I'd like to prove that the product of the roots of two distinct primes $p_1$ and $p_2$ is irrational. That is, $$ \sqrt{p_1 p_2} \notin \mathbb{Q} $$

Would the following be a valid proof?

Suppose $\sqrt{p_1 p_2} \in \mathbb{Q}$.

Then $\sqrt{p_1 p_2} = \frac{a}{b}$ for $a,b \in \mathbb{Z}$ such that $\gcd(a,b) = 1$. $$ p_1 p_2 = \frac{a^2}{b^2} \implies p_1p_2b^2 = a^2 \implies p_1, p_2 \mid a^2 \implies p_1, p_2 \mid a. $$ So $a = p_1 p_2 n$, for nonzero $n \in \mathbb{Z}$.

Let the unique prime factorization of $a = p_1 p_2 (p_{i_1} \cdots p_{i_n})$ and of $b=p_{j_1} \cdots p_{j_m}$.

Then $$ p_1p_2 = \frac{a^2}{b^2} = \frac{p_1^2 p_2^2 (p_{i_1}^2 \cdots p_{i_n}^2)}{p_{j_1}^2 \cdots p_{j_m}^2} = p_1 p_2 \underbrace{\left( \frac{p_1 p_2 (p_{i_1}^2 \cdots p_{i_n}^2)}{p_{j_1}^2 \cdots p_{j_m}^2} \right)}_{=1} $$ But this implies that $p_1 = p_{j_t}$ and $p_2 = p_{j_s}$ for $t, s \in [1,m]$. That is $p_1, p_2 \mid b$.

Therefore, $p_1$ and $p_2$ are divisors of both $a$ and $b$, which contradicts $\gcd(a,b)=1$.

Hence, $\sqrt{p_1 p_2} \notin \mathbb{Q}$.

P.S. Sorry for the double subscripts.

  • 1
    $\begingroup$ The proof looks OK, I would suggest trying to use a factorization to get that $p_1,p_2|b$ instead of dividing. $\endgroup$ – Michael Burr May 7 '17 at 20:48
  • 2
    $\begingroup$ The proof is fine, but a little over complex. Once you show that $p_1$, say, divides $a$ you can show that $p_1$ divides $b$ and that's enough. $\endgroup$ – lulu May 7 '17 at 20:49
  • 4
    $\begingroup$ These number-theoretic proofs are important to know how to do--nothing wrong with them--but know that are a variety of other ways of going about this. For instance, one could consider the polynomial $f(x) = x^2 - p_1p_2$. The irreducibility of $f$ would imply $\sqrt{p_1p_2} \notin \mathbb{Q}$; one can use Eisenstein's criterion or the rational root test to show this. $\endgroup$ – Kaj Hansen May 7 '17 at 20:52
  • $\begingroup$ @KajHansen How irreducibility of $f(x)$ imply $\sqrt{p_1p_2} \notin \mathbb{Q} $? $\endgroup$ – user157835 Jul 4 '17 at 2:29

Once you get at $p_1 p_2 b^2 = a^2$ I'd just point out that, say, $p_1$ appears an odd number times on the left and an even number of times on the right.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.