# Thread: [SOLVED] Archimedian Property proof

1. ## [SOLVED] Archimedian Property proof

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.

2. For k such that $\frac{1}{k} < x$, choose n such that $2^n > k$. Then $\frac{1}{2^n} < \frac{1}{k} < x$.

3. let n>1, so 2^n>n therefore 1/(2^n)<1/n<x.

4. ## Clarification

Originally Posted by icemanfan
For k such that $\frac{1}{k} < x$, choose n such that $2^n > k$. Then $\frac{1}{2^n} < \frac{1}{k} < x$.
So I can arbitrarily choose an n such that 2^n > k?

5. ## Clarification

Originally Posted by chiph588@
let n>1, so 2^n>n therefore 1/(2^n)<1/n<x.
So I can just state that n > 1? And how do I know 2^n > n? That looks familiar. I think we proved something like that before with math induction.

6. actually, n >= 0 (Typo)

2^n > n is proved by induction

base case: 2^0 > 1

inductive case: suppose 2^n > n
then 2*2^n > 2n
2^(n+1) > 2n > n+1

Therefore 2^n > n

7. Originally Posted by ilikedmath
So I can arbitrarily choose an n such that 2^n > k?
The choice is not entirely arbitrary, but since 2^n has no upper bound, for any given k, there must be some value of n for which 2^n > k.

8. ## I think I got it

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.