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Progress:

Since, $p,q$ are irrationals and $p^2$ and $q^2$ are relatively prime, thus, $p^2\cdot{q^2}$ cannot be a proper square, so, $pq$ is also irrational. Suppose, $pq=k$, then: $$\sqrt{k}\cdot\sqrt{k}=\sqrt{pq}\cdot\sqrt{pq}=k$$ Which implies that $\sqrt{pq}$ is irrational, since, $k$ is irrational and that $p^2\cdot{q^2}$ cannot be a proper square being $p^2$ and $q^2$ relatively prime and that concludes the proof.

The above lines are my attempt to prove the assertion. Is the proof correct? If not then how can I improve or disprove it.

Regrads

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  • $\begingroup$ Are $p^2$ and $q^2$ integers? If so, you seemed to have used the facts that (a) the square root of an integer is either an integer or irrational, and (b) the square root of an irrational number is irrational, without explicitly stating them or justifying them $\endgroup$ – Henry Sep 13 '16 at 0:19
  • $\begingroup$ Yes, $p^2$ and $q^2$ are integers, since I stated they are relatively prime. I also used those two facts you mentioned. @Henry $\endgroup$ – user18724 Sep 13 '16 at 0:25
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The first part is correct, but you could clarify it a little bit:

Since $p$ and $q$ are irrational, they are not integers

Since $p$ and $q$ are not integers, $p^2$ and $q^2$ are not perfect squares

Since in addition to that $p^2$ and $q^2$ are relatively prime, $p^2q^2$ is not a perfect square

Since $p^2q^2$ is not a perfect square, $\sqrt{p^2q^2}=pq$ is irrational


The second part is somewhat obscure, but you could simply use the following argument instead:

Since $pq$ is irrational, it is not a multiple of two integers

Since $pq$ is not a multiple of two integers, it is not a perfect square

Since $pq$ is not a perfect square, $\sqrt{pq}$ is irrational

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  • $\begingroup$ @user18724: Thank you :) $\endgroup$ – barak manos Sep 13 '16 at 12:44
  • $\begingroup$ By the way, what I just noticed is that the second argument from the first part is not necessarily true. "Since $p$ and $q$ are not integers, $p^2$ and $q^2$ are not perfect squares", here is one example: $p=\sqrt{8}$ and $q=\sqrt{2}$ where both are not integers, but raising their power to $2$ yields $16=\sqrt{8^2}\cdot{\sqrt{2^2}}=8\cdot{2}$, which is a perfect square, since $4^2=16$ @barak manos $\endgroup$ – user18724 Sep 13 '16 at 15:05
  • $\begingroup$ @user18724: You're wrong. The perfect square that you are talking about is $p^2q^2$, I was referring to $p^2$ AND $q^2$ (separated). $\endgroup$ – barak manos Sep 13 '16 at 15:29
  • $\begingroup$ Ok, got it. @barak manos $\endgroup$ – user18724 Sep 13 '16 at 15:35
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Except that you haven't proved $p^2 * q^2 $ cannot be perfect square, rest of the proof is correct.

Since $p^2 $ and $q^2$ are relatively prime, they do not have common prime factors. And they are not perfect squares. Else p,q would be rational numbers. This is necessary since 25, 4 are relatively prime but their product is perfect square.

Hence atleast one of the prime factors of p^2 and q^2 will be raised to an odd power in prime factorisation and those primes will be distinct. So $p^2 * q^2 $ cannot be perfect square

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You are correct up to the point that $p^2$ is not a square, and $q^2$ is not a square. However, that doesn't at all stop $p^2q^2$ from being a square. You just assumed it, you did not prove it when $p^2$ and $q^2$ are relatively prime. Once that is proved, maybe you can say that $pq$ is irrational.

So how do we prove this missing link? Well, assume the contrary, and suppose that $p^2q^2 = k^2$, where $k$ is an integer. Write $k^2$ as a product of primes, say $k^2 = p_1^{k_1}p_2^{k_2}\ldots p_n^{k_n}$. Note that $2 | k_i$ for all $i$ as $k^2$ is a perfect square.

Now, each $p_i$ divides either $p^2$ or $q^2$. Without loss of generality, if it divides $p^2$, then by a simple result, $p_i^2$ also divides $p^2$. Furthermore, it is not possible that the same $p_i$ divides both $p^2$ and $q^2$, because they are co prime. Hence, each of $p^2$ and $q^2$ is the product of squares of prime numbers, because for any prime dividing these quantities, their square also divides these quantities. Taking the square root on both sides of the previous statement, we get that $p$ and $q$ are products of primes. This is a contradiction, as they are both irrational.

Hence, $p^2q^2$ is not a square. You can use a Euclidean argument to prove that $pq$ is irrational from here on.

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  • $\begingroup$ Thanks for your response. However, you seem to have misread some of my points. First of all I didn't say $p$ and $q$ are relatively prime, because they are irrationals therefore not integers, only integers can be relatively prime. But $p^2$ and $q^2$ are integers and relatively prime to each other, which is enough to imply that their product is not a proper square. @астон вілла олоф мэллбэрг $\endgroup$ – user18724 Sep 13 '16 at 0:47
  • $\begingroup$ Yes, it is enough, but you did not show that in your answer. It is not obvious, it has to be proved. That's what I did. I've edited the answer to make up for my mistake. $\endgroup$ – астон вілла олоф мэллбэрг Sep 13 '16 at 0:57
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Suppose $\sqrt{pq}$ is rational, say $\sqrt{pq} = \dfrac mn$ where $m$ and $m$ are positive, relatively prime integers. Then $p^2 q^2 = \dfrac{m^4}{n^4}$ where $p^2q^2$ is an integer. Since $m$ and $n$ are relatively prime integers, then $m^4$ and $n^4$ are also relatively prime integers. It follows that $n^4=1$ and hence $n=1$. So we find that $p^2q^2 = m^4$ for some positive integer $m$.

Since $p^2$ and $q^2$ are relatively prime integers,then any prime divisor of $m$ cannot be a common divisor of both $p^2$ and $q^2$. It follow that there must be relatively prime positive integers $a$ and $b$ such that $m = ab$, $\;p^2 = a^4$, and $q^2 = b^4$. Hence $p=a^2$ and $q=b^2$, which implies that $p$ and $q$ are rational (integers). By contradiction, $\sqrt{pq}$ must be irrational.

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