $0 \in S$. From the definition of set difference, $0 \notin S'$. So this minimal element of $S'$ has the form $k + 1$ where $k \in \N$. We can consider the von Neumann construction of the natural numbers. By definition of natural number addition, it is noted that $k + 1 \in \N$ is the immediate successor element of $k \in \N$. Thus $k \in S$ but $k + 1 \notin S$. From $(E)$, this contradicts the definition of $S$. Thus if $S' \ne \O$, it has no minimal element. This contradicts the Well-Ordering Principle, and so  $S' = \O$. So $S = N$. Thus we have proved that '''WOP''' implies '''PFI'''.
We have that: {{begin-eqn}} {{eqn | l = \dfrac 5 2       | r = 2 + \dfrac 1 2       | c =  }} {{eqn | ll= \leadsto       | l = 2       | o = \le       | r = \dfrac 5 2       | c =  }} {{eqn | o = <       | r = 3       | c =  }} {{end-eqn}} Hence $2$ is the floor of $\dfrac 5 2$ by definition. {{qed}}
We have that: {{begin-eqn}} {{eqn | l = -\dfrac 5 2       | r = -3 + \dfrac 1 2       | c =  }} {{eqn | ll= \leadsto       | l = -3       | o = \le       | r = -\dfrac 5 2       | c =  }} {{eqn | o = <       | r = -2       | c =  }} {{end-eqn}} Hence $-3$ is the floor of $-\dfrac 5 2$ by definition. {{qed}}
We have that $14$ is an integer. Thus this is a specific example of Real Number is Integer iff equals Floor: $\floor x = x \iff x \in \Z$ {{qed}}
{{begin-eqn}} {{eqn | l = \sqrt 9       | o = \le       | r = \sqrt 10       | c =  }} {{eqn | o = <       | r = \sqrt 16       | c =  }} {{eqn | ll= \leadsto       | l = 3       | o = \le       | r = \sqrt 10       | c =  }} {{eqn | o = <       | r = 4       | c =  }} {{end-eqn}} Hence $3$ is the floor of $\sqrt {10}$ by definition. {{qed}}
Since $\sequence {x_n}$ does not converge to $0$, by Cauchy Sequence Is Eventually Bounded Away From Non-Limit then: :$\exists K \in \N$ and $C \in \R_{>0}: \forall n > K: C <  \norm {x_n}$ or equivalently: :$\exists K \in \N$ and $C \in \R_{>0}: \forall n > K: 1 < \dfrac {\norm {x_n} } C$ By {{NormAxiom|1}}: :$\forall n > K : x_n \ne 0$ Let $\sequence {y_n}$ be the subsequence of $\sequence {x_n}$ defined as: :$y_n = x_{K + n}$ By Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence: :$\sequence {y_n}$ is a Cauchy sequence. So $\sequence { {y_n}^{-1} }$ is well-defined and $\sequence { {y_n}^{-1} } = \sequence {\paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$. Let $\epsilon > 0$ be given. Let $\epsilon' = \epsilon C^2$, then $ \epsilon' > 0$. Similarly, $\sequence {y_n}$ is a Cauchy sequence, we can find $N$ such that: :$\forall n, m > N_2: \norm {y_n - y_m} < \epsilon'$ Thus $\forall n, m > N$: :$(1): \quad 1 < \dfrac {\norm {y_n} } C, \dfrac {\norm {y_m} } C$ :$(2): \quad \norm {y_n - y_m} < \epsilon'$ Hence: {{begin-eqn}} {{eqn | l = \norm { {y_n}^{-1} - {y_m}^{-1} }       | o = <       | r = \dfrac {\norm {y_n} } C \norm { {y_n }^{-1} - {y_m}^{-1} } \dfrac {\norm {y_m} } C       | c = $(1)$ above }} {{eqn | r = \dfrac 1 {C^2} \paren {\norm {y_n} \norm { {y_n }^{-1} - {y_m}^{-1 } } \norm {y_m} } }} {{eqn | r = \dfrac 1 {C^2} \norm {y_n \paren { {y_n }^{-1 } - {y_m }^{-1} } y_m}       | c = {{NormAxiom|2}} }} {{eqn | r = \dfrac 1 {C^2} \norm {\paren {y_n {y_n }^{-1 } - y_n {y_m }^{-1} } y_m }       | c = Ring Axiom $(\text