hcju/proofwikiqa huggingface dataset:
We have: {{begin-eqn}} {{eqn | l = \phi \left({33 \, 817 \, 088}\right)       | r = 16 \, 842 \, 752       | c = {{EulerPhiLink|33,817,088|33 \, 817 \, 088}} }} {{eqn | r = 2^{16} \times 257       | c =  }} {{end-eqn}} Consider the equation: :$(1): \quad \phi \left({x}\right) = 2^{16} \times 257$ Even solutions to $(1)$ are of the form: :$x = 2^{9 - \epsilon_0 - 2 \epsilon_1 - 4 \epsilon_2} \times 3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2$ where each $\epsilon_k$ equals $0$ or $1$. {{explain|Justify the above by showing the working.}} {{AimForCont}} there exists an odd integer $k$ satisfying $(1)$: :$\phi \left({k}\right) = 2^{16} \times 257$ Let $p$ be a prime factor of $k$. Then as Euler Phi Function is Multiplicative, either: :$p - 1 = 2^j$ for some $j \in \Z$ such that $1 \le j \le 16$ or: :$p - 1 = 2^j \times 257$ for some $j \in \Z$ such that $1 \le j \le 16$ But $2^j \times 257 + 1$ is composite for $1 \le j \le 16$: {{explain|Do the arithmetic to demonstrate the above}} So the only possible prime factors of $k$ are the Fermat primes: :$3, 5, 17, 257, 65 \, 537$ such that: : $257$ occurs with multiplicity $2$ : all other prime factors occurs with multiplicity $1$. Because of the size of $257^2 \times 67 \, 537$ it follows that $257$ and $65 \, 537$ cannot appear together. We have that: {{begin-eqn}} {{eqn | l = \phi \left({3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2}\right)       | r = 2^{8 + \epsilon_0 + 2 \epsilon_1 + 4 \epsilon_2} \times 257       | c =  }} {{eqn | o = <       | r = 2^{16} \times 257       | c =  }} {{end-eqn}} {{explain|expand the reasoning of the above}} Thus there can be no odd integer $k$ satisfying $(1)$. It can be established by computer that $33 \, 817 \, 088$ is the smallest such even integer with this property. {{qed}}
From {{EulerPhiLink|83,623,935|83 \, 623 \, 935}}: :$\map \phi {83 \, 623 \, 935} = 41 \, 811 \, 968$ and then: {{begin-eqn}} {{eqn | l = 2 \times 41 \, 811 \, 968       | r = 83 \, 623 \, 936       | c =  }} {{eqn | r = 1 + 83 \, 623 \, 935       | c =  }} {{end-eqn}} {{qed|lemma}} Then we have that $83 \, 623 \, 937$ is the $868 \, 421$st prime number. From Euler Phi Function of Prime: :$\map \phi {83 \, 623 \, 937} = 83 \, 623 \, 936$ {{begin-eqn}} {{eqn | l = \map \phi {83 \, 623 \, 935 \times 83 \, 623 \, 937}       | r = \map \phi {83 \, 623 \, 935} \times \map \phi {83 \, 623 \, 937}       | c = Euler Phi Function is Multiplicative }} {{eqn | r = 41 \, 811 \, 968 \times 83 \, 623 \, 936       | c = from above }} {{end-eqn}} Then we have that: {{begin-eqn}} {{eqn | l = 83 \, 623 \, 935 \times 83 \, 623 \, 937       | r = \paren {83 \, 623 \, 936 - 1} \times \paren {83 \, 623 \, 936 + 1}       | c =  }} {{eqn | r = 83 \, 623 \, 936^2 - 1^2       | c = Difference of Two Squares }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = 2 \times \map \phi {83 \, 623 \, 935 \times 83 \, 623 \, 937}       | r = 2 \times 41 \, 811 \, 968 \times 83 \, 623 \, 936       | c =  }} {{eqn | r = 83 \, 623 \, 936^2       | c =  }} {{eqn | r = 83 \, 623 \, 935 \times 83 \, 623 \, 937 + 1       | c =  }} {{end-eqn}} {{qed}}
