Let $S = { r \in \mathbb{Q} : r \lt 2}$. Prove that $S$ does not have a largest element.

Question

Let $S$ = $[{ r \in \mathbb{Q} : r \lt 2}]$. Prove that $S$ does not have a largest element.

My method:

Assume to the contrary that $S$ does have a largest element, where $S$ = $[{r \in \mathbb{Q} : r \lt 2}]$.

This is the furthest I got and I am unsure of how to proceed next. Would I assume that for $m \in \mathbb{Z}$ $rm \gt S$?

Answer 1

Can you use that the rationals are dense? If yes, the answer is immediate: assume $S$ has a maximum, say $r_0$ and therefore, $r_0<2$. But since the rationals are dense in the real numbers, you can find $r_1\in\mathbb Q$, such that $r_0<r_1<2$. It follows that $r_1\in S$, but that contradicts the assumption that $r_0$ is the maximum of $S$.

Answer 2

Say the largest element is $r$, It is obvious that $\frac{2+r}{2}$ is still en element of $S$, which is a rational number. What's more $\frac{2+r}{2} > r $, that is a contradiction.