Let $a$,$b$ be relatively prime integers which are both greater than or equal to $􏰀2$ . Show that $(a/b)^2$ cannot be an integer.

I have tried to use contradiction to prove this result, but I am not sure that my idea is correct or not. Below is my idea (not a complete proof).

Suppose $(a/b)^2$ be an integer. We have $a/b$ is an integer. Then, we have $a=bk$ where $k$ is an integer since $b$ divides $a$. Since $a,b$ are relatively prime, therefore $b$ must be $1$. Hence, contradiction arises. I am not sure that my idea is correct or not, could anyone help me to check it? If my idea is wrong, could you give me a idea to do this question?

  • $\begingroup$ You already know that $a/b$ is not an integer. It seems to me what is really being asked is to show that if a rational number is not an integer, then neither is its square. $\endgroup$ – MPW Mar 18 '20 at 15:11
  • $\begingroup$ You have to exclude the case that $(a/b)^2=m$ is not a perfect square. $\endgroup$ – Maik Pickl Mar 18 '20 at 15:11
  • $\begingroup$ Your very first line is not quite right. If $(a/b)^2$ is integer, then it does not follow that $a/b$ is integer. Say, what if $(a/b)^2 = 2$? $\endgroup$ – user726394767 Mar 18 '20 at 15:11