Haskell, 40 bytes

(Couldn't resist posting this after seeing Haskell answers with verbose imports – now I see some other language answers are also essentially using this method.)

f takes a pair of integers (numerator and denominator of the fraction; they don't need to be in smallest terms but the denominator must be positive) and returns an integer.

f(n,d)|m<-mod n d,m>0=f(d,m)|0<1=div n d

Try it online!

• Ignoring type mismatch, the fractional part of n/d (assuming d positive) is mod n d/d, so unless mod n d==0, f recurses with a representation of d/mod n d.

Haskell, (削除) 40 (削除ここまで) 34 bytes

Edit:

• -6 bytes: @WheatWizard pointed out the fraction can probably be given as two separate arguments.

(Couldn't resist posting this after seeing Haskell answers with verbose imports – now I see some other language answers are also essentially using this method.)

! takes two integer arguments (numerator and denominator of the fraction; they don't need to be in smallest terms but the denominator must be positive) and returns an integer. Call as 314!100.

n!d|m<-mod n d,m>0=d!m|0<1=div n d

Try it online!

• Ignoring type mismatch, the fractional part of n/d (assuming d positive) is mod n d/d, so unless mod n d==0, ! recurses with a representation of d/mod n d.