hcju/proofwikiqa huggingface dataset:
By Form of Geometric Sequence of Integers from One, the general term of $G_n$ can be expressed as: :$a_j = q^j$ for some $q \in \Z$. Let $k \nmid m$. Then by the Division Theorem there exists a unique $q \in \Z$ such that: :$m = k q + b$ for some $b$ such that $0 < b < k$. Thus: :$a_m = a^{k q} a^b$ which is not a power of $k$. {{qed}} {{Euclid Note|10|IX}}
By Form of Geometric Sequence of Integers from One, the general term of $G_n$ can be expressed as: :$a_j = q^j$ for some $q \in \Z$. Hence the result from Divisors of Power of Prime. {{qed}} {{Euclid Note|11|IX}}
By Form of Geometric Sequence of Integers from One, the general term of $G_n$ can be expressed as: :$a_j = q^j$ for some $q \in \Z$. Thus by hypothesis: :$p \divides q^n$ From Euclid's Lemma for Prime Divisors: General Result: :$p \divides q$ Hence the result. {{qed}} {{Euclid Note|12|IX}}
From Form of Geometric Sequence of Integers from One, the elements of $Q_n$ are given by: :$Q_n = \tuple {1, a, a^2, \ldots, a^n}$ From Divisors of Power of Prime, each of $a_j$ for $j \in \set {1, 2, \ldots, n}$ are the only divisors of $a_n$. {{qed}} {{Euclid Note|13|IX}}
Let the common ratio of $P$ in canonical form be $\dfrac q p$. By Form of Geometric Sequence of Integers in Lowest Terms: :$P = \tuple {p^2, p q, q^2}$ Then: {{begin-eqn}} {{eqn | l = p       | o = \perp       | r = q       | c = {{Defof|Canonical Form of Rational Number}} }} {{eqn | ll= \leadsto       | l = q       | o = \perp       | r = \paren {p + q}       | c = Numbers are Coprime iff Sum is Coprime to Both }} {{eqn | ll= \leadsto       | l = q       | o = \perp       | r = p \paren {p + q}       | c = Integer Coprime to all Factors is Coprime to Whole }} {{eqn | ll= \leadsto       | l = q^2       | o = \perp       | r = p \paren {p + q}       | c = Square of Coprime Number is Coprime }} {{eqn | ll= \leadsto       | l = q^2       | o = \perp       | r = p^2 + p q       | c = factorising }} {{eqn | ll= \leadsto       | l = c       | o = \perp       | r = a + b       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | ll= \leadsto       | l = p^2       | o = \perp       | r = p q + q^2       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = \perp       | r = b + c       | c = factorising }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = p + q       | o = \perp       | r = p       | c = Numbers are Coprime iff Sum is Coprime to Both }} {{eqn | lo= \land       | l = p + q       | o = \perp       | r = q       | c = Numbers are Coprime iff Sum is Coprime to Both }} {{eqn | ll= \leadsto       | l = \paren {p + q}^2       | o = \perp       | r = p q       | c = as above }} {{eqn | ll= \leadsto       | l = p^2 + q^2 + 2 p q       | o = \perp       | r = p q       | c = factorising }} {{eqn | ll= \leadsto       | l = p^2 + q^2       | o = \perp       | r = p q       | c =  }} {{eqn | ll= \leadsto       | l = b       | o = \perp       | r = a + c       | c =  }} {{end-eqn}} {{qed}} {{Euclid Note|15|IX}}
Suppose such a $c$ exists. From Coprime Numbers form Fraction in Lowest Terms, $\dfrac a b$ is in canonical form. From Ratios of Fractions in Lowest Terms: :$a \divides b$ where $\divides$ denotes divisibility. This contradicts the fact that $a$ and $b$ are coprime. Hence such a $c$ cannot exist. {{qed}} {{Euclid Note|16|IX}}
