question on prove by induction that for each n$\in\mathbb{N}_{\ge2}$, $n^2$< $n^3$

Question

I have to prove by induction that for each n$\in\mathbb{N}_{\ge2}$, $n^2$< $n^3$.

If I try to prove for P(1) I end up with 1 < 1. Is this right? Why does it or does it not make sense?

Answer 1

Technically, you can start at $1$ for this sort of proof. Namely, you are proving:

$$\forall n\in\mathbb N(n\geq 2\implies n^3> n^2)$$

So you prove that since $1\geq 2$ is not true, then $1\geq 2\implies 1^3> 1^2$ is true.

Then show that if the statement is true for $n$, then, if $n+1=2$, prove that case, separately, and if $n+1>2$ then $n\geq 2$, so you can use standard induction, because you can assume $n^3>n^2$ when $n\geq 2$.

If $P(n)$ is a statement, and $m$ is a natural number, in order to prove:

$$\forall n(n\geq m\implies P(n))$$

You can prove: $P(m)$ and $$\forall n (n\geq m\implies (P(n)\implies P(n+1))$$

Answer 2

The condition says $n\ge2$

If $n^3>n^2$ for $n=m$ i.e., $m^3>m^2$

$(m+1)^3=m^3+3m^2+3m+1>m^2+3m^2+3m+1$(by inductive hypothesis)

It is sufficient to establish $m^2+3m^2+3m+1\ge(m+1)^2\iff3m^2+m\ge0$ which holds true for $m\ge0$

Answer 3

Let $n=2$ $$2^2 \le 2^3$$ Let n=k. Assume for $n=k$ $$n^2 \le n^3$$ Let k=n+1 $$(k+1)^2 \le (k+1)^3$$ $$k^2+2k+1 \le k^3+3k^2+3k+1$$ $$1 \le k^3+2k^2+k+1$$ $$0 \le k^3+2k^2+k$$ This is true for all $k \ge 0$, thus the inequality holds. Since we've proven the case for n=2, this is true by induction for all $n \ge 2$