Let $S \subset R \sqbrk X$ be the subset of all elements that are linear combinations of monomials. Let $\iota: S \to R \sqbrk X$ denote the inclusion mapping. Suppose for the moment that $S$ is a commutative ring with unity. Then by Universal Property of Polynomial Ring, there exists a ring homomorphism $g: R \sqbrk X \to S$ with $\map g X = X$. By Inclusion Mapping on Subring is Homomorphism, $\iota: S \to R \sqbrk X$ is a ring homomorphism. By Composition of Ring Homomorphisms is Ring Homomorphism, $\iota \circ g: R \sqbrk X \to R \sqbrk X$ is a ring homomorphism. By construction, $\map {\paren {\iota \circ g} } X = X$. By Universal Property of Polynomial Ring, there exists a unique ring homomorphism $h : R \sqbrk X \to R \sqbrk X$ with $\map h X = X$. We have that $\iota \circ g$ is such a ring homomorphism. By Identity Mapping is Ring Automorphism, the identity mapping $I$ on $R \sqbrk X$ is one too. By uniqueness, $\iota \circ g = I$. By Identity Mapping is Surjection and Surjection if Composite is Surjection, $\iota$ is a surjection. By Inclusion Mapping is Surjection iff Identity, $S = R \sqbrk X$. It remains to show that $S$ is a commutative ring with unity. {{finish}}
Note that by Union of Finite Sets is Finite, the union $S \cup T$ is finite. Let $m$ be the cardinality of $S$ and $n$ be the cardinality of $T$. Let $\N_{< m}$ denote an initial segment of the natural numbers. Let $\sigma: \N_{< m} \to S$ and $\tau: \N_{< n} \to T$ be bijections. Let $\alpha: \N_{< n} \to \left[{m \,.\,.\, m + n - 1}\right]$ be the mapping defined as: :$\alpha \left({k}\right) = k + m$ By Translation of Integer Interval is Bijection, $\alpha$ is a bijection. By Composite of Bijections is Bijection and Disjoint Union of Bijections is Bijection, the union: :$\sigma \cup (\tau\circ \alpha) : \N_{< m} \cup \left[{m \,.\,.\, m + n - 1}\right] \to S \cup T$ is a bijection. By Union of Integer Intervals, $\N_{< m} \cup \left[{m \,.\,.\, m + n - 1}\right] = \N_{< m + n}$. We have: {{begin-eqn}} {{eqn | l = \sum_{u \mathop \in S \mathop \cup T} f \left({u}\right)       | r = \sum_{i \mathop = 0}^{m + n - 1} f \circ \left({\sigma \cup \left({\tau \circ \alpha}\right)}\right) \left({i}\right)       | c = {{Defof|Summation}} }} {{eqn | r = \sum_{i \mathop = 0}^{m - 1} f \circ \left({\sigma \cup \left({\tau \circ \alpha}\right)}\right) \left({i}\right) + \sum_{i \mathop = m}^{m + n - 1} f \circ \left({\sigma \cup \left({\tau \circ \alpha}\right)}\right) \left({i}\right)       | c = Indexed Summation over Adjacent Intervals }} {{eqn | r = \sum_{i \mathop = 0}^{m - 1} \left({f \circ \sigma}\right) \left({i}\right) + \sum_{i \mathop = m}^{m + n - 1} \left({g \circ \tau \circ \alpha}\right) \left({i}\right)       | c = Restriction of Union of Mappings to Component of Domain }} {{eqn | r = \sum_{i \mathop = 0}^{m - 1} \left({f \circ \sigma}\right) \left({i}\right) + \sum_{i \mathop = 0}^{n - 1} \left({g \circ \tau}\right) \left({i}\right)       | c = Definition of $\alpha$, Indexed Summation over Translated Interval }} {{eqn | r = \displaystyle \sum_{s \mathop \in S} f \left({s}\right) + \sum_{t \mathop \in T} g \left({t}\right)       | c = {{Defof|Summation}} }} {{end-eqn}} {{qed}}
This is a special case of Change of Variables in Summation over Finite Set. {{qed}} Category:Summations 1posxiyi6afj4e2emhfo71ku7qmtc43
By Inverse of Bijection is Bijection, $h^{-1} : \N_{<n} \to S$ is a bijection. By Composite of Bijections is Bijection, the composition $h^{-1}\circ g$ is a permutation of $\N_{<n}$. By Indexed Summation does not Change under Permutation, we have an equality of indexed summations: :$\displaystyle\sum_{i \mathop = 0}^{n - 1} \map {\paren {f \circ h} } i = \sum_{i \mathop = 0}^{n - 1} \map {\paren {f \circ h} \circ \paren {h^{-1} \circ g} } i$ By Composition of Mappings is Associative and Composite of Bijection with Inverse is Identity Mapping, the {{RHS}} equals $\displaystyle \sum_{i \mathop = 0}^{n - 1} \map f {\map g i}$. {{qed}}
