[seanhunter]

I think precisely the rationals being dense in the reals means that for any two real numbers x and y where x < y there exists a rational number m/n (m and n being integers) such that x < m/n < y.

[tromp]

Yes, that's the more formal equivalent formulation. For n > 1/(y-x), the sequence of rationals ... ,i/n, (i+1)/n, ... must land in (x,y).