Tau Function of 304

Tau Function of 304 0 59029 437827 304244 2019-12-05T13:17:44Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {304} = 10$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$304 = 2^4 \times 19$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {304}       | r = \map \tau {2^4 \times 19^1}       | c =  }} {{eqn | r = \paren {4 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 10       | c =  }} {{end-eqn}} The divisors of $304$ can be enumerated as: :$1, 2, 4, 8, 16, 19, 38, 76, 152, 304$ {{OEIS|A018383}}{{qed}} Category:Tau Function Category:304 4sp5aqq0s67es5to2ipj3cfkzxgnnaa


Sigma Function of 304

Sigma Function of 304 0 59031 437054 318249 2019-11-28T11:59:16Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\map \sigma {304} = 620$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$304 = 2^4 \times 19$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {304}       | r = \frac {2^5 - 1} {2 - 1} \times \frac {19^2 - 1} {19 - 1}       | c = Sigma Function of Integer }} {{eqn | r = \frac {31 - 1} 1 \times \frac {361 - 1} {18}       | c =  }} {{eqn | r = 31 \times 20       | c =  }} {{eqn | r = \paren {2^2 \times 5} \times 31       | c =  }} {{eqn | r = 620       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:304 scg937vsfb5vp31ewr73chnst9ufmdk


Sigma Function of 748

Sigma Function of 748 0 59032 318334 318332 2017-09-16T18:43:23Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({748}\right) = 1512$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$748 = 2^2 \times 11 \times 17$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({748}\right)       | r = \frac {2^3 - 1} {2 - 1} \times \left({11 + 1}\right) \times \left({17 + 1}\right)       | c = Sigma Function of Integer }} {{eqn | r = 7 \times 12 \times 18       | c =  }} {{eqn | r = 7 \times \left({2^2 \times 3}\right) \times \left({2 \times 3^2}\right)       | c =  }} {{eqn | r = 2^3 \times 3^3 \times 7       | c =  }} {{eqn | r = 1512       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:748 bak7x1fyh5pcawhxicsevj5k53a0qix


Tau Function of 748

Tau Function of 748 0 59034 304251 304250 2017-07-10T09:47:44Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({748}\right) = 12$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$748 = 2^2 \times 11 \times 17$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({748}\right)       | r = \tau \left({2^2 \times 11^1 \times 17^1}\right)       | c =  }} {{eqn | r = \left({2 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} The divisors of $748$ can be enumerated as: :$1, 2, 4, 11, 17, 22, 34, 44, 68, 187, 374, 748$ {{OEIS|A018625}}{{qed}} Category:Tau Function Category:748 gy36kbpwjvlwbg7l8qemzf3uswc8ul5


Perfect Number/Examples/6

Perfect Number/Examples/6 0 59047 304290 304286 2017-07-10T13:10:44Z Prime.mover 59 wikitext text/x-wiki == Example of Perfect Number == <onlyinclude> $6$ is a perfect number: :$1 + 2 + 3 = 6$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = 6       | r = 2 \times 3       | c =  }} {{eqn | r = 2^{2 - 1} \left({2^2 - 1}\right)       | c =  }} {{end-eqn}} Thus $6$ is in the form $2^{n - 1} \left({2^n - 1}\right)$. $\left({2^2 - 1}\right) = 3$ is prime. So $6$ is perfect by the Theorem of Even Perfect Numbers. The aliquot parts of $6$ are enumerated at $\tau$ of $6$: :$1, 2, 3$ {{qed}} Category:Perfect Numbers/Examples Category:6 l5ifi04j4tscc66zegdw575qgp0xjl4


Perfect Number/Examples/28

Perfect Number/Examples/28 0 59049 304301 304289 2017-07-10T21:11:53Z Prime.mover 59 wikitext text/x-wiki == Example of Perfect Number == <onlyinclude> $28$ is a perfect number: :$1 + 2 + 4 + 7 + 14 = 28$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = 28       | r = 4 \times 7       | c =  }} {{eqn | r = 2^{3 - 1} \left({2^3 - 1}\right)       | c =  }} {{end-eqn}} Thus $28$ is in the form $2^{p - 1} \left({2^p - 1}\right)$. $\left({2^3 - 1}\right) = 7$ is prime. So $28$ is perfect by the Theorem of Even Perfect Numbers. The aliquot parts of $28$ are enumerated at $\tau$ of $28$: :$1, 2, 4, 7, 14$ {{qed}} Category:Perfect Numbers/Examples Category:28 ko1oaen6qi02suw50mk42r15d4yc14d


Perfect Number/Examples/496

Perfect Number/Examples/496 0 59052 304297 2017-07-10T21:05:03Z Prime.mover 59 Created page with "== Example of Perfect Number == <onlyinclude> $496$ is a perfect number: :$1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248..." wikitext text/x-wiki == Example of Perfect Number == <onlyinclude> $496$ is a perfect number: :$1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = 496       | r = 16 \times 31       | c =  }} {{eqn | r = 2^{5 - 1} \left({2^5 - 1}\right)       | c =  }} {{end-eqn}} Thus $496$ is in the form $2^{p - 1} \left({2^p - 1}\right)$. $\left({2^5 - 1}\right) = 31$ is prime. So $496$ is perfect by the Theorem of Even Perfect Numbers. The aliquot parts of $496$ are enumerated at $\tau$ of $496$: :$1, 2, 4, 8, 16, 31, 62, 124, 248$ {{qed}} Category:Perfect Numbers/Examples Category:496 5caq0phszs33pcx9ubnbtaw1dnhv8f4


Tau Function of 8128

Tau Function of 8128 0 59053 304298 2017-07-10T21:08:43Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({8128}\right) = 10$ </onlyinclude> where $\tau$ denotes the Definition:Tau Fun..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({8128}\right) = 10$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$8128 = 2^6 \times 127$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({8128}\right)       | r = \tau \left({2^6 \times 127^1}\right)       | c =  }} {{eqn | r = \left({6 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 14       | c =  }} {{end-eqn}} The divisors of $496$ can be enumerated as: :$1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064, 8128$ {{OEIS|A133024}}{{qed}} Category:Tau Function Category:8128 hm2zpslcr20akwp3rts5nourkv1yngr


Tau Function of 350

Tau Function of 350 0 59059 304308 2017-07-10T22:22:33Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({350}\right) = 12$ </onlyinclude> where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({350}\right) = 12$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$350 = 2 \times 5^2 \times 7$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({350}\right)       | r = \tau \left({2^1 \times 5^2 \times 7^1}\right)       | c =  }} {{eqn | r = \left({1 + 1}\right) \left({2 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} The divisors of $350$ can be enumerated as: :$1, 2, 5, 7, 10, 14, 25, 35, 50, 70, 175, 350$ {{OEIS|A018406}}{{qed}} Category:Tau Function Category:350 pwa2hv67xd0ko87ic5yzleh47df80ss


Tau Function of 368

Tau Function of 368 0 59061 304310 2017-07-10T22:28:37Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({368}\right) = 10$ </onlyinclude> where $\tau$ denotes the Definition:Tau Func..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({368}\right) = 10$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$368 = 2^4 \times 23$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({368}\right)       | r = \tau \left({2^4 \times 23^1}\right)       | c =  }} {{eqn | r = \left({4 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 10       | c =  }} {{end-eqn}} The divisors of $368$ can be enumerated as: :$1, 2, 4, 8, 16, 23, 46, 92, 184, 368$ {{OEIS|A018416}}{{qed}} Category:Tau Function Category:368 6k9v95xl1u1mmviudnaqyvxwui8t5yl


Tau Function of 464

Tau Function of 464 0 59063 473808 304313 2020-06-12T15:22:36Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {464} = 10$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$464 = 2^4 \times 29$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {464}       | r = \map \tau {2^4 \times 29^1}       | c =  }} {{eqn | r = \paren {4 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 10       | c =  }} {{end-eqn}} The divisors of $464$ can be enumerated as: :$1, 2, 4, 8, 16, 29, 58, 116, 232, 464$ {{OEIS|A018469}}{{qed}} Category:Tau Function Category:464 cm87fogt027tziyuwpzb01bo5l7fiqw


Sigma Function of 464

Sigma Function of 464 0 59074 318279 318277 2017-09-16T18:35:27Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({464}\right) = 930$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$464 = 2^4 \times 29$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({464}\right)       | r = \frac {2^5 - 1} {2 - 1} \times \left({29 + 1}\right)       | c = Sigma Function of Integer }} {{eqn | r = \frac {31 - 1} 1 \times 30       | c =  }} {{eqn | r = 31 \times 30       | c =  }} {{eqn | r = \left({2 \times 3 \times 5}\right) \times 31       | c =  }} {{eqn | r = 930       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:464 pwwssq4d21pngtykzhxl1bdd7lri6rr


Sigma Function of 550

Sigma Function of 550 0 59076 318303 318301 2017-09-16T18:38:48Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({550}\right) = 1116$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$550 = 2 \times 5^2 \times 11$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({550}\right)       | r = \left({2 + 1}\right) \times \frac {5^3 - 1} {5 - 1} \times \left({11 + 1}\right)       | c = Sigma Function of Integer }} {{eqn | r = 3 \times \frac {125 - 1} 4 \times 12       | c =  }} {{eqn | r = 3 \times 31 \times \left({2^2 \times 3}\right)       | c =  }} {{eqn | r = 2^2 \times 3^2 \times 31       | c =  }} {{eqn | r = 1116       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:550 czyhqlmea1f0bdvkvlv8w7iocze8mwu


Sigma Function of 572

Sigma Function of 572 0 59078 318309 318307 2017-09-16T18:39:36Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({572}\right) = 1176$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$572 = 2^2 \times 11 \times 13$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({572}\right)       | r = \frac {2^3 - 1} {2 - 1} \times \left({11 + 1}\right) \times \left({13 + 1}\right)       | c = Sigma Function of Integer }} {{eqn | r = \frac {7 - 1} 1 \times 12 \times 14       | c =  }} {{eqn | r = 7 \times \left({2^2 \times 3}\right) \times \left({2 \times 7}\right)       | c =  }} {{eqn | r = 2^3 \times 3 \times 7^2       | c =  }} {{eqn | r = 1176       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:572 f9klaav8ehkgcafcylzae71k86l4ud0


