Can I always write $\frac{c}{d} = \frac{a}{d'}$ with $d'=b+\epsilon$ when $\frac{a}{b}>\frac{c}{d}$?
Assume $a,b,c,d$ are integers and are $>0$ (since comments provided a counterexample when they can be negative)
(if something is wrong and it is not true than perhaps it is true for $a>b, c>d$, but first lets try the general case)
I realize this is trivial, but it seems to me like it shouldn't necessarily be true (specifically, this seems like I can get from any fraction to another just by changing one of the denominator/numerator?)
Proof attempt: We want to find some $d'$ such that $\frac{c}{d}=\frac{a}{d'}$.
Solving for $d'$ we get $$d' = \frac{d}{c} \cdot a$$
Since $\frac{a}{b}>\frac{c}{d}=\frac{a}{d'}$, it must be that $d' > b$ which implies that $d' = b+\epsilon$, where $\epsilon = d'-b$.
This shows what we wanted to show.
(note: so if $d>c$ then $d'>a$, if $c>d$ then $d'<a$
Sure,
If $0 < \frac cd < \frac ab$ then
$\frac cd = \frac cd*\frac aa = \frac a{a\frac dc}$
Since $\frac cd < \frac ab$ then $\frac cd > \frac ba$ so
$a\frac dc > a\frac ba = b$.
.... In general for an $M,N$ there is an $s $ so that $M = N*s$ and if $t=\frac 1t$ then $M = N \frac 1t$ and if $0 < M < N$ then $0< s < 1$ and $t > 1$.
So... this really shouldn't seem like it shouldn't be true. It should seem that it must be true.
(Of course negatives "turn things around")
So just let $d' = a\frac dc> b$ and let $\epsilon = d' - b>0$.
That's all there is to it.