<s>[INST] Proving certain properties for Babylonian method The Babylonian method for approximating square roots is divided into three steps:

*

*Guess an initial approximation $a$ of $ {\sqrt N}$, where $a$ and $N$ are postive rational numebrs and $N$ is not the square of any rational numbers.

*Let $c=\frac{N-a^2}{2a}$.

*Now $a+c$ is a new approximation of $ {\sqrt N}$.

The question is to show $a+c>\sqrt N$.
I think that since the problem did not inform us of the relationship between $a$ and ${\sqrt N}$, so no matter what number we pick for $a$ does not affect the fact that $a+c>\sqrt N$.
Then we will have
$a+c=a+\frac{N-a^2}{2a}=\frac{a^2+N}{2a}$, if $a$ >$\sqrt N$, then we will have $\frac{a^2+N} {2a} >\frac{2N}{2a}$, $\frac{2N}{2a}<\frac{2N}{2\sqrt N}=\sqrt N$. However, how do I find the relation between $\frac{a^2+N} {2a}  $ and $\frac{2N}{2\sqrt N}$? Any help would be greatly appreciated.
 [/INST] \begin{align*}
a+c &= a + \frac{N - a^2}{2a}
= \frac{N + a^2}{2a}
= \frac{(\sqrt{N} - a)^2 + 2a\sqrt{N}}{2a}\\
&=  \frac{(\sqrt{N} - a)^2}{2a} + \sqrt{N} \geq \sqrt{N}
\end{align*}
</s>