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 October 6th, 2016, 11:30 PM #1 Member   Joined: Sep 2016 From: zambia Posts: 31 Thanks: 0 irrational numbers 1. Show that if p^2 is divisible by 3,then p is also divisible by 3 October 7th, 2016, 06:23 AM #2 Member   Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 37 Thanks: 4 for p^2 to be devisible by 3, p needs to be 3 or composite number that have 3 at least once in his prime factors.. October 7th, 2016, 06:40 AM #3 Senior Member   Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 If p is divisible by 3, then: p=3n Let's make prime factorization of p: $\displaystyle p=a_1a_2a_3...a_m$ Then: $\displaystyle p^2=a_1a_1a_2a_2a_3a_3...a_ma_m$ Let's make prime factorization of 3n: $\displaystyle 3b_1b_2b_3...b_{2m-1}$ 2m-1 is because of unique factorization theorem, and because of the same theorem, one of the $\displaystyle a_1,a_2,a_3,...,a_m$ must equal to 3. So: $\displaystyle p=a_1a_2a_3...\cdot 3\cdot...a_m$ is divisible by 3. Thanks from topsquark October 7th, 2016, 07:05 AM   #4
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Quote:
 Originally Posted by jeho 1. Show that if p^2 is divisible by 3,then p is also divisible by 3
The contrapositive is much clearer for me: if $p$ is not divisible by $3$, then $p^2$ is not divisible by $3$. This is a direct consequence of the unique prime factorisation of integers. October 8th, 2016, 01:12 AM #5 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 If $p$ is divisible by 3 then it will also divisible by $p^2$ and $p^2$ is divisible by 3 then it will also divisible by $p$. October 8th, 2016, 05:59 AM   #6
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Quote:
 Originally Posted by deesuwalka If $p$ is divisible by 3 then > > it < < will also [be] divisible by $p^2$ and $p^2$ is divisible by 3 then it will also divisible by $p$.
This post doesn't make sense, just from your first 13 words/numbers alone, not including the word "be" I inserted.

Because of the ambiguous construction, the pronoun, "it," which I highlighted, could refer to p or to 3.

If "it" refers to p, then part of your statement means "p will also [be] divisible by $\displaystyle \ p^2. \ \$
That isn't true, unless p = $\displaystyle \ p^2 \ \ne \ 0.$

If "it" refers to 3, then that part of your statement means "3 will also [be] divisible by $\displaystyle \ p^2. \ \$
That isn't true unless p is -1 or 1.

Last edited by Math Message Board tutor; October 8th, 2016 at 06:15 AM. Tags irrational, numbers Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Pengkuan Math 61 February 29th, 2016 04:46 PM Albert.Teng Algebra 4 February 12th, 2014 04:55 PM niki500 Number Theory 5 October 7th, 2012 09:10 PM elim Number Theory 1 September 22nd, 2011 12:03 PM Mighty Mouse Jr Algebra 1 October 16th, 2010 07:46 PM

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