D)$: Distributivity }} {{eqn | r = \dfrac 1 {C^2} \norm {y_n {y_n}^{-1} y_m - y_n {y_m}^{-1} y_m}       | c = Ring Axiom $(\text D)$: Distributivity }} {{eqn | r = \dfrac 1 {C^2} \norm {y_m - y_n}       | c = Inverse Property of a Division Ring }} {{eqn | r = \dfrac 1 {C^2} \epsilon'       | o = <       | c = $(2)$ above }} {{eqn | r = \dfrac 1 {C^2} \paren {\epsilon C^2}       | c = Definition of $\epsilon'$ }} {{eqn | r = \epsilon }} {{end-eqn}} So: :$\sequence { { {y_n}^{-1} } }$ is a Cauchy sequence in $R$. {{qed}}
Let $\epsilon > 0$. Since $\sequence {x_n}$ is a Cauchy sequence then: :$\exists N: \forall n,m > N: \norm {x_n - x_m } < \epsilon$ Now let $R = N$. Then from Strictly Increasing Sequence of Natural Numbers: :$\forall r, s > R: n_r \ge r$ and $n_s \ge s$ Thus $n_r, n_s > N$ and so: :$\norm {x_{n_r} - x_{n_s} } < \epsilon$ The result follows. {{qed}}
Let $p$ be a prime factor of $s$. Then for some $n \in \Z_{>0}$: :$p^{2 n} \divides s t$ where $\divides$ denotes the divisibility relation. But we have that $s \perp t$ and so $p \nmid t$. Thus $p^{2 n} \divides s$. This holds for all prime factors of $s$. Thus $s$ is the product of squares of primes. Thus $s$ is square. The same argument holds for $t$. Hence the result. {{qed}}
Let $\sequence {x_n}$ be the constant sequence:  :$\forall n, x_n = a$ Given $\epsilon > 0$: :$\forall n, m \ge 1: \norm {x_n - x_m} = \norm {a - a} = \norm {0} = 0 < \epsilon$ The result follows. {{qed}}
Follows directly from Product Rule for Normed Division Ring Sequences, setting :$\sequence {y_n} := \sequence {x_n}$ and: :$\sequence {x_n} := \tuple {a, a, a, \ldots}$ {{qed}}
From Multiple Rule for Normed Division Ring Sequences: :$\sequence {-y_n} = \sequence {\paren {-1} y_n}$ is a Cauchy sequence.  From Sum Rule for Normed Division Ring Sequences: :$\sequence {x_n - y_n} = \sequence {x_n + \paren {-y_n} }$ is a Cauchy sequence. {{qed}}
From the Multiple Rule for Normed Division Ring Sequences: :$\sequence {a x_n}$ is a Cauchy sequence :$\sequence {b y_n}$ is a Cauchy sequence. The result now follows directly from the Sum Rule for Normed Division Ring Sequences: :$\sequence {a x_n + b y_n}$ is a Cauchy sequence. {{qed}}
By definition of the set of residue classes modulo $m$, $Z_m$ is the quotient set of congruence modulo $m$: :$\Z_m = \dfrac \Z {\mathcal R_m}$ where $\mathcal R_m$ is the congruence relation modulo $m$ on the set of all $a, b \in \Z$: :$\mathcal R_m = \set {\tuple {a, b} \in \Z \times \Z: \exists k \in \Z: a = b + k m}$ By the Fundamental Theorem on Equivalence Relations, $Z_m$ is a partition of $\Z$. Thus: :$\eqclass a m = \eqclass b m$ {{iff}}: :$x \in \eqclass a m \iff x \in \eqclass b m$ {{iff}}: :$a \equiv b \pmod m$ {{qed}}
By definition of the set of residue classes modulo $m$, $\Z_m$ is the quotient set of congruence modulo $m$: :$\Z_m = \dfrac \Z {\mathcal R_m}$ where $\mathcal R_m$ is the congruence relation modulo $m$ on the set of all $a, b \in \Z$: :$\mathcal R_m = \set {\tuple {a, b} \in \Z \times \Z: \exists k \in \Z: a = b + k m}$ By the Fundamental Theorem on Equivalence Relations, $\Z_m$ is a partition of $\Z$. {{qed}}