Let $n$ be the cardinality of $S$ and $T$. Let $\N_{<n}$ be an initial segment of the natural numbers. Let $h : \N_{<n} \to T$ be a bijection. By definition of summation: : $\displaystyle \sum_{t \mathop \in T} f \left({g \left({t}\right)}\right) = \displaystyle \sum_{i \mathop = 0}^{n - 1} f \left({g \left({h \left({i}\right)}\right)}\right)$ By Composite of Bijections is Bijection, the composition $g \circ h : \N_{<n} \to S$ is a bijection. By definition of summation: : $\displaystyle \sum_{s \mathop\in S} f \left({s}\right) = \displaystyle \sum_{i \mathop = 0}^{n - 1} f \left({g \left({h \left({i}\right)}\right)}\right)$ {{qed}}
The proof goes by induction on $b$. === Basis for the Induction === Let $b < a$. Then: :$b + c < a + c$ Thus both indexed summations are zero. This is our basis for the induction. === Induction Step === Let $b \ge a$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a + c}^{b + c} \map f {i - c}       | r = \sum_{i \mathop = a + c}^{b + c - 1} \map f {i - c} + \map f {b + c - c}       | c = {{Defof|Indexed Summation}} }} {{eqn | r = \sum_{i \mathop = a}^{b - 1} \map f i  + \map f b       | c = Induction Hypothesis }} {{eqn | r = \sum_{i \mathop = a}^b \map f i       | c = {{Defof|Indexed Summation}} }} {{end-eqn}} By the Principle of Mathematical Induction, the proof is complete. {{qed}}
The proof goes by induction on $b$. === Basis for the Induction === Let $b = a-1$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a}^b f(i) + \sum_{i \mathop = b+1}^c f(i)       | r = 0 + \sum_{i \mathop = b+1}^c f(i)       | c = {{Defof|Indexed Summation}}, $b=a-1$ }} {{eqn | l =        | r = \sum_{i \mathop = a}^c f(i)       | c = Identity Element of Addition on Numbers }} {{end-eqn}} This is our basis for the induction. === Induction Step === Let $a \leq b \leq c$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a}^b f(i) + \sum_{i \mathop = b+1}^c f(i)       | r = \sum_{i \mathop = a}^{b-1} f(i) + f(b) + \sum_{i \mathop = b+1}^c f(i)       | c = {{Defof|Indexed Summation}} }} {{eqn | r = \sum_{i \mathop = a}^{b-1} f(i) + \sum_{i \mathop = b}^c f(i)       | c = Indexed Summation without First Term }} {{end-eqn}} By the Principle of Mathematical Induction, the proof is complete. {{qed}}
We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a}^{a+1} f(i)       | r = \sum_{i \mathop = a}^a f(i) + f(a+1)       | c = {{Defof|Indexed Summation}} }} {{eqn | l =        | r = f(a) + f(a+1)       | c = Indexed Summation over Interval of Length One }} {{end-eqn}} {{qed}} Category:Summations 1fmazkc5z3mw4wvpz1inhwswolxgv6k
We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a}^{a} f(i)       | r = \sum_{i \mathop = a}^{a-1} f(i) + f(a)       | c = {{Defof|Indexed Summation}} }} {{eqn | l =        | r = 0 + f(a)       | c = {{Defof|Indexed Summation}}, $a-1 < a$ }} {{eqn | l =        | r = f(a)       | c = Identity Element of Addition on Numbers }} {{end-eqn}} {{qed}}
