hcju/proofwikiqa huggingface dataset:
By definition of equivalence relation, $\mathcal R$ is an equivalence relation {{iff}}: :$\Delta_S \cup \mathcal R^{-1} \cup \mathcal R \circ \mathcal R \subseteq \mathcal R$ From Relation is Connected iff Union with Inverse and Diagonal is Trivial Relation: :$\Delta_S \cup \mathcal R^{-1} \cup \mathcal R = S \times S$ Hence the result. {{qed}}
We have the definition of the power to a real number: :$\paren {n + 1}^{1/n} = \map \exp {\dfrac 1 n \, \map \ln {n + 1} }$ For $n >= 1$ then $n + 1 \le 2 n$. Hence: {{begin-eqn}} {{eqn | l = \frac 1 n \, \map \ln {n + 1}       | o = \le       | r = \frac 1 n \, \map \ln {2 n}       | c = Logarithm is Strictly Increasing }} {{eqn | o = }} {{eqn | r = \frac 1 n \paren {\ln 2 + \ln n}       | c = Logarithm on Positive Real Numbers is Group Isomorphism }} {{eqn | o =}} {{eqn | r = \frac {\ln 2} n + \frac 1 n \ln n       | c =  }} {{end-eqn}} By Powers Drown Logarithms: :$\displaystyle \lim_{n \mathop \to \infty} \frac 1 n \ln n = 0$ By Sequence of Reciprocals is Null Sequence: :$\displaystyle \lim_{n \mathop \to \infty} \frac 1 n = 0$ By Combined Sum Rule for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\frac {\ln 2} n + \frac 1 n \ln n} = \ln 2 \cdot 0 + 0 = 0$ By the Squeeze Theorem for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} = 0$ Hence: :$\displaystyle \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} = \exp 0 = 1$ and the result follows. {{qed}} Category:Characterisation of Non-Archimedean Division Ring Norms 4gtwgr743fe4lo5o5j9ncxmmwnpxg30
By Characterisation of Non-Archimedean Division Ring Norms then: :$\norm {\,\cdot\,}$ is non-Archimedean {{iff}}: ::$\sup \set {\norm {n \cdot 1_R}: n \in \N_{\gt 0}} \le 1$ By norm of unity then: :$\norm {1_R} = 1$ The result follows. {{qed}}
By Characterisation of Non-Archimedean Division Ring Norms: :$\norm{\,\cdot\,}$ is Archimedean $\iff \sup \set {\norm {n \cdot 1_R}: n \in \N_{\gt 0}} \gt 1$ By Corollary 2: :$\sup \set {\norm{n \cdot 1_R}: n \in \N_{\gt 0} } \gt 1 \iff \sup \set {\norm {n \cdot 1_R}: n \in \N_{\gt 0}} = +\infty$ {{qed}}
{{AimForCont}} $C > 1$. By Characterizing Property of Supremum of Subset of Real Numbers: :$\exists m \in \N_{> 0}: \norm {m \cdot 1_R} > 1$ Let    :$x = m \cdot 1_R$  :$y = x^{-1}$ By Norm of Inverse: :$\norm y < 1$ By Sequence of Powers of Number less than One: :$\displaystyle \lim_{n \mathop \to \infty} \norm y^n = 0$ By Reciprocal of Null Sequence then: :$\displaystyle \lim_{n \mathop \to \infty} \frac 1 {\norm y^n} = +\infty$ For all $n \in \N_{> 0}$: {{begin-eqn}} {{eqn | l = \dfrac 1 {\norm y^n}       | r = \norm {y^{-1} }^n       | c = Norm of Inverse }} {{eqn | r = \norm x^n       | c = Definition of $y$ }} {{eqn | r = \norm {x^n}       | c = {{NormAxiom|2}} }} {{eqn | r = \norm {\paren {m \cdot 1_R}^n}       | c = Definition of $x$ }} {{eqn | r = \norm {m^n \cdot 1_R} }} {{end-eqn}} So: :$\displaystyle \lim_{n \mathop \to \infty} \norm {m^n \cdot 1_R} = +\infty$ Hence: :$\sup \set {\norm{n \cdot 1_R}: n \in \N_{> 0} } = +\infty$ This contradicts the assumption that $C < +\infty$. {{qed|lemma}} It follows that $C \le 1$. Then: :$\forall n \in \N_{>0}: \norm{n \cdot 1_R} \le 1$ By Characterisation of Non-Archimedean Division Ring Norms, $\norm{\,\cdot\,}$ is non-Archimedean. By Corollary 1: :$\sup \set {\norm {n \cdot 1_R}: n \in \N_{> 0} } = 1$ So $C = 1$. {{qed}}
