Where does the proof of $\sqrt 2$ is irrational break down when trying to prove the same for $\sqrt 4$?

Question

I'm trying find where the common proof by contradiction that $\sqrt 2$ is irrational breaks down when trying to prove $\sqrt 4$ is irrational.

Assume $\left(\frac pq\right)^2=4$ and $\gcd(p,q)=1$. I guess I could just let $p=2, q=1$ and be done, but why is that an adequate failure of the proof? If it's not, then where else does the proof break down?

Continuing the "proof": Then $p^2=4q^2$, so $p$ is even, that is, $p=2r$. That implies that $4r^2=4q^2$, or that $q=r$.

If $p=2r$ and $q=r$ then $\gcd(p,q)=r$ which contradicts $\gcd(p,q)=1$, leaving us with the untenable conclusion that $\sqrt 4$ is irrational.

Answer 1

It does not contradict the relatively prime assumption because we can take $r=1$.

Answer 2

I think this is a good example of how proofs by contradiction are a bad habit. Once you run into a contradiction, you think you're done. But how do you know you haven't just made a mistake, introducing a new contradiction?

Here is a proof that $\sqrt{2}$ is irrational which is not a proof by contradiction and incidentally which tells you exactly what happens with $\sqrt{4}$. The distinction between this proof and a proof by contradiction is that everything I'm about to say is true.

The key idea is that unique prime factorization can be extended to rational numbers: in the same way that any nonzero integer $n$ has a unique prime factorization

$$n = \pm \prod_p p^{\nu_p(n)}$$

where the exponents $\nu_p(n)$ are non-negative integers and the product runs over all primes, any rational number $r$ has a unique prime factorization

$$r = \pm \prod_p p^{\nu_p(r)}$$

where the exponents $\nu_p(r)$ are now allowed to be negative. If we write $r$ as a fraction $\frac{n}{m}$ then the factorization of $r$ is just given by dividing the factorization of $n$ by the factorization of $m$, and the exponents subtract.

The reason it's cool to write the exponents $\nu_p(r)$ explicitly as a function of $r$ is that this makes it easier to state the very important observation that when you multiply two rational numbers together, the exponents in their unique prime factorizations add:

$$\nu_p(rs) = \nu_p(r) + \nu_p(s).$$

In particular,

$$\boxed{ \nu_p(r^2) = 2 \nu_p(r) }.$$

That is, if $r^2$ is the square of a rational number, then the exponents in its prime factorization must all be even. So why can't $2$ be the square of a rational number? Because one of the exponents in its prime factorization - namely the exponent of $2$ - is odd. And why doesn't $4$ succumb to this argument? Because $4 = 2^2$, so all of the exponents in its prime factorization are even.

This argument proves with no further difficulty that lots of other numbers are irrational besides $\sqrt{2}$: namely, if $n$ and $k$ are positive integers, then $\sqrt[k]{n}$ is rational if and only if every exponent $\nu_p(n)$ in the prime factorization of $n$ is divisible by $k$.

Answer 3

Another common proof that $\sqrt{n}$ is irrational if $n$ is not a perfect square is here: $\sqrt{17}$ is irrational: the Well-ordering Principle

It depends on the fact that $n$ is not a perfect square if and only if $\lfloor \sqrt{n} \rfloor < \sqrt{n} $.