hcju/proofwikips huggingface dataset:
Carmichael Number/Examples/1729
== Example of Carmichael Number== $1729$ is a Carmichael number:  :$\forall a \in \Z: a \perp 1729: a^{1729} \equiv a \pmod {1729}$ while $1729$ is composite.  == Proof == We have that: :$1729 = 7 \times 13 \times 19$ and so: {{begin-eqn}} {{eqn | l = 7^2       | o = \nmid       | r = 1729       | c =  }} {{eqn | l = 13^2       | o = \nmid       | r = 1729       | c =  }} {{eqn | l = 19^2       | o = \nmid       | r = 1729       | c =  }} {{end-eqn}} We also have that: {{begin-eqn}} {{eqn | l = 1728       | r = 288 \times 6       | c =  }} {{eqn | r = 144 \times 12       | c =  }} {{eqn | r = 96 \times 18       | c =  }} {{end-eqn}} The result follows by Korselt's Theorem.  == Sources == * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 509,033,161|next = Carmichael Number/Examples/294,409}}: $509,033,161$
Highly Composite Number/Examples/1
== Example of Highly Composite Number ==  $1$ is a highly composite number, being the smallest positive integer with $1$ divisor or more.  == Proof == From Tau Function of 1: :$\map \tau 1 = 1$ The positive integer $1$ has $1$ divisor, that is, $1$ itself. Vacuously, no smaller positive integer has a greater number of divisors. Thus, despite not actually being composite, $1$ is a highly composite number.
Special Highly Composite Number/Examples/1
Special Highly Composite Number/Examples/1 0 61038 312830 2017-08-23T21:58:22Z Prime.mover 59 Created page with "== Example of Special Highly Composite Number ==  $1$ is a Definition:Special Highly Composite Number|special hig..." wikitext text/x-wiki == Example of Special Highly Composite Number ==  $1$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.  == Proof == We have that $1$ is highly composite. From One Divides all Integers, it follows trivially that $1$ is a divisor of all larger highly composite numbers. Thus, by definition, $1$ is a special highly composite number.  == Sources == * {{Citation|title = An Interesting Subset of the Highly Composite Numbers|author = Steven Ratering|journal = Mathematics Magazine|abbr = Math. Mag.|volume = 64|issue = 5|date = Dec. 1991|startpage = 343|endpage = 346|jstor = 2690653}}
Highly Composite Number/Examples/2
== Example of Highly Composite Number ==  $2$ is a highly composite number, being the smallest positive integer with $2$ divisors or more.  == Proof == From Tau of Prime Number: :$\tau \left({2}\right) = 2$ From Tau Function of 1: :$\tau \left({1}\right) = 1$ That is, the only positive integer smaller than $2$ has a smaller number of divisors. Thus, despite not actually being composite, $2$ is a highly composite number.
Special Highly Composite Number/Examples/2
== Example of Special Highly Composite Number ==  $2$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.  == Proof == We have that $2$ is highly composite. Let $n > 2$ be a highly composite number. From Prime Decomposition of Highly Composite Number, the multiplicity of $2$ in $n$ is at least as high as the multiplicity of any other prime $p$ in $n$. Thus if $p \divides n$ it follows that $2 \divides n$. Thus, by definition, $2$ is a special highly composite number.  == Sources == * {{Citation|title = An Interesting Subset of the Highly Composite Numbers|author = Steven Ratering|journal = Mathematics Magazine|abbr = Math. Mag.|volume = 64|issue = 5|date = Dec. 1991|startpage = 343|endpage = 346|jstor = 2690653}}
Special Highly Composite Number/Examples/6
