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 $t^2 > 2$, then $t^2-2>0$ so there must be a positive integer $n>1$ so that $t^2-2>\frac{1}{n}$.
But $$\left(t-\frac{1}{2nt}\right)^2 = t^2-\frac{1}{n} + \frac 1{4t^2n^2} > t^2-\frac{1}{n} > 2$$
So $t-\frac{1}{2nt}$ is smaller than $t$ and it's square is still greater than 2.