Because $R$ has characteristic $p > 0$, the set: :$\set {n \cdot 1_k: n \in \Z}$ has cardinality $p - 1$. Therefore: :$\sup \set {\norm {n \cdot 1_R}: n \in \Z} = \max \set {\norm {1 \cdot 1_R}, \norm {2 \cdot 1_R}, \cdots, \norm {\paren {p - 1} \cdot 1_R} } < +\infty$ By Corollary 2: :$\norm{\,\cdot\,}$ is non-Archimedean and $C = 1$. {{qed}} Category:Characterisation of Non-Archimedean Division Ring Norms qpz1930i9qr4qtkaa39un5u70jdvup9
=== Necessary Condition === {{:Norms Equivalent to Absolute Value on Rational Numbers/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Norms Equivalent to Absolute Value on Rational Numbers/Sufficient Condition}}{{qed}}
{{WLOG}}, let $S = \set {a, b}$. Let $\odot$ be reflexive. By definition of reflexive relation: :$\Delta_S \subseteq \odot$ where $\Delta_S$ is the diagonal relation: :$\Delta_S = \set {\tuple {x, x}: x \in S}$ That is: :$\set {\tuple {a, a}, \tuple {b, b} } \subseteq \odot$ Suppose $\set {\tuple {a, a}, \tuple {b, b} } = \odot$. Then by Diagonal Relation is Equivalence, $\odot$ is transitive. Suppose there exists $\tuple {x, y} \in \odot$ such that $\tuple x \ne y$ and $\tuple {y, x} \notin \odot$. Then: :$x \odot x, x \odot y \implies x \odot y$ and: :$x \odot y, y \odot y \implies x \odot y$ Now suppose there exists $\tuple {x, y} \in \odot$ such that $\tuple x \ne y$ and $\tuple {y, x} \in \odot$. Then we have: :$x \odot y, y \odot x \implies x \odot x$ and: :$y \odot x, x \odot y \implies y \odot y$ which hold because $\odot$ is reflexive Hence in all cases, a reflexive relation on $S$ is also transitive. {{qed}} Category:Reflexive Relations Category:Transitive Relations lhc0zlz6rinxrqyqa85wu1x5pkupht9
{{WLOG}}, let $S = \set {a, b}$. Let $\odot$ not be symmetric. {{AimForCont}} $\odot$ is not transitive. As $\odot$ is not symmetric: :$\exists \tuple {x, y} \in \odot: \tuple {y, x} \notin \odot$ Thus there are two possibilities: :$\exists \tuple {a, b} \in \odot: \tuple {b, a} \notin \odot$ :$\exists \tuple {b, a} \in \odot: \tuple {a, b} \notin \odot$ {{WLOG}}, let $\tuple {a, b} \in \odot$. As $\odot$ is not transitive, either: :$\exists \tuple {x, a} \in \odot: \tuple {x, b} \notin \odot$ or: :$\exists \tuple {b, y} \in \odot: \tuple {a, y} \notin \odot$ In the first case, $x \ne b$ as that would make $\odot$ symmetric. In the second case, $y \ne a$ as that would also make $\odot$ symmetric. Hence $x = a$ and $y = b$, and we have either: :$\exists \tuple {a, a} \in \odot: \tuple {a, b} \notin \odot$ or: :$\exists \tuple {b, b} \in \odot: \tuple {a, b} \notin \odot$ But it has been established that $\tuple {a, b} \in \odot$. From this contradiction it follows that $\odot$ is transitive. Hence it is not possible for $\odot$ to be neither symmetric nor transitive. {{qed}} Category:Symmetric Relations Category:Transitive Relations 8tm3k4p73d1idz37803ov16ps7v5yv1
{{WLOG}}, let $S = \set a$. There are $2$ relations on $S$: $(1): \quad \odot := \O$, which is the null relation on $S$. From Null Relation is Antireflexive, Symmetric and Transitive, $\odot$ is antireflexive, symmetric and transitive. Thus in this case $\odot$ is both symmetric and transitive. $(2): \quad \odot := \set {\tuple {a, a} }$, which is the trivial relation on $S$. From Trivial Relation is Equivalence, $\odot$ is reflexive, symmetric and transitive. Thus in this case also, $\odot$ is both symmetric and transitive. {{qed}} Category:Symmetric Relations Category:Transitive Relations 3mj1jd4zecex4gsih4bo4ufrwlqndfh