Because $g : \closedint c d \to \closedint a b$ is a bijection, these sets are equivalent. By Cardinality of Integer Interval, $\closedint a b$ has cardinality $b - a + 1$. Thus: :$b - a + 1 = d - c + 1$ Thus :$c - a = d - b$ By Indexed Summation over Translated Interval: :$\displaystyle \sum_{i \mathop = c}^d \map f {\map g i} = \sum_{i \mathop = a}^b \map f {\map g {i + c - a} }$ By Translation of Integer Interval is Bijection, the mapping $T : \closedint a b \to \closedint c d$ defined as: :$\map T k = k + c - a$ is a bijection. By Composite of Bijections is Bijection, $g \circ T$ is a permutation of $\closedint a b$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = c}^d \map f {\map g i}       | r = \sum_{i \mathop = a}^b \map f {\map g {i + c - a} }       | c = Indexed Summation over Translated Interval }} {{eqn | r = \sum_{i \mathop = a}^b \map f {\map g {\map T i} }       | c = Definition of $T$ }} {{eqn | r = \sum_{i \mathop = a}^b \map f i       | c = Indexed Summation does not Change under Permutation }} {{end-eqn}} {{qed}}
Note that if $k \in \closedint a b$, then indeed $k + c \in \closedint {a + c} {b + c}$. === Injectivity === Let $k, l \in \closedint a b$ with $k + c = l + c$. By Integer Addition is Cancellable, $k = l$. Thus $T$ is injective. {{qed|lemma}} === Surjectivity === Let $m \in \closedint {a + c} {b + c}$. Then $m - c \in \closedint a b$. Then $\map T {m - c} = m - c + c = m$. Thus $T$ is surjective. {{qed}}
The proof goes by induction on $b$. === Basis for the Induction === Let $b=a$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a}^a f(i)       | r = f(a)       | c = Indexed Summation over Interval of Length One }} {{eqn | r = f(a) + 0       | c = Identity Element of Addition on Numbers }} {{eqn | r = f(a) + \sum_{i \mathop = a+1}^{a} f(i)       | c = {{Defof|Indexed Summation}}, $a+1 > a$ }} {{end-eqn}} This is our basis for the induction. === Induction Step === Let $b \geq a+1$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a}^b f(i)       | r = \sum_{i \mathop = a}^{b-1} f(i) + f(b)       | c = {{Defof|Indexed Summation}}, $b \geq a$ }} {{eqn | r = f(a) + \sum_{i \mathop = a+1}^{b-1} f(i) + f(b)       | c = induction hypothesis }} {{eqn | r = f(a) + \sum_{i \mathop = a+1}^{b} f(i)       | c = {{Defof|Indexed Summation}}, $b \geq a+1$ }} {{end-eqn}} By the Principle of Mathematical Induction, the proof is complete. {{qed}}
By Cardinality of Integer Interval, $\left[{a \,.\,.\, b}\right]$ has cardinality $b - a + 1$. By Translation of Integer Interval is Bijection, the mapping $T : \left[{0 \,.\,.\, b - a}\right] \to \left[{a \,.\,.\, b}\right]$ defined as: :$T \left({k}\right) = k + a$ is a bijection. By definition of summation: :$\displaystyle \sum_{k \mathop \in \left[{a \,.\,.\, b}\right]} f \left({k}\right) = \sum_{k \mathop = 0}^{b - a} f \left({k + a}\right)$ By Indexed Summation over Translated Interval: :$\displaystyle \sum_{k \mathop = 0}^{b - a} f \left({k + a}\right) = \sum_{k \mathop = a}^b f \left({k}\right)$ {{qed}} Category:Summations 1cv3z7q8i4xg9ek6l2wzjw6t8jhuwcb
Let $b < a$. Then $\left[{a \,.\,.\, b}\right]$ is empty. By Empty Set is Finite, $\left[{a \,.\,.\, b}\right]$ is finite. By Cardinality of Empty Set, $\left[{a \,.\,.\, b}\right]$ has cardinality $0$. Let $b \ge a$. By Translation of Integer Interval is Bijection, there exists a bijection between $\left[{a \,.\,.\, b}\right]$ and $\left[{0 \,.\,.\, b - a}\right]$. Thus $\left[{a \,.\,.\, b}\right]$ is finite of cardinality $b - a + 1$. {{qed}} Category:Set Theory tr6bchkgtxm9a40pp9dzcux54f7jo80
