hcju/proofwikiqa huggingface dataset:
=== Definition 1 implies Definition 2 === Let $f^{-1}: T \to S$ be an inverse mapping of $f: S \to T$ by definition 1. From Mapping is Injection and Surjection iff Inverse is Mapping it follows that $f^{-1}$ is a bijection. By Composite of Bijection with Inverse is Identity Mapping: : $f^{-1} \circ f = I_S$ : $f \circ f^{-1} = I_T$ Thus $f^{-1}: T \to S$ is an inverse mapping of $f: S \to T$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $g: T \to S$ be an inverse mapping of $f: S \to T$ by definition 2. From Left and Right Inverse Mappings Implies Bijection it follows that both $f$ and $g$ are bijections. From Mapping is Injection and Surjection iff Inverse is Mapping it follows that $g = f^{-1}$. Thus $g: T \to S$ is an inverse mapping of $f: S \to T$ by definition 1. {{qed}}
From Inverse is Mapping implies Mapping is Injection and Surjection: :$f$ is both an injection and a surjection. Again from Inverse is Mapping implies Mapping is Injection and Surjection: :$g$ is both an injection and a surjection. The result follows by definition of bijection. {{qed}}
=== Necessary Condition === Let $f: S \to T$ be an extension of $g: A \to T$. Then by definition: {{begin-eqn}} {{eqn | lo= \forall x \in A:       | l = \map f x       | r = \map g x       | c =  }} {{eqn | r = \map g {\map {i_A} x}       | c = {{Defof|Inclusion Mapping}} }} {{eqn | r = \map {\paren {g \circ i_A} } x       | c = {{Defof|Composition of Mappings}} }} {{end-eqn}} That is: :$f = g \circ i_A$ {{qed|lemma}} === Sufficient Condition === Let $f = g \circ i_A$ where $i_A$ is the inclusion mapping on $A$. Then: {{begin-eqn}} {{eqn | lo= \forall x \in A:       | l = \map f x       | r = \map {\paren {g \circ i_A} } x       | c = Definition of $f$ }} {{eqn | r = \map g {\map {i_A} x}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map g x       | c = {{Defof|Inclusion Mapping}} }} {{end-eqn}} So, by definition, $f$ is an extension of $g$. {{qed}}
Let $A \subseteq S$ be a subset of $S$. Let $g: A \to T$ be the restriction of $f$ to $A$. By definition of restriction of mapping: :$\forall x \in A: \map g x = \map f x$ and hence there is one mapping $g: A \to T$ such that $g$ is a restriction of $f$. Therefore for each subset of $S$ there exists a unique restriction of $f$ to that subset. Therefore there are as many restriction of $f$ as there are subsets of $S$. That is, by definition of power set, the same as the number of elements of the power set of $S$. From Cardinality of Power Set of Finite Set, that is $2^n$. {{qed}}
Consider the $\beta$th projection. Let $x_\beta \in S_\beta$. Let $\map x \beta = x_\beta$ Suppose $\gamma \in I: \gamma \ne \beta$. As $S_\gamma \ne \O$ it is possible to use the axiom of choice to choose $\map x \gamma \in S_\gamma$. Then: :$\displaystyle x \in \prod_{\alpha \mathop \in I} S_\alpha$ and: :$\map {\pr_\beta} x = \map x \beta$ Hence the result. {{qed}}
This is an instance of the taxicab metric on the cartesian product of $A_{1'}, A_{2'}, \ldots, A_{n'}$. This is proved in Taxicab Metric is Metric. {{qed}}
The taxicab metric on $\R^n$ is: : $\displaystyle d_1 \left({x, y}\right) = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert$ for $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \R^n$. === Proof of $M1$ === {{begin-eqn}} {{eqn | l = d_1 \left({x, x}\right)       | r = \sum_{i \mathop = 1}^n \left\vert{x_i - x_i}\right\vert       | c = Definition of $d_1$ }} {{eqn | r = \sum_{i \mathop = 1}^n 0       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $M1$ holds for $d_1$. {{qed|lemma}} === Proof of $M2$ === {{begin-eqn}} {{eqn | l = d_1 \left({x, y}\right) + d_1 \left({y, z}\right)       | r = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert + \sum_{i \mathop = 1}^n \left\vert{y_i - z_i}\right\vert       | c = Definition of $d_1$ }} {{eqn | r = \sum_{i \mathop = 1}^n \left({\left\vert{x_i - y_i}\right\vert + \left\vert{y_i - z_i}\right\vert}\right)       | c =  }} {{eqn | o = \ge       | r = \sum_{i \mathop = 1}^n \left\vert{x_i - z_i}\right\vert       | c = Triangle Inequality for Real Numbers }} {{eqn | r = d_1 \left({x, z}\right)       | c = Definition of $d_1$ }} {{end-eqn}} So axiom $M2$ holds for $d_1$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = d_1 \left({x, y}\right)       | r = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert       | c = Definition of $d_1$ }} {{eqn | r = \sum_{i \mathop = 1}^n \left\vert{y_i - x_i}\right\vert       | c =  }} {{eqn | r = d \left({y, x}\right)       | c = Definition of $d_1$ }} {{end-eqn}} So axiom $M3$ holds for $d_1$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \implies       | lo= \exists i \in \left\{ {1, 2, \ldots, n}\right\}:       | l = x_i       | o = \ne       | r = y_i       | c =  }} {{eqn | ll= \implies       | l = \left\vert{x_i - y_i}\right\vert       | o = >       | r = 0       | c = Definition of Absolute Value }} {{eqn | ll= \implies       | l = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert       | o = >       | r = 0       | c =  }} {{eqn | ll= \implies       | l = d_1 \left({x, y}\right)       | o = >       | r = 0       | c = Definition of $d_1$ }} {{end-eqn}} So axiom $M4$ holds for $d_1$. {{qed}}
This is an instance of the Chebyshev distance on the cartesian product of metric spaces $A_1, A_2, \ldots, A_3$. This is proved in Chebyshev Distance is Metric. {{qed}}
=== Proof of $M1$ === {{begin-eqn}} {{eqn | l = \map {d_\infty} {x, x}       | r = \max_{i \mathop = 1}^n \size {x_i - x_i}       | c = Definition of $d_\infty$ }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $M1$ holds for $d_\infty$. {{qed|lemma}} === Proof of $M2$ === Let $k \in \closedint 1 n$ such that: {{begin-eqn}} {{eqn | l = \size {x_k - z_k}       | r = \max_{i \mathop = 1}^n \size {x_i - z_i}       | c =  }} {{eqn | r = \map {d_\infty} {x, z}       | c =  }} {{end-eqn}} Then by the Triangle Inequality for Real Numbers: :$\size {x_k - z_k} \le \size {x_k - y_k} + \size {y_k - z_k}$ But by the nature of the $\max$ operation: :$\displaystyle \size {x_k - y_k} \le \max_{i \mathop = 1}^n \size {x_i - y_i}$ and: :$\displaystyle \size {y_k - z_k} \le \max_{i \mathop = 1}^n \size {y_i - z_i}$ Thus: :$\displaystyle \size {x_k - y_k} + \size {y_k - z_k} \le \max_{i \mathop = 1}^n \size {x_i - y_i} + \max_{i \mathop = 1}^n \size {y_i - z_i}$ Hence: :$\map {d_\infty} {x, z} \le \map {d_\infty} {x, y} + \map {d_\infty} {y, z}$ So axiom $M2$ holds for $d_\infty$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = \map {d_\infty} {x, y}       | r = \max_{i \mathop = 1}^n \size {x_i - y_i}       | c = Definition of $d_\infty$ }} {{eqn | r = \max_{i \mathop = 1}^n \size {y_i - x_i}       | c = {{Defof|Absolute Value}} }} {{eqn | r = \map {d_\infty} {y, x}       | c = Definition of $d_\infty$ }} {{end-eqn}} So axiom $M3$ holds for $d_\infty$. {{qed|lemma}} === Proof of $M4$ === Let $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$. {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \leadsto       | lo= \exists k \in \closedint 1 n       | l = x_k       | o = \ne       | r = y_k       | c =  }} {{eqn | ll= \leadsto       | l = \size {x_k - y_k}       | o = >       | r = 0       | c = {{Defof|Absolute Value}} }} {{eqn | ll= \leadsto       | l = \max_{i \mathop = 1}^n \size {x_i - y_i}       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_\infty} {x, y}       | o = >       | r = 0       | c = Definition of $d_\infty$ }} {{end-eqn}} So axiom $M4$ holds for $d_\infty$. {{qed}}
