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}$.
known that $a+c>\sqrt N$.
The question is to show if $N<5a^2$ then $|a^2-N|>(a+c)^2-N.$
My thoughts are since $a+c=\frac{N+a^2}{2a}$, then $(a+c)^2=\frac{N^2+2a^2N+a^4}{4a^2}$. If we let $N=5a^2$, we will get $(a+c)^2-N=4a^2$ which is equal to $|a^2-N|$, then how can you tell the relation when $N<5a^2$ since we have $N$ on both side and an absolute value? Usually, when there is an absolute value and inequality we think about triangular inequality but that doesn't seems to work. Greatly appreciated for the help!
Go a little further:$$ (a+c)^2=\frac{N^2+2a^2N+a^4}{4a^2}\\ (a+c)^2-N=\frac{N^2-2a^2N+a^4}{4a^2}\\ (a+c)^2-N=\frac{(N-a^2)^2}{4a^2} $$
If $N<5a^2$, then $4a^2>N-a^2$. That gives you a lower bound for the denominator, above:$$ \frac{(N-a^2)^2}{4a^2}<\frac{(N-a^2)^2}{N-a^2}=|N-a^2| $$