hcju/proofwikips huggingface dataset:
Condition for Infimum of Subset to equal Infimum of Set
Condition for Infimum of Subset to equal Infimum of Set 0 49120 265908 2016-08-02T09:05:03Z Ivar Sand 2302 Created page with "== Lemma ==  Let $S$ be a real set.  Let $T$ be a subset of $S$.  Let $S$ and $T$ admit Definition:Infimum of Subset of Real..." wikitext text/x-wiki == Lemma == Let $S$ be a real set. Let $T$ be a subset of $S$. Let $S$ and $T$ admit infima. Then: : $\inf S = \inf T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$ == Proof == === Necessary Condition === Let $\inf S = \inf T$. The aim is to establish that $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$. We have: {{begin-eqn}} {{eqn | l = \inf S       | r = \inf T }} {{eqn | ll = \implies       | l = \inf S       | o = \ge       | r = \inf T }} {{eqn | ll = \iff       | l = \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon       | o = >       | r = t       | c = Infima of two Real Sets }} {{end-eqn}} === Sufficient Condition === Let $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$. The aim is to establish that $\inf S = \inf T$. We have: {{begin-eqn}} {{eqn | l = \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon       | o = >       | r = t }} {{eqn | ll = \iff       | l = \inf S       | o = \ge       | r = \inf T       | c = Infima of two Real Sets }} {{eqn | ll = \iff       | l = \inf S       | o = \ge       | r = \inf T \ge \inf S       | c = as $\inf T \ge \inf S$ is true by Infimum of Subset }} {{eqn | ll = \iff       | l = \inf S       | r = \inf T       | c =  }} {{end-eqn}}
Limit of Monotone Real Function/Increasing/Corollary
== Corollary to Limit of Increasing Function ==  Let $f$ be a real function which is increasing on the open interval $\openint a b$. If $\xi \in \openint a b$, then: :$\map f {\xi^-}$ and $\map f {\xi^+}$ both exist and: :$\map f x \le \map f {\xi^-} \le \map f \xi \le \map f {\xi^+} \le \map f y$ provided that $a < x < \xi < y < b$.  == Proof == $f$ is bounded above on $\openint a b$ by $\map f \xi$. By Limit of Increasing Function, the supremum is $\map f {\xi^-}$. So it follows that: :$\forall x \in \openint a \xi: \map f x \le \map f {\xi^-} \le \map f \xi$ A similar argument for $\openint \xi b$ holds for the other inequalities.  == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Limit of Decreasing Function|next = Derivative of Monotone Function}}: $\S 12.5$
Limit of Monotone Real Function/Decreasing/Corollary
== Corollary to Limit of Decreasing Function ==  Let $f$ be a real function which is decreasing on the open interval $\openint a b$. If $\xi \in \openint a b$, then: :$\map f {\xi^-}$ and $\map f {\xi^+}$ both exist and: :$\map f x \ge \map f {\xi^-} \ge \map f \xi \ge \map f {\xi^+} \ge \map f y$ provided that $a < x < \xi < y < b$.  == Proof == $f$ is bounded below on $\openint a \xi$ by $\map f \xi$. By Limit of Decreasing Function, the infimum is $\map f {\xi^-}$. So it follows that: :$\forall x \in \openint a \xi: \map f x \ge \map f {\xi^-} \ge \map f \xi$ A similar argument for $\openint \xi b$ holds for the other inequalities.