=== Proof of $M1$ === {{begin-eqn}} {{eqn | l = \map {d_\infty} {x, x}       | r = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, x_i} }       | c = Definition of $d_\infty$ }} {{eqn | r = 0       | c = as $d_i$ fulfils axiom $M1$ }} {{end-eqn}} So axiom $M1$ holds for $d_\infty$. {{qed|lemma}} === Proof of $M2$ === Let $k \in \closedint 1 n$ such that: {{begin-eqn}} {{eqn | l = \map {d_k} {x_k, z_k}       | r = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, z_i} }       | c =  }} {{eqn | r = \map {d_\infty} {x, z}       | c =  }} {{end-eqn}} Then by application of axiom $M2$ for metric $d_k$: :$\map {d_k} {x_k, z_k} \le \map {d_k} {x_k, y_k} + \map {d_k} {y_k, z_k}$ But by the nature of the $\max$ operation: :$\displaystyle \map {d_k} {x_k, y_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$ and: :$\displaystyle \map {d_k} {y_k, z_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {y_i, z_i} }$ Thus: :$\displaystyle \map {d_k} {x_k, y_k} + \map {d_k} {y_k, z_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} } + \max_{i \mathop = 1}^n \set {\map {d_i} {y_i, z_i} }$ Hence: :$\map {d_\infty} {x, z} \le \map {d_\infty} {x, y} + \map {d_\infty} {y, z}$ So axiom $M2$ holds for $d_\infty$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = \map {d_\infty} {x, y}       | r = \max_{i \mathop = 1}^n \map {d_i} {x_i, y_i}       | c = Definition of $d_\infty$ }} {{eqn | r = \max_{i \mathop = 1}^n \map {d_i} {y_i, x_i}       | c = as $d_i$ fulfils axiom $M3$ }} {{eqn | r = \map {d_\infty} {y, x}       | c = Definition of $d_\infty$ }} {{end-eqn}} So axiom $M3$ holds for $d_\infty$. {{qed|lemma}} === Proof of $M4$ === Let $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$. {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \leadsto       | lo= \exists k \in \closedint 1 n       | l = x_k       | o = \ne       | r = y_k       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_k} {x_k, y_k}       | o = >       | r = 0       | c = as $d_k$ fulfils axiom $M4$ }} {{eqn | ll= \leadsto       | l = \max_{i \mathop = 1}^n \map {d_i} {x_i, y_i}       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_\infty} {x, y}       | o = >       | r = 0       | c = Definition of $d_\infty$ }} {{end-eqn}} So axiom $M4$ holds for $d_\infty$. {{qed}}
=== Proof of $M1$ === {{begin-eqn}} {{eqn | l = d_k \left({x, x}\right)       | r = k \cdot d \left({x, x}\right)       | c = Definition of $d_k$ }} {{eqn | r = 0       | c = as $d$ fulfils axiom $M1$ }} {{end-eqn}} So axiom $M1$ holds for $d_k$. {{qed|lemma}} === Proof of $M2$ === {{begin-eqn}} {{eqn | l = d_k \left({x, y}\right) + d_k \left({y, z}\right)       | r = k \cdot d \left({x, y}\right) + k \cdot d \left({y, z}\right)       | c = Definition of $d_k$ }} {{eqn | r = k \left({d \left({x, y}\right) + d \left({y, z}\right)}\right)       | c =  }} {{eqn | o = \ge       | r = k \cdot d \left({x, z}\right)       | c = as $d$ fulfils axiom $M2$ }} {{eqn | r = d_k \left({x, z}\right)       | c = Definition of $d_k$ }} {{end-eqn}} So axiom $M2$ holds for $d_k$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = d_k \left({x, y}\right)       | r = k \cdot d \left({x, y}\right)       | c = Definition of $d_k$ }} {{eqn | r = k \cdot d \left({y, x}\right)       | c = as $d$ fulfils axiom $M3$ }} {{eqn | r = d_k \left({y, x}\right)       | c = Definition of $d_k$ }} {{end-eqn}} So axiom $M3$ holds for $d_k$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \implies       | l = d \left({x, y}\right)       | o = >       | r = 0       | c = as $d$ fulfils axiom $M4$ }} {{eqn | ll= \implies       | l = k \cdot d \left({x, y}\right)       | o = >       | r = 0       | c = as $k > 0$ }} {{eqn | ll= \implies       | l = d_k \left({x, y}\right)       | o = >       | r = 0       | c = Definition of $d_k$ }} {{end-eqn}} So axiom $M4$ holds for $d_k$. {{qed}}
