hcju/proofwikiqa huggingface dataset:
Let $T$ be the set: :$T = \set {n \in P: \map s n \ne n}$ We will use Axiom $(\text P 5)$ to prove that $T = P$. === Part $1$: $0 \in T$ === From Axiom $(\text P 4)$, a fortiori: :$\map s 0 \ne 0$ Hence $0 \in T$. {{qed|lemma}} === Part $2$: $n \in T \implies \map s n \in T$ === Let $n \in T$. Let $m = \map s n$. Then $n \ne m$, as $\map s n \ne n$. {{AimForCont}} $m \notin T$. That is: :$\map s m = m$ Then that would mean: :$\map s n = \map s m = m$ But from Axiom $(\text P 3)$, $s$ is injective. From this contradiction, it follows that $m \in T$. Hence we have that: :$n \in T \implies \map s n \in T$ {{qed|lemma}} === Conclusion === By Axiom $(\text P 5)$, we conclude that $T = P$. From the definition of $T$, the result follows. {{qed}}
Let $x = \eqclass {a, b} {}$, $y = \eqclass {c, d} {}$ and $z = \eqclass {e, f} {}$ for some $x, y, z\in \Z$. Then: {{begin-eqn}} {{eqn | l = x + z       | r = y + z       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {a, b} {} + \eqclass {e, f} {}       | r = \eqclass {c, d} {} + \eqclass {e, f} {}       | c = }} {{eqn | ll= \leadsto       | l = \eqclass {a + e, b + f} {}       | r = \eqclass {c + e, d + f} {}       | c = {{Defof|Integer Addition}} }} {{eqn | ll= \leadsto       | l = a + e       | r = c + e       | c = {{Defof|Integer}} }} {{eqn | lo= \land       | l = b + f       | r = d + f       | c = {{Defof|Integer}} }} {{eqn | ll= \leadsto       | l = a       | r = c       | c = Natural Number Addition is Cancellable }} {{eqn | lo= \land       | l = b       | r = d       | c = Natural Number Addition is Cancellable }} {{eqn | ll= \leadsto       | l = \eqclass {a, b} {}       | r = \eqclass {c, d} {}       | c = {{Defof|Integer}} }} {{end-eqn}} {{qed}}
Let $x \in \openint a b$. Then by definition of open interval: :$a < x < b$ Thus: :$a \le x \le b$ and so by definition of closed interval: :$x \in \closedint a b$ {{qed}}
{{begin-eqn}} {{eqn | l = \relcomp C {\relcomp B A}       | r = C \setminus \paren {B \setminus A}       | c = {{Defof|Relative Complement}} }} {{eqn | r = \paren {C \setminus B} \cup \paren {C \cap A}       | c = Set Difference with Set Difference is Union of Set Difference with Intersection }} {{eqn | r = \paren {C \setminus B} \cup A       | c = Intersection with Subset is Subset }} {{eqn | r = A \cup \paren {C \setminus B}       | c = Union is Commutative }} {{eqn | r = A \cup \relcomp C B       | c = {{Defof|Relative Complement}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \bigcup_{\alpha \mathop \in I} A_\alpha       | c =  }} {{eqn | lll=\leadsto       | ll= \exists \alpha \in I:       | l = x       | o = \in       | r = A_\alpha       | c = {{Defof|Union of Family}} }} {{eqn | lll=\leadsto       | ll= \exists \alpha \in I:       | l = x       | o = \in       | r = B_\alpha       | c = {{Defof|Subset}} }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \bigcup_{\alpha \mathop \in I} B_\alpha       | c = {{Defof|Union of Family}} }} {{end-eqn}} By definition of subset: :$\displaystyle \bigcup_{\alpha \mathop \in I} A_\alpha \subseteq \bigcup_{\alpha \mathop \in I} B_\alpha$ {{qed}}