Prime Factors of 52 Factorial
== Example of Factorial ==  The prime decomposition of $52!$ is given as: :$52! = 2^{49} \times 3^{23} \times 5^{12} \times 7^8 \times 11^4 \times 13^4 \times 17^3 \times 19^2 \times 23^2 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47$  == Proof == For each prime factor $p$ of $52!$, let $a_p$ be the integer such that: :$p^{a_p} \divides 52!$ :$p^{a_p + 1} \nmid 52!$ Taking the prime factors in turn: {{begin-eqn}} {{eqn | l = a_2       | r = \sum_{k \mathop > 0} \floor {\frac {52} {2^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} 2} + \floor {\frac {52} 4} + \floor {\frac {52} 8 } + \floor {\frac {52} {16} } + \floor {\frac {52} {32} }       | c =  }} {{eqn | r = 26 + 13 + 6 + 3 + 1       | c =  }} {{eqn | r = 49       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_3       | r = \sum_{k \mathop > 0} \floor {\frac {52} {3^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} 3} + \floor {\frac {52} 9} + \floor {\frac {52} {27} }       | c =  }} {{eqn | r = 17 + 5 + 1       | c =  }} {{eqn | r = 23       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_5       | r = \sum_{k \mathop > 0} \floor {\frac {52} {5^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} 5} + \floor {\frac {52} {25} }       | c =  }} {{eqn | r = 10 + 2       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_7       | r = \sum_{k \mathop > 0} \floor {\frac {52} {7^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} 7} + \floor {\frac {52} {49} }       | c =  }} {{eqn | r = 7 + 1       | c =  }} {{eqn | r = 8       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{11}       | r = \sum_{k \mathop > 0} \floor {\frac {52} {11^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} {11} }       | c =  }} {{eqn | r = 4       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{13}       | r = \sum_{k \mathop > 0} \floor {\frac {52} {13^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} {13} }       | c =  }} {{eqn | r = 4       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{17}       | r = \sum_{k \mathop > 0} \floor {\frac {52} {17^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} {17} }       | c =  }} {{eqn | r = 3       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{19}       | r = \sum_{k \mathop > 0} \floor {\frac {52} {19^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} {19} }       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{23}       | r = \sum_{k \mathop > 0} \floor {\frac {52} {23^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {52} {23} }       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} Similarly: :$a_{29} = 1$ :$a_{31} = 1$ :$a_{37} = 1$ :$a_{41} = 1$ :$a_{43} = 1$ :$a_{47} = 1$ Hence the result.
Prime Factors of 13 Factorial
== Example of Factorial ==  The prime decomposition of $13!$ is given as: :$13! = 2^{10} \times 3^5 \times 5^2 \times 7 \times 11 \times 13$  == Proof == For each prime factor $p$ of $13!$, let $a_p$ be the integer such that: :$p^{a_p} \divides 13!$ :$p^{a_p + 1} \nmid 13!$ Taking the prime factors in turn: {{begin-eqn}} {{eqn | l = a_2       | r = \sum_{k \mathop > 0} \floor {\frac {13} {2^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {13} 2} + \floor {\frac {13} 4} + \floor {\frac {13} 8}       | c =  }} {{eqn | r = 6 + 3 + 1       | c =  }} {{eqn | r = 10       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_3       | r = \sum_{k \mathop > 0} \floor {\frac {13} {3^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {13} 3} + \floor {\frac {13} 9}       | c =  }} {{eqn | r = 4 + 1       | c =  }} {{eqn | r = 5       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_5       | r = \sum_{k \mathop > 0} \floor {\frac {13} {5^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {13} 5}       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_7       | r = \sum_{k \mathop > 0} \floor {\frac {13} {7^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {13} 7}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} Similarly: :$a_{11} = 1$ :$a_{13} = 1$ Hence the result.
