Strictly Increasing Sequence of Natural Numbers

Theorem

Let $\N_{>0}$ be the set of natural numbers without zero:

$\N_{>0} = \left\{{1, 2, 3, \ldots}\right\}$

Let $\left \langle {n_r} \right \rangle$ be strictly increasing sequence in $\N_{>0}$.

Then:

$\forall r \in \N_{>0}: n_r \ge r$

Proof

This is to be proved by induction on $r$.

For all $r \in \N_{>0}$, let $P \left({r}\right)$ be the proposition $n_r \ge r$.

Basis for the Induction

When $r = 1$, it follows that $n_1 \ge 1$ as $\N_{>0}$ is bounded below by $1$.

Thus $P(1)$ is true.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So our induction hypothesis is that $n_k \ge k$.

Then we need to show that $n_{k+1} \ge k+1$.

Induction Step

This is our induction step:

Suppose that $n_k \ge k$.

Then as $\left \langle {n_r} \right \rangle$ is strictly increasing:

$n_{k+1} > n_k \ge k$

From Sum with One is Immediate Successor in Naturally Ordered Semigroup, we have:

$n_{k+1} > k \implies n_{k+1} \ge k+1$

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore $\forall r \in \N_{>0}: n_r \ge r$.

$\blacksquare$