The proof proceeds by induction on $b$. For all $b \in \Z_{\ge 0}$, let $\map P b$ be the proposition: :$\displaystyle \sum_{i \mathop = a}^b \map h i = \sum_{i \mathop = a}^b \map f i + \sum_{i \mathop = a}^b \map g i$ === Basis for the Induction === Let $b < a$. Then all indexed summations are zero. Because $0 = 0 + 0$, the result follows. === Basis for the Induction === Let $b = a$. {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a}^b \map h i       | r = \sum_{i \mathop = a}^a \map h i       | c =  }} {{eqn | r = \map h a       | c =  }} {{eqn | r = \map f a + \map g a       | c = {{Defof|Pointwise Addition}} }} {{eqn | r = \sum_{i \mathop = a}^a \map f i + \sum_{i \mathop = a}^a \map g i       | c =  }} {{eqn | r = \sum_{i \mathop = a}^b \map f i + \sum_{i \mathop = a}^b \map g i       | c = as $a = b$ }} {{end-eqn}} Thus $\map P a$ is seen to hold. This is our basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P k$ is true, where $k \ge a$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{i \mathop = a}^k \map h i = \sum_{i \mathop = a}^k \map f i + \sum_{i \mathop = a}^k \map g i$ from which it is to be shown that: :$\displaystyle \sum_{i \mathop = a}^{k + 1} \map h i = \sum_{i \mathop = a}^{k + 1} \map f i + \sum_{i \mathop = a}^{k + 1} \map g i$ === Induction Step === We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a}^{k + 1} \map h i       | r = \sum_{i \mathop = a}^k \map h i + \map h {k + 1}       | c = {{Defof|Indexed Summation}} }} {{eqn | r = \sum_{i \mathop = a}^k \map h i + \map f {k + 1} + \map g {k + 1}       | c = {{Defof|Pointwise Addition}} }} {{eqn | r = \sum_{i \mathop = a}^k \map f i + \sum_{i \mathop = a}^k \map g i + \map f {k + 1} + \map g {k + 1}       | c = Induction Hypothesis }} {{eqn | r = \sum_{i \mathop = a}^k \map f i + \map f {k + 1} + \sum_{i \mathop = a}^k \map g i + \map g {k + 1}       | c = Commutative Law of Addition }} {{eqn | r = \sum_{i \mathop = a}^{k + 1} \map f i + \sum_{i \mathop = a}^{k + 1} \map g i       | c = {{Defof|Indexed Summation}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed}}
Let $n$ be the cardinality of $S$. Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{< n}$ is an initial segment of the natural numbers. By definition of summation, we have to prove the following equality of indexed summations: :$\displaystyle \sum_{i \mathop = 0}^{n - 1} h \left({\sigma \left({i}\right)}\right) = \sum_{i \mathop = 0}^{n - 1} f \left({\sigma \left({i}\right)}\right) + \sum_{i \mathop = 0}^{n - 1} g \left({\sigma \left({i}\right)}\right)$ By Sum of Mappings Composed with Mapping, $h \circ \sigma = f \circ \sigma + g \circ \sigma$. The above equality now follows from Indexed Summation of Sum of Mappings. {{qed}}
The proof goes by induction on $b$. === Basis for the Induction === Let $b < a$. Then all indexed summations are zero. Because $0 = \lambda \cdot 0$, the result follows. This is our basis for the induction. === Induction Step === Let $b \geq a$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a}^b g \left({i}\right)       | r = \sum_{i \mathop = a}^{b - 1} g \left({i}\right) + g \left({b}\right)       | c = {{Defof|Indexed Summation}} }} {{eqn | r = \sum_{i \mathop = a}^{b - 1} g \left({i}\right) + \lambda \cdot f \left({b}\right)       | c = {{Defof|Product of Mapping with Scalar}} }} {{eqn | r = \lambda \cdot \sum_{i \mathop = a}^{b - 1} f \left({i}\right) + \lambda \cdot f \left({b}\right)       | c = Induction Hypothesis }} {{eqn | r = \lambda \cdot \left({\sum_{i \mathop = a}^{b - 1} f \left({i}\right) + f \left({b}\right)}\right)       | c = Multiplication of Numbers Distributes over Addition }} {{eqn | r = \lambda \cdot \sum_{i \mathop = a}^b f \left({i}\right)       | c = {{Defof|Indexed Summation}} }} {{end-eqn}} By the Principle of Mathematical Induction, the proof is complete. {{qed}}
