prove $n_i \ge i$ by induction

Question

$n_i$ is an strictly increasing monotone sequence, and $i \in N$. $n_i \in N$

$P(1) : n_1 \ge 1$

Suppose $P(k) : n_k \ge k$

Then, $P(k+1) : n_{k+1} \ge n_k$. I am stuck in here.

Could you give some hint??

Thank you in advance.

Answer 1

From $$P(k) : n_k \ge k$$

You need to prove that $$P(k+1) : n_{k+1} \ge k+1$$

Note that $$n_{k+1} > n_k \ge k \implies n_{k+1} > k$$

Since $n_{k+1}$ is an integer, $$n_{k+1} > k \implies n_{k+1} \ge k+1$$

Thus $$P(k+1) : n_{k+1} \ge k+1$$