Proof.

Let \(P(n)\) be the statement \(n \lt 100\text{.}\) We will prove \(P(n)\) is true for all \(n \in \N\text{.}\) First we establish the base case: when \(n = 0\text{,}\) \(P(n)\) is true, because \(0 \lt 100\text{.}\) Now for the inductive step, assume \(P(k)\) is true. That is, \(k \lt 100\text{.}\) Now if \(k \lt 100\text{,}\) then \(k\) is some number, like 80. Of course \(80+1 = 81\) which is still less than 100. So \(k +1 \lt 100\) as well. But this is what \(P(k+1)\) claims, so we have shown that \(P(k) \imp P(k+1)\text{.}\) Thus by the principle of mathematical induction, \(P(n)\) is true for all \(n \in \N\text{.}\)