Let $n$ be the cardinality of $S$. Let $\sigma : \N_{< n} \to S$ be a bijection, where $\N_{< n}$ is an initial segment of the natural numbers. By definition of summation, we have to prove the following equality of indexed summations: :$\displaystyle \sum_{i \mathop = 0}^{n - 1} g \left({\sigma \left({i}\right)}\right) = \lambda \cdot \sum_{i \mathop = 0}^{n - 1} f \left({\sigma \left({i}\right)}\right)$ By Multiple of Mapping Composed with Mapping, $g \circ \sigma = \lambda \cdot \left({f \circ \sigma}\right)$. The above equality now follows from Indexed Summation of Multiple of Mapping. {{qed}}
We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a}^b \left( \lambda \cdot f(i) + \mu \cdot g(i) \right)       | r = \sum_{i \mathop = a}^b \left( \lambda \cdot f(i) \right) + \sum_{i \mathop = a}^b \left( \mu \cdot g(i) \right)       | c = Indexed Summation of Sum of Mappings }} {{eqn | r = \lambda \cdot \sum_{i \mathop = a}^b f(i) + \mu \cdot \sum_{i \mathop = a}^b g(i)       | c = Indexed Summation of Multiple of Mapping }} {{end-eqn}} {{qed}}
The proof goes by induction on $b$. === Basis for the Induction === Let $b < a$. Then all indexed summations are zero. Because $|0| \leq |0|$ by definition of the standard absolute value, the result follows. This is our basis for the induction. === Induction Step === Let $b \geq a$. We have: {{begin-eqn}} {{eqn | l = \left\vert \sum_{i \mathop = a}^{b} f(i) \right\vert       | r = \left\vert \sum_{i \mathop = a}^{b-1} f(i) + f(b) \right\vert       | c = {{Defof|Indexed Summation}} }} {{eqn | r = \left\vert \sum_{i \mathop = a}^{b-1} f(i) \right\vert + \left\vert f(b) \right\vert       | c = Triangle Inequality       | o = \leq }} {{eqn | r = \sum_{i \mathop = a}^{b-1} \left\vert f(i) \right\vert + \left\vert f(b) \right\vert       | c = Induction Hypothesis, {{defof|Relation Compatible with Operation}}       | o = \leq }} {{eqn | r = \sum_{i \mathop = a}^{b} \left\vert f(i) \right\vert        | c = {{Defof|Indexed Summation}} }} {{end-eqn}} By the Principle of Mathematical Induction, the proof is complete. {{qed}}
Let $n$ be the cardinality of $S$. Let $\sigma: \N_{< n} \to S$ be a bijection, where $\N_{< n}$ is an initial segment of the natural numbers. By definition of summation, we have to prove the following inequality of indexed summations: :$\displaystyle \left\vert{\sum_{i \mathop = 0}^{n - 1} f \left({\sigma \left({i}\right)}\right)}\right\vert \le \sum_{i \mathop = 0}^{n-1} \left({\left\vert{f}\right\vert \circ \sigma}\right) \left({i}\right) $ By Absolute Value of Mapping Composed with Mapping: : $\left\vert{f}\right\vert \circ \sigma = \left\vert{f \circ \sigma}\right\vert$ The above equality now follows from Triangle Inequality for Indexed Summations. {{qed}}