Prime Factors of 39 Factorial
== Example of Factorial ==  The prime decomposition of $39!$ is given as: :$39! = 2^{35} \times 3^{18} \times 5^8 \times 7^5 \times 11^3 \times 13^3 \times 17^2 \times 19^2 \times 23 \times 29 \times 31 \times 37$  == Proof == For each prime factor $p$ of $39!$, let $a_p$ be the integer such that: :$p^{a_p} \divides 39!$ :$p^{a_p + 1} \nmid 39!$ Taking the prime factors in turn: {{begin-eqn}} {{eqn | l = a_2       | r = \sum_{k \mathop > 0} \floor {\frac {39} {2^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} 2} + \floor {\frac {39} 4} + \floor {\frac {39} 8 } + \floor {\frac {39} {16} } + \floor {\frac {39} {32} }       | c =  }} {{eqn | r = 19 + 9 + 4 + 2 + 1       | c =  }} {{eqn | r = 35       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_3       | r = \sum_{k \mathop > 0} \floor {\frac {39} {3^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} 3} + \floor {\frac {39} 9} + \floor {\frac {39} {27} }       | c =  }} {{eqn | r = 13 + 4 + 1       | c =  }} {{eqn | r = 18       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_5       | r = \sum_{k \mathop > 0} \floor {\frac {39} {5^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} 5} + \floor {\frac {39} {25} }       | c =  }} {{eqn | r = 7 + 1       | c =  }} {{eqn | r = 8       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_7       | r = \sum_{k \mathop > 0} \floor {\frac {39} {7^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} 7}       | c =  }} {{eqn | r = 5       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{11}       | r = \sum_{k \mathop > 0} \floor {\frac {39} {11^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} {11} }       | c =  }} {{eqn | r = 3       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{13}       | r = \sum_{k \mathop > 0} \floor {\frac {39} {13^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} {13} }       | c =  }} {{eqn | r = 3       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{17}       | r = \sum_{k \mathop > 0} \floor {\frac {39} {17^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} {17} }       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_{19}       | r = \sum_{k \mathop > 0} \floor {\frac {39} {19^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {39} {19} }       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} Similarly: :$a_{23} = 1$ :$a_{29} = 1$ :$a_{31} = 1$ :$a_{37} = 1$ Hence the result.
Lucas' Theorem/Corollary
== Corollary to Lucas' Theorem == Let $p$ be a prime number. Let $n, k \in \Z$.  Let the representations of $n$ and $k$ to the base $p$ be given by: :$n = a_r p^r + \cdots + a_1 p + a_0$ :$k = b_r p^r + \cdots + b_1 p + b_0$ Then: :$\displaystyle \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$  where $\dbinom n k$ denotes a binomial coefficient. == Proof == Consider the representations of $n$ and $k$ to the base $p$: :$n = a_r p^r + \cdots + a_1 p + a_0$ :$k = b_r p^r + \cdots + b_1 p + b_0$ Let: :$n_1 = \floor {n / p}$ :$k_1 = \floor {k / p}$ We have that: :$n \bmod p = a_0$ :$k \bmod p = b_0$ :$n_1 = a_r p^{r - 1} + a_{r - 1} p^{r - 2} \cdots + a_1$ :$k_1 = b_r p^{r - 1} + b_{r - 1} p^{r - 2} \cdots + b_1$ It follows from Lucas' Theorem that: :$\dbinom n k \equiv \dbinom {n_1} {k_1} \dbinom {a_0} {b_0} \pmod p$ Now we do the same again to the representation to the base $p$ of $n_1$ and $n_2$. Thus: :$\dbinom n k \equiv \dbinom {\floor {n_1 / p}} {\floor {k_1 / p} } \dbinom {a_1} {b_1} \dbinom {a_0} {b_0} \pmod p$ and so on until: :$\floor {n_r / p}$ and: :$\floor {k_r / p}$ Hence the result.  {{Namedfor|François Édouard Anatole Lucas|cat = Lucas}} == Sources == * {{citation|date = 1947|title = Binomial Coefficients Modulo a Prime|journal = American Mathematical Monthly|abbr = Amer. Math. Monthly|volume = 54|startpage = 589|endpage = 592|author = N.J. Fine|jstor = 2304500}} * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Lucas' Theorem|next = Kummer's Theorem}}: $\S 1.2.6$: Binomial Coefficients: Exercise $10 \ \text{(f)}$
