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.
I guess you could, you would start with something like assuming that if you have 2 rationals x and y, then anything in between x and y must be irrational. So we want to contradict the statement
if $x < y ∈ ℚ$, then $\lnot\exists z∈ℚ,$ $x < z < y$
Then you could say since x and y are rational, you can write them as
$x = \frac{p}{q}$ and $x = \frac{r}{s}$
then find something between them, i.e
one option for z is $z = \frac{x + y}{2}$ since it is between x and y, and this can then be written as
$z = \frac{\frac{p}{q} + \frac{r}{s}}{2}$
which you can then reslove, and since we know that $p, q, r, s$ and $2$ are all rational, z must be rational which is a contradiction.
Is this what you were looking for?