Irrationality of $\sqrt 3$ proof verification

Question

I'm new to proofs and I'd like to check if this proof is valid and if there is anything I'm overlooking.

We have to prove that $\sqrt 3$ is an irrational number.

We can suppose the opposite; suppose $\sqrt 3$ is a rational number. Then there exist integers p and $q \neq 0$ such that $\sqrt 3$ = $\frac{p}{q}$. Without loss of generality, we can assume p and q are coprime numbers.

$\sqrt 3$ = $\frac{p}{q}$ $\implies$ $3q^{2} = p^{2}$ $\implies$ 3 | $p^{2}$ $\implies$ $\exists k \in \Bbb{Z}$ such that $p^{2} = 3k$

When we plug that into $3q^{2} = p^{2}$ we get:

$3q^{2} = (3k)^{2}$ $\implies$ $q^{2} = 3k^{2}$ $\implies$ 3 | $q^{2}$

We assumed that p and q are coprime numbers, meaning they have no common factors other than 1. However, we got that 3 is a factor of both $q^{2}$ and $p^{2}$. This implies that it must hold that q = p = 3. This leads to a contradiction where $\frac{p}{q}$ = 1 = $\sqrt 3$.

As our assumption led to a contradiction, $\sqrt 3$ must be an irrational number.

Answer 1

Many people in the comments are saying you skip one step at the end. You need to show that:

$$3|p^2 \Rightarrow 3|p$$ If this is true, then $3$ is a factor of both $p$ and $q$ and we assumed $p$ and $q$ are coprime, this is a contradiction. Saying that since $p$ and $q$ are coprime and $3$ divides both then $3=p=q$ is logically ok due to the principle of explosion, but for proof purposes you have already finished.

This is a very common error, and it is not hard to rectify (think how prime factorization works). Good luck.

Answer 2

Sorry, but your attempt isn't good. You state that $p^2$ is a multiple of $3$, which is right, but then you write $p^2=3k$ and all you can get is that $$ 3q^2=3k $$ which doesn't lead to contradictions: just to $k=q^2$ and again to $p^2=3q^2$. Dead end.

You're missing the key step, namely that from $3\mid p^2$ you can deduce that $3\mid p$, because $3$ is prime. Thus you can say that $p=3k$ and $$ 3q^2=(3k)^2 $$ so your conclusion works.

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In a more general way, let $n$ be a positive integer. Then $\sqrt{n}$ is rational if and only if $n$ is a square. Indeed, the rational root test applied to $x^2-n$ tells us that a rational root has to be integer.

In particular, the square root of any prime is irrational. But so are also $\sqrt{6}$, $\sqrt{111}$ and so on.