Prove that $2^n>2n$ for all integral values of n greater than 2

Question

Prove $2^n >2n$ for all integral values of n greater than 2.

Let $p_n$ be the statement: $$2^n>2n\ \forall\ n\gt2$$

If the inequality is valid for $n=k$ where $k>2$: $$p_k: 2^k>2k$$

Then for $n=k+1$: $$p_{k+1} = 2^{k+1}>2(k+1)$$

I don't know how to do the inductive step itself, I have only done series/recurrence relations inductions. Have I used the correct layout/notation? Is there more cool notation I could add to improve the mathematical-ness of the proof?

Thanks

Try to use the fact that if $a,b,c$ areal positive numbers and $b>c$, then $ab>ac$. — Git Gud, Aug 31 '13 at 13:53
@GitGud i'm not understanding what to do with the inequalities.
I know that $2^{k+1} = 2^k\cdot2^1$ but how can I use this to my advantage? — salman, Aug 31 '13 at 13:58
@user90771 The comment above and amWhy's answer should make that clear for you now. — Git Gud, Aug 31 '13 at 13:59

score 8 · Accepted Answer · answered Aug 31 '13 at 13:58

8

Don't forget the "base case": $p(3)$. It may seem obviously true, but formally, an inductive proof requires it.

With respect to the inductive step, note that $$2^{k+1} = \underbrace{2 \cdot 2^k \gt 2\cdot 2k}_{\text{inductive step}} = 4k = \underbrace{\color{blue}{\bf 2k+2k\geq 2k+2}}_{\text{for all }\; k \geq 1}=2(k+1)$$

answered Aug 31 '13 at 13:58

amWhy

209,954

Blueish $2+...$ – Mikasa Aug 31 '13 at 14:00
To be completly precise $2\cdot 2^k>2\cdot 2k$ isn't the induction hypothesis, but rather equivalent to it. – Git Gud Aug 31 '13 at 14:02
@amwhy why is $2.2^k > 2 \cdot 2k$? – salman Aug 31 '13 at 14:03
Yes, @GitGud is correct: the inductive hypothesis applies to the fact that $2^k > 2k$ which then implies $2\cdot 2^k > 2\cdot 2k$. – amWhy Aug 31 '13 at 14:03
@user90771 Does the immediately preceding comment help? Simply multiply each side of the inductive hypothesis by $2$: that gives us $2\cdot 2^k > 2\cdot 2k$. – amWhy Aug 31 '13 at 14:05
@amwhy yes it does! So multiplying the inductive hypothesis gives $2^{k+1}>4k$ and we know that $2^{k+1}>2(k+1)$ Why is $4k \ge 2k+2$? – salman Aug 31 '13 at 14:09
Indeed! And don't forget to show $p(3)$ holds too! ;-) – amWhy Aug 31 '13 at 14:11
@amWhy my book strangely does that after proving the inductive hypothesis. – salman Aug 31 '13 at 14:14

score 1 · Answer 2 · edited Aug 31 '13 at 14:18

When you will use mathematical induction, try keep on your mind that you have to follow 3 steps:

Verify that your statement is true for a first case, commonly for 1.
Suppose that your statement is true for a particular $k$ $\in \mathbb{N}$
Then try to prove that you statement is true for $k+1$.

In you problem:

The first number that you statement works is $3$, because $2^3 > 2\cdot3 $
Supose that there exists a $k \in \mathbb{N}$ such that $2^k > 2\cdot k $
Now we will show that $2^{k+1} > 2\cdot(k+1) $:
Note that $2^{k+1} = 2^k\cdot 2 > 2k\cdot2 = 4k = 2k+2k > 2k+2 = 2(k+1)$.

Hence, $2^k > 2\cdot k $ for all $k \in \mathbb{N}$ with $k \geq 3$.

Prove that $2^n>2n$ for all integral values of n greater than 2

2 Answers2