How would you prove that $\sqrt{2}$ is irrational?

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  • Is $n^{th}$ root of $2$ an irrational number? [duplicate]

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You have probably already seen the proof for $n=2$. So, let’s assume that $n \geq 3$.

Seeking a contradiction, suppose that $\sqrt[n]{2}$ is rational. This, together with the positivity of $\sqrt[n]{2}$ imply that there exist $p,q \in \mathbb{N}$ such that $\sqrt[n]{2}=p/q$. Raising both sides to the $n^{th}$ power, we see that
$$2=\frac{p^n}{q^n}.$$
Multiplying through by $q^n$ and using $2q^n = q^n + q^n$, we have
$$q^n + q^n = p^n. $$
This violates Fermat’s Last Theorem, giving a contradiction. Thus $\sqrt[n]{2}$ is irrational.

By the rational root theorem, all rational roots of $x^n – 2$ are one of $\{ \pm 1 , \pm 2 \}$. It’s easy to verify that none of these work if $n \ge 2$.

If you’re willing to use the Fundamental Theorem of Arithmetic, which says that there’s essentially only one way to write an integer as a product of primes, then the proof drops right out.

For, if $2^{1/n}=p/q$ for any integers $p$ and $q$, take the $n$-th power and clear of fractions to get $p^n=2q^n$. How many $2$’s do you see on the left? A number that’s divisible by $n$. How many on the right? A number that’s $1$ more than a multiple of $n$. And there’s your contradiction.

Exactly the same way that you prove $\sqrt{2}$ is irrational — the standard proof generalizes perfectly.

I would argue that the polynomial
$$
p(x)=x^{n}-2
$$

is irreducible over $\mathbb{Q}$ using Eisenstein criterion and in
particular it does not have a root in $\mathbb{Q}$