== Example of Special Highly Composite Number ==  $6$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.  == Proof == By inspection of the sequence of highly composite numbers, $6$ is highly composite.  $n > 6$ is a highly composite number which is not divisible by $6$. We have that $2$ is a special highly composite number. Therefore $2$ is a divisor of $n$. As $6$ is not a divisor of $n$, it follows that $3$ is also not a divisor of $n$. By Prime Decomposition of Highly Composite Number, that means $n = 2^k$ for some $k \ge 3$. Then: {{begin-eqn}} {{eqn | l = 2^{k - 2} \times 3       | o = <       | r = 2^k       | c = because $3 < 2^2 = 4$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^{k - 2} \times 3}\right)       | o = <       | r = \tau \left({2^k}\right)       | c = as $2^k$ is highly composite }} {{eqn | ll= \leadsto       | l = \left({k - 1}\right) \left({1 + 1}\right)       | o = <       | r = k + 1       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = k       | o = <       | r = 3       | c = after algebra }} {{end-eqn}} But this contradicts our deduction that $n = 2^k$ where $k \ge 3$. The result follows by Proof by Contradiction.  == Sources == * {{Citation|title = An Interesting Subset of the Highly Composite Numbers|author = Steven Ratering|journal = Mathematics Magazine|abbr = Math. Mag.|volume = 64|issue = 5|date = Dec. 1991|startpage = 343|endpage = 346|jstor = 2690653}}
Special Highly Composite Number/Examples/12
== Example of Special Highly Composite Number ==  $12$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.  == Proof == By inspection of the sequence of highly composite numbers, $12$ is highly composite.  $n > 12$ is a highly composite number which is not divisible by $12$. We have that $6$ is a special highly composite number. Therefore $6$ is a divisor of $n$. As $12$ is not a divisor of $n$, it follows that the multiplicity of $2$ in $n$ is $1$. From Prime Decomposition of Highly Composite Number, that means: : $n = 2 \times 3 \times 5 \times r$ where $r$ is a possibly vacuous square-free product of prime numbers strictly greater than $5$. Then: {{begin-eqn}} {{eqn | l = 2^3 \times 3 \times r       | o = <       | r = 2 \times 3 \times 5 \times r       | c = because $4 = 2^2 < 5$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^3 \times 3 \times r}\right)       | o = <       | r = \tau \left({2 \times 3 \times 5 \times r}\right)       | c = as $2 \times 3 \times 5 \times r$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^3 \times 3}\right) \times \tau \left({r}\right)       | o = <       | r = \tau \left({2 \times 3 \times 5}\right) \times \tau \left({r}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^3 \times 3}\right)       | o = <       | r = \tau \left({2 \times 3 \times 5}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({3 + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({1 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 8       | o = <       | r = 8       | c = which is a falsehood }} {{end-eqn}} The result follows by Proof by Contradiction.  == Sources == * {{Citation|title = An Interesting Subset of the Highly Composite Numbers|author = Steven Ratering|journal = Mathematics Magazine|abbr = Math. Mag.|volume = 64|issue = 5|date = Dec. 1991|startpage = 343|endpage = 346|jstor = 2690653}}
Sigma Function of 2620
== Example of Sigma Function of Integer ==  :$\sigma \left({2620}\right) = 5544$  where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$2620 = 2^2 \times 5 \times 131$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({2620}\right)       | r = \frac {2^3 - 1} {2 - 1} \times \left({5 + 1}\right) \times \left({131 + 1}\right)       | c =  }} {{eqn | r = \frac 7 1 \times 6 \times 132       | c =  }} {{eqn | r = 7 \times \left({2 \times 3}\right) \times \left({2^2 \times 3 \times 11}\right)       | c =  }} {{eqn | r = 2^3 \times 3^2 \times 7 \times 11       | c =  }} {{eqn | r = 5544       | c =  }} {{end-eqn}}
