Showing $\mathbb{Q}$ does not have the L.U.B. property I am trying to prove in a different way than how it was already proved on this website (another question). So yes, this is sort of a duplicate.

Claim: $\mathbb{Q}$ does not have the least upper bound property

Let $S = \{r \in \mathbb{Q} : r > 0~~ \text {and} ~~r^2 < 2 \}$. Clearly $S$ is bounded above by 2 and $S$ is nonempty since $1 \in S$.
To prove S has no least upper bound it suffices to prove that if $t$ is an upperbound for $S$ then there exists a upper bound $t' \in \mathbb Q$ such that $t' < t$.
It can then be shown that for $n \in \mathbb{N}$ is sufficiently large then
$$\left(t-\frac{1}{n} \right)^2 > 2$$ given that $t^2 > 2$
But I get stuck here. How would I go about proving this is true? any ideas would help. I appreciate it.
Thankyou

If you substitute 4 for 2 in your proof, then the conclusion will be false: $\{r \in \mathbb{Q} : r > 0$ and $r^2 < 4 \}$ does have a least upper bound in $\mathbb Q$ (because 4 is a perfect square). So somewhere you have to use the fact that if $r \in \mathbb Q$, then $r^2 \ne 2$.  
So your proof should have two parts:


*

*Show that if $r \in \mathbb Q$, then $r^2 \ne 2$.

*Show that if $r \in \mathbb Q$ and $r^2 > 2$, then $\exists t \in \mathbb Q$ such that $2 < t^2 < r^2$.  


If you can prove these (can you?), then the result follows: any upper bound $r$ will satisfy $r^2 > 2$ (because of 1), so it can't be a least upper bound (because of 2).
Edited to reply to comment: OK, if $r^2 > 2$, then $r^2 = 2 + \epsilon$ for some rational $\epsilon > 0$. Now look at $t = r - \epsilon/2r$.