Let $x \in \bigcup \mathbb S$. By the definition of union, there exists an $S \in \mathbb S$ such that $x \in S$. By the premise, there exists a $T \in \mathbb T$ such that $S \subseteq T$. By the definition of Subset: :$x \in T$ Thus by the definition of union: :$x \in \bigcup \mathbb T$ {{qed}} Category:Set Union Preserves Subsets p76bex02v5vodwdiwzu81ezytmoz9dx
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \bigcap_{\alpha \mathop \in I} A_\alpha       | c = }} {{eqn | lll=\leadsto       | ll= \forall \alpha \in I:       | l = x       | o = \in       | r = A_\alpha       | c = Intersection is Subset }} {{eqn | lll=\leadsto       | ll= \forall \alpha \in J:       | l = x       | o = \in       | r = A_\alpha       | c = {{Defof|Subset}}: $J \subseteq I$ }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \bigcap_{\alpha \mathop \in J} A_\alpha       | c = {{Defof|Intersection of Family}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \bigcup_{\alpha \mathop \in J} A_\alpha       | c = }} {{eqn | lll=\leadsto       | ll= \exists \alpha \in J:       | l = x       | o = \in       | r = A_\alpha       | c = {{Defof|Union of Family}} }} {{eqn | lll=\leadsto       | ll= \exists \alpha \in I:       | l = x       | o = \in       | r = A_\alpha       | c = {{Defof|Subset}}: $J \subseteq I$ }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \bigcup_{\alpha \mathop \in I} A_\alpha       | c = Set is Subset of Union }} {{end-eqn}} {{qed}}
Let $\displaystyle A = \bigcap_{x \mathop \in \R_{> 0} } A_x$. {{AimForCont}} $A \ne \O$. Then: :$\exists y \in \R_{> 0}: y \in A$ By definition of open interval: :$y \notin \openint 0 y = A_y$ and so by definition of intersection of family: :$y \notin A$ From this contradiction it follows that $A$ has no elements. That is: :$\displaystyle \bigcap_{x \mathop \in \R_{> 0} } A_x = \O$ {{qed}}
Let $\displaystyle A = \bigcap_{x \mathop \in \R_{> 0} } A_x$. Let $y \in A$. Then by definition of union of family: :$\exists x \in \R_{> 0}: y \in A_x$ As $A_x \subseteq \R_{> 0}$ it follows by definition of subset that: :$y \in \R_{> 0}$ So: :$\displaystyle \bigcap_{x \mathop \in \R_{> 0} } A_x \subseteq \R_{> 0}$ {{qed|lemma}} Let $y \in \R_{> 0}$. By the Archimedean Property:  :$\exists z \in \N: z > y$ and so: :$y \in A_z$ That is by definition of union of family: :$y \in \displaystyle \bigcap_{x \mathop \in \R_{> 0} } A_x$ So by definition of subset: :$\displaystyle \R_{> 0} \subseteq \bigcap_{x \mathop \in \R_{> 0} } A_x$ {{qed|lemma}} By definition of set equality: :$\displaystyle \bigcup_{x \mathop \in \R_{> 0} } A_x = \R_{> 0}$ {{qed}}
Let $\displaystyle B = \bigcap_{x \mathop \in \R_{> 0} } B_x$. We have that: :$\forall x \in \R_{> 0}: 0 \in \closedint 0 x$ So by definition of intersection: :$0 \in B$ and so by Singleton of Element is Subset: :$\displaystyle \set 0 \subseteq \bigcap_{x \mathop \in \R_{> 0} } B_x$ {{AimForCont}} $\exists y \in \R_{> 0}: y \in B$. By definition of $B_x$: :$y \notin \closedint 0 {\dfrac y 2 = B_{y/2}$ and so by definition of intersection of family: :$y \notin B$ From this contradiction it follows that there can be no elements in $B$ apart from $0$. That is: :$\displaystyle \bigcap_{x \mathop \in \R_{> 0} } A_x \subseteq \set 0$ By definition of set equality: :$\displaystyle \bigcap_{x \mathop \in \R_{> 0} } B_x = \set 0$ {{qed}}
