This is a question to which the answer is of course intuitively obvious. We are asked to prove that there is an infinite amount of rationals $t$ with $0
Definition Suppose $r,q\in \mathbb{Q}$. We can write both $r$ and $q$ with the same denominator $b\in \mathbb{N}^*$, $r=\frac{a}{b}$ and $q=\frac{c}{b}$, for some $a,c\in\mathbb{Z}$. Then $r< q \iff a< b$.
Definition Suppose $r,q\in \mathbb{Q}$. Suppose also $r=\frac{a}{b}$ and $q=\frac{c}{d}$ for some $a,b,c,d\in \mathbb{Q}$. Then $r=q\iff ad=bc$.
Now here's my idea for a proof:
Suppose we have a finite subset $S\subset Q$ of rational numbers with every so that for every $s\in S, 0.
If $S$ is empty, we have $\frac{1}{2}\in \mathbb{Q}$ and $0<\frac{1}{2}<1$, so $S\neq\{q\in \mathbb{Q}|0 so the interval contains at least one element.
Suppose $S$ contains one element $s$. Then we have that $1-s$ is also in the interval so the interval contains more than one element.
Now suppose $S$ contains more than one element. Take two distinct elements $r,q, r< q$ such that there is no element $t$ in $S$ with $r and we can conclude that $\{q\in \mathbb{Q}|0
is infinite.
While I think this is a good approach, I am not sure and would very much appreciate your help and comments. I also haven't used the definition of $=$, so maybe someone could suggest how I would do that.
Thank you
EDIT: In the comments, Hagen von Eitzen remarks that I haven't proven that $\frac{d}{2b}\notin S$. I have edited the text above in such a way that I think I have gotten around this. I understand the biggest mistake in this proof is the argument that if $|S|=1$ then there is a rational number which satisfies the conditions which is not in $S$.
PS. I don't really have the time to look at the answer but when I get home later I will carefully look through it and upvote/select as answer if it was helpful (don't worry, I'm not that strict :P).