{{AimForCont}} there exists $b$ such that $\dfrac {a_0} {a_1} = \dfrac {a_n} b$. Then: :$\dfrac {a_0} {a_n} = \dfrac {a_1} b$ By Ratios of Fractions in Lowest Terms: :$a_0 \divides a_1$ where $\divides$ denotes divisibility. From Divisibility of Elements in Geometric Sequence of Integers: :$a_0 \divides a_n$ But $a_0 \perp a_n$. From this contradiction it follows that there can be no such $b$. {{qed}} {{Euclid Note|17|IX}}
Let $P = \tuple {a, b, c}$ be a geometric sequence. Then by definition their common ratio is: :$\dfrac b a = \dfrac c b$ From Two Coprime Integers have no Third Integer Proportional it cannot be the case that $a$ and $b$ are coprime. Thus condition $(1)$ is satisfied. From Form of Geometric Sequence of Integers, $P$ is in the form: :$\tuple {k p^2, k p q, k q^2}$ from which it can be seen that: :$k p^2 \divides k^2 p^2 q^2$ demonstrating that condition $(2)$ is satisfied. {{qed}} {{Euclid Note|18|IX}}
Let $S = \left\{{r_1, r_2, \ldots, r_n}\right\}$ be a set of $n$ odd numbers, where $n = 2 m$. By definition of odd number, this can be expressed as: :$S = \left\{{2 s_1 + 1, 2 s_2 + 1, \ldots, 2 s_n + 1}\right\}$ where: :$\forall k \in \left[{1 \,.\,.\, n}\right]: r_k = 2 s_k + 1$ Then: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n r_k       | r = \sum_{k \mathop = 1}^{2 m} 2 s_k + 1       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^{2 m} 2 s_k + \sum_{k \mathop = 1}^{2 m} 1       | c =  }} {{eqn | r = 2 \times 2 m \sum_{k \mathop = 1}^{2 m} s_k + 2 m       | c =  }} {{eqn | r = 2 \left({2 m \sum_{k \mathop = 1}^{2 m} s_k + m}\right)       | c =  }} {{end-eqn}} Thus, by definition, $\displaystyle \sum_{k \mathop = 1}^n r_k$ is even. {{qed}} {{Euclid Note|22|IX}}
Let $S = \left\{{r_1, r_2, \ldots, r_n}\right\}$ be a set of $n$ odd numbers, where $n = 2 m + 1$. By definition of odd number, this can be expressed as: :$S = \left\{{2 s_1 + 1, 2 s_2 + 1, \ldots, 2 s_n + 1}\right\}$ where: :$\forall k \in \left[{1 \,.\,.\, n}\right]: r_k = 2 s_k + 1$ Then: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n r_k       | r = \sum_{k \mathop = 1}^{2 m + 1} \left({2 s_k + 1}\right)       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^{2 m} \left({2 s_k + 1}\right) + 2 s_n + 1       | c = extracting the $n$th element from the summation }} {{eqn | r = 2 q + 2 s_n + 1       | c = where $q \in \N$: Sum of Even Number of Odd Numbers is Even }} {{eqn | r = 2 \left({q + s_n}\right) + 1       | c =  }} {{end-eqn}} Thus, by definition, $\displaystyle \sum_{k \mathop = 1}^n r_k$ is odd. {{qed}} {{Euclid Note|23|IX}}
Let $a$ and $b$ be even numbers. Then by definition of even number: :$\exists c \in \Z: a = 2 c$ :$\exists d \in \Z: b = 2 d$ So: {{begin-eqn}} {{eqn | l = a - b       | r = 2 c - 2 d       | c =  }} {{eqn | r = 2 \left({c - d}\right)       | c =  }} {{end-eqn}} Hence the result by definition of even number. {{qed}} {{Euclid Note|24|IX}}
Let $a$ be even and $b$ be odd. Then by definition of even number: :$\exists c \in \Z: a = 2 c$ and by definition of odd number: :$\exists d \in \Z: b = 2 d + 1$ So: {{begin-eqn}} {{eqn | l = a - b       | r = 2 c - \left({2 d + 1}\right)       | c =  }} {{eqn | r = 2 \left({c - d}\right) - 1       | c =  }} {{eqn | r = 2 \left({c - d - 1}\right) + 1       | c =  }} {{end-eqn}} Hence the result by definition of odd number. {{qed}} {{Euclid Note|25|IX}}