De Polignac's Formula/Examples/2 in 1000
== Example of Use of De Polignac's Formula ==  The prime factor $2$ appears in $1000!$ to the power of $994$. That is: :$2^{994} \divides 1000!$ but: :$2^{995} \nmid 1000!$  == Proof == Let $\mu$ denote the power of $2$ which divides $1000!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {2^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} 2} + \floor {\frac {1000} 4} + \floor {\frac {1000} 8} + \floor {\frac {1000} {16} } + \floor {\frac {1000} {32} }       | c =  }} {{eqn | o =        | ro= +       | r = \floor {\frac {1000} {64} } + \floor {\frac {1000} {128} } + \floor {\frac {1000} {256} } + \floor {\frac {1000} {512} }       | c =  }} {{eqn | r = 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1       | c =  }} {{eqn | r = 994       | c =  }} {{end-eqn}}
De Polignac's Formula/Examples/5 in 1000
== Example of Use of De Polignac's Formula ==  The prime factor $5$ appears in $1000!$ to the power of $249$. That is: :$5^{249} \divides 1000!$ but: :$5^{250} \nmid 1000!$  == Proof == Let $\mu$ denote the power of $5$ which divides $1000!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {5^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} 5} + \floor {\frac {1000} {25} } + \floor {\frac {1000} {125} } + \floor {\frac {1000} {625} }       | c =  }} {{eqn | r = 200 + 40 + 8 + 1       | c =  }} {{eqn | r = 249       | c =  }} {{end-eqn}}
De Polignac's Formula/Examples/7 in 1000
== Example of Use of De Polignac's Formula ==  The prime factor $7$ appears in $1000!$ to the power of $164$. That is: :$7^{164} \divides 1000!$ but: :$7^{165} \nmid 1000!$  == Proof == Let $\mu$ denote the power of $7$ which divides $1000!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {7^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} 7} + \floor {\frac {1000} {49} } + \floor {\frac {1000} {343} }       | c =  }} {{eqn | r = 142 + 20 + 2       | c =  }} {{eqn | r = 164       | c =  }} {{end-eqn}}
De Polignac's Formula/Examples/11 in 1000
== Example of Use of De Polignac's Formula ==  The prime factor $11$ appears in $1000!$ to the power of $98$. That is: :$11^{98} \divides 1000!$ but: :$11^{99} \nmid 1000!$  == Proof == Let $\mu$ denote the power of $11$ which divides $1000!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {11^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} {11} } + \floor {\frac {1000} {121} }       | c =  }} {{eqn | r = 90 + 8       | c =  }} {{eqn | r = 98       | c =  }} {{end-eqn}}
De Polignac's Formula/Examples/13 in 1000
== Example of Use of De Polignac's Formula ==  The prime factor $13$ appears in $1000!$ to the power of $81$. That is: :$13^{81} \divides 1000!$ but: :$13^{82} \nmid 1000!$  == Proof == Let $\mu$ denote the power of $13$ which divides $1000!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {1000} {13^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {1000} {13} } + \floor {\frac {1000} {169} }       | c =  }} {{eqn | r = 76 + 5       | c =  }} {{eqn | r = 81       | c =  }} {{end-eqn}}
Integral Form of Gamma Function equivalent to Euler Form/Lemma