By definition of the set of residue classes modulo $m$, $Z_m$ is the quotient set of congruence modulo $m$: :$\Z_m = \dfrac \Z {\mathcal R_m}$ where $\mathcal R_m$ is the congruence relation modulo $m$ on the set of all $a, b \in \Z$: :$\mathcal R_m = \set {\tuple {a, b} \in \Z \times \Z: \exists k \in \Z: a = b + k m}$ Thus by definition $\Z_m$ is the '''set of residue classes modulo $m$''': :$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$ Hence: {{begin-eqn}} {{eqn | l = \card {\Z_m}       | r = \card {\set {0, 1, \ldots, m - 1} }       | c =  }} {{eqn | r = m       | c =  }} {{end-eqn}} {{qed}}
By Structure Induced by Ring Operations is Ring then $\struct {R^S, +', \circ'}$ is a ring. We have from Induced Structure Identity that the constant mapping $f_{1_R}: S \to R$ defined as: :$\forall x \in S: \map {f_{1_R} } x = 1_R$ is the identity for $\struct {R^S, \circ'}$. The result follows by definition of ring with unity and unity of ring. {{qed}} Category:Rings of Mappings Category:Rings with Unity 5rqmmpykqgm8c2kdn94fo2bus367hci
The Subring Test used to prove the result. By Constant Rule for Cauchy sequences: :the constant sequence $\tuple {1, 1, 1, \dotsc}$ is a Cauchy sequences. Hence: :$\CC \neq \O$ Let $\sequence {x_n}, \sequence {y_n} \in \CC$. By definition of pointwise addition: :$\sequence {x_n} + \paren {-\sequence {y_n}} = \sequence {x_n - y_n}$. By Difference Rule for Normed Division Ring Sequences: :$\sequence {x_n - y_n}$ is a Cauchy sequence. Hence: :$\sequence {x_n} + \paren {- \sequence {y_n}} \in CC$. By definition of pointwise product: :$\sequence {x_n} \circ \sequence {y_n} = \sequence {x_n \circ y_n}$. By Product Rule for Normed Division Ring Sequences: :the sequence $\sequence {x_n \circ y_n}$ is a Cauchy sequence. Hence: :$\sequence {x_n} \circ \sequence {y_n} \in \CC$. By the Subring Test the result follows. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === We have that for all $X \in S$: :$X \cap A = X \cap A$ That is: :$X \mathrel \alpha X$ Thus $\alpha$ is seen to be reflexive. {{qed|lemma}} === Symmetry === {{begin-eqn}} {{eqn | l = X       | o = \alpha       | r = Y       | c =  }} {{eqn | ll= \implies       | l = X \cap A       | r = Y \cap A       | c =  }} {{eqn | ll= \implies       | l = Y \cap A       | r = X \cap A       | c =  }} {{eqn | ll= \implies       | l = Y       | o = \alpha       | r = X       | c =  }} {{end-eqn}} Thus $\alpha$ is seen to be symmetric. {{qed|lemma}} === Transitivity === Let $X \mathrel \alpha Y$ and $Y \mathrel \alpha Z$. Then by definition: {{begin-eqn}} {{eqn | l = X \cap A       | r = Y \cap A }} {{eqn | l = Y \cap A       | r = Z \cap A }} {{end-eqn}} Hence: :$X \cap A = Z \cap A$ and so by definition of $\alpha$: :$X \mathrel \alpha Z$ Thus $\alpha$ is seen to be transitive. {{qed|lemma}} $\alpha$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
By Structure Induced by Ring Operations is Ring then $\struct {R^S, +', \circ'}$ is a ring. From Structure Induced by Commutative Operation is Commutative, so is the pointwise operation $\circ$ induces on $R^S$. The result follows by definition of commutative ring. {{qed}} Category:Rings of Mappings d9gcno26rkxm8oz0gfxy8wkr83wucul