Let $S_m = \set {m, m + 1, \ldots, m + n - 1}$ be a set containing $n$ successive integers. The proof proceeds by induction on $m$, the smallest number in $S_m$. === Basis for the Induction === The case $m = 1$ is verified as follows: :$S_1 = \set {1, 2, \ldots, n}$ This contains a multiple of $n$, namely, $n$ itself. If $S_1$ is a singleton then this element is trivially unique. If there are any other integers in $S_1$, they are less than $n$ and more than $0$, and so cannot be a multiple of $n$. This is the basis for the induction. === Induction Hypothesis === Fix $m \in \N$ with $m \ge 1$. Assume that: :$S_m = \set {m, m + 1, \ldots, m + n - 1}$ contains a unique multiple of $n$. This is our induction hypothesis. === Induction Step === This is our induction step: Consider the set: :$S_{m + 1} = \set {m + 1, m + 2, \ldots, m + n - 1, m + n}$ As $n \equiv 0 \pmod n$, $m + n$ is a multiple of $n$ {{iff}} $m$ is a multiple of $n$. That means that we can replace $m + n$ with $m$ and instead analyze: :${S_{m + 1} }' = \set {m + 1, m + 2, \ldots, m + n - 1, m}$ {{handwaving}} But ${S_{m + 1} }' = S_m$, which contains a unique multiple of $n$ by the induction hypothesis. The result follows by the Principle of Mathematical Induction. {{qed}} Category:Number Theory g2ax9b3le2fkujs6qx0wil67129zywp
By definition of the Chebyshev distance on $\R^n$, we have: :$\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$ where $x = \left({x_1, x_2, \ldots, x_n}\right)$ and $y = \left({y_1, y_2, \ldots, y_n}\right)$. Let $j$ be chosen so that: :$\displaystyle \left\vert{x_j - y_j}\right\vert = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$ Then: {{begin-eqn}} {{eqn | l = d_\infty \left({x, y}\right)       | r = \left\vert{x_j - y_j}\right\vert       | c = Definition of Chebyshev distance }} {{eqn | o = \le       | r = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert }} {{eqn | r = d_1 \left({x, y}\right) }} {{eqn | o = \le       | r = n \left\vert{x_j - y_j}\right\vert }} {{eqn | r = n \cdot d_\infty \left({x, y}\right) }} {{end-eqn}} {{qed}}
=== Proof of $M1$ === {{begin-eqn}} {{eqn | l = d \left({f, f}\right)       | r = \int_a^b \left\vert{f \left({t}\right) - f \left({t}\right)}\right\vert \ \mathrm d t       | c = Definition of $d$ }} {{eqn | r = \int_a^b 0 \ \mathrm d t       | c = Definition of Absolute Value }} {{eqn | r = 0       | c = Definite Integral of Constant }} {{end-eqn}} So axiom $M1$ holds for $d$. {{qed|lemma}} === Proof of $M2$ === {{begin-eqn}} {{eqn | l = \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert + \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert       | o = \ge       | r = \left\vert{f \left({t}\right) - h \left({t}\right)}\right\vert       | c = Triangle Inequality for Real Numbers }} {{eqn | ll= \implies        | l = \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t + \int_a^b \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert \ \mathrm d t       | o = \ge       | r = \int_a^b \left\vert{f \left({t}\right) - h \left({t}\right)}\right\vert \ \mathrm d t       | c = Relative Sizes of Definite Integrals }} {{eqn | ll= \implies        | l = d \left({f, g}\right) + d \left({g, h}\right)       | o = \ge       | r = d \left({f, h}\right)       | c = Definition of $d$ }} {{end-eqn}} So axiom $M2$ holds for $d$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = d \left({f, g}\right)       | r = \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t       | c = Definition of $d$ }} {{eqn | r = \int_a^b \left\vert{g \left({t}\right) - f \left({t}\right)}\right\vert \ \mathrm d t       | c = Definition of Absolute Value }} {{eqn | r = d \left({g, f}\right)       | c = Definition of $d$ }} {{end-eqn}} So axiom $M3$ holds for $d$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | lo= \forall t \in \left[{a \,.