Sigma Function of 650

Sigma Function of 650 0 59080 435882 318324 2019-11-20T19:00:14Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\map \sigma {650} = 1302$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$650 = 2 \times 5^2 \times 13$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {650}       | r = \paren {2 + 1} \times \frac {5^3 - 1} {5 - 1} \times \paren {13 + 1}       | c = Sigma Function of Integer }} {{eqn | r = 3 \times \frac {125 - 1} 4 \times 14       | c =  }} {{eqn | r = 3 \times 31 \times \paren {2 \times 7}       | c =  }} {{eqn | r = 2 \times 3 \times 7 \times 31       | c =  }} {{eqn | r = 1302       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:650 bm6g40cfm16s71b0igpb06njswgr2vd


Tau Function of 770

Tau Function of 770 0 59567 305915 305909 2017-07-22T08:34:49Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({770}\right) = 16$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$770 = 2 \times 5 \times 7 \times 11$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({770}\right)       | r = \tau \left({2^1 \times 5^1 \times 7^1 \times 11^1}\right)       | c =  }} {{eqn | r = \left({1 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | c =  }} {{eqn | r = 16       | c =  }} {{end-eqn}} The divisors of $770$ can be enumerated as: :$1, 2, 5, 7, 10, 11, 14, 22, 35, 55, 70, 77, 110, 154, 385, 770$ {{OEIS|A018636}}{{qed}} Category:Tau Function Category:770 6jgxiy0fxo1q0xqw1xnjov11x3mhfz9


Multiplicative Persistence/Examples/25

Multiplicative Persistence/Examples/25 0 60112 308235 308234 2017-08-02T06:02:20Z Prime.mover 59 wikitext text/x-wiki == Examples of Multiplicative Persistence == <onlyinclude> $25$ is the smallest positive integer which has a multiplicative persistence of $2$. </onlyinclude> == Proof == Trivially: {{begin-eqn}} {{eqn | n = 1       | l = 2 \times 5       | r = 10 }} {{eqn | n = 2       | l = 1 \times 0       | r = 0 }} {{end-eqn}} All positive integers between $10$ and $19$ are seen to have a multiplicative persistence of $1$: :$1 \times n = n$ where $n$ is a single digit. Then for 2-digit positive integers starting with $2$: :$2 \times n > 9 \implies n > 4$ by inspection. Hence the result. {{qed}} Category:Multiplicative Persistence Category:25 q4xeibo3s0mj1uxi8vb11276c4hz6vy


De Polignac's Formula/Examples/2 in 720 Factorial 0 60227 391883 308633 2019-02-09T23:45:11Z Prime.mover 59 wikitext text/x-wiki == Example of Use of De Polignac's Formula == <onlyinclude> The prime factor $2$ appears in $720!$ to the power of $716$. That is: :$2^{716} \divides 720!$ but: :$2^{717} \nmid 720!$ </onlyinclude> == Proof == Let $\mu$ denote the power of $2$ which divides $720!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {720} {2^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {720} 2} + \floor {\frac {720} 4} + \floor {\frac {720} 8} + \floor {\frac {720} {16} } + \floor {\frac {720} {32} }       | c =  }} {{eqn | o =        | r = + \floor {\frac {720} {64} } + \floor {\frac {720} {128} } + \floor {\frac {720} {256} } + \floor {\frac {720} {512} }       | c =  }} {{eqn | r = 360 + 180 + 90 + 45 + 22 + 11 + 5 + 2 + 1       | c =  }} {{eqn | r = 716       | c =  }} {{end-eqn}} {{qed}} Category:De Polignac's Formula maxgk7rg0b1l1jnx4zh0pbqnazmfcbw


De Polignac's Formula/Examples/3 in 720 Factorial 0 60228 391885 334099 2019-02-09T23:46:07Z Prime.mover 59 wikitext text/x-wiki == Example of Use of De Polignac's Formula == <onlyinclude> The prime factor $3$ appears in $720!$ to the power of $356$. That is: :$3^{356} \divides 720!$ but: :$3^{357} \nmid 720!$ </onlyinclude> == Proof == Let $\mu$ denote the power of $3$ which divides $720!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {720} {3^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {720} 3} + \floor {\frac {720} 9} + \floor {\frac {720} {27} } + \floor {\frac {720} {81} } + \floor {\frac {720} {243} }       | c =  }} {{eqn | r = 240 + 80 + 26 + 8 + 2       | c =  }} {{eqn | r = 356       | c =  }} {{end-eqn}} {{qed}} Category:De Polignac's Formula aku8yebg8tzo3oxc1cpymz97023rxyq


De Polignac's Formula/Examples/5 in 720 Factorial 0 60229 391886 358683 2019-02-09T23:47:00Z Prime.mover 59 wikitext text/x-wiki == Example of Use of De Polignac's Formula == <onlyinclude> The prime factor $5$ appears in $720!$ to the power of $178$. That is: :$5^{178} \divides 720!$ but: :$5^{179} \nmid 720!$ </onlyinclude> == Proof == Let $\mu$ denote the power of $5$ which divides $720!$ {{begin-eqn}} {{eqn | l = \mu       | r = \sum_{k \mathop > 0} \floor {\frac {720} {5^k} }       | c = De Polignac's Formula }} {{eqn | r = \floor {\frac {720} 5} + \floor {\frac {720} {25} } + \floor {\frac {720} {125} } + \floor {\frac {720} {625} }       | c =  }} {{eqn | r = 144 + 28 + 5 + 1       | c =  }} {{eqn | r = 178       | c =  }} {{end-eqn}} {{qed}} Category:De Polignac's Formula 45qodtu6gqlwu69yvt7btcb8fpudk4c


Sigma Function of 1184

Sigma Function of 1184 0 60464 454635 317765 2020-03-15T15:40:51Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\map \sigma {1184} = 2394$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == From Sigma Function of Integer :$\displaystyle \map \sigma n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$ where $n = \displaystyle \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$. We have that: :$1184 = 2^5 \times 37$ Hence: {{begin-eqn}} {{eqn | l = \map \sigma {1184}       | r = \frac {2^6 - 1} {2 - 1} \times \frac {37^2 - 1} {37 - 1}       | c =  }} {{eqn | r = \frac {63} 1 \times \frac {38 \times 36} {36}       | c =  }} {{eqn | r = 63 \times 38       | c =  }} {{eqn | r = \paren {3^2 \times 7} \times \paren {2 \times 19}       | c =  }} {{eqn | r = 2 \times 3^2 \times 7 \times 19       | c =  }} {{eqn | r = 2394       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:1184 ganlt3wy11purs9zxg4h9t2r84dq0mg


Sigma Function of 1210

Sigma Function of 1210 0 60465 317791 317789 2017-09-16T13:01:46Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({1210}\right) = 2394$ </onlyinclude> 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: :$1210 = 2 \times 5 \times 11^2$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({1210}\right)       | r = \left({2 + 1}\right) \times \left({5 + 1}\right) \times \frac {11^3 - 1} {11 - 1}       | c =  }} {{eqn | r = 3 \times 6 \times \frac {1330} {10}       | c =  }} {{eqn | r = 3 \times 6 \times 133       | c =  }} {{eqn | r = 3 \times \left({2 \times 3}\right) \times \left({7 \times 19}\right)       | c =  }} {{eqn | r = 2 \times 3^2 \times 7 \times 19       | c =  }} {{eqn | r = 2394       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:1210 6eiwqik7ft4fpi17l6dtmvwllrjpuuv


Sigma Function of 1638

Sigma Function of 1638 0 60492 318129 318127 2017-09-16T14:56:56Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({1638}\right) = 4368$ </onlyinclude> where $\sigma$ denotes the $\sigma$ function. == Proof == We have that: :$1638 = 2 \times 3^2 \times 7 \times 13$ Hence: {{begin-eqn}} {{eqn | l = \sigma \left({1638}\right)       | r = \left({2 + 1}\right) \frac {3^3 - 1} {3 - 1} \times \left({7 + 1}\right) \times \left({13 + 1}\right)       | c = Sigma Function of Integer }} {{eqn | r = 3 \times \frac {26} 2 \times 8 \times 14       | c =  }} {{eqn | r = 3 \times 13 \times 2^3 \times \left({2 \times 7}\right)       | c =  }} {{eqn | r = 2^4 \times 3 \times 7 \times 13       | c =  }} {{eqn | r = 4368       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function of Integer Category:1638 piry1g4q0g9krzjvna8gqdbqybd7dko


Tau Function of 1638

Tau Function of 1638 0 60493 445384 309876 2020-01-31T11:05:00Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {1638} = 24$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$1638 = 2 \times 3^2 \times 7 \times 13$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {1638}       | r = \map \tau {2^1 \times 3^2 \times 7^1 \times 13^1}       | c =  }} {{eqn | r = \paren {1 + 1} \paren {2 + 1} \paren {1 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 24       | c =  }} {{end-eqn}} The divisors of $1638$ can be enumerated as: :$1, 2, 3, 6, 7, 9, 13, 14, 18, 21, 26, 39, 42, 63, 78, 91, 117, 126, 182, 234, 273, 546, 819, 1638$ {{qed}} Category:Tau Function Category:1638 s9kup1bhls2umalmu3n3pig5qpa3a4y


