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Given the number a:

$\ a^2\ = H$, where H is a perfect square. Let :

$\ a= c*d\ $, where c are it's prime factors.

How do we prove that:

$\ a^2\ $ = $\ (c * d)^2\ $ = $\ c^2 * d^2\ $

mathmaniage
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    if $a,c,d$ are naturals which i assume they are in this case, you essentially want to prove $c \cdot d = d \cdot c $ which is almost axiomatic – mdave16 Jul 31 '17 at 11:23
  • @mdave16, I want to prove (14)'s square is equal to the individual square of 7 and 2 – mathmaniage Jul 31 '17 at 11:28
  • stity's answer should be enough, unless you're looking to prove that multiplication is associative and commutative on the naturals, that is a slightly longer affair. – mdave16 Jul 31 '17 at 11:31

1 Answers1

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The multiplication on the naturals is associative and commutative :

  • Associative : $$a*(b*c)=(a*b)*c$$
  • Commutative : $$a*b=b*a$$

So you get $$(c*d)*(c*d)=c*d*c*d=c*c*d*d=c^2*d^2$$

stity
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