Bernoulli's Inequality/Proof 1

Theorem
Let $X$ be one of the sets of numbers $\N$, $\Z$, $\Q$, or $\R$.

Let $n \in \N$, $x \in X$, $x \ge -1$.

Then:
 * $\left({1 + x}\right)^n \ge 1 + nx$

Proof
For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\left({1 + x}\right)^n \ge 1 + nx$

We use the Principle of Mathematical Induction.

Basis for the Induction
$P \left({0}\right)$ is the case:
 * $\left({1 + x}\right)^0 \ge 1$

so $P \left({0}\right)$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\left({1 + x}\right)^k \ge 1 + kx$

We need to show that:
 * $\left({1 + x}\right)^{k+1} \ge 1 + \left({k + 1}\right) x$

Induction Step
This is our induction step:

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