Let $\displaystyle B = \bigcap_{x \mathop \in \R_{> 0} } B_x$. Let $y \in B$. Then by definition of union of family: :$\exists x \in \R_{> 0}: y \in B_x$ As $B_x \subseteq \R_{> 0}$ it follows by definition of subset that: :$y = 0$ or :$y \in \R_{> 0}$ In either case, $y \in \R_{\ge 0}$ So: :$\displaystyle \bigcap_{x \mathop \in \R_{> 0} } B_x \subseteq \R_{\ge 0}$ {{qed|lemma}} Let $y \in \R_{\ge 0}$. If $y = 0$ then $y \in \R_{\ge 0}$ by definition. Otherwise, by the Archimedean Property:  :$\exists z \in \N: z > y$ and so: :$y \in B_z$ That is by definition of union of family: :$\displaystyle y \in \bigcap_{x \mathop \in \R_{\ge 0} } B_x$ So by definition of subset: :$\displaystyle \R_{\ge 0} \subseteq \bigcap_{x \mathop \in \R_{> 0} } B_x$ {{qed|lemma}} By definition of set equality: :$\displaystyle \bigcup_{x \mathop \in \R_{> 0} } B_x = \R_{\ge 0}$ {{qed}}
From Set with Relative Complement forms Partition: :$A = \set {X \mid \relcomp A X}$ :$B = \set {Y \mid \relcomp B Y}$ and so by definition of partition: :$A = X \cup \relcomp A X$ :$B = Y \cup \relcomp B Y$ By Cartesian Product of Unions: :$A \times B = \paren {X \times Y} \cup \paren {\relcomp A X \times \relcomp B Y} \cup \paren {X \times \relcomp B Y} \cup \paren {\relcomp A X \times Y}$ and so: {{begin-eqn}} {{eqn | l = \paren {A \times B} \setminus \paren {X \times Y}       | r = \paren {\relcomp A X \times \relcomp B Y} \cup \paren {X \times \relcomp B Y} \cup \paren {\relcomp A X \times Y}       | c =  }} {{eqn | r = \paren {X \times \relcomp B Y} \cup \paren {\relcomp A X \times \relcomp B Y} \cup \paren {\relcomp A X \times Y} \cup \paren {\relcomp A X \times \relcomp B Y}       | c =  }} {{eqn | r = \paren {\paren {X \cup \relcomp A X} \times \relcomp B Y} \cup \paren {\relcomp A X \times \paren {Y \cup \relcomp B Y} }       | c =  }} {{eqn | r = \paren {A \times \relcomp B Y} \cup \paren {\relcomp A X \times B}       | c =  }} {{end-eqn}} {{qed}} {{MissingLinks|Justification for the steps above}}
Let $f$ be such that: :$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$ In particular, it holds for all subsets of $A$ which are singletons. Now, consider any $x, y \in A$. We have: {{begin-eqn}} {{eqn | l = \map f x       | r = \map f y       | c = }} {{eqn | ll= \leadsto       | l = \set {\map f x}       | r = \set {\map f y}       | c =  }} {{eqn | ll= \leadsto       | l = \map {f^\to} {\set x}       | r = \map {f^\to} {\set y}       | c = {{Defof|Direct Image Mapping of Mapping}} }} {{eqn | ll= \leadsto       | l = \set x       | r = \map {f^\gets} {\map {f^\to} {\set x} }       | c = by hypothesis: $A = \map {\paren {f^\gets \circ f^\to} } A$ }} {{eqn | r = \map {f^\gets} {\map {f^\to} {\set y} }       | c =  }} {{eqn | r = \set y       | c = by hypothesis: $A = \map {\paren {f^\gets \circ f^\to} } A$ }} {{eqn | ll= \leadsto       | l = x       | r = y       | c =  }} {{end-eqn}} So $f$ is an injection. {{qed}}
Suppose that $f$ is not an injection. Then two elements of $S$ map to the same one element of $T$. That is: :$\exists a_1, a_2 \in S, b \in T: \map f {a_1} = \map f {a_2} = b$ Let $A = \set {a_1}$. Then: {{begin-eqn}} {{eqn | l = \map {f^\to} A       | r = \set b       | c =  }} {{eqn | ll= \leadsto       | l = \map {f^\gets} {\map {f^\to} A}       | o = \supseteq       | r = \set {a_1, a_2}       | c = {{Defof|Preimage of Subset under Mapping}} }} {{eqn | ll= \leadsto       | l = \map {f^\gets} {\map {f^\to} A}       | o = \supsetneqq       | r = A       | c = as $A = \set {a_1}$ }} {{end-eqn}} So by the Rule of Transposition: :$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$ implies that $f$ is an injection. {{qed}}
