
#1
Nov210, 02:39 AM

P: 5

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. 



#2
Nov210, 05:58 AM

P: 617

For your example I would first show: 3^{n} ≥ 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: 3^{n} ≥ 3n. Then our goal is to show: 3^{n+1}≥3(n+1) To do that I would prove the following: 3^{n} ≥ 3n ⇒ 3+3^{n} ≥ 3(n+1) Then i would prove: 3^{n+1}≥3+3^{n} for n>1 Putting these together: 3^{n+1}≥3+3^{n}≥3(n+1) This step shows our goal! Thus by the principle of induction: 3^{n} ≥ 3n for Natural n Then you know: 3^{n} ≥ 3n ⇒3^{n} + 2 ≥ 3n or 3^{n} ≥ 3n ⇒3*3*3^{n} =3^{n+2} ≥ 3n from the properties of inequalities. It's hard to tell which of these you were trying to prove how you wrote it. 



#3
Nov210, 10:41 AM

P: 5

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? 



#4
Nov210, 09:14 PM

P: 617

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



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