Tau Function of 945

Tau Function of 945 0 60606 425091 310276 2019-09-13T12:10:32Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {945} = 16$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$945 = 3^3 \times 5 \times 7$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {945}       | r = \map \tau {3^3 \times 5^1 \times 7^1}       | c =  }} {{eqn | r = \paren {3 + 1} \paren {1 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 16       | c =  }} {{end-eqn}} The divisors of $945$ can be enumerated as: :$1, 3, 5, 7, 9, 15, 21, 27, 35, 45, 63, 105, 135, 189, 315, 945$ {{OEIS|A018736}}{{qed}} Category:Tau Function Category:945 87e3zg4t9qjr3unoyfn33x5h0mnybkn


Tau Function of 910

Tau Function of 910 0 60608 439494 310279 2019-12-14T10:31:38Z Prime.mover 59 wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\map \tau {910} = 16$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$910 = 2 \times 5 \times 7 \times 13$ Thus: {{begin-eqn}} {{eqn | l = \map \tau {910}       | r = \map \tau {2^1 \times 5^1 \times 7^1 \times 13^1}       | c =  }} {{eqn | r = \paren {1 + 1} \paren {1 + 1} \paren {1 + 1} \paren {1 + 1}       | c =  }} {{eqn | r = 16       | c =  }} {{end-eqn}} The divisors of $910$ can be enumerated as: :$1, 2, 5, 7, 10, 13, 14, 26, 35, 65, 70, 91, 130, 182, 455, 910$ {{OEIS|A018717}}{{qed}} Category:Tau Function Category:910 ccz2ot8ebhj08bja1wyq1810eollxc4


Tau Function of 1184

Tau Function of 1184 0 60611 310282 2017-08-10T06:33:21Z Prime.mover 59 Created page with "== Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({1184}\right) = 12$ </onlyinclude> where $\tau$ denotes the Definition:Tau Fun..." wikitext text/x-wiki == Example of Use of $\tau$ Function == <onlyinclude> :$\tau \left({1184}\right) = 12$ </onlyinclude> where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$1184 = 2^5 \times 37$ Thus: {{begin-eqn}} {{eqn | l = \tau \left({1184}\right)       | r = \tau \left({2^5 \times 37^1}\right)       | c =  }} {{eqn | r = \left({5 + 1}\right) \left({2 + 1}\right)       | c =  }} {{eqn | r = 12       | c =  }} {{end-eqn}} The divisors of $1184$ can be enumerated as: :$1, 2, 4, 8, 16, 32, 37, 74, 148, 296, 592, 1184$ {{qed}} Category:Tau Function Category:1184 hp5q548fdculppmqc91vdbj1uf8rot5


Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999/Examples 0 60630 478390 478377 2020-07-16T06:53:03Z Prime.mover 59 wikitext text/x-wiki == Examples of Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999 == <onlyinclude> {{begin-eqn}} {{eqn | l = 4 \times 999       | r = 3996       | c =  }} {{eqn | ll= \leadsto       | l = 3 + 996       | r = 999       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 15 \times 999       | r = 14 \, 985       | c =  }} {{eqn | ll= \leadsto       | l = 14 + 985       | r = 999       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 47 \times 999       | r = 46 \, 953       | c =  }} {{eqn | ll= \leadsto       | l = 46 + 953       | r = 999       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 57 \times 999       | r = 56 \, 943       | c =  }} {{eqn | ll= \leadsto       | l = 56 + 943       | r = 999       | c =  }} {{end-eqn}} </onlyinclude> Category:Integer whose Digits when Grouped in 3s add to Multiple of 999 is Divisible by 999 k75tr675sa2cztel9d4thg7069dew4m


Carmichael Number/Examples/1729

Carmichael Number/Examples/1729 0 60981 323299 312511 2017-10-26T20:16:49Z Prime.mover 59 wikitext text/x-wiki == Example of Carmichael Number== $1729$ is a Carmichael number: <onlyinclude> :$\forall a \in \Z: a \perp 1729: a^{1729} \equiv a \pmod {1729}$ while $1729$ is composite. </onlyinclude> == 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. {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 509,033,161|next = Carmichael Number/Examples/294,409}}: $509,033,161$ Category:Carmichael Numbers Category:1729 qgour7zwj5qk61w91iqa7mzf20wlqhx


Highly Composite Number/Examples/1

Highly Composite Number/Examples/1 0 61035 492964 312820 2020-10-06T18:06:40Z Prime.mover 59 wikitext text/x-wiki == Example of Highly Composite Number == <onlyinclude> $1$ is a highly composite number, being the smallest positive integer with $1$ divisor or more. </onlyinclude> == 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. {{qed}} Category:Highly Composite Numbers 0p8650qep457aueoerbrrp7pp93pyns


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 == <onlyinclude> $1$ is a Definition:Special Highly Composite Number|special hig..." wikitext text/x-wiki == Example of Special Highly Composite Number == <onlyinclude> $1$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers. </onlyinclude> == 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. {{qed}} == 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}} Category:Special Highly Composite Numbers i3t0aihfktpa57wuu91qmdax7ttfb3j


Highly Composite Number/Examples/2

Highly Composite Number/Examples/2 0 61039 312834 312833 2017-08-23T22:04:28Z Prime.mover 59 wikitext text/x-wiki == Example of Highly Composite Number == <onlyinclude> $2$ is a highly composite number, being the smallest positive integer with $2$ divisors or more. </onlyinclude> == 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. {{qed}} Category:Highly Composite Numbers opsdh13u7m4uhchr7v0zb1rvxf5e8ar


Special Highly Composite Number/Examples/2 0 61065 412551 313064 2019-07-12T16:23:03Z Prime.mover 59 wikitext text/x-wiki == Example of Special Highly Composite Number == <onlyinclude> $2$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers. </onlyinclude> == 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. {{qed}} == 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}} Category:Special Highly Composite Numbers Category:2 3m6blgzzkw9e3zqsh0mzkup9sr9v5cc


Special Highly Composite Number/Examples/6 0 61067 313065 312947 2017-08-26T06:11:36Z Prime.mover 59 wikitext text/x-wiki == Example of Special Highly Composite Number == <onlyinclude> $6$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers. </onlyinclude> == Proof == By inspection of the sequence of highly composite numbers, $6$ is highly composite. {{AimForCont}} $n > 6$ 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. {{qed}} == 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}} Category:Special Highly Composite Numbers Category:6 85yg2crs8eimf36fbfbd8u16gudlxwr


Special Highly Composite Number/Examples/12 0 61069 313066 312950 2017-08-26T06:13:05Z Prime.mover 59 wikitext text/x-wiki == Example of Special Highly Composite Number == <onlyinclude> $12$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers. </onlyinclude> == Proof == By inspection of the sequence of highly composite numbers, $12$ is highly composite. {{AimForCont}} $n > 12$ 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. {{qed}} == 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}} Category:Special Highly Composite Numbers Category:12 ne9hzwbf2tx192vwctgj5lqk14clq3z


Special Highly Composite Number/Examples/60 0 61070 313067 312953 2017-08-26T06:17:38Z Prime.mover 59 wikitext text/x-wiki == Example of Special Highly Composite Number == <onlyinclude> $60$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers. </onlyinclude> == Proof == By inspection of the sequence of highly composite numbers, $60$ is highly composite. {{AimForCont}} $n > 60$ is a highly composite number which is not divisible by $60$. We have that $12$ is a special highly composite number. Therefore $12$ is a divisor of $n$. As $60$ is not a divisor of $n$, it follows that while $3$ is a divisor of $n$, $5$ is not. From Prime Decomposition of Highly Composite Number, no prime number greater than $5$ is a divisor of $n$. Thus: :$n = 2^a \times 3^b$ where $a \ge b \ge 1$. This will be investigated on a case-by-case basis. :$(1): \quad b = 1$ That is, $n = 2^a \times 3$. We have that $n > 60$. Therefore: :$(1 \text a): \quad a \ge 5$ as $2^4 \times 3^1 = 48$. Then: {{begin-eqn}} {{eqn | l = 2^{a - 3} \times 3 \times 5       | o = <       | r = 2^a \times 3       | c = because $5 < 2^3 = 8$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 3} \times 3 \times 5}\right)       | o = <       | r = \tau \left({2^a \times 3}\right)       | c = as $2^a \times 3$ is highly composite }} {{eqn | ll= \leadsto       | l = \left({\left({a - 3}\right) + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({a + 1}\right) \left({1 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 4 \left({a - 2}\right)       | o = <       | r = 2 \left({a + 1}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = a       | o = <       | r = 5       | c = which is a contradiction of $(1 \text a)$ }} {{end-eqn}} It follows by Proof by Contradiction that $b \ne 1$. {{qed|lemma}} :$(2): \quad b = 2$ That is, $n = 2^a \times 3^2$. We have that $n > 60$. Therefore: :$(2 \text a): \quad a \ge 3$ as $2^2 \times 3^2 = 36$. Then: {{begin-eqn}} {{eqn | l = 2^{a - 1} \times 3 \times 5       | o = <       | r = 2^a \times 3^2       | c = because $5 < 2 \times 3 = 6$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 1} \times 3 \times 5}\right)       | o = <       | r = \tau \left({2^a \times 3^2}\right)       | c = as $2^a \times 3^2$ is highly composite }} {{eqn | ll= \leadsto       | l = \left({\left({a - 1}\right) + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({a + 1}\right) \left({2 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 4 a       | o = <       | r = 3 \left({a + 1}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = a       | o = <       | r = 3       | c = which is a contradiction of $(2 \text a)$ }} {{end-eqn}} It follows by Proof by Contradiction that $b \ne 2$. {{qed|lemma}} :$(3): \quad b \ge 3$ By Prime Decomposition of Highly Composite Number we have that $a \ge 3$. Then: {{begin-eqn}} {{eqn | l = 2^{a - 1} \times 3^{b - 1} \times 5       | o = <       | r = 2^a \times 3^b       | c = because $5 < 2 \times 3 = 6$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 1} \times 3^{b - 1} \times 5}\right)       | o = <       | r = \tau \left({2^a \times 3^b}\right)       | c = as $2^a \times 3^b$ is highly composite }} {{eqn | ll= \leadsto       | l = \left({\left({a - 1}\right) + 1}\right) \left({\left({b - 1}\right) + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({a + 1}\right) \left({b + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 2 a b       | o = <       | r = \left({a + 1}\right) \left({b + 1}\right)       | c =  }} {{eqn | ll= \leadsto       | l = a b       | o = <       | r = a + b + 1       | c =  }} {{eqn | ll= \leadsto       | l = 3 a       | o = <       | r = a + b + 1       | c = as $b \ge 3$ }} {{eqn | ll= \leadsto       | l = 2 a       | o = <       | r = b + 1       | c =  }} {{eqn | ll= \leadsto       | l = 2 b       | o = <       | r = b + 1       | c = as $a \ge b$ }} {{eqn | ll= \leadsto       | l = b       | o = <       | r = 1       | c = which is a contradiction of $(3)$ }} {{end-eqn}} {{qed|lemma}} By Proof by Cases it is seen that the existence of a highly composite $n$ not divisible by $60$ leads to a contradiction. The result then follows by Proof by Contradiction. {{qed}} == 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}} Category:Special Highly Composite Numbers Category:60 2scpj1d25yf7x7ls4qlytolch75s6en