\,.\, b}\right]:       | l = \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert       | o = \ge       | r = 0       | c = Definition of Absolute Value }} {{eqn | ll= \implies       | l = \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t       | o = \ge       | r = 0       | c = Sign of Function Matches Sign of Definite Integral }} {{end-eqn}} From Zero Definite Integral of Nowhere Negative Function implies Zero Function we have that: :$d \left({f, g}\right) = 0 \implies f = g$ on $\left[{a \,.\,.\, b}\right]$. So axiom $M4$ holds for $d$. {{qed}}
The Euclidean metric on $\mathcal A$ is a special case of the $p$-product metric. The result follows from $p$-Product Metric is Metric. {{qed}}
We have that the Euclidean metric on $\mathcal A$ is defined as: :$\displaystyle \map {d_2} {x, y} = \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2}$ where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \mathcal A$. === Proof of $M1$ === {{begin-eqn}} {{eqn | l = \map {d_2} {x, x}       | r = \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, x_i} }^2}^{\frac 1 2}       | c = Definition of $d_2$ }} {{eqn | r = \paren {\sum_{i \mathop = 1}^n 0^2}^{\frac 1 2}       | c = as $d_{i'}$ fulfils axiom $M1$ }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $M1$ holds for $d_2$. {{qed|lemma}} === Proof of $M2$ === Let: :$(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$ :$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$ :$(3): \quad \map {d_{i'} } {x_i, y_i} = r_i$ :$(4): \quad \map {d_{i'} } {y_i, z_i} = s_i$. Thus we need to show that: :$\displaystyle \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^2}^{\frac 1 2} \ge \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^2}^{\frac 1 2}$ We have: {{begin-eqn}} {{eqn | l = \map {d_2} {x, y} + \map {d_2} {y, z}       | r = \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^2}^{\frac 1 2}       | c = Definition of $d_2$ }} {{eqn | r = \paren {\sum r_i^2}^{\frac 1 2} + \paren {\sum s_i^2}^{\frac 1 2}       | c =  }} {{eqn | o = \ge       | r = \paren {\sum \paren {r_i + s_i}^2}^{\frac 1 2}       | c = Minkowski's Inequality for Sums: index $2$ }} {{eqn | r = \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} + \map {d_{i'} } {y_i, z_i} }^2}^{\frac 1 2}       | c = Definition of $r_i$ and $s_i$ }} {{eqn | o = \ge       | r = \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^2}^{\frac 1 2}       | c = as $d_{i'}$ fulfils axiom $M2$ }} {{eqn | r = \map {d_2} {x, z}       | c = Definition of $d_2$ }} {{end-eqn}} So axiom $M2$ holds for $d_2$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = \map {d_2} {x, y}       | r = \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2}       | c = Definition of $d_2$ }} {{eqn | r = \paren {\sum \paren {\map {d_{i'} } {y_i, x_i} }^2}^{\frac 1 2}       | c = as $d_i$ fulfils axiom $M3$ }} {{eqn | r = \map {d_2} {y, x}       | c = Definition of $d_2$ }} {{end-eqn}} So axiom $M3$ holds for $d_2$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = \exists k \in \closedint 1 n: x_k       | o = \ne       | r = y_k       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_k} {x_k, y_k}       | o = >       | r = 0       | c = as $d_k$ fulfils axiom $M4$ }} {{eqn | ll= \leadsto       | l = \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2}       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_2} {x, y}       | o = >       | r = 0       | c = Definition of $d_2$ }} {{end-eqn}} So axiom $M4$ holds for $d_2$. {{qed}}