Let $f$ be an injection. From Subset of Domain is Subset of Preimage of Image, we have that: :$\forall A \subseteq S: A \subseteq \paren {f^{-1} \circ f} \sqbrk A$ by dint of $f$ being a relation. So what we need to do is show that: :$\forall A \subseteq S: \paren {f^{-1} \circ f} \sqbrk A \subseteq A$ Take any $A \subseteq S$. Let $x \in A$. We have: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \paren {f^{-1} \circ f} \sqbrk A       | c =  }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = f^{-1} \sqbrk {f \sqbrk A}       | c = {{Defof|Composition of Mappings}} }} {{eqn | lll=\leadsto       | l = \map f x       | o = \in       | r = f \sqbrk A       | c = {{Defof|Inverse of Mapping}} }} {{eqn | lll=\leadsto       | ll= \exists y \in A:       | l = \map f x       | r = \map f y       | c = {{Defof|Image of Subset under Mapping}} }} {{eqn | lll=\leadsto       | l = x       | r = y       | c = {{Defof|Injection}} }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = A       | c = as $y \in A$ }} {{end-eqn}} Thus we see that: :$\paren {f^{-1} \circ f} \sqbrk A \subseteq A$ and hence the result: :$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$ {{qed}}
Let $f$ be such that: :$\forall B \subseteq T: B = \paren {f \circ f^{-1} } \sqbrk B$ In particular, it holds for $T$ itself. Hence: {{begin-eqn}} {{eqn | l = T       | r = \paren {f \circ f^{-1} } \sqbrk T       | c =  }} {{eqn | l = T       | r = f \sqbrk {f^{-1} \sqbrk T}       | c = {{Defof|Composition of Mappings}} }} {{eqn | o = \subseteq       | r = f \sqbrk S       | c = Image of Subset under Relation is Subset of Image: Corollary 2 }} {{eqn | r = \Img f       | c = {{Defof|Image of Mapping}} }} {{eqn | o = \subseteq       | r = T       | c = Image is Subset of Codomain: Corollary 1 }} {{end-eqn}} So: :$T \subseteq \Img f \subseteq T$ and so by definition of set equality: :$\Img f = T$ So, by definition, $f$ is a surjection. {{qed}}
Suppose $f$ is not a surjection. $T$ must have at least two elements for this to be the case. Let one of these two elements not be the image of any element of $S$. That is, let $b_1, b_2 \in T$ such that: :$\exists a \in S: f \paren a = b_1$ :$\nexists x \in S: f \paren x = b_2$ Let $B = \set {b_1, b_2}$. Then: {{begin-eqn}} {{eqn | l = f^{-1} \sqbrk B       | r = A       | c = for some $A \subseteq S$ }} {{eqn | ll= \leadsto       | l = f \sqbrk {f^{-1} \sqbrk B}       | r = \set {b_1}       | c = as $b_2$ has no preimage }} {{eqn | ll= \leadsto       | l = f \sqbrk {f^{-1} \sqbrk B}       | o = \ne       | r = B       | c = as $B = \set {b_1, b_2}$ }} {{end-eqn}} So by the Rule of Transposition: :$\forall B \subseteq S: B = \paren {f \circ f^{-1} } \sqbrk B$ implies that $f$ is an surjection. {{qed}}