Special Highly Composite Number/Examples/2520 0 61071 322270 321092 2017-10-22T12:17:49Z Prime.mover 59 wikitext text/x-wiki == Example of Special Highly Composite Number == <onlyinclude> $2520$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers. </onlyinclude> == Proof == By inspection of the sequence of highly composite numbers, $2520$ is highly composite. For reference, the prime decomposition of $2520$ is: :$2520 = 2^3 \times 3^2 \times 5 \times 7$ {{AimForCont}} $n > 2520$ is a highly composite number which is not divisible by $2520$. We have that $60$ is a special highly composite number. Therefore $60$ is a divisor of $n$. It follows that $3$, $4$ and $5$ are all divisors of $n$. But as $2520$ is not a divisor of $n$, it follows that at least one of $7$, $8$ and $9$ is not a divisor of $n$. These will be investigated on a case-by-case basis. :$(1): \quad 7$  is not a divisor of $n$. By Prime Decomposition of Highly Composite Number we have that: :$n = 2^a \times 3^b \times 5^c$ where $a \ge b \ge c \ge 1$. Suppose that $a < 3$. Then: :$n \le 2^2 \times 3^2 \times 5^2 = 900$ which is too small. So we have that $a \ge 3$. Then: {{begin-eqn}} {{eqn | l = 2^{a - 3} \times 3^b \times 5^c \times 7       | o = <       | r = 2^a \times 3^b \times 5^c       | c = because $7 < 2^3 = 8$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 3} \times 3^b \times 5^c \times 7}\right)       | o = <       | r = \tau \left({2^a \times 3^b \times 5^c}\right)       | c = as $2^a \times 3^b \times 5^c$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 3} \times 7}\right) \times \tau \left({3^b \times 5^c}\right)       | o = <       | r = \tau \left({2^a}\right) \times \tau \left({3^b \times 5^c}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 3} \times 7}\right)       | o = <       | r = \tau \left({2^a}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({\left({a - 3}\right) + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({a + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 2 \left({a - 2}\right)       | o = <       | r = a + 1       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = <       | r = 5       | c =  }} {{end-eqn}} Now suppose that $b < 2$. Then: :$n \le 2^5 \times 3 \times 5 = 480$ which is too small. So we have that $b \ge 2$. Then: {{begin-eqn}} {{eqn | l = 2^a \times 3^{b - 2} \times 5^c \times 7       | o = <       | r = 2^a \times 3^b \times 5^c       | c = because $7 < 3^2 = 9$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^a \times 3^{b - 2} \times 5^c \times 7}\right)       | o = <       | r = \tau \left({2^a \times 3^b \times 5^c}\right)       | c = as $2^a \times 3^b \times 5^c$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^a \times 5^c}\right) \times \tau \left({3^{b - 2} \times 7}\right)       | o = <       | r = \tau \left({2^a \times 5^c}\right) \times \tau \left({3^b}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({3^{b - 2} \times 7}\right)       | o = <       | r = \tau \left({3^b}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({\left({b - 2}\right) + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({b + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 2 \left({b - 1}\right)       | o = <       | r = b + 1       | c =  }} {{eqn | ll= \leadsto       | l = b       | o = <       | r = 3       | c =  }} {{end-eqn}} But $c \le b$ and so $c < 3$ as well. Thus we have upper bounds on $a$, $b$ and $c$. Since $2^a \times 3^b \times 5^c > 2520$, it must be the case that: :$n = 2^4 \times 3^2 \times 5^2$ which gives that $n = 3600$. But: : from {{TauLink|3600}} we have that $\tau \left({3600}\right) = 45$ : from {{TauLink|2520}} we have that $\tau \left({2520}\right) = 48$ This contradicts our hypothesis that $3600$ is highly composite. By Proof by Contradiction it follows that $7$ must be a divisor of $n$. {{qed|lemma}} :$(2): \quad 9$ is not a divisor of $n$, but $7$ is. By Prime Decomposition of Highly Composite Number we have that: :$n = 2^a \times 3^1 \times 5^1 \times 7^1 \times 11^e \times r$ where: :$e$ is either $0$ or $1$ :$r$ is a possibly vacuous square-free product of prime numbers strictly greater than $11$. Suppose $e = 1$. Then: {{begin-eqn}} {{eqn | l = 2^a \times 3^3 \times 5 \times 7 \times r       | o = <       | r = 2^a \times 3 \times 5 \times 7 \times 11 \times r       | c = because $9 = 3^2 < 11$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^a \times 3^3 \times 5 \times 7 \times r}\right)       | o = <       | r = \tau \left({2^a \times 3 \times 5 \times 7 \times 11 \times r}\right)       | c = as $2^a \times 3 \times 5 \times 7 \times 11 \times r$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^a \times 5 \times 7 \times r}\right) \times \tau \left({3^3}\right)       | o = <       | r = \tau \left({2^a \times 5 \times 7 \times r}\right) \times \tau \left({3 \times 11}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({3^3}\right)       | o = <       | r = \tau \left({3 \times 11}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = 3 + 1       | o = <       | r = \left({1 + 1}\right) \left({1 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 4       | o = <       | r = 4       | c = which is an absurdity }} {{end-eqn}} So $e = 0$ and so by Prime Decomposition of Highly Composite Number $r = 1$. Thus: :$n = 2^a \times 3 \times 5 \times 7$ We have that $n > 2520$, so: :$(2 \text a): a \ge 5$ Then: {{begin-eqn}} {{eqn | l = 2^{a - 2} \times 3^2 \times 5 \times 7       | o = <       | r = 2^a \times 3 \times 5 \times 7       | c = because $3 < 2^2 = 4$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 2} \times 3^2 \times 5 \times 7}\right)       | o = <       | r = \tau \left({2^a \times 3 \times 5 \times 7}\right)       | c = as $2^a \times 3 \times 5 \times 7$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 2} \times 3^2}\right) \times \tau \left({5 \times 7}\right)       | o = <       | r = \tau \left({2^a \times 3}\right) \times \tau \left({5 \times 7}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 2} \times 3^2}\right)       | o = <       | r = \tau \left({2^a \times 3}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({\left({a - 2}\right) + 1}\right) \left({2 + 1}\right)       | o = <       | r = \left({a + 1}\right) \left({1 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 3 \left({a - 1}\right)       | o = <       | r = 2 \left({a + 1}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = a       | o = <       | r = 5       | c = which is a contradiction of $(2 \text a)$ }} {{end-eqn}} It follows by Proof by Contradiction that $9$ is a divisor of $n$. {{qed|lemma}} :$(3): \quad 8$ is not a divisor of $n$, but $7$ and $9$ both are. By Prime Decomposition of Highly Composite Number we have that: :$n = 2^2 \times 3^2 \times 5^c \times 7^d \times 11^e \times r$ where $r$ is a possibly vacuous product of prime numbers strictly greater than $11$. Suppose: :$(3 \text a): \quad e > 0$ Then: {{begin-eqn}} {{eqn | l = 2^5 \times 3^2 \times 5^c \times 7^d \times 11^{e - 1} \times r       | o = <       | r = 2^2 \times 3^2 \times 5^c \times 7^d \times 11^e \times r       | c = because $8 = 2^3 < 11$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 3^2 \times 5^c \times 7^d \times 11^{e - 1} \times r}\right)       | o = <       | r = \tau \left({2^2 \times 3^2 \times 5^c \times 7^d \times 11^e \times r}\right)       | c = as $2^2 \times 3^2 \times 5^c \times 7^d \times 11^e \times r$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 11^{e - 1} }\right) \times \tau \left({3^2 \times 5^c \times 7^d \times r}\right)       | o = <       | r = \tau \left({2^2 \times 11^e}\right) \times \tau \left({3^2 \times 5^c \times 7^d \times r}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^5 \times 11^{e - 1} }\right)       | o = <       | r = \tau \left({2^2 \times 11^e}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({5 + 1}\right) \left({\left({e - 1}\right) + 1}\right)       | o = <       | r = \left({2 + 1}\right) \left({e + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 6 e       | o = <       | r = 3 \left({e + 1}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = e       | o = <       | r = 1       | c = which is a contradiction of $(3 \text a)$ }} {{end-eqn}} So $e = 0$ and so Prime Decomposition of Highly Composite Number $r = 1$. Thus: :$n = 2^2 \times 3^2 \times 5^c \times 7^d$ Suppose $c = 2$. Then: {{begin-eqn}} {{eqn | l = 2^4 \times 3^2 \times 5^1 \times 7^d       | o = <       | r = 2^2 \times 3^2 \times 5^2 \times 7^d       | c = because $4 = 2^2 < 5$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^4 \times 3^2 \times 5^1 \times 7^d}\right)       | o = <       | r = \tau \left({2^2 \times 3^2 \times 5^2 \times 7^d}\right)       | c = as $2^2 \times 3^2 \times 5^2 \times 7^d$ is highly composite }} {{eqn | ll= \leadsto       | l = \tau \left({2^4 \times 5}\right) \times \tau \left({3^2 \times 7^d}\right)       | o = <       | r = \tau \left({2^2 \times 5^2}\right) \times \tau \left({3^2 \times 7^d}\right)       | c = Tau Function is Multiplicative }} {{eqn | ll= \leadsto       | l = \tau \left({2^4 \times 5}\right)       | o = <       | r = \tau \left({2^2 \times 5^2}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = \left({4 + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({2 + 1}\right) \left({2 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 10       | o = <       | r = 9       | c = which is an absurdity }} {{end-eqn}} The remaining possibility is that $c = 1$ and $d = 1$: Thus: :$n = 2^2 \times 3^2 \times 5 \times 7 = 1260$ But this is a contradiction of our supposition that $n > 2520$. It follows by Proof by Contradiction that $8$ is a divisor of $n$. {{qed|lemma}} By Proof by Cases it is seen that the existence of a highly composite $n$ not divisible by $2520$ leads to a contradiction. The result then follows by Proof by Contradiction. {{qed}} == 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}} Category:Special Highly Composite Numbers Category:2520 h5re8uupp5lyxmk06qhw4oe3ena9n0u