We prove the norm axioms. === Positive Definiteness === As $\mathbf 0 = \tuple {0, \ldots, 0}$, it follows that: :$\norm {\mathbf 0} = \sqrt {0^2 + \ldots + 0^2} = 0$ Suppose instead that $\norm {\mathbf v} = 0$ for some $\mathbf v = \tuple {v_1, \ldots, v_n}$ with $v_1, \ldots, v_n \in \R$. As $\displaystyle \sqrt {\sum_{i \mathop = 1}^n v_n^2 } = 0$, it follows from squaring both sides that: :$\displaystyle \sum_{i \mathop = 1}^n v_n^2 = 0$ From Even Power is Non-Negative, it follows that for all $i$: $v_i = 0$. Hence $\mathbf v = \mathbf 0$. {{qed|lemma}} === Positive homogeneity === {{ProofWanted|I refuse to believe this is not covered somewhere, but cannot find it}} === Triangle Inequality === Follows from Triangle Inequality for Vectors in Euclidean Space. {{qed}} Category:Examples of Norms Category:Euclidean Space h0t1y2y2hps3keus5ajty6mep8p8nd5
By the Division Theorem: :$(1): \quad n = q m + r$ where $0 \le r < q$. As $r < q$, it follows that the greatest multiple of $m$ up to $n$ is $q m$. So all the multiples of $m$ up to $n$ are: :$m, 2 m, 3 m, \ldots, q m$ Dividing both sides of $(1)$ by $q$: :$(2): \quad \dfrac n m = q + \dfrac r m$ Taking the floor of $(2)$: :$\floor {\dfrac n m} = \floor {q + \dfrac r m}$ But as $0 \le \dfrac r m < 1$: :$\floor {q + \dfrac r m} = q$ Recall that all the multiples of $m$ up to $n$ are $m, 2 m, 3 m, \ldots, q m$. It follows that the number of multiples of $m$ up to $n$ is: :$q = \floor {\dfrac n m}$ {{qed}}
Proof by Counterexample: Let $f: \R \to \R$ be given by: :$f \left({x}\right) = \begin{cases} 0 & : x \in \Q \\ 1 & : x \in \R \end{cases}$ where $\Q$ is the set of rational numbers. Then $f {\restriction_\Q}: \Q \to \R$ is the constant function $f_0$ with value $0$. By Constant Mapping is Continuous, $f {\restriction_\Q}$ is continuous at every point. However, $f$ is not continuous at any point on $\R$. {{qed}}
The interval is an instance of a set. Hence Supremum Metric on Bounded Real-Valued Functions is Metric can be directly applied.
Let $f \in \mathscr C \left[{a \,.\,.\, b}\right]$. Then by Image of Closed Real Interval is Bounded, $f$ is bounded on $\left[{a \,.\,.\, b}\right]$. Thus $f \in \mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right)$ and the result follows. {{qed}}
{{AimForCont}} that: :$\exists c \in \left[{a \,.\,.\, b}\right]: h \left({c}\right) > 0$ As $h$ is continuous, there exists some closed real interval $\left[{r \,.\,.\, s}\right] \subseteq \left[{a \,.\,.\, b}\right]$ where $r < s$ such that: :$\exists \epsilon \in \R_{>0}: \forall x \in \left[{r \,.\,.\, s}\right]: h \left({x}\right) > \dfrac {h \left({c}\right)} 2$ From Sign of Function Matches Sign of Definite Integral: :$\displaystyle \int_r^s h \left({x}\right) \rd x > 0$ This makes a strictly positive contribution to the integral. $h$ is still continuous on $\left[{a \,.\,.\, r}\right]$ and $\left[{s \,.\,.\, b}\right]$. So, again from Sign of Function Matches Sign of Definite Integral: :$\forall x \in \left [{a \,.\,.\, r} \right]: f \left({x}\right) \ge 0 \implies \displaystyle \int_a^r f \left({x}\right) \rd x \ge 0$ :$\forall x \in \left [{s \,.\,.\, b} \right]: f \left({x}\right) \ge 0 \implies \displaystyle \int_s^b f \left({x}\right) \rd x \ge 0$ Thus as: :$\displaystyle \int_a^b h \left({x}\right) \rd x = \int_a^r f \left({x}\right) \rd x + \int_r^s f \left({x}\right) \rd x + \int_s^b f \left({x}\right) \rd x$ it follows that: :$\displaystyle \int_a^b h \left({x}\right) \rd x > 0$ But by hypothesis: :$\displaystyle \int_a^b h \left({x}\right) \rd x = 0$ Thus by contradiction, there can be no $c \in \left[{a \,.\,.\, b}\right]$ such that $h \left({c}\right) > 0$. Hence the result. {{qed}}