# In relation to the Babylonian method for computing a square root, if N/b = c, will c always be less than b?

I'm trying to wrap my head around the Babylonian method/algorithm for computing the square root of a number N. I can't seem to explain in words why, when a is too large i.e a^2 > N, then why you divide N/a to get b.

For example.

Compute the square root of 9409, where N = 9409.

If you guess a = 100, you get 10,000, which is too large as a^2 > N

so you then decide N/a = b and you say that b is an under estimate.

Why would b, be an underestimate?

In the wikipedia article for this: https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method it relates to the sentence "The basic idea is that if x is an overestimate to the square root of a non-negative real number S then S/x will be an underestimate, or vice versa".

I hope I'm making sense, if not, please let me know.

If there are any resources where I may be able to get a better understanding please feel free to link them, including any practise problems.

If $$a^2>n$$, then $$\frac{n}{a^2}<1$$. Let $$b=\frac{n}a$$; then $$b^2=\frac{n^2}{a^2}=n\cdot\frac{n}{a^2} so $$b<\sqrt{n}$$.
• @David: It came from the inequality in the first line: $\frac{n}{a^2}<1$. Jan 15, 2021 at 20:08
• @David: Just multiply the inequality $\frac{n}{a^2}<1$ by $n$; $n$ is positive, so the directly of the inequality is unchanged, and you get $\frac{n^2}{a^2}<n$. Jan 15, 2021 at 20:15