0

I've proven the case where only one of $a$, $b$ is zero. But this is the proof for both $a$ and $b$ nonzero. This is what I have:

Suppose $a\sqrt3 + b\sqrt5 = X$ for $X$ rational. Squaring,

$X^{2}$ = $3a^{2} + 2(ab\sqrt3 \sqrt5$) + $5b^{2}$

The first and last expression on the right hand side will be rational, and since by assumption $a, b$ are nonzero we can get a new rational number $X' = \frac{X^{2} -3a^{2} -5b^{2}}{2ab}$ such that

$X' = \sqrt3 \sqrt5$ = $\sqrt{15}$.

Hence the problem comes down to showing that this is a contradiction. I.e., that $\sqrt{15}$ is irrational.

I guess my question is whether or not my reasoning is sound so far, and whether there is an easier way to prove this that I'm missing. If it's all good, then how do we prove that $\sqrt{15}$ is irrational?

Robert Shore
  • 23,332
Wilson Brians
  • 747
  • 1
    if $\sqrt{15}$ is rational then $15=\frac{a^2}{b^2}$ hence $3\times 5 b^2=a^2$ if you look at the factorization of $a^2$ you see that every prime appears an even number of times and the same for $b^2$ and so you get a contradiction – Spotty Oct 09 '16 at 21:25
  • The line of proof is correct so far. For the last step, an alternative way to show that $\sqrt{15}$ is irrational is to note that it is a root of the polynomial equation $x^2-15=0$. But by the rational root theorem... – dxiv Oct 09 '16 at 21:28

2 Answers2

2

Your argument is fine so far, and if you can modify your proof for $\sqrt 3$ being irrational you are home. Assume $\sqrt{15}=\frac ab$ in lowest terms. Square and find that $15\mid a$, then argue that $15^2 \mid a$ and so on. Note that because you squared you do not have logical equivalence-there are irrational numbers like $\sqrt 3$ that square to rational numbers. If you find $X'$ is rational it could still be that $X$ is irrational. That is not a problem here, because $X'$ is in fact irrational.

Ross Millikan
  • 374,822
-1

An alternative method: Denote $$X := a \sqrt{3} + b \sqrt{5}. $$ By rearranging, squaring, rearranging, and squaring again, we find that $$X^4 - 2 (3 a^2 + 5 b^2) X^2 + (3 a^2 - 5 b^2)^2 .$$ So, $X^2$ is a root of a quadratic with discriminant $240 a^2 b^2 = 15 (4 a b)^2$, which cannot be a square (except when $a = 0$ or $b = 0$) because $15$ is not. So $X^2$ is not rational, hence $X$ is not. If $a = 0$ or $b = 0$, rearranging gives that $\sqrt{5}$, resp., $\sqrt{3}$, are rational, a contradiction.

Travis Willse
  • 99,363