I have a proof of that if $q\in \mathbb{Q}$ then $ \sqrt{q}$ is rational if and only if $q$ is a perfect square (it can be written in the form $q={p_1}^{a_1}...{p_n}^{a_n}$ where integers $a_j$, which may be positive or negative are even). I just need an explanation why in this proof, when we squared $r$ there is $r^2=\pm p_1^{2a_1}...p_k^{2a_n}$ not just $r^2=p_1^{2a_1}...p_k^{2a_n}$? (Proof in the photo)

  • $\begingroup$ What photo? ${}$ $\endgroup$
    – Wojowu
    Feb 17, 2019 at 18:34
  • $\begingroup$ Hmm there must be a photo, I dont know, why it doesnt show. Here link: i.stack.imgur.com/hHQBR.png @Wojowu $\endgroup$
    – Bambeil
    Feb 17, 2019 at 18:36
  • 5
    $\begingroup$ This is a typo. $\endgroup$
    – user65203
    Feb 17, 2019 at 18:39

1 Answer 1


Hypothesis: the author copy-pasted the expression of $r$ and forgot about the $\pm$ symbol, because, you are correct: when you square a positive or negative integer, it should be positive.


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