That $\alpha$ is an equivalence relation is proved in Equivalence Relation on Power Set induced by Intersection with Subset. We have that: :$\eqclass \O \alpha = \set {X \in S: X \cap A = \O \cap A = \O}$ Thus: {{begin-eqn}} {{eqn | l = X       | o = \in       | r = \eqclass \O \alpha       | c =  }} {{eqn | ll= \leadsto       | l = X \cap A       | r = \O       | c = {{Defof|Equivalence Class}} }} {{eqn | ll= \leadsto       | l = X \cap A       | o = \subseteq       | r = \relcomp T A       | c = Empty Intersection iff Subset of Complement }} {{eqn | ll= \leadsto       | l = X       | o = \subseteq       | r = T \setminus A       | c = {{Defof|Relative Complement}} }} {{eqn | ll= \leadsto       | l = X       | o = \in       | r = \powerset {T \setminus A}       | c = {{Defof|Power Set}} }} {{eqn | ll= \leadsto       | l = \eqclass \O \alpha       | o = \subseteq       | r = \powerset {T \setminus A}       | c = {{Defof|Subset}} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = X       | o = \in       | r = \powerset {T \setminus A}       | c =  }} {{eqn | ll= \leadsto       | l = X \cap A       | r = \varnothing       | c = reversing the above argument }} {{eqn | ll= \leadsto       | l = X       | o = \alpha       | r = \varnothing       | c =  }} {{eqn | ll= \leadsto       | l = X       | o = \in       | r = \eqclass \O \alpha       | c =  }} {{eqn | ll= \leadsto       | l = \powerset {T \setminus A}       | o = \subseteq       | r = \eqclass \O \alpha       | c = {{Defof|Subset}} }} {{end-eqn}} {{qed}}
That $\alpha$ is an equivalence relation is proved in Equivalence Relation on Power Set induced by Intersection with Subset. Let $U \subseteq A$. Then $U$ is in its own $\alpha$-equivalence class. Now suppose $U' \subseteq A$ such that $U \ne U'$. From Intersection with Subset is Subset: :$U \cap A = U$ and: :$U' \cap A = U'$ So: :$U \ne U' \implies \eqclass U \alpha \ne \eqclass {U'} \alpha$ and so there are as many $\alpha$-equivalence classes as there are subsets of $A$. That is: :$\card {\set {\eqclass X \alpha: X \in S} } = \powerset A$ The result follows from Cardinality of Power Set of Finite Set. {{qed}}
Proof by Counterexample: Let $S = \set {a, b, c}$. Let: : $\alpha = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {b, a}, \tuple {b, c}, \tuple {c, b} }$ By inspection it is seen that $\alpha$ is both reflexive and symmetric. However, we have: :$a \mathrel \alpha b$ and $b \mathrel \alpha c$ but it is not the case that $a \mathrel \alpha c$. Hence $\alpha$ is both reflexive and symmetric but not transitive. {{qed}}
Proof by Counterexample: Let $S = \Z$ be the set of integers. Let $n \in \Z$ such that $n > 1$. Let $\alpha$ be the relation on $S$ defined as: :$\forall x, y \in S: x \mathrel \alpha y \iff \size {x - y} < n$ where $\size x$ denotes the absolute value of $x$. It is seen that: :$\forall x \in \Z: \size {x - x} = 0 < n$ and so: :$\forall x \in \Z: x \mathrel \alpha x$ Thus $\alpha$ is reflexive Then it is seen that: :$\forall x, y \in \Z: \size {x - y} = \size {y - x}$ and so if $\size {x - y} < n$ then so is $\size {y - x} < n$. That is, if $x \mathrel \alpha y$ then $y \mathrel \alpha x$. Thus $\alpha$ is symmetric. Now let $x = -\paren{n - 1}, y = 0, z = n - 1$. We have: {{begin-eqn}} {{eqn | l = \size {x - y}       | r = n - 1 }} {{eqn | l = \size {y - z}       | r = n - 1 }} {{end-eqn}} and so both $x \mathrel \alpha y$ and $y \mathrel \alpha z$. But: {{begin-eqn}} {{eqn | l = \size {x - z}       | r = \size {-\paren {n - 1} - \paren {n - 1} } }} {{eqn | r = 2 n - 2 }} {{eqn | o = \ge       | r = n + 2 - 2       | c = as $n \ge 2$ }} {{eqn | r = n       | c = as $n \ge 2$ }} {{eqn | ll= \leadsto       | l = \neg \size {x - z}       | o = <       | r = n       | c =  }} {{end-eqn}} Thus: :$\neg x \mathrel \alpha z$ and so $\alpha$ is not transitive. Hence $\alpha$ is both reflexive and symmetric but not transitive. {{qed}}
Proof by Counterexample: Let $S = \set {a, b, c}$. Let: : $\alpha = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {b, c}, \tuple {a, c} }$ By inspection it is seen that $\alpha$ is both reflexive and transitive. However, we have: :$a \mathrel \alpha b$ but it is not the case that $b \mathrel \alpha a$. Hence $\alpha$ is both reflexive and transitive but not symmetric. {{qed}}
Proof by Counterexample: Let $S = \Z$ be the set of integers. Let $\alpha$ be the relation on $S$ defined as: :$\forall x, y \in S: x \mathrel \alpha y \iff x \le y$ It is seen that: :$\forall x \in \Z: x \le x$ and so: :$\forall x \in \Z: x \mathrel \alpha x$ Thus $\alpha$ is reflexive. Then it is seen that: :$\forall x, y, z \in \Z: x \le y, y \le z \implies x \le z$ Thus $\alpha$ is transitive. Now let $x = 1$ and $y = 2$. Then: : $x \le y$ but it is not the case that $y \le x$ and so $\alpha$ is not symmetric. Hence $\alpha$ is both reflexive and transitive but not symmetric. {{qed}}
Proof by Counterexample: Let $S = \Z$ be the set of integers. Let $\alpha$ be the relation on $S$ defined as: :$\forall x, y \in S: x \mathrel \alpha y \iff x = y = 0$ {{begin-eqn}} {{eqn | l = x       | o = \alpha       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = y = 0       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = x = 0       | c =  }} {{eqn | ll= \leadsto       | l = y       | o = \alpha       | r = x       | c =  }} {{end-eqn}} Thus $\alpha$ is symmetric. Now let $x \mathrel \alpha y$ and $y \mathrel \alpha z$ Then: :$x = y = 0, y = z = 0$ and so :$x \mathrel \alpha z$ Now let $x = \Z$ such that $x \ne 0$. Then it is not the case that: :$x \mathrel \alpha x$ and so $\alpha$ is not reflexive. Hence $\alpha$ is both symmetric and transitive but not reflexive. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === We have that for all $x \in \N$: :$x = 2^0 x$ and as $0 \in \Z$ it follows that: :$x \mathrel \alpha x$ Thus $\alpha$ is seen to be reflexive. {{qed|lemma}} === Symmetry === {{begin-eqn}} {{eqn | l = x       | o = \alpha       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = x        | r = 2^n y       | c = for some $n \in \Z$ }} {{eqn | ll= \leadsto       | l = y       | r = 2^{-n} x       | c = dividing both sides by $2^{-n}$ }} {{eqn | ll= \leadsto       | l = y       | o = \alpha       | r = x       | c = as $-n \in \Z$ }} {{end-eqn}} Thus $\alpha$ is seen to be symmetric. {{qed|lemma}} === Transitivity === Let $x \mathrel \alpha y$ and $y \mathrel \alpha z$. Then by definition: {{begin-eqn}} {{eqn | l = y       | r = 2^m z       | c = for some $m \in \Z$ }} {{eqn | l = x       | r = 2^n y       | c = for some $n \in \Z$ }} {{eqn | r = 2^n \paren {2^m z}       | c =  }} {{eqn | r = 2^{n + m} z       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = \alpha       | r = z       | c = as $n + m \in \Z$ }} {{end-eqn}} Thus $\alpha$ is seen to be transitive. {{qed|lemma}} $\alpha$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
