# Proving irrationality of $\sqrt 3$

So I understand how to prove that $\sqrt 3$ is irrational. However, I think I'm missing something, conceptually.

So we assume that $\sqrt 3$ is rational, and thus can be expressed as $\frac{p}q$, $p \in \mathbb Z, q \in \mathbb Z^*$.

Thus, $(\frac{p}q)^2$ = 3, which means that $p^2 = 3q^2$, which means that $q^2$ is a multiple of $3$, which means that $q$ is a multiple of 3. How do we know this? I've read that it comes from the Fundamental Theorem of Arithmetic, but I just don't see how it follows?

We then go on and express $q$ as $3r, r \in \mathbb Z$, thus $3r = q^2$. Therefore, $p$ and $q$ are multiples of each other, and we have a contradiction.

My question is, why coudln't we use the exact same logic for proving that $\sqrt 4$ is irrational (which of course it isn't)? Does the Fundamental Theorem of Arthmetic imply something for our $\sqrt 3$ that it doesn't for the $\sqrt 4$ proof? This feels like a stupid questions seeing as $\sqrt 4$ is obviously not irrational. I just don't understand how the logic doesn't tranpose.

Any help is appreciated, thanks.

• Well you know p, not q, is a multiple of 3. Thats what $p^2=3q^2$ is saying. You have a few typos. – CogitoErgoCogitoSum Apr 21 '18 at 0:45
• The square root of three satisfies $(\sqrt{3})^\color{red}{2}=3$. On the other hand the cube root of three satisfies $(\sqrt{3})^\color{red}{3}=3$ – JMoravitz Apr 21 '18 at 0:45
• As for how the fundamental theorem of arithmetic implies that $3\mid q^3\implies 3\mid q$, remember that $3$ is prime and that if a prime divides evenly into a product of two numbers it must be true that it divides evenly into at least one of the numbers individually. Here $3\mid q^3$ and remembering $q^3=q^2\cdot q$ implies that $3\mid q^2$ or that $3\mid q$. In the second case we are done, and in the first case we apply the same logic again noting that $q^2=q\cdot q$ to again arrive at $3\mid q$ – JMoravitz Apr 21 '18 at 0:47
• As for why it doesn't work for $\sqrt{4}$, recognize that $3$ is prime and that $4$ is not prime. – JMoravitz Apr 21 '18 at 0:50
• @JMoravitz Thanks for your help. However, there's one thing I'm still confused about: we can use this logic to prove for $\sqrt 6$, and that's not a prime number? What's the difference between 4 and 6? Thanks. – iaskdumbstuff Apr 21 '18 at 1:02

First - I think you meant to write $$\left( \frac{p}{q} \right)^2 = 3$$ - squared, not cubed.
The argument then continues by claiming that if $p^2 = 3q^2$ then $3$ must divide $p$. That follows from the fact that if a prime (in this case $3$) divides a product it must divide one of the factors (here they are $p$ and $p$).
You can't assert that if $p^2 = 4q^2$ then $4$ must divide $p$. If $p$ were just singly even then $4$ would still divide $p^2$.
• @EthanBolker Thanks for your answer. So when once we say that $p^2 = 4q^2$, we can't go any further, since we can't make any conclusions on what values $p$ is a multiple of? – iaskdumbstuff Apr 21 '18 at 1:01
• @user494405 .If we have $p^2=4q^2$ we can conclude that $p$ is even (because $p^2$ is even ) so let $p=2p'$. Then $4p'^2=4q^2,$ so $p'^2=q^2.$ In other words $\sqrt 4\in \Bbb Q$ if there exist non-zero integers $p',q$ such that $p'^2=q^2.$ And there are, e.g. $p'=q=1.$ – DanielWainfleet Apr 21 '18 at 9:22