Let $a$ and $b$ be odd numbers. Then by definition of odd number: :$\exists c \in \Z: a = 2 c + 1$ :$\exists d \in \Z: b = 2 d + 1$ So: {{begin-eqn}} {{eqn | l = a - b       | r = 2 c + 1 - \left({2 d + 1}\right)       | c =  }} {{eqn | r = 2 \left({c - d}\right)       | c =  }} {{end-eqn}} Hence the result by definition of even number. {{qed}} {{Euclid Note|26|IX}}
Let $a$ be odd and $b$ be even. Then by definition of odd number: :$\exists c \in \Z: a = 2 c + 1$ and by definition of even number: :$\exists d \in \Z: b = 2 d$ So: {{begin-eqn}} {{eqn | l = a - b       | r = 2 c + 1 - 2 d       | c =  }} {{eqn | r = 2 \paren {c - d} + 1       | c =  }} {{end-eqn}} Hence the result by definition of odd number. {{qed}} {{Euclid Note|27|IX}}
Let $a$ be odd and $b$ be even. Then by definition of odd number: :$\exists c \in \Z: a = 2 c + 1$ and by definition of even number: :$\exists d \in \Z: b = 2 d$ So: {{begin-eqn}} {{eqn | l = a \times b       | r = \left({2 c + 1}\right) \times 2 d       | c =  }} {{eqn | r = 2 \left({2 c d + d}\right)       | c =  }} {{end-eqn}} Hence the result by definition of even number. {{qed}} {{Euclid Note|28|IX}}
Let $a$ be odd and $b$ be even. Then by definition of odd number: :$\exists c \in \Z: a = 2 c + 1$ :$\exists d \in \Z: b = 2 d + 1$ So: {{begin-eqn}} {{eqn | l = a \times b       | r = \paren {2 c + 1} \times \paren {2 d + 1}       | c =  }} {{eqn | r = 4 c d + 2 c + 2 d + 1       | c =  }} {{eqn | r = 2 \paren {2 c d + c + d} + 1       | c =  }} {{end-eqn}} Hence the result by definition of odd number. {{qed}} {{Euclid Note|29|IX}}
By definition of an even number: :$\exists r \in \Z: b = 2 r$ By definition of an odd number: :$\exists s \in \Z: a = 2 s + 1$ Thus: :$a \divides 2 \iff a = \pm 1$ in which case from One Divides all Integers: :$a \divides \dfrac b 2$ We have that: :$a \divides 2 r$ and so by Euclid's Lemma: :$a \divides r = \dfrac b 2$ {{qed}} {{Euclid Note|30|IX}}
By definition of odd number: :$a \perp 2$ The result follows from Integer Coprime to all Factors is Coprime to Whole. {{qed}} {{Euclid Note|31|IX}}
As $a$ is a power of $2$ greater than $2$: :$\exists k \in \Z_{>1}: a = 2^k$ Thus: :$a = 2^2 2^{k - 2}$ and so has $2^2 = 4$ as a divisor. Let $b$ be an odd number. By definition: :$b \nmid 2$ The result follows by Integer Coprime to all Factors is Coprime to Whole. {{qed}} {{Euclid Note|32|IX}}
By definition: :$a = 2 r$ where $r$ is an odd integer. Thus: :$a$ has an even divisor and: :$a$ has an odd divisor. Hence the result by definition of even-times odd integer. As $r$ is an odd integer it follows that $2 \nmid r$. Thus $a$ is not divisible by $4$. Hence $a$ is not even-times even. {{qed}} {{Euclid Note|33|IX}}
As $\dfrac a 2$ is an even integer it follows that: :$\dfrac a 2 = 2 r$ for some $r \in \Z$. That is: :$a = 2^2 r$ and so $a$ is even-times even by definition. {{qed}} Suppose $a$ is not even-times odd. Then $a$ does not have an odd divisor. Thus in its prime decomposition there are no odd primes. Thus $a$ is in the form: :$a = 2^k$ for some $k \in \Z_{>0}$. That is, $a$ is a power of $2$. This contradicts condition $(1)$. From this contradiction it is deduced that $a$ is even-times odd. {{qed}} {{Euclid Note|34|IX}}
