I recently came across an interesting problem in number theory, namely,
Given distinct integers $a, b, c, d < N$ ( with the condition that $a<b$ and $c<d$ ) , where $N$ is also an integer :
Prove that $\frac{a}{b}$ and $\frac{c}{d}$ are at least $\smash[t]{\frac{1}{N^2}}$ apart.
I started by considering the difference between the fractions:
$$
\frac{a}{b} - \frac{c}{d} = \frac{ad-bc}{bd}
> \frac{ad-bc}{N^2}
$$
Now, WLOG if $b<d$, then we have 2 cases: either $a<c$ or $a>c$. Now, if $a>c$, then the numerator of the RHS becomes greater than $1$, so the theorem is proven.
However , I cannot solve the case where $a<c$ , since now I have to consider to what extent $a$ can be lesser than $c$ to make the numerator of my fraction closer to $1$ , and I got stuck there.
Any help would be appreciated!
Issue is Direct Consequence of Farey Sequence :
"In mathematics, the Farey sequence of order n is the sequence of completely reduced fractions , ... between 0 and 1 , ... , which when in lowest terms have denominators less than or equal to n , arranged in order of increasing size."
We can move the terms around to get fractions between $0$ & $1$ , to match the Farey Sequence.
Here $n=N-1$
We want to check the gap between Farey neighbours (skipping over neighbours will increase the over-all gap) & then get the minimum gap.
It is well-known that Farey neighours $a/b$ & $c/d$ have this :
$bc-ad=1$
Hence the gap between Farey nieghbours is $a/b-c/d=-1/bd$
We know that $b$ & $d$ can be maximum $N-1$ & $N-2$
Hence minimum gap is $1/(N-1)(N-2)$
We can not get smaller gap. We can not even get $1/N^2$.