How Does the Inequality 2*3^k >= 3 Arise in Mathematical Induction?

[beatem]

Hi,

I'm trying to learn mathematical induction for proving inequalities, but there is just one step I cannot get past: finding another inequality that is added to the inductive hypothesis.

For example, in this problem:

Prove for all positive integers (n >= 1), prove 3^n + 2 >= 3n.

I understand the basis step and in general how to do induction, but for some reason, the example says that that after I get the hypothesis, 3^k + 2 >= 3k (for some arbitrary k), it can generate the inequality 2*3^k >= 3 for all k >= 1. Where does this come from? I can follow how it adds this inequality to the hypothesis, but what is this, and how would I go about getting this?

This isn't just a generic problem by the way: I've looked at many examples, but I can't figure out what this is when dealing with inequalities and induction.

 

[JonF]

For your example I would first show: 3ⁿ ≥ 3n which is easier.
You said you can do the basic step. So let's move on to the induction.

To do the induction we suppose n, then we prove if n is true, n+1 is true.
So first suppose: 3ⁿ ≥ 3n. Then our goal is to show: 3ⁿ⁺¹≥3(n+1)

To do that I would prove the following:
3ⁿ ≥ 3n ⇒ 3+3ⁿ ≥ 3(n+1)
Then i would prove: 3ⁿ⁺¹≥3+3ⁿ for n>1
Putting these together: 3ⁿ⁺¹≥3+3ⁿ≥3(n+1) This step shows our goal!
Thus by the principle of induction: 3ⁿ ≥ 3n for Natural n

Then you know: 3ⁿ ≥ 3n ⇒3ⁿ + 2 ≥ 3n or 3ⁿ ≥ 3n ⇒3*3*3ⁿ =3ⁿ⁺² ≥ 3n from the properties of inequalities. It's hard to tell which of these you were trying to prove how you wrote it.

 

[beatem]

Thanks for the reply!

Sorry: I meant (3^n)+2

So is there no need for the extra inequality of 2*3^n >= 3? Or am I just missing something?

 

[JonF]

No need for the other inequality, which i think you typed incorrectly.