Let $\displaystyle \lim_{x \mathop \to +\infty} \map f x = +\infty$. From the definition of infinite limit at infinity, this means that: :$\forall M > 0: \exists N > 0: x > N \implies \map f x > M$. Suppose $\lambda > 0$. Then $M > 0 \iff \lambda^{-1} M > 0$. Also, $\map f x > \lambda^{-1} M \iff \lambda f\map f x > M$. So: :$\forall M > 0: \exists N > 0: x > N \implies \lambda \map f x > M$ From the definition of infinite limit at infinity: :$\displaystyle \lim_{x \mathop \to +\infty} \lambda \map f x = +\infty$. The proof for $\displaystyle \lim_{x \mathop \to -\infty} \map f x = +\infty$ is analogous. Now, suppose $\lambda < 0$. Then $-\lambda f > 0$, and so $-\lambda f \to + \infty$, from above. Write $\lambda f = -\paren {-\lambda f}$ and the result follows from Negative of Real Function that Increases Without Bound. {{qed}}
Let $\displaystyle \lim_{x \mathop \to +\infty} \map f x = -\infty$. From the definition of infinite limits at infinity, this means that: :$\forall M < 0: \exists N > 0: x > N \implies \map f x < M$ Suppose $\lambda > 0$. Then $M < 0 \iff \lambda^{-1}M < 0$. Also, $\map f x < \lambda^{-1}M \iff \lambda \map f x < M$ So: :$\forall M < 0: \exists N > 0: x > N \implies \lambda \map f x < M$ From the definition of infinite limits at infinity: :$\displaystyle \lim_{x \mathop \to +\infty} \lambda \map f x = -\infty$ The proof for $\displaystyle  \lim_{x \mathop \to -\infty} \map f x = -\infty$ is analogous. Now, suppose $\lambda < 0$. Then $-\lambda f < 0$, and so $-\lambda f \to -\infty$, from above. Write $\lambda f = -\paren {-\lambda f}$ and the result follows from Negative of Real Function that Decreases Without Bound. {{qed}}
:400px Let $AB$ and $C$ be two unequal magnitudes such that $AB > C$. From {{EuclidDefLink|V|4|Existence of Ratio}} there exists a multiple of $C$ which is greater than $AB$. Thus let $DE$ be a multiple of $C$ which is greater than $AB$. Let $DE$ be divided into parts $DF$, $FG$ and $GE$ (for example) which are equal to $C$. From $AB$ let $BH$ be subtracted which is greater than half $AB$. From $AH$ let $HK$ be subtracted which is greater than half $AH$. Let this process be continued until the divisions of $AB$ are equal in number to the divisions of $DE$. Let $AK, KH, HB$ be divisions which are equal in number to $DF, FG, GE$. We have that $DE > AB$. From $DE$ there has been subtracted $EG$ which is less than half $DE$. From $AB$ there has been subtracted $BH$ which is greater than half $AB$. Therefore the remainder $GD$ of $DE$ is greater than the remainder $HA$ of $AB$. From $GD$ there has been subtracted $GF$ which is half $GD$. From $HA$ there has been subtracted $HK$ which is greater than half $HA$. Therefore the remainder $DF$ of $GD$ is greater than the remainder $AK$ of $HA$. But $DF = C$. Therefore $C > AK$. Therefore there is left of the magnitude $AB$ the magnitude $AK$ which is less than the lesser magnitude set out, which is $C$. {{qed}} {{Euclid Note|1|X}}
:500px Let $AB$ and $CD$ be unequal magnitudes such that $AB < CD$ which fulfil the condition of the statement. Suppose $AB$ and $CD$ are commensurable magnitudes. Then by definition some magnitude will measure them both. Let $E$ be such a magnitude that measures both $AB$ and $CD$. Let $AB$ measure $FD$ and leave $CF$ from $CD$. Let $CF$ measure $BG$ and leave $AG$ from $AB$. By hypothesis, this process can be repeated indefinitely. Let it be repeated until some magnitude be left that is less than $E$. Since: : $E$ measures $AB$ and: : $AB$ measures $DF$ then: : $E$ measures $FD$. But $E$ also measures $CD$. Therefore $E$ also measures $CF$. But $CF$ measures $BG$. Therefore $E$ also measures $BG$. But $E$ also measures the whole of $AB$. Therefore $E$ will also measures $AG$. But $AG$ is less than $E$. That means the greater magnitude measures the lesser magnitude. From this contradiction it follows that no magnitude can measure both $AB$ and $CD$. Therefore, by definition, $AB$ and $CD$ are incommensurable. {{qed}} {{Euclid Note|2|X}}
Let $AB$ and $CD$ be two commensurable magnitudes such that $AB < CD$. $AB$ either measures $CD$ or it does not. Let $AB$ measure $CD$. As $AB$ also measures itself, it follows that $AB$ is a common measure for $AB$ and $CD$. As no magnitude can measure a smaller magnitude, it follows that $AB$ is the greatest common measure of $AB$ and $CD$. :500px Let $AB$ not measure $CD$. Let the less be continually subtracted in turn from the greater. We are given that $AB$ and $CD$ are commensurable. From Incommensurable Magnitudes do not Terminate in Euclid's Algorithm, that which is left over will eventually measure the one before it. Let $AB$ measure $ED$ and leave $EC$ less than $AB$. Let $EC$ measure $FB$ and leave $AF$ less than $EC$. Let $AF$ measure $CE$. Since: : $AF$ measure $CE$ and: : $CE$ measure $FB$ it follows that: : $AF$ measure $FB$. But $AF$ also measures itself. Therefore $AF$ measures the whole of $AB$. But $AB$ also measures $DE$. Therefore $AF$ also measures $DE$. But $AB$ also measures $CE$. Therefore $AF$ also measures the whole of $CD$. Therefore $AF$ is a common measure for $AB$ and $CD$. Suppose there were some magnitude $G$ which is greater than $AF$ which is also a common measure for $AB$ and $CD$. Since: :$G$ measures $AB$ and: :$AB$ measures $ED$ it follows that: :$G$ measures $ED$. But $G$ also measures the whole of $CD$. Therefore $G$ also measures the remainder $CE$. But $CE$ measures $FB$. Therefore $G$ also measures $FB$. But $G$ also measures the whole of $AB$. Therefore it also measures the remainder $AF$. But $G$ is greater than $AF$, and so cannot measure it. From this contradiction it follows that $G$ cannot be a common measure for $AB$ and $CD$. That is, $AF$ is the greatest common measure of $AB$ and $CD$. {{qed}} {{Euclid Note|3|X}}
Let $V \subseteq M_2$ be an open set with respect to the subspace topology of $M_2$. By definition of the subspace topology, $V = U \cap M_2$ for an open set $U \in \tau_2$. We have that: {{begin-eqn}} {{eqn | l = \left({f {\restriction_{M_1 \times M_2} } }\right)^{-1} \left[{V}\right]       | r = \left\{ {x \in M_1 : f {\restriction_{M_1 \times M_2} } \left({x}\right) \in V} \right\}       | c = Definition of Preimage }} {{eqn | r = \left\{ {x \in M_1 : f {\restriction_{M_1 \times M_2} } \left({x}\right) \in U}\right\}       | c = $f \left({M_1}\right) \subseteq M_2$ }} {{eqn | r = \left\{ {x \in M_1 : f \left({x}\right) \in U}\right\}       | c = Definition of Restriction of Mapping }} {{eqn | r = \left\{ {x \in S_1 : f \left({x}\right) \in U}\right\} \cap M_1       | c = Definition of Set Intersection }} {{eqn | r = f^{-1} \left[{U}\right] \cap M_1       | c = Definition of Preimage }} {{end-eqn}} By definition of continuous mapping, $f^{-1} \left[{U}\right]$ is open in $T_1$. By definition of subspace topology, $f^{-1} \left[{U}\right] \cap M_1$ is open with respect to the subspace topology of $M_1$. As $f^{-1} \left[{U}\right] \cap M_1 = {f {\restriction_{M_1 \times M_2}}}^{-1} \left[{V}\right]$ it follows that ${f {\restriction_{M_1 \times M_2}}}^{-1} \left[{V}\right]$ is open with respect to the subspace topology of $M_1$. Since $V \subseteq M_2$ was an arbitrary open set, $f$ is continuous. {{qed}} Category:Topology Category:Continuous Mappings s06ioeumwuwzpcvqj6g9ux7ciieh01e