Let $g$ be a surjection. Let $B \subseteq T$. Let $b \in B$. Then: {{begin-eqn}} {{eqn | ll= \exists a \in S:       | l = b       | r = \map f a       | c = {{Defof|Surjection}} }} {{eqn | lll=\leadsto       | l = a        | o = \in       | r = f^{-1} \sqbrk B       | c = {{Defof|Preimage of Subset under Mapping}} }} {{eqn | lll=\leadsto       | l = b       | o = \in       | r = f \sqbrk {f^{-1} \sqbrk B}       | c = {{Defof|Image of Subset under Mapping}} }} {{eqn | lll=\leadsto       | l = B       | o = \subseteq       | r = f \sqbrk {f^{-1} \sqbrk B}       | c = {{Defof|Subset}} }} {{eqn | lll=\leadsto       | l = B       | o = \subseteq       | r = \paren {f \circ f^{-1} } \sqbrk B       | c = {{Defof|Composition of Mappings}} }} {{end-eqn}} From Subset of Codomain is Superset of Image of Preimage, we already have that: :$\paren {f \circ f^{-1} } \sqbrk B \subseteq B$ So: :$B \subseteq \paren {f \circ f^{-1} } \sqbrk B \subseteq B$ and by definition of set equality: :$B = \paren {f \circ f^{-1} } \sqbrk B$ {{qed}}
=== Sufficient Condition === Suppose $S_k = \set {s_k}$ for all $k \in \set {1, 2, \dotsc, n}$ where $k \ne j$. Let $\map {\pr_j} x = \map {\pr_j} y = z$ for $x, y \in S$. Then by definition of $j$th projection, $x, y \in S$ are given by: :$x = \tuple {s_1, s_2, \dotsc, s_{j - 1}, z, s_{j + 1}, \dotsc, s_n}$ :$y = \tuple {s_1, s_2, \dotsc, s_{j - 1}, z, s_{j + 1}, \dotsc, s_n}$ and so $x = y$. Thus $\pr_j$ is an injection by definition. {{qed|lemma}} === Necessary Condition === Let $\pr_j$ be an injection. Thus: :$\forall x, y \in S: \map {\pr_j} x = \map {\pr_j} y \implies x = y$ Let $x, y \in S$ be such that: :$x = \tuple {x_1, x_2, \dotsc, x_{j - 1}, z, x_{j + 1}, \dotsc, x_n}$ :$y = \tuple {y_1, y_2, \dotsc, y_{j - 1}, z, y_{j + 1}, \dotsc, y_n}$ Then: :$\map {\pr_j} x = \map {\pr_j} y = z$ For $x = y$ it is necessary that: :$x_1 = y_1, x_2 = y_2, \dotsc, x_{j - 1} = y_{j - 1}, x_{j + 1} = y_{j + 1}, \dotsc, x_n = y_n$ That is: :$\forall k \in \set {1, 2, \dotsc, n}, k \ne j: x_k = y_k$ That is, that $S_k$ is a singleton for all $k \in \set {1, 2, \dotsc, n}$ where $k \ne j$. {{qed}}
From the definition of factorial: :$n! = 1 \times 2 \times \cdots \times \left({n-1}\right) \times n$ Thus every number less than $n$ appears as a divisor of $n!$. The result follows from definition of congruence. {{qed}} Category:Factorials Category:Divisors 3p1tjojrrsp4gejtpfuz94uxurucs8o
Let $a - b \in P$ and $b - c \in P$. By condition $(2)$: :$\paren {a - b} + \paren {b - c} \in P$ Simplifying: :$a - c \in P$ The result follows by definition of transitive relation. {{qed}}
Proof by induction: For all $n \in \N_{\ge 0}$, let $\map P n$ be the proposition: :$f^n$ is an injection. $\map P 0$ is true, as this is the case Identity Mapping is Injection. $\map P 1$ is true, as this is the assertion that $f$ is an injection. === Basis for the Induction === $\map P 2$ is the assertion that $f^2$ is an injection. {{begin-eqn}} {{eqn | l = \map {f^2} x       | r = \map f {\map f x}       | c = Definition of $f^n$ }} {{eqn | r = \map {\paren {f \circ f} } x       | c = {{Defof|Composition of Mappings}} }} {{end-eqn}} But $f \circ f$ is an injection by Composite of Injections is Injection. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$f^k$ is an injection. Then we need to show: :$f^{k + 1}$ is an injection. === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \map {f^{k + 1} } x       | r = \map f {\map {f^k} x}       | c = Definition of $f^n$ }} {{eqn | r = \map {\paren {f \circ f^k} } x       | c =  }} {{end-eqn}} By the induction hypothesis $f^k$ is an injection. By Composite of Injections is Injection it follows that $f^{k + 1}$ is an injection. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore for all $n \in \N$: :$f^n$ is an injection. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === By definition, $f^0$ is the identity mapping. So by definition of $f^0$: :$\forall a \in S: a = \map {f^0} a$ That is: :$a \mathop {\mathcal R} a$ So $\mathcal R$ has been shown to be reflexive. {{qed|lemma}} === Symmetry === Let $a \mathop {\mathcal R} b$. That is: :$b = \map {f^k} a$ for some $k \in \Z$. Let $g_k$ be the restriction of $f^k$ to its image. From Injection to Image is Bijection, $g^k$ is a bijection. From Inverse of Bijection is Bijection: :$\map {\paren {g^k}^{-1} } b = a$ Consider the extension $\paren {f^k}^{-1}$ of $\paren {g^k}^{-1}$ to its codomain which is $S$. From Inverse of Injection is Many-to-One Relation, $\map {\paren {f^k}^{-1} } b = a$ is well-defined. By interpreting $\map {\paren {f^k}^{-1} } b = a$ as $\map {f^{-k} } b = a$, it follows that we can set $j = -k$ and so: :$\exists j \in \Z: \map {f^j} b = a$ That is: :$b \mathop {\mathcal R} a$ So $\mathcal R$ has been shown to be symmetric. {{qed|lemma}} === Transitivity === Let $a \mathop {\mathcal R} b$ and $b \mathop {\mathcal R} c$. That is: :$b = \map {f^{k_1} } a$ :$c = \map {f^{k_2} } b$ where $k_1, k_2 \in \Z$. That is: :$c = \map {f^{k_2} } {\map {f^{k_1} } a}$ By Composition of Repeated Compositions of Injections: :$c = \map {f^{k_2} } {\map {f^{k_1} } a} = \map {f^{k_1 + k_2} } a$ and so: :$a \mathop {\mathcal R} c$ So $\mathcal R$ has been shown to be transitive. {{qed|lemma}} $\mathcal R$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
Proof by induction: Let $m \in \Z_{\ge 0}$ be given. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$f^n \circ f^m = f^{m + n}$ $P \left({0}\right)$ is true, as this is the case: {{begin-eqn}} {{eqn | l = f^0 \circ f^m       | r = I_S \circ f^m       | c = where $I_S$ is the identity mapping }} {{eqn | r = f^m       | c = Identity Mapping is Left Identity }} {{end-eqn}} === Basis for the Induction === $P \left({1}\right)$ is true, as this is the case: :$f^{m+1} \left({x}\right) = f \left({f^m \left({x}\right)}\right)$ by definition. 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: :$f^k \circ f^m = f^{m + k}$ Then we need to show: :$f^{k+1} \circ f^m = f^{m + k + 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \left({f^{k+1} \circ f^m}\right) \left({x}\right)       | r = \left({f \left({f^k}\right) \circ f^m}\right) \left({x}\right)       | c = Definition of $f^{k+1}$ }} {{eqn | r = \left({\left({f \circ f^k}\right) \circ f^m}\right) \left({x}\right)       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \left({f \circ \left({f^k \circ f^m}\right)}\right) \left({x}\right)       | c = Composition of Mappings is Associative }} {{eqn | r = \left({f \circ f^{m+k} }\right) \left({x}\right)       | c = Induction Hypothesis }} {{eqn | r = f^{m+k+1} \left({x}\right)       | c = Basis for the Induction }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m, n \in \Z_{\ge 0}: f^n \circ f^m = f^{m + n}$ {{qed}} Category:Injections Category:Composite Mappings l10y9a8b2nveuifhsttjiyzw45bbi2n
