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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.

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