Show that $\{x\in\mathbb Q\mid x^2\leq 2\}$ has no supremum in $\mathbb Q$. Why my argument is not correct?

Question

Let $A=\{x\in\mathbb Q\mid x^2\leq 2\}$. Prove that $A$ has no supremum in $\mathbb Q$. I had that to an exam, and I had a grade of $0/15$. Could someone explain me why ?

Proof : Let $M\in\mathbb Q$ s.t. $M=\sup(A)$. Suppose $M<\sqrt 2$. Then, by density of $\mathbb Q$ in $\mathbb R$ there is $x\in \mathbb Q$ s.t. $M<x<\sqrt 2$. Since $x^2\leq 2$, we have that $x\in A$ which is a contradiction.

If $\sqrt 2<M$. Still by density of $\mathbb Q$ in $\mathbb R$, there is $y\in \mathbb Q$ s.t. $\sqrt 2<y<M$. They $y$ is an upper bound of $A$ which contradict that $M$ is the smallest upper bound.

Question : What's wrong in my argument ? (I'll normally see the corrector next week, but I would like to know why this is wrong because to me it looks completely correct).

Answer 1

I imagine that full marks would have been avoiding talking about $\mathbb R$ and $\sqrt 2$ and actually constructing rational numbers in the ranges you suggest. What I mean is talking about $m\over n$ as your rational number and saying things like $\left({m\over n}\right)^2<2$ instead of ${m\over n}<\sqrt2$.

Still, I'll take a stand and say that getting 0 marks for this is harsh. A fully correct proof would look very much like this:

Assume $M={m\over n}\in\mathbb Q$ is the supremum of $A$, for the purposes of contradiction. If $\left({m\over n}\right)^2<2$, [insert proof by construction that $M$ is not an upper bound of $A$]. If $\left({m\over n}\right)^2>2$, [insert proof by construction that $M$ is not the least upper bound of $A$]. Obviously, $\left({m\over n}\right)^2=2$ is not achievable in the rational numbers. Therefore, our assumption that $A$ has a supremum in $\mathbb Q$ was erroneous.

The proof shown here is a fair skeleton of that proof. At least one or two points would have recognized that.