1. ## Induction Proof n

Hi, so I need to prove that 2^n > n for all positive numbers. I'm not sure if this is the right place to post this (couldnt find anywhere for basic proofs, but this is for my college linear algebra/intro proof class).

My attempt:

2^(n+1) > n + 1
Step 1:Since 2^(n+1) > n + 1 if 2^(n + 1) > 2(n+1) then 2^(n+1) > n + 1
Step 2: 2^(n+1) > 2(n+1)
Step 3: 2^n * 2 >2(n+1)
Step 4: 2^n > n + 1
Step 5: Since 2^n > n + 1. 2^n will be greater than n.

Is this a good way to prove this?? I Feel like it's wrong/terrible verbose. Any suggestions or better ways to do so (without having to to do all the replacements)?

2. ## Re: Induction Proof n

I expect you need to use induction for this. You are trying to prove \displaystyle \displaystyle \begin{align*} 2^n > n \end{align*} for all

Base step:

\displaystyle \displaystyle \begin{align*} n&= 1 \\ \\ 2^1 &= 2 \\ \\ 2^1 &> 1 \end{align*}

The statement is proved for the base step.

Inductive step: Assume that \displaystyle \displaystyle \begin{align*} 2^k > k \end{align*}, then we have

\displaystyle \displaystyle \begin{align*} 2^{k + 1} &= 2 \cdot 2^k \\ &> 2k \textrm{ since } 2^k > k \\ &> k + 1 \textrm{ since } 2k > k + 1 \textrm{ for all } k > 1 \end{align*}

So the inductive step is proven. Therefore we have now proven \displaystyle \displaystyle \begin{align*} 2^n > n \end{align*} for all integer \displaystyle \displaystyle \begin{align*} n > 0 \end{align*}.

3. ## Re: Induction Proof n

Thanks for the prompt reply. Didn't even think of that.