That $\alpha$ is an equivalence relation is proved in Equivalence Relation on Natural Numbers such that Quotient is Power of Two. Let $p$ be a prime number whose $\alpha$-equivalence class is $\eqclass p \alpha$. {{AimForCont}} $\eqclass p \alpha$ contains a prime number $q$ such that $q \ne p$. Then: :$p = 2^n q$ for some $n \in \Z$ If $n = 0$ then $p = q$ which contradicts $q \ne p$. Thus $n \ne 0$. Let $n > 0$. Then: :$p = r q$ for $r \in \Z$ such that $r = 2^n$. Thus $p$ is a composite number with divisors $q$ and $r$ (and possibly more). This contradicts the supposition that $p$ is prime. Let $n < 0$. Then: :$p = \dfrac q {2^m}$ where $m = -n$ and so $m > 0$. Thus: :$q = r p$ for $r \in \Z$ such that $r = 2^m$.   Thus $q$ is a composite number with divisors $p$ and $r$ (and possibly more). This contradicts the supposition that $q$ is prime. Thus all cases of $n$ lead to a contradiction. It follows that no $\alpha$-equivalence class can contain more than one prime number. {{qed}}
That $\alpha$ is an equivalence relation is proved in Equivalence Relation on Natural Numbers such that Quotient is Power of Two. Suppose $r$ is even. Then $r = 2 s$ for some $s \in \N$. Thus $s \in \eqclass r \alpha$ such that $s < r$. This contradicts the supposition that $r$ is the smallest such natural number. Thus $r$ has to be the smallest odd composite number, and so: :$r = 3^2 = 9$ {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === We have that for all $\mathbf A \in S$: :$\mathbf A^r = \mathbf A^r$ for all $r \in \N$. It follows by definition of $\alpha$ that: :$\mathbf A \mathrel \alpha \mathbf A$ Thus $\alpha$ is seen to be reflexive. {{qed|lemma}} === Symmetry === {{begin-eqn}} {{eqn | l = \mathbf A       | o = \alpha       | r = \mathbf B       | c =  }} {{eqn | ll= \implies       | l = \mathbf A^r        | r = \mathbf B^s       | c = for some $r, s \in \N$ }} {{eqn | ll= \implies       | l = \mathbf B^s       | r = \mathbf A^r       | c =  }} {{eqn | ll= \implies       | l = \mathbf B       | o = \alpha       | r = \mathbf A       | c =  }} {{end-eqn}} Thus $\alpha$ is seen to be symmetric. {{qed|lemma}} === Transitivity === Let: : $\mathbf A \mathrel \alpha \mathbf B$ and $\mathbf B \mathrel \alpha \mathbf C$ for square matrices of order $n$ $\mathbf A, \mathbf B, \mathbf C$. Then by definition: {{begin-eqn}} {{eqn | l = \mathbf A^r       | r = \mathbf B^s       | c = for some $r, s \in \N$ }} {{eqn | l = \mathbf B^u       | r = \mathbf C^v       | c = for some $u, v \in \N$ }} {{eqn | ll= \leadsto       | l = \mathbf A^{r u}       | r = \mathbf B^{s u}       | c = raising both sides to $u$th power }} {{eqn | l = \mathbf B^{s u}       | r = \mathbf C^{s v}       | c = raising both sides to $s$th power }} {{eqn | ll= \leadsto       | l = \mathbf A^{r u}       | r = \mathbf C^{s v}       | c =  }} {{end-eqn}} Thus $\alpha$ is seen to be transitive. {{qed|lemma}} $\alpha$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === We have that for all $x \in \Z$: :$x^2 \equiv x^2 \pmod 