Yahoo Answers: Answers and Comments for How would you prove that 2^n > n? [Mathematics]
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From Dan P
enSG
Wed, 29 Oct 2008 07:05:25 +0000
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Yahoo Answers: Answers and Comments for How would you prove that 2^n > n? [Mathematics]
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From Puggy: If induction is involved, I'm assuming you...
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Wed, 29 Oct 2008 07:11:20 +0000
If induction is involved, I'm assuming you mean for all natural numbers n. Let's start with the base case of n = 1.
If n = 1, then 2^n = 2^1 = 2, and
n = 1, so LHS > RHS and the inequality holds true for n = 1.
For our induction hypothesis, we assume 2^k > k for some k > 1.
(We want to prove that 2^(k + 1) > (k + 1) )
However,
2^(k + 1) = (2^k)(2^1)
= (2^k)(2)
which, by our induction hypothesis, is
> k(2)
= 2k
= k + k
And since k > 1
> k + 1
Showing that
2^(k + 1) > k + 1
Confusing? Here's the entire chain all over again.
2^(k + 1) = 2*(2^k) > 2*(k) = k + k > k + 1
So
2^(k + 1) > (k + 1)
Which means the inequality holds true for n = k + 1.
Thus by the Principal of Mathematical Induction,
2^n > n for all natural numbers n.

From xtempore: Good answer Puggy.
And of course it holds f...
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Wed, 29 Oct 2008 07:15:30 +0000
Good answer Puggy.
And of course it holds for all negative numbers too, because
0 < 2 ^ n < 1, for n < 0, and therefore 2 ^ n < n.

From Anonymous: N 2 N
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Mon, 09 Jan 2017 19:15:33 +0000
N 2 N

From Anonymous: This Site Might Help You.
RE:
How would you p...
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Mon, 10 Aug 2015 13:19:09 +0000
This Site Might Help You.
RE:
How would you prove that 2^n > n?
I'm having a few problems going through this exercise even though it's kinda easy. I have to use mathematical induction to prove that 2 ^ n > n.

From kumorifox: Assume the statement is true for n=k
Theref...
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Wed, 29 Oct 2008 07:14:43 +0000
Assume the statement is true for n=k
Therefore, 2^k > k
Now prove it is true for n=k+1
2^k+1 > k+1?
2^k × 2^1 > k+1?
2^k × 2 > k+1 (we had assumed that it is true for n=k, so this is true)
Therefore, if the statement is true for n=k, it is also true for n=k+1
Is it true for n=1?
2¹ > 1? Yes, 2>1
So the statement is true.

From Anonymous: Let P(n) be 2^n>n, then
P(1): 2>1 is true.
A...
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Wed, 29 Oct 2008 07:25:08 +0000
Let P(n) be 2^n>n, then
P(1): 2>1 is true.
Assume that P(k) is true for +ve integer k>1, then
2^(k+1)(k+1)=2*2^kk1=
2^kk+2^k1>0=>
P(k+1): 2^(k+1)>(k+1) is true.
So, the statement P(n) is true for any +ve integer n.
Mistake: next time, you should specify what n is first.