Sigma Function of 2620

Sigma Function of 2620 0 61119 318211 318209 2017-09-16T18:24:46Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({2620}\right) = 5544$ </onlyinclude> 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}} {{qed}} Category:Sigma Function of Integer Category:2620 oo4wii402j4f183zvi1wjki1geo26ro


Sigma Function of 2924

Sigma Function of 2924 0 61123 318232 318230 2017-09-16T18:27:48Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({2924}\right) = 5544$ </onlyinclude> 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}} {{qed}} Category:Sigma Function of Integer Category:2924 lpi8bmft2jg3kv0jl0f7apaxhn8xjbk


Sigma Function of 5020

Sigma Function of 5020 0 61419 318288 318286 2017-09-16T18:36:38Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({5020}\right) = 10 \, 584$ </onlyinclude> 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}} {{qed}} Category:Sigma Function of Integer Category:5020 3fvyay6mnnqrsutdmzfx7aen7y3uq12


Sigma Function of 5564

Sigma Function of 5564 0 61422 449562 318246 2020-02-17T10:32:33Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\map \sigma {5564} = 10 \, 584$ </onlyinclude> 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}} {{qed}} Category:Sigma Function of Integer Category:5564 dkj32zjugpiuvxaye2xd0q0esa16uhu


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 == <onlyinclude> :$\phi \left({5186}\right) = 2592$ </onlyinclude> where $\phi$ denotes the Definition:Euler Phi Fu..." wikitext text/x-wiki == Example of Euler Phi Function of 2 times Odd Prime == <onlyinclude> :$\phi \left({5186}\right) = 2592$ </onlyinclude> 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}} {{qed}} Category:Euler Phi Function of 2 times Odd Prime Category:5186 6u42hodcge8arxocxxfdwq9byioakzg


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 == <onlyinclude> :$\phi \left({5187}\right) = 2592$ </onlyinclude> whe..." wikitext text/x-wiki == Example of Euler $\phi$ Function of Square-Free Integer == <onlyinclude> :$\phi \left({5187}\right) = 2592$ </onlyinclude> 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}} {{qed}} Category:Euler Phi Function of Square-Free Integer Category:5187 pfzyjagdkn9eh3xfjg84vlul4jf99nt


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 == <onlyinclude> :$\phi \left({5188}\right) = 2592$ </onlyinclude> where $\phi$ denotes the Defi..." wikitext text/x-wiki == Example of Use of Euler $\phi$ Function == <onlyinclude> :$\phi \left({5188}\right) = 2592$ </onlyinclude> 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}} {{qed}} Category:Euler Phi Function Category:5188 ksc0vo75vdfeh9ceq0pw48wdpg33m9c


Hurwitz's Theorem (Number Theory)/Lemma 1 0 61447 408768 314012 2019-06-17T20:24:04Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> 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: :$\size {\xi - \dfrac p q} < \dfrac 1 {A \, q^2}$ </onlyinclude> == 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. {{AimForCont}} 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 {{LHS}} of $(1)$ becomes less than $1$. At the same time, the {{RHS}} 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. {{qed}} {{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 $194$ Category:Hurwitz's Theorem (Number Theory) 3jryiijqyoi0w0w4np98gsgoqlqeckx


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 == <onlyinclude> 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 == <onlyinclude> 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: :$\left|{\xi - \dfrac p q}\right| < \dfrac 1 {A \, q^2}$ </onlyinclude> == 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. {{qed}} == 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 $195$ Category:Hurwitz's Theorem (Number Theory) mtf3n0u2276ibpmkdowb12f8np3vhc4


Sigma Function of 6232

Sigma Function of 6232 0 62161 318239 318237 2017-09-16T18:29:08Z Prime.mover 59 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({6232}\right) = 12 \, 600$ </onlyinclude> 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}} {{qed}} Category:Sigma Function of Integer Category:6232 3wiisvb0sskiv79patt7x501veejja0


Sigma Function of 6368

Sigma Function of 6368 0 62162 318240 318236 2017-09-16T18:29:20Z Prime.mover 59 Prime.mover moved page Sigma of 6368 to Sigma Function of 6368 wikitext text/x-wiki == Example of Sigma Function of Integer == <onlyinclude> :$\sigma \left({6368}\right) = 12 \, 600$ </onlyinclude> 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}} {{qed}} Category:Sigma Function of Integer Category:6368 iidrztpqoybupgrji3pcirgmyhsiy7p


Completion Theorem (Measure Spaces)/Lemma 0 62230 317250 317249 2017-09-13T21:08:39Z Lord Farin 560 wikitext text/x-wiki == 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: :$\bar \mu \left({E \cup N}\right) := \mu \left({E}\right)$ <onlyinclude> The mapping $\bar \mu$ is well-defined, i.e.: :$\forall E, F \in \Sigma: \forall N, M \in \mathcal N: E \cup N = F \cup M \implies \mu \left({E}\right) = \mu \left({F}\right)$ </onlyinclude> == 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: :$\mu \left({E}\right) = \mu \left({F}\right)$ {{qed}} Category:Measure Theory q3ci6501bmbncb84lqewf3oofmb6nkt


Language of Propositional Logic has Unique Parsability/Lemma 0 62976 352583 319218 2018-05-01T21:04:07Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $\mathcal L_0$ be the language of propositional logic. <onlyinclude> 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 $\mathbf C = \mathbf E$. </onlyinclude> == 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. {{qed}} == 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 $1.4.2$ Category:Language of Propositional Logic j1eji49p7l5w74u7w2cl9j60e14lacs


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 == <onlyinclude> Let $\Gamma$ be a finished branch of a Definition:Proposit..." wikitext text/x-wiki == Corollary to Finished Branch Lemma == <onlyinclude> 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. </onlyinclude> == Proof == {{Finish|Branch not contradictory. See Finished Set Lemma}} Category:Propositional Tableaus 74txt8zkki3759oge83rh2k2wi40qnn


Pythagorean Triangle/Examples/4485-5852-7373 0 63453 478778 478748 2020-07-18T12:16:53Z Prime.mover 59 wikitext text/x-wiki == Example of Primitive Pythagorean Triangle == <onlyinclude> The triangle whose sides are of length $4485$, $5852$ and $7373$ is a primitive Pythagorean triangle. </onlyinclude> :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. {{qed}} Category:Examples of Pythagorean Triangles rsjb3kgh1ab6emmlccixjpdz13jyq52


Pythagorean Triangle/Examples/3059-8580-9109 0 63454 478782 478746 2020-07-18T12:20:06Z Prime.mover 59 wikitext text/x-wiki == Example of Primitive Pythagorean Triangle == <onlyinclude> The triangle whose sides are of length $3059$, $8580$ and $9109$ is a primitive Pythagorean triangle. </onlyinclude> :700px It has generator $\tuple {78, 55}$. == Proof == We have: {{begin-eqn}} {{eqn | l = 78^2 - 55^2       | r = 6084 - 3025       | c =  }} {{eqn | r = 3059       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 2 \times 78 \times 55       | r = 8580       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 78^2 + 55^2       | r = 6084 + 3025       | c =  }} {{eqn | r = 9109       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 3059^2 + 8580^2       | r = 9 \, 357 \, 481 + 73 \, 616 \, 400       | c =  }} {{eqn | r = 82 \, 973 \, 881       | c =  }} {{eqn | r = 9109^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $3059$, $8580$ and $9109$  form a Pythagorean triple. We have that: {{begin-eqn}} {{eqn | l = 3059       | r = 7 \times 19 \times 23       | c =  }} {{eqn | l = 8580       | r = 2^2 \times 3 \times 5 \times 11 \times 13       | c =  }} {{end-eqn}} It is seen that $3059$ and $8580$ share no prime factors. That is, $3059$ and $8580$ are coprime. Hence, by definition, $3059$, $8580$ and $9109$ form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle. {{qed}} Category:Examples of Pythagorean Triangles s2evzbxc685l7qigf75rmq6fh6k46es