The proof proceeds by induction on $d$. === Basis for the Induction === Let $d < c$. Then the indexed summation in the {{RHS}} is zero. By Indexed Summation of Zero, so is the {{LHS}}. This is our basis for the induction. === Induction Step === Let $d \ge c$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = a}^b \sum_{j \mathop = c}^d \map f {i, j}        | r = \sum_{i \mathop = a}^b \paren {\sum_{j \mathop = c}^{d - 1} \map f {i, j} + \map f {i, d} }       | c = {{Defof|Indexed Summation}} }} {{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop = c}^{d - 1} \map f {i, j} + \sum_{i \mathop = a}^b \map f {i, d}       | c = Indexed Summation of Sum of Mappings }} {{eqn | r = \sum_{j \mathop = c}^{d - 1} \sum_{i \mathop = a}^b \map f {i, j} + \sum_{i \mathop = a}^b \map f {i, d}       | c = Induction Hypothesis }} {{eqn | r = \sum_{j \mathop = c}^d \sum_{i \mathop = a}^b \map f {i, j}       | c = {{Defof|Indexed Summation}} }} {{end-eqn}} By the Principle of Mathematical Induction, the proof is complete. {{qed}}
Note that by Subset of Finite Set is Finite, $T$ is indeed finite. By Set is Disjoint Union of Subset and Relative Complement, $S$ is the disjoint union of $S \setminus T$ and $T$. The result now follows from Sum over Disjoint Union of Finite Sets. {{qed}}
Note that by Subset of Finite Set is Finite, $B$ consists of finite sets. The result now follows from Sum over Disjoint Union of Finite Sets. {{qed}}
Follows from: :Mapping Defines Additive Function of Subalgebra of Power Set :Power Set is Algebra of Sets :Inclusion-Exclusion Principle {{qed}}
Note that by Subset of Finite Set is Finite, $\operatorname{Supp} \left({f}\right)$ is indeed finite. The result now follows from: * Sum over Complement of Finite Set * Sum of Zero over Finite Set * Identity Element of Addition on Numbers {{qed}}
At least three proofs are possible: * by induction, using Identity Element of Addition on Numbers * using Indexed Summation of Multiple of Mapping * using Indexed Summation of Sum of Mappings {{ProofWanted}} Category:Summations 3drcgu3jy45li9v3a1ivxrq7qiai7y4
At least three proofs are possible: * Using the definition of summation and Indexed Summation of Zero * using Indexed Summation of Sum of Mappings * using Summation of Multiple of Mapping on Finite Set {{ProofWanted}} Category:Summations oqey1cchdp8n9xxz38hcng5j4dierj5
By Support of Zero Mapping, the support of $0$ is empty. By Empty Set is Finite, the support of $0$ is indeed finite. By Summation over Empty Set, $\displaystyle \sum_{s \mathop \in S} \map 0 s = \sum_{s \mathop \in \O} \map 0 s = 0$ {{qed}} Category:Summations 0vnmuzgyu2aw8ll79sxek547d0ijrwv
Define an extension $\overline f$ of $f$ to $S \times T$ by: :$\overline f \left({s, t}\right) = \begin{cases} f \left({s, t}\right) & : \left({s, t}\right) \in D \\ 0  & : \left({s, t}\right) \notin D \end{cases}$ Then for all $s \in S$, by: : Preimage of Disjoint Union is Disjoin Union : Sum over Disjoint Union of Finite Sets : Summation over Finite Set of Zero: :$\displaystyle \sum_{t \mathop \in \pi_2 \left({\pi_1^{-1} \left({s}\right)}\right)} f \left({s, t}\right) = \sum_{t \mathop \in T} \overline f \left({s, t}\right)$ Thus: :$\displaystyle \sum_{s \mathop \in S} \sum_{t \mathop \in \pi_2 \left({\pi_1^{-1} \left({s}\right)}\right)} f \left({s, t}\right) = \sum_{s \mathop \in S} \sum_{t \mathop \in T} \overline f \left({s, t}\right)$ Similarly: :$\displaystyle \sum_{t \mathop \in T} \sum_{s \mathop \in \pi_1 \left({\pi_2^{-1} \left({t}\right)}\right)} f \left({s, t}\right) = \sum_{t \mathop \in T} \sum_{s \mathop \in S} \overline f \left({s, t}\right)$ By Exchange of Order of Summation over Cartesian Product of Finite Sets, the result follows. {{qed}} Category:Summations 1dvrkgudjt280sscptdkqnz44ldqhi3