Sigma Function of 2924
== Example of Sigma Function of Integer ==  :$\sigma \left({2924}\right) = 5544$  where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$2924 = 2^2 \times 17 \times 43$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({2924}\right)       | r = \frac {2^3 - 1} {2 - 1} \times \left({17 + 1}\right) \times \left({43 + 1}\right)       | c =  }} {{eqn | r = \frac 7 1 \times 18 \times 44       | c =  }} {{eqn | r = 7 \times \left({2 \times 3^2}\right) \times \left({2^2 \times 11}\right)       | c =  }} {{eqn | r = 2^3 \times 3^2 \times 7 \times 11       | c =  }} {{eqn | r = 5544       | c =  }} {{end-eqn}}
Sigma Function of 5020
== Example of Sigma Function of Integer ==  :$\sigma \left({5020}\right) = 10 \, 584$  where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$5020 = 2^2 \times 5 \times 251$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({5020}\right)       | r = \frac {2^3 - 1} {2 - 1} \times \left({5 + 1}\right) \times \left({251 + 1}\right)       | c =  }} {{eqn | r = \frac 7 1 \times 6 \times 252       | c =  }} {{eqn | r = 7 \times \left({2 \times 3}\right) \times \left({2^2 \times 3^2 \times 7}\right)       | c =  }} {{eqn | r = 2^3 \times 3^3 \times 7^2       | c =  }} {{eqn | r = 10 \, 584       | c =  }} {{end-eqn}}
Sigma Function of 5564
== Example of Sigma Function of Integer ==  :$\map \sigma {5564} = 10 \, 584$  where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\map \sigma n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$5564 = 2^2 \times 13 \times 107$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {5020}       | r = \frac {2^3 - 1} {2 - 1} \times \paren {13 + 1} \times \paren {107 + 1}       | c =  }} {{eqn | r = \frac 7 1 \times 14 \times 108       | c =  }} {{eqn | r = 7 \times \paren {2 \times 7} \times \paren {2^2 \times 3^3}       | c =  }} {{eqn | r = 2^3 \times 3^3 \times 7^2       | c =  }} {{eqn | r = 10 \, 584       | c =  }} {{end-eqn}}
Euler Phi Function of 5186
Euler Phi Function of 5186 0 61432 313964 2017-08-29T07:04:25Z Prime.mover 59 Created page with "== Example of Euler Phi Function of 2 times Odd Prime ==  :$\phi \left({5186}\right) = 2592$  where $\phi$ denotes the Definition:Euler Phi Fu..." wikitext text/x-wiki == Example of Euler Phi Function of 2 times Odd Prime ==  :$\phi \left({5186}\right) = 2592$  where $\phi$ denotes the Euler $\phi$ Function. == Proof == We have that: :$5186 = 2 \times 2593$ Thus: {{begin-eqn}} {{eqn | l = \phi \left({5186}\right)       | r = 2593 - 1       | c = Euler Phi Function of 2 times Odd Prime }} {{eqn | r = 2592       | c =  }} {{end-eqn}}
Euler Phi Function of 5187
Euler Phi Function of 5187 0 61433 313968 2017-08-29T07:09:59Z Prime.mover 59 Created page with "== Example of Euler $\phi$ Function of Square-Free Integer ==  :$\phi \left({5187}\right) = 2592$  whe..." wikitext text/x-wiki == Example of Euler $\phi$ Function of Square-Free Integer ==  :$\phi \left({5187}\right) = 2592$  where $\phi$ denotes the Euler $\phi$ Function. == Proof == From Euler Phi Function of Square-Free Integer: :$\displaystyle \phi \left({n}\right) = \prod_{\substack {p \mathop \backslash n \\ p \mathop > 2} } \left({p - 1}\right)$ where $p \mathop \backslash n$ denotes the primes which divide $n$. We have that: :$5187 = 3 \times 7 \times 13 \times 19$ and so is square-free. Thus: {{begin-eqn}} {{eqn | l = \phi \left({5187}\right)       | r = \left({2 - 1}\right) \left({7 - 1}\right) \left({13 - 1}\right) \left({19 - 1}\right)       | c =  }} {{eqn | r = 2 \times 6 \times 12 \times 18       | c =  }} {{eqn | r = 2 \times \left({2 \times 3}\right) \times \left({2^2 \times 3}\right) \times \left({2 \times 3^2}\right)       | c =  }} {{eqn | r = 2^5 \times 3^4       | c =  }} {{eqn | r = 2592       | c =  }} {{end-eqn}}
