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.

You have $p^2=3k$ but when you plug in the next step, you have $3q^2=p^2 \implies 3q^2=\color{red}{(3k)^2}$. You need to show that if $3 \mid p^2$, then $3 \mid p$. — Anurag A, Dec 21 '23 at 20:00
Did you need to use the property: if $3 \mid q^2$ then $3 \mid q$. To see the problem, repeat your proof for the similar claim that $\sqrt{4}$ is irrational. — GEdgar, Dec 21 '23 at 20:01
For that part, wouldn't we use the fact that three is prime, while four is just $2^2$? — Someone, Dec 21 '23 at 20:07
It is worth to compare your proof with the ones posted here already, e.g., this one and others. Even though each proof is different, common errors appear again and again. — Dietrich Burde, Dec 21 '23 at 20:15
@Someone ... Yes, $3$ is prime, which the OP never mentioned. — GEdgar, Dec 21 '23 at 20:20
Euclid's lemma: if the prime $p$ divides $ab$, then $p|a$ or $p|b$. — Jean-Claude Arbaut, Dec 22 '23 at 22:17
This is the infinite descent way of proving $\sqrt{3}$ is irrational. Another approach is using the Rational Root Theorem. — John, Dec 23 '23 at 02:36

Robert Murray · Answer 1 · 2023-12-22T21:08:15.373

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.

edited Dec 22 '23 at 21:08

answered Dec 21 '23 at 20:35

Robert Murray

1,788

How does $3|p$ and $3|q$ imply $3=p=q$? – Gonçalo Dec 21 '23 at 23:02
Oops! I took it from the author without thinking! I’ll fix it thank you! – Robert Murray Dec 21 '23 at 23:32

score 1 · Answer 2 · answered Dec 22 '23 at 22:49

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.

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.

Irrationality of $\sqrt 3$ proof verification

2 Answers2