Pythagorean Triangle/Examples/1380-19,019-19,069 0 63455 478798 478740 2020-07-18T12:30:57Z Prime.mover 59 wikitext text/x-wiki == Example of Primitive Pythagorean Triangle == <onlyinclude> The triangle whose sides are of length $1380$, $19 \, 019$ and $19 \, 069$ is a primitive Pythagorean triangle. </onlyinclude> :700px It has generator $\left({138, 5}\right)$. == Proof == We have: {{begin-eqn}} {{eqn | l = 138^2 - 5^2       | r = 19 \, 044 - 25       | c =  }} {{eqn | r = 19 \, 019       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 2 \times 138 \times 5       | r = 1380       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 138^2 + 5^2       | r = 19 \, 044 + 25       | c =  }} {{eqn | r = 19 \, 069       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 1380^2 + 19 \, 019^2       | r = 1 \, 904 \, 400 + 361 \, 722 \, 361       | c =  }} {{eqn | r = 363 \, 626 \, 761       | c =  }} {{eqn | r = 19 \, 069^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $1380$, $19 \, 019$ and $19 \, 069$  form a Pythagorean triple. We have that: {{begin-eqn}} {{eqn | l = 1380       | r = 2^2 \times 3 \times 5 \times 23       | c =  }} {{eqn | l = 19 \, 019       | r = 7 \times 11 \times 13 \times 19       | c =  }} {{end-eqn}} It is seen that $1380$ and $19 \, 019$ share no prime factors. That is, $1380$ and $19 \, 019$ are coprime. Hence, by definition, $1380$, $19 \, 019$ and $19 \, 069$ form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle. {{qed}} Category:Examples of Pythagorean Triangles k7a305zpkonaow2uzak0kn3stqp6v1m


Carmichael Number/Examples/294,409

Carmichael Number/Examples/294,409 0 64010 392221 333769 2019-02-12T16:07:42Z Prime.mover 59 wikitext text/x-wiki == Example of Carmichael Number == $294 \, 409$ is a Carmichael number: <onlyinclude> :$\forall a \in \Z: a \perp 294 \, 409: a^{294 \, 409} \equiv a \pmod {294 \, 409}$ while $294 \, 409$ is composite. </onlyinclude> == Proof == We have that: :$294 \, 409 = 37 \times 73 \times 109$ First note that $294 \, 409$ is square-free. Hence the square of none of its prime factors is a divisor of $294 \, 409$: :$\forall p \divides 294 \, 409: p^2 \nmid 294 \, 409$ We also see that: {{begin-eqn}} {{eqn | l = 294 \, 408       | r = 2^3 \times 3^3 \times 29 \times 47       | c =  }} {{eqn | r = 8178 \times 36       | c =  }} {{eqn | r = 4089 \times 72       | c =  }} {{eqn | r = 2726 \times 108       | c =  }} {{end-eqn}} Thus $294 \, 409$ is a Carmichael number by Korselt's Theorem. {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Carmichael Number/Examples/1729|next = Smallest Integer which is Sum of 2 Fourth Powers in 2 Ways}}: $509,033,161$ Category:Carmichael Numbers Category:294,409 78fchhna1ilh2bjp5ltl9itndk8c6wg


Euler Phi Function of 6

Euler Phi Function of 6 0 64278 391878 324134 2019-02-09T23:41:24Z Prime.mover 59 wikitext text/x-wiki == Example of Euler $\phi$ Function of Square-Free Integer == <onlyinclude> :$\map \phi 6 = 2$ </onlyinclude> where $\phi$ denotes the Euler $\phi$ Function. == Proof == From Euler Phi Function of Square-Free Integer: :$\displaystyle \map \phi n = \prod_{\substack {p \mathop \divides n \\ p \mathop > 2} } \paren {p - 1}$ where $p \divides n$ denotes the primes which divide $n$. We have that: :$6 = 2 \times 3$ and so is square-free. Thus: {{begin-eqn}} {{eqn | l = \map \phi 6       | r = \paren {3 - 1}       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} {{qed}} Category:Euler Phi Function of Square-Free Integer Category:6 gi0ykexw3gggcps7d1ep2kk89o3valv


Odd Amicable Pair/Examples/29,912,035,725-34,883,817,075 0 64360 467815 324454 2020-05-14T06:00:22Z Prime.mover 59 wikitext text/x-wiki == Example of Odd Amicable Pair == <onlyinclude> $29 \, 912 \, 035 \, 725$ and $34 \, 883 \, 817 \, 075$ are an odd amicable pair: :$\map \sigma {29 \, 912 \, 035 \, 725} = \map \sigma {34 \, 883 \, 817 \, 075} = 64 \, 795 \, 852 \, 800 = 29 \, 912 \, 035 \, 725 + 34 \, 883 \, 817 \, 075$ </onlyinclude> == Proof == By definition, $m$ and $n$ form an amicable pair {{iff}}: :$\map \sigma m = \map \sigma n = m + n$ where $\map \sigma n$ denotes the $\sigma$ function of $n$. Thus: {{begin-eqn}} {{eqn | l = \map \sigma {29 \, 912 \, 035 \, 725}       | r = 64 \, 795 \, 852 \, 800       | c = {{SigmaLink|29,912,035,725|29 \, 912 \, 035 \, 725}} }} {{eqn | r = 29 \, 912 \, 035 \, 725 + 34 \, 883 \, 817 \, 075       | c =  }} {{eqn | r = \map \sigma {34 \, 883 \, 817 \, 075}       | c = {{SigmaLink|34,883,817,075|34 \, 883 \, 817 \, 075}} }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 64,795,852,800|next = Odd Amicable Pair/Examples/31,695,652,275-33,100,200,525}}: $64,795,852,800$ Category:Amicable Pairs sl6nmndx3i6chz3kyt71ni5m1qdfbg6


Odd Amicable Pair/Examples/31,695,652,275-33,100,200,525 0 64361 324455 2017-11-03T07:11:08Z Prime.mover 59 Created page with "== Example of Odd Amicable Pair == <onlyinclude> $31 \, 695 \, 652 \, 275$ and $33 \, 100 \, 200 \, 525$ are an Defin..." wikitext text/x-wiki == Example of Odd Amicable Pair == <onlyinclude> $31 \, 695 \, 652 \, 275$ and $33 \, 100 \, 200 \, 525$ are an odd amicable pair: :$\sigma \left({31 \, 695 \, 652 \, 275}\right) = \sigma \left({33 \, 100 \, 200 \, 525}\right) = 64 \, 795 \, 852 \, 800 = 31 \, 695 \, 652 \, 275 + 33 \, 100 \, 200 \, 525$ </onlyinclude> == Proof == By definition, $m$ and $n$ form an amicable pair {{iff}}: :$\sigma \left({m}\right) = \sigma \left({n}\right) = m + n$ where $\sigma \left({n}\right)$ denotes the $\sigma$ function. Thus: {{begin-eqn}} {{eqn | l = \sigma \left({31 \, 695 \, 652 \, 275}\right)       | r = 64 \, 795 \, 852 \, 800       | c = {{SigmaLink|31,695,652,275|31 \, 695 \, 652 \, 275}} }} {{eqn | r = 31 \, 695 \, 652 \, 275 + 33 \, 100 \, 200 \, 525       | c =  }} {{eqn | r = \sigma \left({33 \, 100 \, 200 \, 525}\right)       | c = {{SigmaLink|33,100,200,525|33 \, 100 \, 200 \, 525}} }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Odd Amicable Pair/Examples/29,912,035,725-34,883,817,075|next = Odd Amicable Pair/Examples/32,129,958,525-32,665,894,275}}: $64,795,852,800$ Category:Amicable Pairs l41u780ot9j3ot1lx6zoqja2377ajfo


Odd Amicable Pair/Examples/32,129,958,525-32,665,894,275 0 64362 324456 2017-11-03T07:15:34Z Prime.mover 59 Created page with "== Example of Odd Amicable Pair == <onlyinclude> $32 \, 129 \, 958 \, 525$ and $32 \, 665 \, 894 \, 275$ are an Defin..." wikitext text/x-wiki == Example of Odd Amicable Pair == <onlyinclude> $32 \, 129 \, 958 \, 525$ and $32 \, 665 \, 894 \, 275$ are an odd amicable pair: :$\sigma \left({32 \, 129 \, 958 \, 525}\right) = \sigma \left({32 \, 665 \, 894 \, 275}\right) = 64 \, 795 \, 852 \, 800 = 32 \, 129 \, 958 \, 525 + 32 \, 665 \, 894 \, 275$ </onlyinclude> == Proof == By definition, $m$ and $n$ form an amicable pair {{iff}}: :$\sigma \left({m}\right) = \sigma \left({n}\right) = m + n$ where $\sigma \left({n}\right)$ denotes the $\sigma$ function. Thus: {{begin-eqn}} {{eqn | l = \sigma \left({32 \, 129 \, 958 \, 525}\right)       | r = 64 \, 795 \, 852 \, 800       | c = {{SigmaLink|32,129,958,525|32 \, 129 \, 958 \, 525}} }} {{eqn | r = 32 \, 129 \, 958 \, 525 + 32 \, 665 \, 894 \, 275       | c =  }} {{eqn | r = \sigma \left({32 \, 665 \, 894 \, 275}\right)       | c = {{SigmaLink|32,665,894,275|32 \, 665 \, 894 \, 275}} }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Odd Amicable Pair/Examples/31,695,652,275-33,100,200,525|next = Multiply Perfect Number of Order 6}}: $64,795,852,800$ Category:Amicable Pairs h8hkza7rcn54glac3ff8jchw53909h2