Checking in turn each of the criteria for equivalence: === Reflexivity === Let $\triangle A$ be a triangle. By definition, by Triangle Side-Side-Side Equality, $\triangle A$ is trivially congruent to itself. Thus $\cong$ is seen to be reflexive. {{qed|lemma}} === Symmetry === Let $\triangle A \cong \triangle B$. Then: :all the sides of $\triangle A$ are equal to the sides of $\triangle B$ :all the angles contained by the sides of $\triangle A$ are equal to the angles contained by the sides of $\triangle B$. It follows directly that: :all the sides of $\triangle B$ are equal to the sides of $\triangle A$ :all the angles contained by the sides of $\triangle B$ are equal to the angles contained by the sides of $\triangle A$. That is: : $\triangle B \cong \triangle A$ Thus $\cong$ is seen to be symmetric. {{qed|lemma}} === Transitivity === Let: :$\triangle A \cong \triangle B$ :$\triangle B \cong \triangle C$ Then: :all the sides of $\triangle A$ are equal to the sides of $\triangle B$ :all the angles contained by the sides of $\triangle A$ are equal to the angles contained by the sides of $\triangle B$. and: :all the sides of $\triangle B$ are equal to the sides of $\triangle C$ :all the angles contained by the sides of $\triangle B$ are equal to the angles contained by the sides of $\triangle C$. From Equality is Equivalence Relation, it follows that: :all the sides of $\triangle A$ are equal to the sides of $\triangle C$ :all the angles contained by the sides of $\triangle A$ are equal to the angles contained by the sides of $\triangle C$. That is: : $\triangle A \cong \triangle C$ Thus $\cong$ is seen to be transitive. {{qed|lemma}} $\cong$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
First suppose $T = \O$ there are no elements in $f$ From Null Relation is Mapping iff Domain is Empty Set, there are no elements in $f$. Hence in this case $\card f = 0$, whatever $\card S$ may be, By definition of mapping, $f$ is a set of ordered pairs $\tuple {s, t}$ where $s \in S$ and $t \in T$, such that: :$(1): \quad \forall s \in S: \exists \tuple {s, t} \in f$ and: :$(2): \quad \forall x \in S: \tuple {s, t_1} \in f \land \tuple {s, t_2} \in f \implies t_1 = t_2$ From $(1)$ it follows that $\card f \ge n$. {{AimForCont}} that $\card f > n$. Then by the Pigeonhole Principle: :$\exists x \in f: \exists y_1, y_2 \in T, y_1 \ne y_2: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f$ and so $f$ is not a mapping. From this contradiction it follows that $\card f = n$. {{qed}}
=== $(1)$ {{iff}} $(2)$ === By definition of divisor, $n$ is divisible by $2$ {{iff}}: :$n = 2 r$ where $r \in \Z$. Thus definition 1 is logically equivalent to definition 2. {{qed|lemma}} === $(2)$ {{iff}} $(3)$ === By definition of congruence modulo $2$: :$x \equiv y \pmod 2 \iff \exists r \in \Z: x - y = 2 r$ Setting $y = 0$: :$x \equiv 0 \pmod 2 \iff \exists r \in \Z: x = 2 r$ Thus definition 2 is logically equivalent to definition 3. {{qed}} Category:Even Integers egqqpgs1q6kfmxnv4cgwqatztjhscxh
Let $x$ be an even integer. Then by definition: :$x = 2 n$ for some integer $n$. {{AimForCont}} $y = x + 5 = 2 m$ for some integer $m$. Then: {{begin-eqn}} {{eqn | l = x       | r = 2 m - 5       | c =  }} {{eqn | r = \paren {2 m - 6} + 1       | c =  }} {{eqn | r = 2 \paren {m - 3} + 1       | c =  }} {{eqn | r = 2 r + 1       | c = where $r = m - 3 \in \Z$ }} {{end-eqn}} Hence $x$ is odd. But this contradicts our premise that $x$ is even. Hence the result by Proof by Contradiction. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^1 \paren {2 i - 1}^2       | r = \paren {2 \times 1 - 1}^2       | c =  }} {{eqn | r = 1^2       | c =  }} {{eqn | r = \dfrac {\paren {4 \times 1 - 1}^3} 3       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{i \mathop = 1}^k \paren {2 i - 1}^2 = \frac {4 k^3 - k} 3$ from which it is to be shown that: :$\displaystyle \sum_{i \mathop = 1}^{k + 1} \paren {2 i - 1}^2 = \frac {4 \paren {k + 1}^3 - \paren {k + 1} } 3$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^{k + 1} \paren {2 i - 1}^2       | r = \sum_{i \mathop = 1}^k \paren {2 i - 1}^2 + \paren {2 \paren {k + 1} - 1}^2       | c =  }} {{eqn | r = \frac {4 k^3 - k} 3 + \paren {2 k + 1}^2       | c = Induction Hypothesis }} {{eqn | r = \frac {4 k^3 - k + 12 k^2 + 12 k + 3} 3       | c =  }} {{eqn | r = \frac {\paren {k + 1} \paren {4 k^2 + 8 k + 3} } 3       | c =  }} {{eqn | r = \frac {\paren {k + 1} \paren {4 \paren {k + 1}^2  - 1} } 3       | c =  }} {{eqn | r = \frac {4 \paren {k + 1}^3 - \paren {k + 1} } 3       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{> 0}: \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^1 j \paren {j + 2}       | r = 1 \times 3       | c =  }} {{eqn | r = 3       | c =  }} {{eqn | r = \frac {1 \paren {1 + 1} \paren {2 \times 1 + 7} } 6       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{j \mathop = 1}^k j \paren {j + 2} = \frac {k \paren {k + 1} \paren {2 k + 7} } 6$ from which it is to be shown that: :$\displaystyle \sum_{j \mathop = 1}^{k + 1} j \paren {j + 2} = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 7} } 6$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^{k + 1} j \paren {j + 2}       | r = \sum_{j \mathop = 1}^k j \paren {j + 2} + \paren {k + 1} \paren {k + 3}       | c =  }} {{eqn | r = \frac {k \paren {k + 1} \paren {2 k + 7} } 6 + \paren {k + 1} \paren {k + 3}       | c = Induction Hypothesis }} {{eqn | r = \paren {k + 1} \paren {\frac {k \paren {2 k + 7} + 6 \paren {k + 3} } 6}       | c =  }} {{eqn | r = \paren {k + 1} \paren {\frac {2 k^2 + 13 k + 18} 6 }       | c =  }} {{eqn | r = \frac {\paren {k + 1} \paren {k + 2} \paren {2 k + 9} } 6       | c =  }} {{eqn | r = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 7} } 6       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{>0}: \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$ {{qed}} </onlyinclude>
=== Lemma 1 === {{:Arcsin as an Integral/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Arcsin as an Integral/Lemma 2}}{{qed|lemma}} :$\displaystyle \map \arcsin x = \map {\arcsin_A} x = \map {\arcsin_G} x = \int_0^\Theta \frac {\d x} {\sqrt {1 - x^2} }$ {{qed}} hd79ptfc8vvlm8f8hye4haku88a0b3m
Proof by induction: Let $L_n$ be the Lucas number defined as the sum of two Fibonacci numbers: :$L_n = F_{n - 1} + F_{n + 1}$ For all $n \in \N$, let $\map P n$ be the proposition: :$L_n = \begin{cases} 2 & : n = 0 \\ 1 & : n = 1 \\ L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$ === Basis for the Induction === We have that: :$L_0 = F_{-1} + F_1 = 1 + 1 = 2$ :$L_1 = F_0 + F_2 = 0 + 1 = 1$ :$L_2 = F_1 + F_3 = 1 + 2 = 3$ Thus $\map P 0$, $\map P 1$ and $\map P 2$ hold. $\map P 3$ is the case: : $L_3 = F_2 + F_4 = 1 + 3 = 4$ So $\map P 3$, as $L_3 = L_1 + L_2$. This is our basis for the induction. === Induction Hypothesis === Let us make the supposition that, for some $k \in \N: k \ge 1$, the proposition $\map P j$ holds for all $j \in \N: 1 \le j \le k$. We shall show that it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\forall 1 \le j \le k: L_j = L_{j - 1} + L_{j - 2}$ Then we need to show: :$L_{k + 1} = L_k + L_{k - 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = L_{k + 1}       | r = F_k + F_{k + 2}       | c = by hypothesis }} {{eqn | r = F_{k - 2} + F_{k - 1} + F_k + F_{k + 1}       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = \paren {F_{k - 2} + F_k} + \paren {F_{k - 1} + F_{k + 1} }       | c =  }} {{eqn | r = L_{k - 1} + L_k       | c = by hypothesis }} {{end-eqn}} Hence $L_n = L_{n - 2} + L_{n - 1}$ follows by the Second Principle of Mathematical Induction. That is: $\sequence {L_n}$ is the sequence defined as: :$L_n = \begin{cases} 2 & : n = 0 \\ 1 & : n = 1 \\ L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$ {{qed}}