== Lemma for Integral Form of Gamma Function equivalent to Euler Form ==  Let $0 \le t \le m$. Then: :$0 \le e^{-t} - \left({1 - \dfrac t m}\right)^m \le t^2 \dfrac {e^{-t} } m$  == Proof == From Exponential of x not less than 1+x: :$1 + x \le e^x$ Let $x = \pm \dfrac t m$. Then: {{begin-eqn}} {{eqn | l = 1 - \dfrac t m       | o = \le       | r = e^{-t/m} }} {{eqn | ll= \leadsto       | l = \left({1 - \dfrac t m}\right)^m       | o = \le       | r = e^{-t}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = 1 + \dfrac t m       | o = \le       | r = e^{t/m} }} {{eqn | ll= \leadsto       | l = \left({1 + \dfrac t m}\right)^m       | o = \le       | r = e^t       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = e^{-t}       | o = \ge       | r = \left({1 - \dfrac t m}\right)^m }} {{eqn | r = e^t \left({1 - \dfrac t m}\right)^m e^{-t}       | c =  }} {{eqn | o = \ge       | r = e^{-t} \left({1 - \dfrac t m}\right)^m \left({1 + \dfrac t m}\right)^m       | c =  }} {{eqn | r = e^{-t} \left({1 - \dfrac {t^2} {m^2} }\right)^m       | c = Difference of Two Squares }} {{eqn | o = \ge       | r = e^{-t} \left({1 - \dfrac {t^2} {m^2} }\right)       | c = Corollary to Bernoulli's Inequality }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = e^{-t} - \left({1 - \dfrac t m}\right)^m }} {{eqn | o = \le       | r = t^2 \dfrac {e^{-t} } m       | c =  }} {{end-eqn}}  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Partial Gamma Function expressed as Integral/Lemma|next = Integral Form of Gamma Function equivalent to Euler Form/Proof 2}}: $\S 1.2.5$: Permutations and Factorials: Exercise $20$
Faà di Bruno's Formula/Example/0
== Example of use of Faà di Bruno's Formula: $n = 0$ == Consider Faà di Bruno's Formula: {{:Faà di Bruno's Formula}} When $n = 0$ we have:  :$D_x^0 w = w$  == Proof == {{:Faà di Bruno's Formula/Example/0/Proof}}
Faà di Bruno's Formula/Example/1
== Example of use of Faà di Bruno's Formula: $n = 1$ == Consider Faà di Bruno's Formula: {{:Faà di Bruno's Formula}} When $n = 1$ we have:  :$D_x^1 w = D_u^1 w D_x^1 u$  == Proof == {{:Faà di Bruno's Formula/Example/1/Proof}}
Faà di Bruno's Formula/Example/2
Faà di Bruno's Formula/Example/2 0 49374 267042 2016-08-17T20:45:57Z Prime.mover 59 Created page with "== Example of use of Faà di Bruno's Formula: $n = 2$ ==  Consider Faà di Bruno's Formula: {{:Faà di Bruno's Formula}}  When $n = 2$ we have:  :$D_x^2 w..." wikitext text/x-wiki == Example of use of Faà di Bruno's Formula: $n = 2$ == Consider Faà di Bruno's Formula: {{:Faà di Bruno's Formula}} When $n = 2$ we have:  :$D_x^2 w = D_u^2 w \left({D_x^1 u}\right)^2 + D_u^1 w D_x^2 u$  == Proof == {{:Faà di Bruno's Formula/Example/2/Proof}}
Ceiling of x+m over n/Corollary
Ceiling of x+m over n/Corollary 0 49513 316257 267646 2017-09-08T16:29:41Z Caliburn 3218 cat wikitext text/x-wiki == Corollary to Ceiling of $\frac {x + m} n$ ==  Let $n \in \Z$ such that $n > 0$. Let $x \in \R$. Then: :$\left \lceil{\dfrac x n}\right \rceil = \left \lceil{\dfrac {\left \lceil{x}\right \rceil} n}\right \rceil$ where $\left\lceil{x}\right\rceil$ denotes the ceiling of $x$.  == Proof == This is a special case of Ceiling of $\dfrac {x + m} n$: :$\left \lceil{\dfrac {x + m} n}\right \rceil = \left \lceil{\dfrac {\left \lceil{x}\right \rceil + m} n}\right \rceil$ where $m = 0$. == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Ceiling of x+m over n|next = Ceiling of x+m over n/Proof 1}}: $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$
Floor of x+m over n/Corollary
== Corollary to Floor of $\frac {x + m} n$ ==  Let $n \in \Z$ such that $n > 0$. Let $x \in \R$. Then: :$\floor {\dfrac x n} = \floor {\dfrac {\floor x} n}$ where $\floor x$ denotes the floor of $x$.  == Proof == This is a special case of Floor of $\dfrac {x + m} n$: :$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$ where $m = 0$. == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Floor of x+m over n|next = Floor of x+m over n/Proof 1}}: $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$