Aurifeuillian Factorization/Examples/2^4n+2 + 1 0 65177 385072 385071 2018-12-28T15:20:25Z Prime.mover 59 wikitext text/x-wiki == Example of Aurifeuillian Factorizations == <onlyinclude> :$2^{4 n + 2} + 1 = \paren {2^{2 n + 1} - 2^{n + 1} + 1} \paren {2^{2 n + 1} + 2^{n + 1} + 1}$ </onlyinclude> == Proof == From Sum of Squares as Product of Factors with Square Roots: :$x^2 + y^2 = \paren {x + \sqrt {2 x y} + y} \paren {x - \sqrt {2 x y} + y}$ Let $x = 2^{2 n + 1}$ and $y = 1$. Then: {{begin-eqn}} {{eqn | l = 2^{4 n + 2} + 1       | r = \paren {2^{2 n + 1} }^2 + 1^2       | c =  }} {{eqn | r = \paren {2^{2 n + 1} + \sqrt {2 \times \paren {2^{2 n + 1} } \times 1} + 1} \paren {2^{2 n + 1} - \sqrt {2 \times \paren {2^{2 n + 1} } \times 1} + 1}       | c =  }} {{eqn | r = \paren {2^{2 n + 1} + \sqrt {2^{2 n + 2} } + 1} \paren {2^{2 n + 1} - \sqrt {2^{2 n + 2} } + 1}       | c =  }} {{eqn | r = \paren {2^{2 n + 1} - 2^{n + 1} + 1} \paren {2^{2 n + 1} + 2^{n + 1} + 1}       | c =  }} {{end-eqn}} {{qed}} == Historical Note == {{:Aurifeuillian Factorization/Examples/2^4n+2 + 1/Historical Note}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Prime Decomposition of 2^58+1/Historical Note|next = Aurifeuillian Factorization/Examples/2^4n+2 + 1/Historical Note}}: $2^{58} + 1$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Prime Decomposition of 2^58+1/Historical Note|next = Aurifeuillian Factorization/Examples/2^4n+2 + 1/Historical Note}}: $2^{58} + 1$ Category:Aurifeuillian Factorizations 0c3oxexmsom4rnbzn48k0x3fupkqo47


Tutte's Wheel Theorem/Lemma

Tutte's Wheel Theorem/Lemma 0 66822 340403 340082 2018-01-28T07:02:36Z Stixme 3311 wikitext text/x-wiki {{MissingLinks}} {{tidy}} == Lemma == If a graph $G$ is 3-connected with $|V\left({G}\right)| > 4$ then $\exists e, \: e \in E\left({G}\right)$ such that $G \thinspace / \thinspace e$ is also 3-connected. == Proof == Suppose that no such edge $e$ exists.  Then $\forall e, \: e = xy \in E\left({G}\right)$, $\: G \thinspace / \thinspace e$ contains a vertex cut $S, \: |S| ≤ 2$. Since $\kappa\left({G}\right) \geq 3$, the contracted vertex $v_{x,y}$ of $G \thinspace / \thinspace e$ lies in $S$ (i.e. $\exists z, \: z \in G, \: z \notin \{x,y\}$.)   Let $S = \{v_{x,y}, z\}$. Then $T = \{x, y, z\}$ is a vertex cut of $G$.  Thus, every vertex in $T$ has an edge to every component of $G'=G-T$.  Let $C$ be the smallest component of $G'$.  Let $u \in N\left({z}\right) \cap C$ where $N\left({z}\right)$ is the set of all neighbours of the vertex $z$. By assumption, $G \thinspace / \thinspace zu$ is again not 3-connected, so again $\exists w$ such that $\{w, z, u\}$ is a vertex cut of $G$.  It also follows that every vertex in $\{w, z, u\}$ has an edge to every component of $G'' = G - \{w, z, u\}$.  Since $x,y$ are connected, $\exists D$, $D$ is a component of $G''$ and $D \cap \{x,y\} = \emptyset$.  Then, $D \subseteq N\left({z}\right) \cap V\left({C}\right) = \emptyset$.  Hence $D \varsubsetneqq C$ by the choice of $D$, which contradicts the assumption that $C$ was the smallest component.  {{qed|lemma}} Category:Graph Theory b4if14xx8b5hx14s014gee025jvwj56


Completion Theorem (Metric Space)/Lemma 1 0 66833 453813 340173 2020-03-10T15:50:34Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $M = \struct {A, d}$ be a metric space. Let $\CC \sqbrk A$ denote the set of all Cauchy sequences in $A$. Define the equivalence relation $\sim$ on $\CC \sqbrk A$ by: :$\displaystyle \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$ Denote the equivalence class of $\sequence {x_n} \in \CC \sqbrk A$ by $\sqbrk {x_n}$. Denote the set of equivalence classes under $\sim$ by $\tilde A$. Define $\tilde d: \tilde A \to \R_{\ge 0}$ by: :$\displaystyle \map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$ Then: <onlyinclude> :$\tilde d$ is well-defined on $\tilde A$. </onlyinclude> == Proof == Let $\sequence {x_n}$, $\sequence {\hat x_n}$, $\sequence {y_n}$, $\sequence {\hat y_n} \in \CC \sqbrk A$ be such that: :$\sequence {x_n} \sim \sequence {\hat x_n}$ :$\sequence {y_n} \sim \sequence {\hat y_n}$ We have: {{begin-eqn}} {{eqn | l = \map d {x_n, y_n} - \map d {\hat x_n, \hat y_n}       | o = \le       | r = \map d {x_n, \hat x_n} + \map d {\hat x_n, y_n} - \map d {\hat x_n, \hat y_n}       | c = Triangle Inequality }} {{eqn | o = \le       | r = \map d {x_n, \hat x_n} + \map d {\hat x_n, \hat y_n} + \map d {\hat y_n, y_n} - \map d {\hat x_n, \hat y_n}       | c = Triangle Inequality }} {{eqn | r = \map d {x_n, \hat x_n} + \map d {\hat y_n, y_n}       | c =  }} {{end-eqn}} By an identical argument, we can also show that: :$\map d {\hat x_n, \hat y_n} - \map d {x_n, y_n} \le \map d {x_n, \hat x_n} + \map d {\hat y_n, y_n}$  and therefore: :$\displaystyle 0 \le \size {\map d {x_n, y_n} - \map d {\hat x_n, \hat y_n} } \le \map d {x_n, \hat x_n} + \map d {\hat y_n, y_n}$ Passing to the limit $n \to \infty$ and using the Combination Theorem for Sequences we have shown that: $\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \lim_{n \mathop \to \infty} \map d {\hat x_n, \hat y_n}$ But this precisely means that: :$\map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} } = \map {\tilde d} {\sqbrk {\hat x_n}, \sqbrk {\hat y_n} }$ {{qed}} Category:Completion Theorem 20d0c7shgcq6vlnumi2nzcsg5cx4w0g


Completion Theorem (Metric Space)/Lemma 2 0 66834 364362 340176 2018-09-04T21:11:34Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $M = \left({A, d}\right)$ be a metric space. Let $\mathcal C \left[{A}\right]$ denote the set of all Cauchy sequences in $A$. Define the equivalence relation $\sim$ on $\mathcal C \left[{A}\right]$ by: :$\displaystyle \left\langle{x_n}\right\rangle \sim \left\langle{y_n}\right\rangle \iff \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right) = 0$ Denote the equivalence class of $\left\langle{x_n}\right\rangle \in \mathcal C \left[{A}\right]$ by $\left[{x_n}\right]$. Denote the set of equivalence classes under $\sim$ by $\tilde A$. Define $\tilde d: \tilde A \to \R_{\ge 0}$ by: :$\displaystyle \tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right)$ Then: <onlyinclude> :$\tilde d$ is a metric on $\tilde A$. </onlyinclude> == Proof == To prove $\tilde d$ is a metric, we verify that it satisfies the axioms $M1$, $M2$, $M3$ and $M4$. === Proof of $M4$ === Let $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = \infty$. Then $\left\langle{x_n}\right\rangle$ and $\left\langle y_n \right\rangle$ cannot both be Cauchy. So $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) < \infty$ for $\left[{x_n}\right], \left[{y_n}\right] \in \tilde A$. By the definition of $\tilde d$, for any $\left[{x_n}\right], \left[{y_n}\right] \in \tilde A$, $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right)$ must be a limit point of $R_{\ge 0}$. The closure of $\R_{\ge 0}$ is $\R_{\ge 0}$, so $\tilde d: \tilde A \times \tilde A \to \R_{\ge 0}$. So axiom $M4$ holds for $\tilde d$. === Proof of $M1$ === Let $\tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = 0$, which means that: :$\displaystyle \lim_{n \mathop \to \infty} d\left({x_n, y_n}\right) = 0$ So by definition: :$\left\langle{x_n}\right\rangle \sim \left\langle{y_n}\right\rangle$ and: :$\left[{x_n}\right] = \left[{y_n}\right]$ As $d$ is a metric, we also find immediately: :$\tilde d \left({\left[{x_n}\right], \left[{x_n}\right]}\right) = 0$ So axiom $M1$ holds for $\tilde d$. {{qed|lemma}} === Proof of $M3$ === We have that: {{begin-eqn}} {{eqn | l = \tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right)       | r = \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right)       | c = }} {{eqn | r = \lim_{n \mathop \to \infty} d \left({y_n, x_n}\right)       | c = $d$ is a Metric }} {{eqn | r = \tilde d \left({\left[{y_n}\right], \left[{x_n}\right]}\right)       | c = }} {{end-eqn}} So axiom $M3$ holds for $\tilde d$. {{qed|lemma}} === Proof of $M2$ === We have that: {{begin-eqn}} {{eqn | l = \tilde d \left({\left[{x_n}\right], \left[{z_n}\right]}\right)       | r = \lim_{n \mathop \to \infty} d \left({x_n, z_n}\right)       | c = }} {{eqn | o = \le       | r = \lim_{n \mathop \to \infty} \left\{ {d \left({x_n, y_n}\right) + d \left({y_n, z_n}\right)}\right\}       | c = $d$ is a metric, and using elementary properties of limits (Reference?) }} {{eqn | r = \lim_{n \mathop \to \infty} d\left({x_n, y_n}\right) +  \lim_{n \to \infty} d \left({y_n, z_n}\right)       | c = Sum Rule for Real Sequences }} {{eqn | r = \tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) + \tilde d \left({\left[{y_n}\right], \left[{z_n}\right]}\right) }} {{end-eqn}} So axiom $M2$ holds for $\tilde d$. {{qed|lemma}} Thus $\tilde d$ satisfies all the metric space axioms and so is a metric. {{qed}} Category:Completion Theorem irpehes3ivyz25mgvi3adr78b3pr3ht


