What is different ways to prove that $\sqrt p$ is irrational when $p$ is prime?

Question

I know this fact can be proved by contradiction (reductio ad absurdum) but please give proofs by different methods.

Answer 1

By Eisenstein's criterion, the polynomial $x^2-p$ is irreducible.

(To answer WimC's complaint below: Consider the splitting field of $x^2-p$ over $\mathbf Q_p$. It contains an element of valuation $1/2$, so it is a proper over-field of $\mathbf Q_p$, and therefore $x^2-p$ is irreducible over $\mathbf Q_p$ (and a fortiori over $\mathbf Q$)).

Answer 2

The idea behind the proof is that if $\sqrt p$ is rational then it should be integer. Until here that is not a proof by contradiction.

However afterward, we have to assume that if $\sqrt p$ is rational then it is integer and then it is impossible for $\sqrt p$ to be integer and hence contradiction. So it is still a proof by contradiction.

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Now the proof:

First of all observe that because $p$ is prime there is no natural number $x>1$ such that $x^2=p$ otherwise $x\mid p$ for $x<p$.

Secondly if there are $x,y\in\mathbb N$, such that $(x,y)=1$, and $x^2=py^2$, then we have: $$ \frac xy=\sqrt p\implies \frac{py}x=\sqrt p $$ Note that $py> x$, $y>1$. Write $x=q_0y+r_0$ where $y>r_0>0$. $r_0$ cannot be zero otherwise $\sqrt p=q_0$ which we saw that is not possible.

Now we can write: $$ \frac xy= \frac{py}x=\sqrt p\implies\frac{py-q_0x}{x-q_0y}=\sqrt p $$

Now we found another fraction with $0<r_0=x-q_0y<y$, which means a fraction with a smaller denominator. Now take $\frac{y_0}{r_0}=\frac{py-q_0x}{r_0}$ and $\frac xy$, and write $y=q_1r_0+r_1$. Again $r_1\neq 0$ because otherwise $r_0\mid y$ and $r_0\mid x$ and hence $\frac{y_0}{r_0}$ would be a natural number which is not possible. Therefore $r_1>0$.

Using the same procedure we find $y_1$ and $r_1<r_0$ such that $\frac{y_1}{r_1}=\sqrt p$. By continuing like this we find sequence of $y_n$ and $r_n$ such that $r_0>r_1>\dots>r_n>\dots>0$ which means that eventually we should arrive to $r_n=1$ for some $n$ and hence $\sqrt p=y_n$ which is not possible.

Answer 3

If $\sqrt p$ is rational then there are whole numbers $a$ and $b$ such that $\sqrt p = \frac{a}{b}$, where $a > 0$ and $b > 0$ and they are co-primes, i.e. there is no prime that divides both. (If there is, then divide $a$ and $b$ by that number and get a new $a$ and $b$ that meet this criterion).

Then $\large p = \frac{a^2}{b^2}$ and then $\large\frac{pb^2}{a^2} = 1$.
Since $a$ cannot divide $b$ then $p$ is divisible by $a$ which makes $p$ non-prime.

Answer 4

By Gauss's lemma, any rational root of $x^2-p$ must be an integer.

But by unique factorization, $p$ cannot be the square of an integer.

Answer 5

By Dirichlet's theorem on primes in arithmetic progressions and by quadratic reciprocity (!!), there exists a prime $\ell$ such that $$\left(\frac{p}{\ell}\right) = -1.$$

By Gauss's lemma, any rational root of $x^2−p$ must be an integer. But if $p$ were the square of an integer, it would certainly be a square mod $\ell$ also. Thus, $x^2-p$ has no rational root.