# Irrationality of $\sqrt{2}$: why can we assume lowest terms?

In all textbook proofs of the irrationality of $$\sqrt{2}$$, one will inevitably see something to the effect of "we assume that $$\frac{p}{q}$$ is in lowest terms", or that numerator and denominator do not have 2 as a common factor etc.

But why can we do this? I'm starting to self study math but given that I am embarking on a journey of rigor, this seems a decidedly un-rigorous introduction to the subject. Is there more behind this just being an assumption, or does it necessarily follow from the way we've set it up? If the latter, I honestly do not see it. Please advise.

• Every time you cancel a factor, the numerator gets smaller. So it has to stop in less than $p$ steps. Commented May 31, 2021 at 5:49
• Note that you can choose whether you want to assume lowest terms, and reach a contradiction the first time you simplify, or not assume lowest terms and show that you can keep simplifying the fraction ad infinitum, and reach a contradiction that way. I used to prefer the latter, personally, but I like the former now. Commented May 31, 2021 at 5:59
• No need to assume lowest terms. See math.stackexchange.com/a/812443/589
– lhf
Commented May 31, 2021 at 17:49

Here are approaches:

1. Two positive integers always have a greatest common divisor. Just because a rational number is not initially written in reduced form does not stop you from passing immediately to that reduced form. It is there whether you use it or not. Why be concerned about using it? Here is why there is a reduced form. When a positive rational number $$r$$ is written as $$m/n$$ for some positive integers $$m$$ and $$n$$, there will be a greatest common factor $$d$$ of $$m$$ and $$n$$. Then $$m = dp$$ and $$n = dq$$ for some positive integers $$p$$ and $$q$$. The numbers $$p$$ and $$q$$ don't have a common factor greater than $$1$$ (otherwise $$d$$ is not the greatest common factor of $$m$$ and $$n$$, right?), so $$r = m/n = p/q$$ and this shows each positive rational has a representation as a ratio of positive integers that have no common factor bigger than $$1$$. (If you have a fussy rigor issue about rational numbers having a reduced form, then I'm surprised you don't express concern about why $$\sqrt{2}$$ exists or what real numbers are -- that has a lot more depth to it than the existence of a reduced form for fractions and how to prove properties of divisibility among positive integers.)

2. Find a proof that does not use the condition of $$r$$ having a reduced form representation. If $$r$$ is a positive rational number where $$r^2 = 2$$ and we write $$r = m/n$$ where $$m$$ and $$n$$ are positive integers, then $$2n^2 = m^2$$, so $$m^2$$ is even, which implies $$m$$ is even and then $$n$$ is even by the usual argument you've seen (typically assuming $$m$$ and $$n$$ are relatively prime, but that's irrelevant up to this point so we don't need to assume it). Since $$m$$ and $$n$$ are even, $$m = 2m_1$$ and $$n = 2n_1$$. Thus $$r = m/n = m_1/n_1$$. Run through the argument with $$m_1$$ and $$n_1$$ in place of $$m$$ and $$n$$ to see $$m_1$$ and $$n_1$$ are even: $$m_1 = 2m_2$$ and $$n_1 = 2n_2$$. Thus $$r = m_1/n_1 = m_2/n_2$$ and $$m = 2m_1 = 4m_2$$ and $$n = 2n_1 = 4n_2$$. We can do this arbitrarily often, thus getting for each integer $$k \geq 1$$ that $$r = m/n = m_k/n_k$$ where $$m = 2^km_k$$ and $$n = 2^kn_k$$ for positive integers $$m_k$$ and $$n_k$$. Hence $$m \geq 2^k$$ and $$n \geq 2^k$$: $$m$$ and $$n$$ are greater than or equal to each power of $$2$$. That's a contradiction, so $$\sqrt{2}$$ is irrational.

• (+1) Nice answer! The second point appeals so much! Commented May 31, 2021 at 6:23

If the argument was fleshed out more, then it would go something like this:

Assume, for the sake of contradiction, that $$\sqrt{2}$$ is rational. This means that it can be expressed in the form $$\frac{a}{b}$$, where $$a$$ and $$b$$ are integers, and $$b\neq0$$. If $$a$$ and $$b$$ have any common factors, then cancel them, and denote the lowest terms fraction as $$\frac{p}{q}$$.