By definition of Lucas numbers: :$L_n = F_{n - 1} + F_{n + 1}$ Hence: :$F_n L_n = F_n \paren {F_{n - 1} + F_{n + 1} }$ From Fibonacci Number in terms of Smaller Fibonacci Numbers: :$\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$ The result follows by setting $m = n$. {{qed}}
Let $n \in \Z$. Let $m \in \Z_{\ge 0}$ be such that: :$3^m + 1 \le \size {2 n} \le 3^{m + 1} - 1$ where $\size {2 n}$ denotes the absolute value of $2 n$. As $2 n$ is even, this is always possible, because $3^r$ is always an odd integer for non-negative $r$. Let $d = \dfrac {3^{m + 1} - 1} 2$. Let $k = n + d$. We have that: {{begin-eqn}} {{eqn | l = \size {2 n}       | o = \le       | r = 3^{m + 1} - 1       | c =  }} {{eqn | ll= \leadsto       | l = \size n       | o = \le       | r = d       | c = Definition of $d$ }} {{eqn | ll= \leadsto       | l = -d       | o = \le       | r = n \le d       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = \le       | r = n + d \le 3^{m + 1} - 1       | c =  }} {{end-eqn}} Let $k = n + d \in \Z$ be represented in ternary notation: :$k = \displaystyle \sum_{j \mathop = 0}^m s_j 3^j$ where $s_j \in \set {0, 1, 2}$. By the Basis Representation Theorem, this expression for $k$ is unique. Now we have: {{begin-eqn}} {{eqn | l = d       | r = \dfrac {3^{m + 1} - 1} {3 - 1}       | c = by definition }} {{eqn | r = \sum_{j \mathop = 0}^m 3^j       | c = Sum of Geometric Sequence }} {{end-eqn}} Hence we see: {{begin-eqn}} {{eqn | l = n       | r = k - d       | c = by definition }} {{eqn | r = \sum_{j \mathop = 0}^m s_j 3^j - \sum_{j \mathop = 0}^m 3^j       | c =  }} {{eqn | r = \sum_{j \mathop = 0}^m \paren {s_j - 1} 3^j       | c =  }} {{eqn | r = \sum_{j \mathop = 0}^m r_j 3^j       | c = where $r_j \in \set {-1, 0, 1}$ }} {{end-eqn}} Hence $n$ has a representation in balanced ternary. The representation for $k$ in ternary notation is unique, as established. Hence the representation in balanced ternary for $n$ is also unique. {{qed}} {{Proofread|Not completely sure that uniqueness has been properly proved.}}
By Complex Modulus is Norm then the complex modulus satisfies the norm axioms on the set of complex numbers $\C$. Since the real numbers $\R$ is a subset of the complex numbers $\C$ then the complex modulus satisfies the norm axioms on the real numbers $\R$. By Complex Modulus of Real Number equals Absolute Value then the absolute value satisfies the norm axioms on set of real numbers $\R$. {{qed}}
As none of the coefficients $a_j$ in $\displaystyle \sum_{j \mathop = 1}^s a_j k^j$ is (strictly) negative, the summation itself likewise cannot be negative Thus: :$0 \le n$ The equality is satisfied when $a_j = 0$ for all $j$. We then have: {{begin-eqn}} {{eqn | l = n       | r = \sum_{j \mathop = 1}^s a_j k^j       | c =  }} {{eqn | o = \le       | r = \paren {k - 1} \sum_{j \mathop = 1}^s k^j       | c = as $a_j \le k - 1$ for all $j$ }} {{eqn | r = \paren {k - 1} \dfrac {k^{s + 1} - 1} {k - 1}       | c = Sum of Geometric Sequence }} {{eqn | r = k^{s + 1} - 1       | c =  }} {{eqn | o = <       | r = k^{s + 1}       | c =  }} {{end-eqn}} {{qed}}
