# Proof an inequality by mathematical induction

I have a problem that I have to solve using mathematical induction but I'm stuck from a part. The problem is:

Proof that $\large n<2^n$ is true for $\large n \in \mathbb{N}\$

So, I did that steps:

1) $n = 1$, $\qquad$ $1<2^1$ $\rightarrow$ $1 < 2$ $\qquad$ THIS IS TRUE

2) $n = k$, $\qquad$ $k < 2^k$

3) $n = k+1$, $\qquad$ $k+1 < 2^{k+1}\$

$\qquad$$\qquad$$\qquad$$\quad k < 2^k \qquad$$\qquad$$\qquad$$\quad$ $2 * k < 2^k * 2$ $\qquad$ I multiplied by $2^1$ in both sides to get

$\qquad$$\qquad$$\qquad$$\quad 2k < 2^{k+1} I'm stuck in this part, what should I do next?? Please be specific • If k>1 then k+1<k+k=2k, so ... – DiegoMath Sep 7 '14 at 6:38 ## 3 Answers When you have k<2^k (2) then k+\color{red}1<2^k+\color{red}1. But$$\color{red}1< \color{green}{2^k},~~\forall k\in\mathbb N.$$so you get$$k+\color{red}1<2^k+\color{red}1<2^k+\color{green}{2^k}.$$For n=1 we have$$1 < 2.$$From n=k to n=k+1 note that k < 2^{k} is equivalent to k+1 \leq 2^{k}, so that$$k+1 \leq 2^{k} < 2^{k+1}$\$ follows.

I think you should have done it like this

By assumption:

          k+1<2^k

2(k+1)<2*2^k            * by 2 both sides


since 2*2^k>2(k+1)

and 2(k+1)>k+1

  for example if n>10

therefore     n>5


Therefore k+1<2*2^k

         =2^(k+1)


LHS < RHS

Then statement n<2^n is valid for all Natural Numbers