Euler Phi Function of 5188
Euler Phi Function of 5188 0 61434 313969 2017-08-29T07:12:17Z Prime.mover 59 Created page with "== Example of Use of Euler $\phi$ Function ==  :$\phi \left({5188}\right) = 2592$  where $\phi$ denotes the Defi..." wikitext text/x-wiki == Example of Use of Euler $\phi$ Function ==  :$\phi \left({5188}\right) = 2592$  where $\phi$ denotes the Euler $\phi$ Function. == Proof == From Euler Phi Function of Integer: :$\displaystyle \phi \left({n}\right) = n \prod_{p \mathop \backslash n} \left({1 - \frac 1 p}\right)$ where $p \mathop \backslash n$ denotes the primes which divide $n$. We have that: :$5188 = 2^2 \times 1297$ Thus: {{begin-eqn}} {{eqn | l = \phi \left({5188}\right)       | r = 5188 \left({1 - \dfrac 1 2}\right) \left({1 - \dfrac 1 {1297} }\right)       | c =  }} {{eqn | r = 5188 \times \frac 1 2 \times \frac {1296} {1297}       | c =  }} {{eqn | r = 2 \times 1 \times 1296       | c =  }} {{eqn | r = 2592       | c =  }} {{end-eqn}}
Hurwitz's Theorem (Number Theory)/Lemma 1
== Lemma ==  Let $\xi$ be an irrational number. Let $A \in \R$ be a real number strictly greater than $\sqrt 5$. Then there may exist at most a finite number of relatively prime integers $p, q \in \Z$ such that: :$\size {\xi - \dfrac p q} < \dfrac 1 {A \, q^2}$  == Proof == We will take as our example of such an irrational number: :$\xi = \dfrac {\sqrt 5 - 1} 2$ This is equal to $1 - \phi$, where $\phi$ is the Golden mean.  that there exist an infinite number of $p, q$ with $p \perp q$ such that: :$\size {\xi - \dfrac p q} < \dfrac 1 {A \, q^2}$ Then there exist an infinite number of $p, q$ with $p \perp q$ such that: :$\xi = \dfrac p q + \dfrac \delta {q^2}$ where: :$\size \delta < \dfrac 1 A < \dfrac 1 {\sqrt 5}$ Hence: {{begin-eqn}} {{eqn | l = \dfrac \delta q       | r = q \xi - p       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac \delta q - \dfrac {q \sqrt 5} 2       | r = -\dfrac q 2 - p       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = \dfrac {\delta^2} {q^2} - \delta \sqrt 5       | r = \paren {\dfrac q 2 - p}^2 - \dfrac {5 q^2} 4       | c =  }} {{eqn | r = p^2 + p q - q^2       | c =  }} {{end-eqn}} When $q$ is large, the  of $(1)$ becomes less than $1$. At the same time, the  is always an integer. Thus: :$p^2  + p q - q^2 = 0$ or: :$\paren {2 p + q}^2 = 5 q^2$ which would lead to: :$p = 2 q$ which contradicts the stipulation that $p$ and $q$ are coprime. Hence by Proof by Contradiction there cannot be an infinite number of such $p, q$. Hence the result.  {{Namedfor|Adolf Hurwitz}} == Sources == * {{BookReference|An Introduction to the Theory of Numbers|1979|G.H. Hardy|author2 = E.M. Wright|ed = 5th|edpage = Fifth Edition}}: $11.8$: The measure of the closest approximation to an arbitrary irrational: Theorem $194$
Hurwitz's Theorem (Number Theory)/Lemma 2
