# Proof that if $a/b$ is a perfect square then $a\cdot b$ is also a perfect square

It's my first question here, my name is Massimiliano. How can I prove that if $a/b$ is a perfect square then $ab$ is also a perfect square without using the unicity of prime decomposition?

My work: If $a/b$ is a perfect square then there exists $c$ such that $c^2 = a/b$ so $c^2b^2 = (cb)^2 = ab$, and therefore $ab$ is a perfect square.

But (to the contrary) if $c^2 = ab$ and I divide by $b^2$ I obtain $(c/b)^2 = a/b$. But I don't know if $c/b$ is actually an integer.

Thank you for help!

• Do you mean to ask how to show that if $a/b$ is a square then $ab$ is a square? That's not what you have written. But anyway $ab=(a/b)(b^2)$. Oct 2 '15 at 9:42
• I have helped format your question. Please check if what I wrote is correct.
– Ian
Oct 2 '15 at 9:44
• $a/b$ is not an integer, necessarily. But for rationals it makes sense to ask whether it is a perfect square as well. Oct 2 '15 at 9:47
• I'm using only N (natural number 0,1,2,...) and my problem is to demonstrate that if ab is a square then a/b is a square too; but i'm able only to demonstrate that if a/b is a square then ab is a square. Oct 2 '15 at 10:02
• do we've to show bothway? Oct 2 '15 at 10:09

If $\frac{a}{b}=q^2$, $a=bq^2$, then $a\cdot b = bq^2\cdot b = (bq)^2$.