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Let $x=\frac{A}{B}$ be a positive rational number in lowers terms (i.e., $A, B\in\mathbb{N}$ and $hcf(A,B)=1$). Prove that $\sqrt{x}$ is rational if and only if $A$ and $B$ are both perfect squares. (Remember that your proof should work for cases like $A=40$, where $A$ is not a perfect square but has a factor that is a perfect square.)

I know prime factorization is involved but that's basically it. Proving that $\sqrt{y}$ (for some number y) is rational if and only if $y$ is a perfect square I can do. Applying that that to the fraction mentioned above? Not so much.

Any and all help would be appreciated. Thank you.

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    $\begingroup$ Write $\displaystyle\sqrt x = {p \over q}$. Then $\displaystyle{A \over B} = {p^2 \over q^2}$, and $Aq^2 = Bp^2$. Now use uniqueness of prime factorizations. $\endgroup$ Commented Mar 31, 2016 at 7:57

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First of all, you have an if and only if statement, so you need to prove both directions.

One of them should be trivial. If $A,B$ are perfect squares, then it shuold be easy to show that $\sqrt x$ is rational.

Now, for the other direction, follow the same steps you are used to:

  • Assume that $x$ is rational.
  • Therefore, $x=\frac pq$ for some $p,q$ such that $hcf(p,q)=1$.
  • Now, take a look at the equality $\frac AB = \frac{p^2}{q^2}$. Multiply the equality by $Bq^2$. What do you get?
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