Let $\sequence {z_n}$ be a sequence in $S$ such that: :$\displaystyle \lim_{n \mathop \to \infty} z_n = \alpha$ Since $\sequence {z_n}$ and $\alpha$ both lie in $S$: :$\displaystyle \lim_{n \mathop \to \infty} \map f {z_n} \to \alpha$ But: :$\forall n \in \N: \map g {z_n} = \map f {z_n}$ and also: :$\map g \alpha = \map f \alpha$ The result follows from Limit of Function by Convergent Sequences. {{qed}} Category:Metric Spaces Category:Continuous Mappings on Metric Spaces dznt78ygvoyxfoho264sgdkxf28je6l
Let $y \in f \left[{S}\right]$. By the definition of image: :$\exists z \in S : f \left({z}\right) = y$ Since $f$ is a bijection, $f^{-1}$ is a mapping. Let $x = f^{-1} \left({y}\right)$. Suppose that $x \notin S$. Suppose that $x = z$. Then $x \in S$. Thus $x \ne z$. By the definition of inverse mapping: : $f\left({x}\right) = y$ By Equality is Transitive: : $f \left({x}\right) = f \left({z}\right)$ Thus, $f$ is not an injection. Thus, $f$ is not a bijection. This contradicts the assumption. Therefore, $x \in S$. By the definition of restriction of mapping: : $f {\restriction_{S \times f \left[{S}\right]}} \left({x}\right) = f\left({x}\right)$ By Equality is Transitive: : $f {\restriction_{S \times f \left[{S}\right]}} \left({x}\right) = y$ Suppose that there exists a $w \in S$ such that $w \ne x$ and $f {\restriction_{S \times f \left[{S}\right]}} \left({w}\right) = y$. By the definition of restriction of mapping: : $f {\restriction_{S \times f \left[{S}\right]}} \left({w}\right) = f\left({w}\right)$ By Equality is Transitive: : $f \left({w}\right) = y$ By Equality is Transitive: : $f \left({w}\right) = f\left({x}\right)$ Thus, $f$ is not an injection. Thus, $f$ is not a bijection. This contradicts the assumption. Therefore, there exists a unique $x \in S$ such that: : $f {\restriction_{S \times f\left[{S}\right]}} \left({x}\right) = y$ Since $y \in f\left[{S}\right]$ was arbitrary, $f {\restriction_{S \times f \left[{S}\right]}}$ is a bijection. Thus, ${\left(f {\restriction_{S \times f \left[{S}\right]}}\right)}^{-1}$ is a mapping. Let $x' = {\left(f {\restriction_{S \times f \left[{S}\right]}}\right)}^{-1}\left({y}\right)$. By the definition of inverse mapping: : $f {\restriction_{S \times f \left[{S}\right]}} \left({x'}\right) = y$ By the definition of restriction of mapping, we have $f \left({x'}\right) = y$. Suppose that $x' \ne x$. By Equality is Transitive: : $f\left({x'}\right) = f\left({x}\right)$ Thus, $f$ is not an injection. Thus, $f$ is not a bijection. This contradicts the assumption. Therefore, $x' = x$. By Equality is Transitive: : $f^{-1}\left({y}\right) = (f {\restriction_{S \times f \left[{S}\right]}})^{-1} \left({y}\right)$ By the definition of restriction of mapping: : $f^{-1} {\restriction_{f \left[{S}\right] \times S}} \left({y}\right) = f^{-1} \left({y}\right)$ By Equality is Transitive: : ${\left(f {\restriction_{S \times f \left[{S}\right]}}\right)}^{-1} \left({y}\right) = f^{-1} {\restriction_{f\left[{S}\right] \times S}} \left({y}\right)$ Since $y \in f\left[{S}\right]$ was arbitrary: : ${\left(f {\restriction_{S \times f \left[{S}\right]}}\right)}^{-1} = f^{-1} {\restriction_{f \left[{S}\right] \times S}}$ {{qed}} Category:Restrictions Category:Inverse Mappings ge779b0r8bkxb2vrgidb7cpyz5v4qid
