I am trying to prove in a different way then 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