For k such that , choose n such that . Then .
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.
I think I got it [I have a proof within a proof]:
My proof: Let x > 0. By induction 2^n > n
(Proof: Base case: for n = 1, 2^1 > 1. Check. Inductive step: Assume true for n = k, k a natural number. That is 2^k > k is true. To show this true for k + 1, we need to show 2^(k+1) > k +1 which implies 2*2^k > k +1. By assumption that 2^k > k, then 2*2^k > 2k. But for all k ≥ 1, 2k ≥ k + 1, so
2^(k+1) > 2k ≥ k+1. Therefore 2^(k+1) > k + 1, and 2^n > n. QED)
So since 2^n > n, then 1/2^n < 1/n. By the Archimedian Property, 1/n < x. Therefore
1/2^n < 1/n < x which implies 1/2^n < x. QED.