1
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

If $a^2>n$, then $\frac{n}{a^2}<1$. Let $b=\frac{n}a$; then $$b^2=\frac{n^2}{a^2}=n\cdot\frac{n}{a^2}<n\cdot1=n\,,$$ so $b<\sqrt{n}$.

$\endgroup$
4
  • $\begingroup$ in the second line, where did the inequality come from? $\endgroup$
    – David
    Jan 15, 2021 at 20:06
  • $\begingroup$ @David: It came from the inequality in the first line: $\frac{n}{a^2}<1$. $\endgroup$ Jan 15, 2021 at 20:08
  • $\begingroup$ I don't understand where it was substituted or how it cancels out? Am I missing something with regard to how inequalitites work? $\endgroup$
    – David
    Jan 15, 2021 at 20:13
  • $\begingroup$ @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 $\frac{n^2}{a^2}<n$. $\endgroup$ Jan 15, 2021 at 20:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .