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The Babylonian method for approximating square roots is divided into three steps:

  1. 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.
  2. Let $c=\frac{N-a^2}{2a}$.
  3. 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.

user19170731
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1 Answers1

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\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*}

adfriedman
  • 3,641