Alternative proof for the irrationality of $\sqrt{2}$; questions

Question

So I started reading Conjecture and Proof by Miklos Laczkovich and one of the first proofs he provides is that of the irrationality of the square root of two. I am aware there are alternative proofs (one of which is geometric and another that uses the fundamental theorem of arithmetic) but I have a few questions about this one.

The Proof:

Suppose $\sqrt2 = p/q$, where $p, q$ are positive integers and let $q$ be the smallest such number. Then $2q^2=p^2$ and thus $p^2$ is even. Then p itself must be even; let $p=2p_1$. Then $2p^2 = (2p_1)^2=4p_{1}^{2}$ and thus q is also even. If $q=2q_1$ then $\sqrt{2} =p/q=p_{1}/q_{1}$. Since $q_{1}<q$, this contradicts the minimality of $q$.

My questions:

Why do we let $q$ be the smallest such number? I understand that this creates the contradiction at the end of the proof but I don't know why this is a fundamental need.

Also, if we are picking from the set of positive integers and $q$ is the smallest wouldn't that make $q=1$ if our set is all positive integers if we exlude $0$ in the set, and $q=0$ if we do include $0$ in the set? So in the first case $\sqrt{2}=p$ and the second case $\sqrt{2}= undefinded$. I am unsure of where I am going wrong here

Answer 1

The reason for assuming that $q$ is minimal IS that without this assumption we wouldn't reach a contradiction at the end, so the proof wouldn't work.

Writing a proof is not a mechanical process where each step follows with necessity from what comes before it. Instead, very often the reason to do something is that it will enable a step later in the proof to work.

It's like explaining the right path through a maze: The answer to "why should I turn right here?" is not of the form "because we're following a rule that says we must turn right after we have turned left a prime number of times" (or whatever) -- the answer is, "because that's the way that leads to the exit!"

The person who created the walkthrough of the maze has probably explored more of if than the instructions say, lots of blind ends and so forth. But he doesn't chronicle all of those false starts in his finial explanation of how to get through.

A proof is like a cheat sheet for a maze: All that's required of it is that it leads to the exit; it doesn't need to catalogue all of the blind alleys, or all of the possible steps that don't turn out to help us prove our goal.

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wouldn't that make $q=1$ if our set is all positive integers if we exlude $0$ in the set, and $q=0$ if we do include $0$ in the set?

No. The proof step doesn't say "let $q$ be the smallest integer", but instead "let $q$ be the smallest integer such that there is a $p$ with $p/q=\sqrt2$". Even though at this point in the proof we're assuming that some $p/q$ equals $\sqrt 2$, so there is some $q$ we can choose, it doesn't immediately follow from this that $1$ is among the possible $q$s we can choose.

Answer 2

Saying "q is the smallest such number" is the same as saying the fraction is "reduced to lowest terms". Essentially the proof is showing that you always have factors of 2 in both numerator and denominator so that a fraction equal to square root of 2 cannot be reduced to lowest terms, a contradiction.

Answer 3

The proof assumes that you know what fractions are and how to compute with them, and also that they have more than one representation: if $\frac{p}{q}$ represents a fraction, then $\frac{2p}{2q}$ represents the same fraction. We use this fact to add fractions ($\frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{1+2}{4} = \frac{3}{4}$, etc.), among other things.

Now we make the argument very formal, if we'd like:

For every fraction $\frac{p}{q}$ we can consider denominators that can occur in representations of it: $D_{\frac{p}{q}} := \{ b \in \mathbb{N}: \exists a \in \mathbb{Z}: \frac{p}{q} = \frac{a}{b} \}$. This set is non-empty ($q$ is in it), so has a minimal member in $\mathbb{N}$. Now start arguing as in your proof, and you get a proper contradiction: we get a stricly smaller number in $D_{\frac{p}{q}}$. And $1$ is not a consideration, as it will not be in $D_{\frac{p}{q}}$ at all.