Hurwitz's Theorem (Number Theory)/Lemma 2 0 61450 314015 2017-08-29T21:47:57Z Prime.mover 59 Created page with "{{MissingLinks|Much of the notation used below needs to be explained in its context.}} == Lemma ==  Let $\xi$ be an Definition:Irrational Number|irrational numb..." wikitext text/x-wiki {{MissingLinks|Much of the notation used below needs to be explained in its context.}} == Lemma ==  Let $\xi$ be an irrational number. Let there be $3$ consecutive convergents of the continued fraction to $\xi$. Then at least one of them, $\dfrac p q$, say, satisfies: :$\left|{\xi - \dfrac p q}\right| < \dfrac 1 {A \, q^2}$  == Proof == Let $\dfrac {p_k} {q_k}$ be an arbitrary convergent to $\xi$. Let: :$\dfrac {q_{n - 1} } {q_n} = b_{n + 1}$ Then: {{begin-eqn}} {{eqn | l = \left\lvert{\dfrac {p_n} {q_n} - \xi}\right\rvert       | r = \dfrac 1 {q_n q'_{n + 1} }       | c =  }} {{eqn | r = \dfrac 1 { {q_n}^2} \dfrac 1 {a'_{n + 1} + b_{n + 1} }       | c =  }} {{end-eqn}} It is sufficient to prove that: :$(1): \quad a'_i + b_i \le 5$ cannot be true for all of $n - 1$, $n$ and $n + 1$ of $i$. Suppose $(1)$ is true for $i = n - 1$ and $i = n$. Then: :$a'_{n - 1} = a_{n - 1} + \dfrac 1 {a'_n}$ and: :$\dfrac 1 {b_n} = \dfrac {q_{n - 1} } {q_{n - 2} } = a_{n - 1} + b_{n - 1}$ Hence: :$\dfrac 1 {a'_n} + \dfrac 1 {b_n} = a'_{n - 1} + b_{n - 1} \le \sqrt 5$ and: :$1 = a'_n \dfrac 1 {a'_n} \le \left({\sqrt 5 - b_n}\right) \left({\sqrt 5 - \dfrac 1 {b_n} }\right)$ or: :$b_n + \dfrac 1 {b_n} \le \sqrt 5$ As $b_n$ is rational, the equality cannot happen. We also have that $b_n < 1$. Thus: :${b_n}^2 -b_n \sqrt 5 + 1 < 0$ :$\left({\dfrac {\sqrt 5} 2 - b_n}\right)^2 < \dfrac 1 4$ and so: :$b_n > \dfrac {\sqrt 5 - 1} 2$ If $(1)$ were also true for $i = n + 1$, it could be proved similarly that: :$b_{n + 1} > \dfrac {\sqrt 5 - 1} 2$ and we would then be able to substitute $n + 1$ for $n$ in the above equations, to get: :$a_n = \dfrac 1 {b_{n + 1} } - b_n < \dfrac {\sqrt 5 + 1} 2 - \dfrac {\sqrt 5 - 1} 2 = 1$ from which a contradiction is apparent.  == Sources == * {{BookReference|An Introduction to the Theory of Numbers|1979|G.H. Hardy|author2 = E.M. Wright|ed = 5th|edpage = Fifth Edition}}: $11.8$: The measure of the closest approximation to an arbitrary irrational: Theorem $195$
Sigma Function of 6232
== Example of Sigma Function of Integer ==  :$\sigma \left({6232}\right) = 12 \, 600$  where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$6232 = 2^3 \times 19 \times 41$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({6232}\right)       | r = \frac {2^4 - 1} {2 - 1} \times \left({19 + 1}\right) \times \left({41 + 1}\right)       | c =  }} {{eqn | r = \frac {15} 1 \times 20 \times 42       | c =  }} {{eqn | r = \left({3 \times 5}\right) \times \left({2^2 \times 5}\right) \times \left({2 \times 3 \times 7}\right)       | c =  }} {{eqn | r = 2^3 \times 3^2 \times 5^2 \times 7       | c =  }} {{eqn | r = 12 \, 600       | c =  }} {{end-eqn}}
Sigma Function of 6368
== Example of Sigma Function of Integer ==  :$\sigma \left({6368}\right) = 12 \, 600$  where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$6368 = 2^5 \times 199$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({6368}\right)       | r = \frac {2^6 - 1} {2 - 1} \times \left({199 + 1}\right)       | c =  }} {{eqn | r = \frac {63} 1 \times 200       | c =  }} {{eqn | r = \left({3^2 \times 7}\right) \times \left({2^3 \times 5^2}\right)       | c =  }} {{eqn | r = 2^3 \times 3^2 \times 5^2 \times 7       | c =  }} {{eqn | r = 12 \, 600       | c =  }} {{end-eqn}}
Completion Theorem (Measure Spaces)/Lemma
