Why is the proof that $\sqrt2$ is irrational must $p$ and $q$ be coprime?

Question

I understand most of the proof except how it starts with that $p$ and $q$ in $\frac pq$ where $p$ and $q$ are integers and $q$ can’t equal $0$, that $p$ and $q$ must be coprime? Why must they be coprime? I don’t think it’s part of the definition of a rational number.

Answer 1

They don't have to be coprime. It's just an observation one can make if $\sqrt{2}=\frac{P}{Q}$ for some $P, Q$ such that $d=\gcd(P, Q)>1$, that if $P=pd$ and $Q=qd$, then, necessarily, $p$ and $q$ are coprime. This coprimality is then used later on in the proof.

Answer 2

It's not that they must necessarily be coprime. However, for any rational number, we can always find coprime integers $p$ and $q$ such that $\frac pq$ is that rational number.

If that rational number is $\sqrt 2$, then the two coprime integers we found are both even. Which is a contradiction.

Answer 3

It's not part of what it means to be a rational number (assuming you're using one of the most common definitions of $\mathbb{Q}$, as equivalence classes of $\mathbb{Z} \times \mathbb{N}^{> 0}$ or similar). But it is a fact that you can always find such $p$ and $q$. (For example, $\frac{4}{6} = \frac{2}{3}$.)

If it turned out by some fluke that $\sqrt{2}$ were equal to $\frac{1000}{225}$, then you could always simplify that down to $\frac{40}{9}$ before you started, and then proceed with the usual proof by contradiction to show that $\sqrt{2}$ is not in fact equal to $\frac{1000}{225}$.

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Prove that $\sqrt{2} \not = \frac{1000}{225}$.

If $\sqrt{2} = \frac{1000}{225}$, then $\sqrt{2} = \frac{40}{9}$. Now, the right hand side is something even, squared, divided by something odd; so we've got two powers of $2$ on the right hand side but only one on the left-hand side. Contradiction.

To show that $\sqrt{2}$ is not rational, you basically just have to do this again for the $\frac{\mathrm{odd}}{\mathrm{odd}}$ and $\frac{\mathrm{odd}}{\mathrm{even}}$ cases.

Answer 4

You don't need to assume they are coprime. In fact, if $\sqrt{2}=\frac{p}{q}$, then the proof actually concludes that both $p$ and $q$ are even. Then you cancel the twos and end up with another fraction $\frac{\text{half of }p}{\text{half of }q}$ equal to $\sqrt{2}$. In which you can cancel the twos again ... and again ... ad infinitum, which is clearly impossible.

Answer 5

A rational number is the ratio of any two integers. So in principle we really do have to consider the possibility that we could get $\sqrt2$ by dividing $p/q$ where $p$ and $q$ are not necessarily coprime.

But if you just say that, and proceed to use the same symbols $p$ and $q$ the exact same way you have seen in a proof, eventually you get to the step where you say “but then $p$ and $q$ would both have to be even, which is a contradiction” (or whatever words you use at that point), and you would be wrong. It would not be a contradiction, because you just said $p$ and $q$ could be any integers, so they could both be even.

So if we start with any old $p$ and $q$ we need some extra equations. We let $d$ be the GCD of $p$ and $q$ and let $p'=p/d$ and $q'=q/d.$ Now we can proceed with the rest of the proof, except we have to write $p'$ and $q'$ instead of $p$ and $q,$ and then we really get a contradiction when we want it, because we made it so $p'$ and $q'$ cannot both be even.

But look how cumbersome that was. We had names for two numbers, we got two other names of numbers from them, then we never used the first set of names again. So many words, so many numbers to name.

So what we do instead is we figure out that no matter what integers we start with, we can always do the GCD trick and get a ratio of two coprime integers. Then we skip the part where we give names to the original integers and their GCD, and we just name the coprime integers. It saves a lot of words and symbols. Also, it’s a thing we can use again and again in proofs about rational and irrational numbers, not just for $\sqrt2.$

So put this fact in your bag of tricks. Any two integers define a rational number, but there is always a way to write that exact same number as a ratio of coprime integers.

Answer 6

Well, the goal of this famous proof is to eventually conclude that if such a rational number exists, then the integers $p,q$ have infinitely many factors of $2$ in common. This is why it starts out saying you should simplify the fraction to lowest terms to make sure there are no common factors. Then the proof continues to show that although we've purportedly done this, the integers still have a common factor of $2.$ Then the cycle begins again, ad infinitum, whence it follows that there cannot be any such integers.

So, you see that it's because of the path the proof wants to go through -- to show that we cannot express such rational $p/q$ in lowest terms (assuming it exists) -- that it starts out assuming we've in fact just ensured this.