If $P_n$ is the statement $3^n>n,$ prove $P_{n+1}$ is true whenever $P_n$ is true.

Question

If $3^n>n$ then $3^{n+1}>n+1$

I could not solve this problem

What have you tried so far? Is at least the result intuitively clear to you? Do you know proving statements by induction? — Imago, Jan 26 '17 at 16:09
@Imago I know that first we must show P(1) is true, 2nd step we must assume P(k). Then final step we must show it is true for P(k+1). Then only it will be proved. But i could not understand this question can you explain me. — bappy, Jan 27 '17 at 02:12

Bram28 · Accepted Answer · 2017-01-27T17:53:43.613

2

If $n=0$, then the truth of this statement is easily verifiable, since

$$3^0 = 1 > 0.$$

If $n>0$, i.e. $n\ge1$, then:

$$3^{n+1} = (by \: basic \: arithmetic)$$

$$3\cdot3^n > (by \: assumption \: 3^n > n)$$

$$3n = (by \: basic \: arithmetic)$$

$$2n + n \ge (by \: n \ge 1)$$

$$2n + 1 > (by \: n \ge 1)$$

$$n + 1$$

edited Jan 27 '17 at 17:53

answered Jan 26 '17 at 16:14

Bram28

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Can you please explain the answer by writing each line in such a way which would be easy to understand. – bappy Jan 27 '17 at 17:36
@bappy Sure .. give me a sec – Bram28 Jan 27 '17 at 17:50

score 0 · Answer 2 · answered Jan 26 '17 at 16:18

Presumably, $n\in\Bbb N$ since we're talking about an induction proof (you could also include $n=0$ but I won't here). Thus, $$n\geq 1\implies n+n\geq n+1\implies n+n+n>n+1\implies 3n>n+1$$ You also have $3^n>n$ by assumption. I'll leave the rest to you.

If $P_n$ is the statement $3^n>n,$ prove $P_{n+1}$ is true whenever $P_n$ is true.

2 Answers2