Prove that if m is a perfect square and n is a perfect square, then their product is a perfect square as well

Question

anonymous · Answer

The key here is$$m^2\cdot n^2=(m\cdot n)^2$$

lgbasallote · Answer

hmm

anonymous · Answer

nice...

anonymous · Answer

It has to do with the fact that exponents distribute over multiplication.

lgbasallote · Answer

why do you square m and n?

swissgirl · Answer

Suppose m and n  are perfect squares then by definition m=x^2 and n=y^2
The product of m*n=x^2*y^2 Which are perfect squres
$\sqrt{x^2*y^2}$=x*y
Hence The product of two perfect squraes is a perfect square

anonymous · Answer

oops, ummm

anonymous · Answer

Well maybe what I should have said was $$m=j^2$$$$n=k^2$$$$j^2\cdot k^2=(j\cdot k)^2$$

lgbasallote · Answer

yes that;s what i thought too @swissgirl and @wio but for some reason...it's not a valid proof...

anonymous · Answer

maybe they want you to show an intermediate step like:
$$j^2\cdot k^2=j\cdot j \cdot k \cdot k = j\cdot k \cdot j \cdot k =(j\cdot k)(j \cdot k)=(j \cdot k)^2 $$

swissgirl · Answer

Suppose m and n are perfect squares then by definition m=x^2 and n=y^2 
The product of m*n=x^2*y^2=(x*y)^2 Which is perfect squares 
Hence The product of two perfect squraes is a perfect square

lgbasallote · Answer

ah i got it. i have to substitude jk into x so it will be the definition of a perfect square
$$j^2k^2\implies (jk)^2 \implies x^2$$

swissgirl · Answer

Yessss

anonymous · Answer

You really don't have to do the $x$ think though.

lgbasallote · Answer

well the definition of a perfect square is n = k^2

anonymous · Answer

I mean it doesn't make the fact that $j\cdot k \in \mathbb{Z}$

lgbasallote · Answer

still..since this is proving that nm is a perfect square...the definition of a perfect square has to be shown..otherwise it's not a valid proof.