Euclidean Algorithm/Demonstration
== Example of use of Euclidean Algorithm ==  Using the '''Euclidean Algorithm''', we can investigate in detail what happens when we apply the Division Theorem repeatedly to $a$ and $b$. {{begin-eqn}} {{eqn | l = a       | r = q_1 b + r_1       | c =  }} {{eqn | l = b       | r = q_2 r_1 + r_2       | c =  }} {{eqn | l = r_1       | r = q_3 r_2 + r_3       | c =  }} {{eqn | l = \cdots       | o =        | c =  }} {{eqn | l = r_{n - 2}       | r = q_n r_{n - 1} + r_n       | c =  }} {{eqn | l = r_{n - 1}       | r = q_{n + 1} r_n + 0       | c =  }} {{end-eqn}} From the Division Theorem, we know that the remainder is always strictly less than the divisor. That is, in $a = q b + r$: :$0 \le r < \size b$ So we know that: :$b > r_1 > r_2 > \ldots > r_{n - 2} > r_{n - 1} > r_n > 0$ So the algorithm ''has'' to terminate.   == Sources == * {{BookReference|Number Theory|1971|George E. Andrews|prev = Euclidean Algorithm|next = Existence of Greatest Common Divisor/Proof 4}}: $\text {2-2}$ Divisibility: Theorem $\text {2-2}$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Euclidean Algorithm|next = Definition:Integer Combination}}: $\S 12$: Highest common factors and Euclid's algorithm * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Euclidean Algorithm|next = Definition:Linearly Independent Set}}: $\text{A}.1$: Number theory
Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijection
Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijection 0 49666 268458 2016-08-30T05:32:36Z Prime.mover 59 Created page with "== Theorem ==  Let $\left({S, \circ}\right)$ be a finite semigroup.  Let $a \in S$ be Definition:Right Cancellabl..." wikitext text/x-wiki == Theorem ==  Let $\left({S, \circ}\right)$ be a finite semigroup. Let $a \in S$ be right cancellable. Then the right regular representation $\rho_a$ of $\left({S, \circ}\right)$ with respect to $a$ is a bijection.  == Proof == By Right Cancellable iff Right Regular Representation Injective, $\rho_a$ is an injection. By hypothesis, $S$ is finite. From Injection from Finite Set to Itself is Surjection, $\rho_a$ is a surjection. Thus $\rho_a$ is injective and surjective, and therefore a bijection.
Euler Phi Function/Examples/9
== Example of Use of Euler $\phi$ Function ==  :$\map \phi 9 = 6$  where $\phi$ denotes the Euler $\phi$ function. == Proof == From Euler Phi Function of Prime Power: :$\map \phi {3^k} = 2 \times 3^{k - 1}$ Thus: :$\map \phi 9 = \map \phi {3^2} = 2 \times 3 = 6$ They can be enumerated as: :$1, 2, 4, 5, 7, 8$  == Sources == * {{BookReference|Introduction to the Theory of Finite Groups|1964|Walter Ledermann|ed = 5th|edpage = Fifth Edition|prev = Definition:Euler Phi Function|next = Euler Phi Function of Prime}}: Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$
Group of Rotation Matrices Order 4/Cayley Table
== Cayley Table for Group of Rotation Matrices Order $4$ == Consider the group of rotation matrices order $4$ :$R_4 = \set {\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} }$ $R_4$ can be described completely by showing its Cayley table. Let: :$r_0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ :$r_1 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ :$r_2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ :$r_3 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ Then we have:  :$\begin{array}{r|rrrr} \times & r_0 & r_1 & r_2 & r_3 \\ \hline r_0 & r_0 & r_1 & r_2 & r_3 \\ r_1 & r_1 & r_2 & r_3 & r_0 \\ r_2 & r_2 & r_3 & r_0 & r_1 \\ r_3 & r_3 & r_0 & r_1 & r_2 \\ \end{array}$