Statement to prove:
If x > 0, show there exists n in N (the set of all natural numbers) such that 1/(2^n) < x.
My work on the proof so far:
Let x > 0. By the Archimedian Property, we know if ε > 0, there exists an n in N such that 1/n < ε.
Take x = ε . So there exists an n in N such that 1/n < x.
That is as far as I've gotten. I am stuck as to how I can algebraically manipulate the inequality to get the 2 in there somehow and to get the final form of 1/(2^n) < x.