Approximate the plastic number
Python 2, 49 bytes



n=x=input()
while n**3/x/x&lt;n+x:n+=1
print n,'/',x


Try it online!

The idea is to express the ρ with ρ³=ρ+1 as a fraction n/x whose denominator x is the input accuracy parameter. We take (n/x)³=n/x+1 and clear denominators to get n³=x²(x+n). 

Since the LHS increases in n faster than the RHS, we can approximate the equality point n as the smallest with n³≥x²(x+n). The code counts up n until this is the case, starting at x which is smaller. 

A small byte save is to divide both sides by x² to write n³/x²≥x+n (negated in the while condition). This is floor division in the code, but the fractional part lost is negligible.

A same-length alternative instead puts x as the numerator:

Python 2, 49 bytes



n=x=input()
while x**3/n/n&lt;n+x:n-=1
print x,'/',n


Try it online!