On a proof that the square root of prime numbers is irrational

Question

I know the proof but wanted to understand it a little better. It is proved using proof by contradiction and starts with "Suppose $k$ is a prime number. Let us assume $\sqrt{k}$ is a rational number. Therefore, $\sqrt{k}=s/t$; $s\in\mathbb{Z}$, $t\in\mathbb{N}$, $s$ and $t$ are co-prime. On squaring both sides, it becomes $k=(s/t)^2$."

My doubt is why doesn't it stop here. As $k$ is a prime number, we know that $k=(s/t)^2$ can not be true as a square number is never prime. (Please don't be brutal.)

Answer 1

"as a square number is never prime": this is true in the integers, but we are now in the rationals, this is why we need to look further. (If you stop here, you have only proven that the square root of a prime is not integer.)

Answer 2

If the square root of a prime $p$ would be rational, then $\sqrt p = \frac{s}{t}$ for some integers $s,t\geq 1$. Squaring gives

$pt^2 = s^2$.

Consider the prime factorization of $s$ and $t$. These factorizations are unique. In the product $pt^2$, the multiplicity of $p$ is odd, while in the factorization of $s^2$, the multiplicity of $p$ is even. Contradiction.

Answer 3

No need for $p$ being a prime, only that $\sqrt{p}$ isn't an integer. Suppose $\sqrt p=A/B$ in lowest terms, then $A/B=pB/A$. Now the fractional part of $A/B$ is $b/B$ with $b<B$ and the fractional part of $pB/A$ is $a/A$ with $a<A$. Those fractional parts must be equal, that is, $a/A=b/B$ and so $a/b=A/B=\sqrt{p}$ contrary to $A/B$ being in lowest terms.

Answer 4

To show that there may be an issue - that assumptions need to be proved - we can look at the system of integers of the form $4n+1$ for $n$ an integer. This is closed under multiplication, and has $1$ as the (multiplicative) identity. We don't use the additive structure.

Think about the number $49$ - this is of the form $4n+1$ but the only factors it has of this form are $1$ and $49$. It is clear that when we expand the system to include all odd numbers $49$ becomes a square. So the phenomenon can happen.

Incidentally, this system also demonstrates the need to take care about the difference between irreducible elements in a system, and primes. $49$ is irreducible. But $49$ is a factor of $21\times 77=1617=49\times 33$. When primes and irreducibles part company then unique factorisation can fail (with a prime if $p|ab$ then $p|a$ or $p|b$ and $49$ fails this test in the system under consideration).

This also illustrates that the proof of unique factorisation of integers into irreducible elements (which are in fact prime) depends on the additive structure as well as the multiplicative one.