Prove $n < 2^n$ for all $n \geq 0$ using induction.

Question

Please verify for me?

Base case: $n = 0 $

$0 < 2^0$

$0 < 1$. This is true.

Inductive step: Suppose $n \geq 0$. Assume $P(k)$ is true if $k = n$. We must deduce that $P$ holds for $k+1$.

$n < 2^n$

$n - 2^n < 0$

$(n+1) - 2^{n+1}$

$n+1 - (2^n \times 2)$

$n+1 - (2^n + 2^n)$

$n - (2^n + 2^n) + 1$

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

$(n - 2^n) - (2^n-1)$

We know $(n - 2^n) < 0$, that is to say $(n-2^n)$ is negative.

For all values of $n \geq 0, (2^n-1)$ is always a non-negative number.

A negative number minus a non-negative number is always negative, so

$(n - 2^n) - (2^n-1) < 0$ is true.

$(n - 2^n) < (2^n-1)$.

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

$(n+1) < 2^{n+1}$

So $n < 2^n$ for all $n \geq 0$.

Answer 1

There is an easier way. If $n<2^n$ for some $n$, then $$n+1<2^n+1 \leq 2^n+2^n=2^{n+1}$$.