5$ It follows by definition of $\beta$ that: :$x \mathrel \beta x$ Thus $\beta$ is seen to be reflexive. {{qed|lemma}} === Symmetry === {{begin-eqn}} {{eqn | l = x       | o = \beta       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = x^2       | o = \equiv        | r = y^2       | rr= \pmod 5       | c =  }} {{eqn | ll= \leadsto       | l = y^2       | o = \equiv        | r = x^2       | rr= \pmod 5       | c =  }} {{eqn | ll= \leadsto       | l = y       | o = \beta       | r = x       | c =  }} {{end-eqn}} Thus $\beta$ is seen to be symmetric. {{qed|lemma}} === Transitivity === Let: :$x \mathrel \beta y$ and $y \mathrel \beta z$ for $x, y, z \in \Z$. Then by definition: {{begin-eqn}} {{eqn | l = x^2       | o = \equiv        | r = y^2       | rr= \pmod 5       | c =  }} {{eqn | l = y^2       | o = \equiv        | r = z^2       | rr= \pmod 5       | c =  }} {{eqn | ll= \leadsto       | l = x^2       | o = \equiv        | r = z^2       | rr= \pmod 5       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = \beta       | r = z       | c =  }} {{end-eqn}} Thus $\beta$ is seen to be transitive. {{qed|lemma}} $\beta$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
That $\beta$ is an equivalence relation is proved in Equivalence Relation on Integers Modulo 5 induced by Squaring. The set of residue classes modulo $5$ is: :$\set {\eqclass 0 5, \eqclass 1 5, \eqclass 2 5, \eqclass 3 5, \eqclass 4 5}$ Then: {{begin-eqn}} {{eqn | l = 0 \times 0       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass 0 5 \times_5 \eqclass 0 5       | r = \eqclass 0 5       | c =  }} {{eqn | l = 1 \times 1       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass 1 5 \times_5 \eqclass 1 5       | r = \eqclass 1 5       | c =  }} {{eqn | l = 2 \times 2       | r = 4       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass 2 5 \times_5 \eqclass 2 5       | r = \eqclass 4 5       | c =  }} {{eqn | l = 3 \times 3       | r = 9       | c =  }} {{eqn | o = \equiv       | r = 4       | rr= \pmod 5       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass 3 5 \times_5 \eqclass 3 5       | r = \eqclass 4 5       | c =  }} {{eqn | l = 4 \times 4       | r = 16       | c =  }} {{eqn | o = \equiv       | r = 1       | rr= \pmod 5       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass 4 5 \times_5 \eqclass 4 5       | r = \eqclass 1 5       | c =  }} {{end-eqn}} Thus we have that: :$\eqclass 1 5^2 = \eqclass 4 5^2 = \eqclass 1 \beta$ :$\eqclass 2 5^2 = \eqclass 3 5^2 = \eqclass 2 \beta$ Hence the result. {{qed}}
That $\beta$ is an equivalence relation is proved in Equivalence Relation on Integers Modulo 5 induced by Squaring. From Number of Equivalence Classes we have: We have: {{begin-eqn}} {{eqn | l = \eqclass 1 \beta       | r = \eqclass 4 \beta       | c =  }} {{eqn | l = \eqclass 2 \beta       | r = \eqclass 3 \beta       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \eqclass 0 \beta       | r = \eqclass 5 \beta       | c =  }} {{eqn | r = \eqclass {1 + 4} \beta       | c =  }} {{eqn | r = \eqclass 1 \beta + \eqclass 4 \beta       | c =  }} {{eqn | r = \eqclass 1 \beta + \eqclass 1 \beta       | c =  }} {{eqn | r = \eqclass 2 \beta       | c =  }} {{end-eqn}} Hence the result. {{qed}}