Prove by mathematical induction: $n < 2^n$

Question

Step 1: prove for $n = 1$

1 < 2

Step 2:

$n+1 < 2 \cdot 2^n$

$n < 2 \cdot 2^n - 1$

$n < 2^n + 2^n - 1$

The function $2^n + 2^n - 1$ is surely higher than $2^n - 1$ so if

$n < 2^n$ is true (induction step), $n < 2^n + 2^n - 1$ has to be true as well.

Is this valid argumentation?

This question is similar: Prove by mathematical induction that $2n ≤ 2^n$, for all integer $n≥1$? — Martin Sleziak, Oct 21 '13 at 17:08

score 7 · Answer 1 · answered Oct 21 '13 at 15:56

7

$$n+1<n+n<2^n+2^n=2^{n+1}$$

answered Oct 21 '13 at 15:56

bateman

4,000

Beaten by 35 seconds! – Ahaan S. Rungta Oct 21 '13 at 15:57
@AhaanRungta :-) – bateman Oct 21 '13 at 15:58
1

Thank you. I'll give the answer to @AhaanRungta though, because he answered the question. – Oct 21 '13 at 16:00
1

Thanks! Bateman beat me to it, so I'll give him an upvote. :) Also, Grant, you didn't give me the check mark (accepted answer). :P – Ahaan S. Rungta Oct 21 '13 at 16:01
@Grant: bateman answered the question in the title, which should be arguably more important than anything you ask in the text (unless explictly said otherwise). – Marek Oct 21 '13 at 16:02
@Marek Very true. I suggest not making a big distinction between our posts, Grant. ;) – Ahaan S. Rungta Oct 21 '13 at 16:03
@Ahaan: it would be nice to be able to mark multiple answers as correct because this can definitely happen. But that's discussion for another time and place I suppose :) – Marek Oct 21 '13 at 16:04
Yes, more likely the Meta section. :P – Ahaan S. Rungta Oct 21 '13 at 16:05

score 4 · Accepted Answer · answered Oct 21 '13 at 15:57

4

Yes. Your method works. Alternate: $$ n + 1 < 2n < 2 \cdot 2^n = 2^{n+1}, $$ as desired.

answered Oct 21 '13 at 15:57

Ahaan S. Rungta

7,706

1

You state that n+1<2n. But this isn't true for n=0. What am I missing? – aidandeno Jul 24 '15 at 15:30
This proves the results for all $ n > 0 $. – Ahaan S. Rungta Jul 24 '15 at 18:10

score 2 · Answer 3 · edited Mar 12 '15 at 06:21

2

We want to show that $k+1 < 2^{k+1}$, from the original equation, replacing $n$ with $k$: $$k + 1 < 2^k + 1$$ Thus, one needs to show that: $$2^k + 1 < 2^{k + 1}$$ to complete the proof.

We know that $1 < 2^k$ for $k ≥ 1$. Adding $2^k$ to both sides: $$2^k + 1 < 2^k + 2^k = 2^{k+1},$$ which shows that $n < 2^n$.

edited Mar 12 '15 at 06:21

daOnlyBG

2,711

answered Mar 12 '15 at 05:58

Alex

21

Prove by mathematical induction: $n < 2^n$

3 Answers3