Hero's Method/Lemma 1

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Lemma for Hero's Method

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:

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

$\blacksquare$


Sources

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