First suppose that $r \ne s$. {{WLOG}}, suppose $r > s$. Then from Bounds for Integer Expressed in Base k: {{begin-eqn}} {{eqn | l = a_r k^r       | o = >       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = >       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = \ne       | r = b       | c =  }} {{end-eqn}} Otherwise $r = s$. Let $l$ be the largest such that $a_l \ne b_l$. {{WLOG}}, suppose $a_l > b_l$. For all $j > l$, let $a_1 = a - a_j k_j$ and $b_1 = b - b_j k_j$. As $a_j = b_j$ in this range, $a_j k_j = b_j k_j$ and so the same amount is being subtracted from both. So consider $\paren {a_l - b_l} k_l$. From Bounds for Integer Expressed in Base k: :$\paren {a_l - b_l} k_l > \displaystyle \sum_{j \mathop = 0}^{l - 1} a_j k^j$ and: :$\paren {a_l - b_l} k_l > \displaystyle \sum_{j \mathop = 0}^{l - 1} b_j k^j$ and so $a_1 > b_1$. Hence: :$a_1 + a_j k_j > b_1 + b_j k_j$ Hence $a \ne b$ and the result. {{qed}} {{Proofread}}
Let $x, y \in \R$ be such that $x < y$. Since $0 < a < 1$, we have that: :$\dfrac 1 a > 1$ Then we have that: {{begin-eqn}} {{eqn | l = \paren {\dfrac 1 a}^x        | o = <       | r = \paren {\dfrac 1 a}^y       | c = Real Power Function on Base Greater than One is Strictly Increasing }} {{eqn | ll= \leadstoandfrom       | l = \dfrac 1 {a^x}       | o = <       | r = \dfrac 1 {a^y}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = a^x       | o = >       | r = a^y       | c = Reciprocal Function is Strictly Decreasing }} {{end-eqn}} The result follows. {{qed}} Category:Power Function on Base between Zero and One is Strictly Decreasing 5b3907fc75dupddxgephkodb55l5y9t
Let $J \subseteq \Z$ fulfil the conditions of an integral ideal: :$(1): \quad n, m \in J \implies m + n \in J, m - n \in J$ :$(2): \quad n \in J, r \in \Z \implies r n \in J$ First note that $J$ is non-empty by definition. Then from $(1)$ we have in particular: :$n, m \in J \implies m - n \in J$ Thus by the One-Step Subgroup Test: :$\struct {J, +}$ is a subgroup of the additive group of integers $\struct {\Z, +}$. Then from $(2)$ we have: :$n \in J, r \in \Z \implies r n \in J$ and by Integer Multiplication is Commutative it follows that: :$n \in J, r \in \Z \implies n r \in J$ Thus it is seen that the conditions are fulfilled for $J$ to be an ideal of the ring of integers $\struct {\Z, +, \times}$. {{qed|lemma}} Suppose $J \subseteq \Z$ such that: :$(3): \quad \struct {J, +}$ is a subgroup of $\struct {\Z, +}$ :$(4): \quad \forall n \in J: \forall r \in R: n r \in J \land r n \in J$ From $(4)$, $(2)$ immediately follows. From $(3)$ it follows from the One-Step Subgroup Test that: :$\forall n, m \in J: m - n \in J$ which is a restatement of $(1)$. Thus the conditions are fulfilled for $J$ to be an integral ideal. {{qed}} Category:Integral Ideals Category:Ideal Theory 1gbbe9q46i0bmjvtwjlszsur1l5oy0a
First note that $m \times 0 \in m \Z$ whatever $m$ may be. Thus $m \Z \ne \O$. Let $a, b \in m \Z$. Then: {{begin-eqn}} {{eqn | l = a + b       | r = m j + m k       | c = for some $j, k \in \Z$ by definition of $m \Z$ }} {{eqn | r = m \paren {j + k}       | c =  }} {{eqn | o = \in       | r = m \Z       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = a - b       | r = m j - m k       | c = for some $j, k \in \Z$ by definition of $m \Z$ }} {{eqn | r = m \paren {j - k}       | c =  }} {{eqn | o = \in       | r = m \Z       | c =  }} {{end-eqn}} Let $r \in \Z$. Then: {{begin-eqn}} {{eqn | l = r a       | r = r \paren {m j}       | c = for some $j \in \Z$ by definition of $m \Z$ }} {{eqn | r = m \paren {r j}       | c =  }} {{eqn | o = \in       | r = m \Z       | c =  }} {{end-eqn}} Thus the conditions for $m \Z$ to be an integral ideal are fulfilled. Hence the result. {{qed}}