Hero's Method/Lemma 1

Theorem
Let $a \in \R$ be a real number such that $a > 0$.

Let $x_1 \in \R$ be a real number such that $x_1 > 0$.

Let $\sequence {x_n}$ be the sequence in $\R$ defined recursively by:


 * $\forall n \in \N_{>0}: x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$

Then:
 * $\forall n \in \N_{>0}: x_n > 0$

Proof
The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
 * $x_n > 0$

Basis for the Induction
$\map P 1$ is the case:
 * $x_1 > 0$

which is assumed.

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $x_k > 0$

from which it is to be shown that:
 * $x_{k + 1} > 0$

Induction Step
This is the induction step:

We have that:
 * $x_{k + 1} = \dfrac {x_k + \dfrac a {x_k} } 2$

But as $x_k > 0$ and $a > 0$, it follows that:


 * $\dfrac a {x_k} > 0$

Then as $x_k > 0$ and $\dfrac a {x_k} > 0$, it follows that:
 * $\dfrac 1 2 \paren {x_k + \dfrac a {x_k} } > 0$

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

Therefore:
 * $\forall n \in \N_{>0}: x_n > 0$