The phrasing 'assume that $$\frac{p}{q}$$ is in lowest terms' is a little misleading, in my opinion. It is true that if $$x$$ is a rational number, and $$p$$ and $$q$$ are integers such that $$\frac{p}{q}=x$$, then there is no reason to think that $$\frac{p}{q}$$ is in lowest terms.

What the authors of the proof are trying to say is that any rational number has a unique lowest terms representation. And from this point onwards we will choose to denote that lowest terms representation as $$\frac{p}{q}$$.

• +1 This is the only answer so far that seems to correctly understand and address what OP's confusion is, which is about not the existence of lowest terms representations but the meaning of "assume" in this context. Commented May 31, 2021 at 6:33
• @EricWofsey: Thank you Eric. Mathematicians use words like 'prove', 'assume', etc. in subtly different ways to how people use them in everyday life, inevitably leading to confusion for beginners.
– Joe
Commented May 31, 2021 at 6:54
• @EricWofsey indeed. thank you! Commented May 31, 2021 at 7:18
• @user25849: I'm glad to hear that I've helped. Feel free to accept an answer by clicking the green arrow below the upvote and downvote buttons.
– Joe
Commented May 31, 2021 at 10:20

I also found this a bit nonrigorous at first. But to someone who is just starting out with proofs, it's likely best not to bog them down with details - going into the formalities of why we can make that assumption draws attention away from the actual proof.

Here's a way to think about this (somewhat) more rigorously. One way to define the set of rational numbers $$\mathbb{Q}$$ is as the set of equivalence classes of $$\mathbb{Z}\times(\mathbb{Z}\setminus\{0\})$$ under the equivalence relation $$(a,b)\sim(c,d)\iff ad=bc$$, and write a representative $$(a,b)$$ of this equivalence class as $$\frac{a}{b}$$. (If you're not familiar with this terminology, it essentially means that we define rational numbers by taking pairs of integers, then declaring two pairs $$(a,b)$$ and $$(c,d)$$ to be equal if they satisfy the relation $$ad=bc$$.) This is motivated by the following computation (when $$b$$ and $$d$$ are nonzero): $$\frac{a}{b}=\frac{c}{d}\iff ad=bc.$$

Now it's easier to see why we can assume without loss of generality that $$p$$ and $$q$$ have no common factors. Suppose that $$p=ap'$$ and $$q=aq'$$ for some integers $$a,p',q'$$. Then we have: $$\frac{p}{q}=\frac{ap'}{aq'}.$$ We claim that the fraction $$\frac{ap'}{aq'}=\frac{p'}{q'}$$; but this follows from our definition of the rational numbers, since $$(ap')q'=(aq')p'$$ from the commutativity of integer multiplication. So if $$p$$ and $$q$$ share a common factor $$a$$, then the fraction $$p/q$$ is equal to the fraction $$p'/q'$$, where the common factor of $$a$$ has been "divided out".

The two integers $$p$$ and $$q$$ can only share a finite number of factors (other than $$1$$). This is simply because an integer can itself have only finitely many factors (other than $$1$$). Note that if $$p$$ has infinitely many factors other than $$1$$, then you can write $$p=a_1p'=a_1a_2p''=\cdots$$, where each $$a_j$$ is strictly greater than $$1$$. Thus each $$a_j$$ is greater than or equal to $$2$$ (as each is an integer), and as each of $$p',p'',\cdots\geq1$$, we have $$p\geq 2^n$$ for every $$n\in\mathbb{N}$$, contradicting the assumption that $$p$$ is an integer. Thus $$p$$ and $$q$$ can only have finitely many non-$$1$$ factors, and so we can repeat the process above a finite number of times to cancel the finite number of shared factors of $$p$$ and $$q$$.

We can assume this because all rationals can be written in lowest terms. Suppose you had a rational that you couldn't write in lowest terms, say "$$\frac{p}{q}$$". Then we could keep cancelling out factors forever, i.e. $$p$$ and $$q$$ are infinite, which doesn't make much sense either, so it must be that no such integers exist.

• i'm sorry, yes, then "can" be. but how do i know that it is necessarily so in this case? Commented May 31, 2021 at 5:54