== Lemma == Let $\left({X, \Sigma, \mu}\right)$ be a measure space. Let $\mathcal N$ and $\Sigma^*$ be defined as: :$\mathcal N := \left\{{N \subseteq X: \exists M \in \Sigma: \mu \left({M}\right) = 0, N \subseteq M}\right\}$ :$\Sigma^* := \left\{{E \cup N: E \in \Sigma, N \in \mathcal N}\right\}$ Next, define $\bar \mu: \Sigma^* \to \overline{\R}_{\ge 0}$ by: :$\bar \mu \left({E \cup N}\right) := \mu \left({E}\right)$  The mapping $\bar \mu$ is well-defined, i.e.: :$\forall E, F \in \Sigma: \forall N, M \in \mathcal N: E \cup N = F \cup M \implies \mu \left({E}\right) = \mu \left({F}\right)$  == Proof == Let $N_0, M_0 \in \Sigma$ be null sets such that $N \subseteq N_0, M \subseteq M_0$. Then: :$E \subseteq E \cup N = F \cup M \subseteq F \cup M_0$ so that: :$\mu \left({E}\right) \le \mu \left({F \cup M_0}\right) \le \mu \left({F}\right) + \mu \left({M_0}\right) = \mu \left({F}\right) + 0$ Analogously: :$F \subseteq F \cup M = E \cup N \subseteq E \cup N_0$ so that: :$\mu \left({F}\right) \le \mu \left({E \cup N_0}\right) \le \mu \left({E}\right) + \mu \left({N_0}\right) = \mu \left({E}\right) + 0$ In total: :$\mu \left({E}\right) = \mu \left({F}\right)$
Language of Propositional Logic has Unique Parsability/Lemma
== Lemma == Let $\mathcal L_0$ be the language of propositional logic.  Let $\mathbf A$ be a WFF. Suppose that $\mathbf A = \left({B \circ C}\right) = \left({D * E}\right)$. Then $\mathbf B = \mathbf D$, ${\circ} = {*}$, and $\mathbf C = \mathbf E$.  == Proof == The WFFs $\mathbf B$ and $\mathbf D$ are strings which both start in the same place, right after the first left bracket in $\mathbf A$. By Initial Part of WFF of PropLog is not WFF, neither $\mathbf B$ nor $\mathbf D$ can be an initial part of the other. Therefore $\mathbf B = \mathbf D$. It follows that $* = \circ$ and $\mathbf C = \mathbf E$. Hence the result.  == Sources == * {{BookReference|Mathematical Logic and Computability|1996|H. Jerome Keisler|author2 = Joel Robbin|prev = Definition:Main Connective/Propositional Logic/Definition 2|next = Definition:Abbreviation of WFFs of Propositional Logic}}: $\S 1.4$: Main Connective: Theorem $1.4.2$
Finished Branch Lemma/Corollary
Finished Branch Lemma/Corollary 0 63153 319762 2017-10-04T20:59:11Z Lord Farin 560 Created page with "== Corollary to Finished Branch Lemma ==  Let $\Gamma$ be a finished branch of a Definition:Proposit..." wikitext text/x-wiki == Corollary to Finished Branch Lemma ==  Let $\Gamma$ be a finished branch of a propositional tableau $\left({T, \mathbf H, \Phi}\right)$. Then $\Phi \left[{\Gamma}\right]$, the image of $\Gamma$ under $\Phi$, is satisfiable for boolean interpretations.  == Proof == {{Finish|Branch not contradictory. See Finished Set Lemma}}
Pythagorean Triangle/Examples/4485-5852-7373
== Example of Primitive Pythagorean Triangle ==  The triangle whose sides are of length $4485$, $5852$ and $7373$ is a primitive Pythagorean triangle.  :700px It has generator $\tuple {77, 38}$. == Proof == We have: {{begin-eqn}} {{eqn | l = 77^2 - 38^2       | r = 5929 - 1444       | c =  }} {{eqn | r = 4485       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 2 \times 77 \times 38       | r = 5852       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 77^2 + 38^2       | r = 5929 + 1444       | c =  }} {{eqn | r = 7373       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = 4485^2 + 5852^2       | r = 20 \, 115 \, 225 + 34 \, 245 \, 904       | c =  }} {{eqn | r = 54 \, 361 \, 129       | c =  }} {{eqn | r = 7373^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $4485$, $5852$ and $7373$  form a Pythagorean triple. We have that: {{begin-eqn}} {{eqn | l = 4485       | r = 3 \times 5 \times 13 \times 23       | c =  }} {{eqn | l = 5852       | r = 2^2 \times 7 \times 11 \times 19       | c =  }} {{end-eqn}} It is seen that $4485$ and $5852$ share no prime factors. That is, $4485$ and $5852$ are coprime. Hence, by definition, $693$, $1924$ and $2045$ form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle.