Is it possible to prove the above statement via contradiction? I initially tried solving the statement with that method, but all the solutions I see online use a direct proof. Here is what I had (any criticism would be great, too):
$\phi: (\forall x,y\in Q)[(x <y)\Rightarrow(\exists z \in Q)[(x<z)\land(z<y)]]$ $\lnot\phi: (\exists x,y\in Q)[(x < y)\land(\forall z \in Q)[(x\ge z)\lor(z\ge y)]]$ which is false. (Negation of $p\Rightarrow q$ is equivalent to $p \land \lnot q$).
I'm not very sure when I should give up on one method of proving something in favor of another way.