Relations; aRb if ab=k^2

Question

Hi I am supposed to prove that R is an equivalence relation on N (natural number)

We define the relation R on N as aRb if $ab=k^2$ for some k that belongs to N.

I know that I should prove that the relation is reflexive, symmetric and transitive.

The relation is reflexive if aRa for some a that belongs to N. a^2 is a natural number and therefore a^2 = k^2 and aRa.

The relation is symmetric if aRb then bRa for all a and b that belongs to N.

The relation is symmetric because if aRb then $ab=k^2$ for some k that belongs to Z. This is equivalent to $ba= k^2$ and therefore bRa.

I'm not sure about Transitive tho. I must prove that if aRb and bRc then aRc. How can I do that?

Answer 1

Hint. Try to prove transitivity when $a$ and $b$ are both powers of $2$. A few examples should show you how. Then think about the fundamental theorem of arithmetic.

Answer 2

Since you know that aRb and bRc, you can conclude that there are natural numbers $m$ and $n$ such that $ab=m^2$ and $bc=n^2$. Your job is to prove that $ac$ is also the square of some natural number involving $m$ and $n$. (Hint: multiply those two equations together.)