(New page: Category:MA453Spring2009WaltherIf <math>\scriptstyle p</math> is a prime and <math>\scriptstyle p\ \mid\ a_1a_2\cdots a_n</math>, then <math>\scriptstyle p\ \mid\ a_1</math> or <math>\...) |
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[[Category:MA453Spring2009Walther]]If <math>\scriptstyle p</math> is a prime and <math>\scriptstyle p\ \mid\ a_1a_2\cdots a_n</math>, then <math>\scriptstyle p\ \mid\ a_1</math> or <math>\scriptstyle p\ \mid\ a_2a_3\cdots a_n</math>. If <math>\scriptstyle p\ \mid\ a_1</math>, we are done. Otherwise, <math>\scriptstyle p\ \mid\ a_2</math> or <math>\scriptstyle p\ \mid\ a_3a_4\cdots a_n</math>. This continues iteratively until <math>\scriptstyle p\ \mid\ a_{n-1}a_n</math>. Then <math>\scriptstyle p\ \mid\ a_{n-1}</math> or <math>\scriptstyle p\ \mid\ a_n</math>, and it is clear that <math>\scriptstyle p\ \mid\ a_i</math> for some <math>\scriptstyle1\ \le\ i\ \le\ n</math>. <math>\scriptstyle\Box</math> | [[Category:MA453Spring2009Walther]]If <math>\scriptstyle p</math> is a prime and <math>\scriptstyle p\ \mid\ a_1a_2\cdots a_n</math>, then <math>\scriptstyle p\ \mid\ a_1</math> or <math>\scriptstyle p\ \mid\ a_2a_3\cdots a_n</math>. If <math>\scriptstyle p\ \mid\ a_1</math>, we are done. Otherwise, <math>\scriptstyle p\ \mid\ a_2</math> or <math>\scriptstyle p\ \mid\ a_3a_4\cdots a_n</math>. This continues iteratively until <math>\scriptstyle p\ \mid\ a_{n-1}a_n</math>. Then <math>\scriptstyle p\ \mid\ a_{n-1}</math> or <math>\scriptstyle p\ \mid\ a_n</math>, and it is clear that <math>\scriptstyle p\ \mid\ a_i</math> for some <math>\scriptstyle1\ \le\ i\ \le\ n</math>. <math>\scriptstyle\Box</math> | ||
| − | :--[[User:Narupley| | + | :--[[User:Narupley|Nick Rupley]] 00:27, 22 January 2009 (UTC) |
Latest revision as of 19:28, 21 January 2009
If $ \scriptstyle p $ is a prime and $ \scriptstyle p\ \mid\ a_1a_2\cdots a_n $, then $ \scriptstyle p\ \mid\ a_1 $ or $ \scriptstyle p\ \mid\ a_2a_3\cdots a_n $. If $ \scriptstyle p\ \mid\ a_1 $, we are done. Otherwise, $ \scriptstyle p\ \mid\ a_2 $ or $ \scriptstyle p\ \mid\ a_3a_4\cdots a_n $. This continues iteratively until $ \scriptstyle p\ \mid\ a_{n-1}a_n $. Then $ \scriptstyle p\ \mid\ a_{n-1} $ or $ \scriptstyle p\ \mid\ a_n $, and it is clear that $ \scriptstyle p\ \mid\ a_i $ for some $ \scriptstyle1\ \le\ i\ \le\ n $. $ \scriptstyle\Box $
- --Nick Rupley 00:27, 22 January 2009 (UTC)