By Restriction of Continuous Mapping is Continuous, $f {\restriction_{S \times f \left[{S}\right]}}$ is continuous. By Restriction of Inverse is Inverse of Restriction, $\left({f {\restriction_{S \times f \left[{S}\right]}}}\right)^{-1}$ is well-defined and equal to $f^{-1} {\restriction_{f \left[{S}\right] \times S}}$. By Restriction of Continuous Mapping is Continuous, $f^{-1} {\restriction_{f \left[{S}\right] \times S}}$ is continuous. Since $\left({f {\restriction_{S \times f \left[{S}\right]}}}\right)^{-1} = f^{-1} {\restriction_{f \left[{S}\right] \times S}}$, $\left({f {\restriction_{S \times f \left[{S}\right]}}}\right)^{-1}$ is continuous. Hence, $f {\restriction_{S \times f \left[{S}\right]}}$ is a homeomorphism. {{qed}} Category:Homeomorphisms 99a6bp3gqgbkt7cucgrah6bato7al5k
{{explain|Has it been demonstrated for $i {{=}} 1$? Best if this were put into the house style of induction proofs.}} We use the Principle of Mathematical Induction. Let it be assumed that the result holds for $i = n - 1$. We show that the result holds for $i = n$.  Define: :$q_n := \dfrac {p_n} {p_n-1}$ and for $i = 1, \ldots, n - 1$, define: :$r_i := p_i \cdot \paren {1 - \dfrac 1 {p_n} }$ Then: :$\dfrac 1 {p_n} + \dfrac 1 {q_n} = 1$ :$\displaystyle \sum_{i \mathop = 1}^{n-1} \dfrac 1 {r_i} = 1$ and: :$q_n \cdot r_i = p_i$ Applying Hölder's Inequality to $\displaystyle f := \prod_{i \mathop = 1}^{n-1} f_i$ and $g := f_n$, we find: {{begin-eqn}} {{eqn | l = \int \size {\prod_{i \mathop = 1}^n f_i} \rd \mu       | o = \le       | r = \norm {f_n}_{p_n} \cdot \norm {\prod_{i \mathop = 1}^{n - 1} f_i}_{q_n}       | c =  }} {{eqn | r = \norm {f_n}_{p_n} \cdot \paren {\int \prod_{i \mathop = 1}^{n - 1} \size { {f_i}^{q_n} } \rd \mu}^\frac 1 {q_n}       | c =  }} {{eqn | o = \le       | r = \norm {f_n}_{p_n} \cdot \paren {\prod_{i \mathop = 1}^{n-1} \norm { {f_i}^{q_n} }_{r_i} }^\frac 1 {q_n}       | c =  }} {{eqn | r = \prod_{i \mathop = 1}^n \norm {f_i}_{p_i}       | c =  }} {{end-eqn}} {{qed}} {{Namedfor|Otto Ludwig Hölder}} Category:Lebesgue Spaces pgq1ogkfb0jnvkjt0g7kkhxhtwzcmtz
Suppose $2^n - 1$ is prime. Let $a = 2^{n - 1} \paren {2^n - 1}$. Then $n \ge 2$ which means $2^{n - 1}$ is even and hence so is $a = 2^{n - 1} \paren {2^n - 1}$. Note that $2^n - 1$ is odd. Since all divisors (except $1$) of $2^{n - 1}$ are even it follows that $2^{n - 1}$ and $2^n - 1$ are coprime. Let $\map \sigma n$ be the sigma function of $n$, that is, the sum of all divisors of $n$ (including $n$). From Sigma Function is Multiplicative, it follows that $\map \sigma a = \map \sigma {2^{n - 1} } \map \sigma {2^n - 1}$. But as $2^n - 1$ is prime, $\map \sigma {2^n - 1} = 2^n$ from Sigma Function of Prime Number. Then we have that $\map \sigma {2^{n - 1} } = 2^n - 1$ from Sigma Function of Power of Prime. Hence it follows that $\map \sigma a = \paren {2^n - 1} 2^n = 2 a$. Hence from the definition of perfect number it follows that $2^{n - 1} \paren {2^n - 1}$ is perfect.