I'm new to induction and trying to prove $2^n > n$ for all natural numbers.

I've seen a couple of examples but am confused about the the case going from $k = 1$ to $k =2$.

So I show $2^1 > 1$ as the base case.

Then I assume $2^k > k$

Meaning that

$2.2^k > 2k$


$2^{k+1} > 2k$


$2^{k+1} > k + k > k + 1$

So it is considered proven.

But when $k = 1$, $k + k \not> k + 1$

What am I missing please?

Do I need a special case for going from $k=1$ to $k=2$?

  • 1
    $\begingroup$ Adding a special case when $k=1$ is a perfectly valid solution. Saying $k+k\geq k+1$ for $k\geq1$ is probably nicer though. $\endgroup$ Dec 12, 2018 at 9:45

2 Answers 2


Technically, the statement $$2^{k+1} > k+k>k+1$$ is wrong, precisely because $k$ could be $1$. You can, however, write $$2^{k+1} > k+k \geq k+1$$

and from that, you can still conclude that $2^{k+1}>k+1$. No special case needed, since $a>b$ and $b\geq c$ always implies $a>c$.


No, you don't need a special case. $2^{k+1} >2k $ and $2k \geq k+1$ together impliy $2^{k+1} >k+1 $


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