{{questionable|This is not enough. It might superficially be that e.g. the factors of $2$ and $6$ coincide. Effectively this argument reduces to the theorem; it's almost circular}} Let $m \in \N_{\ge 1}$. Consider the set: :$S = \{ {m, m + 1, m + 2, \ldots, m + n - 1 }\}$ Note $S$ has $n$ elements. By Set of Successive Numbers contains Unique Multiple: :$\left\{ {m} \right\}$ contains a factor of $1$, :$\left\{ {m, m+1} \right\}$ contains a factor of $2$, and in general, :$\left\{ {m, m+1, \ldots, m + j - 1} \right\}$ contains a factor of $j$. It follows that $S$ contains factors of $1, 2, 3, \ldots, n$. Multiplying all elements of $S$ gives: {{begin-eqn}} {{eqn | l = m \left({m+1}\right)\ldots\left({m+n-1}\right)       | r = k_1 2 k_2 3 k_3 \ldots \left({n-1}\right)k_{n-1} n k_n       | c = for some $k_1, k_2, \ldots, k_n \in N$ }} {{eqn | r = \left({k_1 k_2 \ldots k_{n-1}k_n}\right)\left({1 \times 2 \times \ldots \times \left({n - 1}\right) \times n}\right) }} {{eqn | ll = \implies       | l = m^{\overline n}       | r = K n!       | c = for some $K \in \N$ }} {{eqn | ll = \implies       | l = m^{\overline n}       | o = \equiv       | r = 0 \bmod n! }} {{end-eqn}} {{qed}} Category:Number Theory Category:Factorials 0bxzl1xm1nz2t6qgk3dpamjhu03p06r
If it is not the case that $0 \le k \le n$, then the result holds trivially. So let $0 \le k \le n$. By the definition of binomial coefficients: {{begin-eqn}} {{eqn | l = \binom n k       | r = \frac {n!} {k!\left({n - k}\right)!} }} {{eqn | r = \frac {n \left({n - 1}\right) \left({n - 2}\right) \ldots \left({n - k + 1}\right)} {k!} }} {{end-eqn}} The numerator is a product of $k$ successive integers. From Factorial Divides Product of Successive Numbers, $k!$ divides it. {{qed}}
Follows directly from Derivative Function on Set of Functions induces Equivalence Relation. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === :$\forall f \in E: \map f b = \map f b$ That is: :$\forall f \in E: \tuple {f, f} \in \mathcal R$ Thus $\mathcal R$ is seen to be reflexive. {{qed|lemma}} === Symmetry === {{begin-eqn}} {{eqn | l = \tuple {f, g}       | o = \in       | r = \mathcal R       | c =  }} {{eqn | ll= \leadsto       | l = \map g b       | r = \map f b       | c =  }} {{eqn | ll= \leadsto       | l = \map g b       | r = \map f b       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {g, f}       | o = \in       | r = \mathcal R       | c =  }} {{end-eqn}} Thus $\mathcal R$ is seen to be symmetric. {{qed|lemma}} === Transitivity === Let $f, g, h \in E$. {{begin-eqn}} {{eqn | l = \tuple {f, g}       | o = \in       | r = \mathcal R       | c =  }} {{eqn | lo= \land       | l = \tuple {g, h}       | o = \in       | r = \mathcal R       | c =  }} {{eqn | ll= \leadsto       | l = \map f b       | r = \map g b       | c =  }} {{eqn | lo= \land       | l = \map g b       | r = \map h b       | c =  }} {{eqn | ll= \leadsto       | l = \map f b       | r = \map h b       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {f, h}       | o = \in       | r = \mathcal R       | c =  }} {{end-eqn}} Thus $\mathcal R$ is seen to be transitive. {{qed|lemma}} $\mathcal R$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}