Completion Theorem (Metric Space)/Lemma 3 0 66835 340181 340180 2018-01-27T11:17:19Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $M = \left({A, d}\right)$ be a metric space. Let $\mathcal C \left[{A}\right]$ denote the set of all Cauchy sequences in $A$. Define the equivalence relation $\sim$ on $\mathcal C \left[{A}\right]$ by: :$\displaystyle \left\langle{x_n}\right\rangle \sim \left\langle{y_n}\right\rangle \iff \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right) = 0$ Denote the equivalence class of $\left\langle{x_n}\right\rangle \in \mathcal C \left[{A}\right]$ by $\left[{x_n}\right]$. Denote the set of equivalence classes under $\sim$ by $\tilde A$. Define $\tilde d: \tilde A \to \R_{\ge 0}$ by: :$\displaystyle \tilde d \left({\left[{x_n}\right], \left[{y_n}\right]}\right) = \lim_{n \mathop \to \infty} d \left({x_n, y_n}\right)$ Then: <onlyinclude> :$\tilde M = \left({\tilde A, \tilde d}\right)$ is a completion of $M$. </onlyinclude> == Proof == We are to show that: : $(1): \quad \tilde M$ is a complete metric space : $(2): \quad A \subseteq \tilde A$ : $(3): \quad A$ is dense in $\tilde M$ : $(4): \quad \forall x, y \in A : \tilde d \left({x, y}\right) = d \left({x, y}\right)$ For $x \in A$, let $\hat x = \left({x, x, x, \ldots}\right)$ be the constant sequence with value $x$. Let $\phi: A \to \tilde A: x = \left[{\hat x}\right]$. We first demonstrate that $(2)$ holds, by showing that $A \subseteq \tilde A$. {{begin-eqn}} {{eqn | l = \phi \left({x}\right)       | r = \phi \left({y}\right) }} {{eqn | ll= \implies       | l = \left[{\hat x}\right]       | r = \left[{\hat y}\right] }} {{eqn | ll= \implies       | l = \lim_{n \mathop \to \infty} d \left({x, y}\right)       | r = 0 }} {{eqn | ll= \implies       | l = d \left({x, y}\right)       | r = 0 }} {{eqn | ll= \implies       | l = x       | r = y }} {{end-eqn}} Thus: :$A \subseteq \tilde A$ {{qed|lemma}} Henceforth we identify $A$ with its isomorphic copy in $\tilde A$ when it is convenient. Now we demonstrate that $(4)$ holds, by showing that $\phi$ is an injection from $A$ into $\tilde A$. For any $x, y \in A$: {{begin-eqn}} {{eqn | l = \tilde d \left({\left[{\hat x}\right], \left[{\hat y}\right]}\right)       | r = \lim_{n \mathop \to \infty} d \left({x, y}\right) }} {{eqn | r = d \left({x, y}\right) }} {{end-eqn}} That is: :$\forall x, y \in A : \tilde d \left({x, y}\right) = d \left({x, y}\right)$ {{qed|lemma}} Now we demonstrate that $(3)$ holds, by showing that $A$ is dense in $\tilde A$. Recall that the closure of $A$ is the union of $A$ and the limit points of $A$. Let $\left[{x_n}\right] \in \tilde A$ and $\epsilon > 0$ be arbitrary. If we can find $x \in A$ such that $\tilde d \left({\left[{\hat x}\right], \left[{x_n}\right]}\right) < \epsilon$ then we have shown that $A$ is dense in $\tilde A$. Since $\left\langle{x_n}\right\rangle$ is Cauchy, there exists $N \in \N$ such that: :$\forall m, n \ge N: d \left({x_m, x_n}\right) < \epsilon$ Then we have: {{begin-eqn}} {{eqn | l = \tilde d \left({\left[{\hat x_N}\right], \left[{x_n}\right]}\right)       | r = \lim_{n \mathop \to \infty} d \left({x_N, x_n}\right) }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} and therefore $A$ is dense in $\tilde A$. {{qed|lemma}} Finally we demonstrate that $(1)$ holds, by showing that $\left({\tilde A, \tilde d}\right)$ is complete. By the completeness criterion it is sufficient to show that every Cauchy sequence in $\phi \left({A}\right)$ converges in $\tilde A$. Let $\left\langle{\hat w_n}\right\rangle$ be a Cauchy sequence in $\phi \left({A}\right)$, so each $\hat w_n$ has the form $\left\langle{w_n, w_n, w_n, \ldots}\right\rangle$. Since $\phi$ is an isometry: :$\forall m, n \in \N: \tilde d \left({\hat w_n, \hat w_m}\right) = d \left({w_n, w_m}\right)$ Therefore, $\left\langle{w_1, w_2, w_3,\ldots}\right\rangle$ is Cauchy in $A$. Let $W = \left[{\left\langle{w_1, w_2, w_3, \ldots}\right\rangle}\right] \in \tilde A$. We claim that $\left\langle{\hat w_n}\right\rangle$ converges to $W$ in $\tilde A$. Let $\epsilon > 0$ be arbitrary. Since $\left\langle{w_1, w_2, w_3, \ldots}\right\rangle$ is Cauchy in $A$, there exists $N \in \N$ such that for all $m, n \ge N$, we have $d \left({w_n, w_m}\right) < \epsilon$. Thus for all $n > N$: :$\displaystyle \tilde d \left({w_n, W}\right) = \lim_{n \mathop \to \infty} d \left({w_n, W}\right) < \epsilon$  Therefore, $\left\langle{\hat w_n}\right\rangle \to W$ as $N \to \infty$, and $\tilde A$ is complete. {{qed}} {{ACC|Completeness Criterion (Metric Spaces)}} Category:Completion Theorem iunu09pstzvf3vvstpmluzt6nz0bbt7


Heine-Borel Theorem/Euclidean Space/Necessary Condition 0 67083 342472 2018-02-12T08:09:32Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> For any natural number $n \ge 1$, a closed and Definition:Bounded Metric Spac..." wikitext text/x-wiki == Theorem == <onlyinclude> For any natural number $n \ge 1$, a closed and bounded subspace of the Euclidean space $\R^n$ is compact. </onlyinclude> == Proof 1 == {{:Heine-Borel Theorem/Euclidean Space/Necessary Condition/Proof 1}} == Proof 2 == {{:Heine-Borel Theorem/Euclidean Space/Necessary Condition/Proof 2}} {{Namedfor|Heinrich Eduard Heine|name2 = Émile Borel}} Category:Heine-Borel Theorem d74r4b7lk9ga1i4go85xka0z7cy390g


Separated Subsets of Linearly Ordered Space under Order Topology/Lemma 0 67102 471130 342563 2020-05-28T07:05:25Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space. Let $A$ and $B$ be separated sets of $T$. Let $A^*$ and $B^*$ be defined as: :$A^* := \displaystyle \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$ :$B^* := \displaystyle \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$ where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$. Then: :$(1): \quad A \subseteq A^*$ :$(2): \quad B \subseteq B^*$ :$(3): \quad A^* \cap B^* = \O$ == Proof == Let $a \in A$. Then: {{begin-eqn}} {{eqn | l = \closedint a a       | r = \set a       | c =  }} {{eqn | ll= \leadsto       | l = \closedint a a \cap B^-       | r = \O       | c = {{Defof|Separated Sets}} }} {{eqn | ll= \leadsto       | l = \closedint a a       | o = \subseteq       | r = A^*       | c = Definition of $A^*$ }} {{eqn | ll= \leadsto       | l = a       | o = \in       | r = A^*       | c = Definition of $\closedint a a$ }} {{eqn | ll= \leadsto       | l = A       | o = \subseteq       | r = A^*       | c = {{Defof|Subset}} }} {{end-eqn}} Similarly, $B \subseteq B^-$. {{qed|lemma}} {{AimForCont}} $A^* \cap B^* \ne \O$. Then: :$\exists p: p \in A^* \cap B^*$ Hence: :$\exists a, b \in A, c, d \in B: p \in \closedint a b \cap \closedint c d$ But because $A$ and $B$ are separated sets: :$c, d \notin A$ and: :$a, b \notin B$ and so: :$\closedint a b \cap \closedint c d = \O$ Thus $p \notin \closedint a b \cap \closedint c d$ It follows by Proof by Contradiction that $A^* \cap B^* = \O$. {{qed}} == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Definition:Convex Component|next = Separated Subsets of Linearly Ordered Space under Order Topology}}: Part $\text {II}$: Counterexamples: $39$. Order Topology: $3$ Category:Linearly Ordered Spaces i2v65eccpzydkuytk9n5k95db573kxl


Power Series Expansion for Logarithm of 1 + x/Corollary 0 67448 363669 349899 2018-08-26T11:27:37Z Prime.mover 59 wikitext text/x-wiki == Corollary to Power Series Expansion for $\ln \paren {1 + x}$ == <onlyinclude> {{begin-eqn}} {{eqn | l = \ln \paren {1 - x}       | r = -\sum_{n \mathop = 1}^\infty \frac {x^n} n       | c =  }} {{eqn | r = -x - \frac {x^2} 2 - \frac {x^3} 3 - \frac {x^4} 4 - \cdots       | c =  }} {{end-eqn}} valid for $-1 < x < 1$. </onlyinclude> == Proof == By Power Series Expansion for $\ln \paren {1 + x}$: :$\displaystyle \ln \paren {1 + x} = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n$ Then: {{begin-eqn}} {{eqn | l = \ln \paren {1 - x}       | r = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\paren {-x}^n} n       | c = substituting $x \to -x$ }} {{eqn | r = -\sum_{n \mathop = 1}^\infty \paren {-1}^{2 n} \frac {x^n} n  }} {{eqn | r = -\sum_{n \mathop = 1}^\infty \frac {x^n} n  }} {{end-eqn}} {{qed}} Category:Examples of Power Series Category:Logarithms jjoh7jh41havb8b3zshple7iibi2yvy