# Is this a valid proof that $\sqrt 3$ is irrational?

I saw that questions about this have been posted before, but I don't want to spoil things for myself by prematurely looking up an answer. I just want to know if this proof I came up with is valid:

"Suppose $\sqrt 3$ is rational. Then $\sqrt 3 = \frac p q$, where $p,q$ are integers and $q \ne 0$. But then that means $q\sqrt 3 = p$, and we know that no integers satisfy this, so $\sqrt 3$ must be irrational."

Is this good? If not, why not?

• How do you know no integer satisfies $\sqrt 3q=p$? – Bernard Oct 24 '15 at 23:21
• Yes, you need to go further. You can deduce from where you have reached, that $3q^{2} = p^{2}$. This is indeed not possible for relatively prime integers $p$ and $q$, but you need to explain why. – Geoff Robinson Oct 24 '15 at 23:24
• Yes, I don't want to sound discouraging as you are on a good start but "no integers satisfy this" is simply another way of saying "this is irrational". So basically all you've done is say "square root of three is irrational because the square root of three is irrational". – fleablood Oct 24 '15 at 23:28
• @GeoffRobinson Ah, I think I see now! If $p$ and $q$ are relatively prime then $q^2$ and $p^2$ are relatively prime too, which means their only common factor is $1$. If $3q^2=p^2$ then that means that $q^2$ and $p^2$ share all factors except 3, which means that $q=1$, which means that $p$ is $\sqrt 3$. But $3$ is not a perfect square, so $p=\sqrt 3$ can't be an integer, which is a contradiction! So $\sqrt 3$ can't be rational. Thanks for the help! (Seeing the phrase "relatively prime" just set off alarm bells in my head, lol) – Asker Oct 24 '15 at 23:34

It would be very instructive to compare the classic proof that $\sqrt{2}$ is irrational, which starts off very similar to what you have. But there is the crucial additional requirement that $\gcd(p, q) = 1$, that is, they have no prime factors in common. If they do have prime factors in common, divide them out so the fraction is in lowest terms.
Then you square both sides to obtain $$2 = \frac{p^2}{q^2}.$$ Moving stuff around, we get $p^2 = 2q^2$, which means that $p^2$ is even. From this we derive a contradiction of $\gcd(p, q) = 1$.
I think that if you fully understand this proof, you should be able to carry the same concepts over to proving $\sqrt{3}$ is also irrational.
A little housekeeping note: $\sqrt{3} q$ could cause confusion with $\sqrt{3q}$. It would be clearer to write $q\sqrt{3}$.