Digit-by-digit calculation of square root

  • Thread starter jd12345
  • Start date
Let's call the initial number a. The part already determined form the square root is p.

The basic idea is like that: For each step, find the next digit such that the squared value (consisting of p plus the new digit) is smaller than a, but the same number with one more as next digit would (squared) give a value larger than a.
Now, how do we find this digit?

We already know that p^2<a and the difference d=a-p^2 can be calculated.
Now we add a small value x/10 to p. This gives (p+x/10)^2=p^2+2px/10+x^2/100 and should be a better approximation to the square root.

Therefore, we want to find an x with 2px/10+x^2/100 <= d.
To avoid decimal digits, shift everything by two places, which is equivalent to a multiplication of 100 on both sides:
20 p x + x^2 <= 100d.
100d is now called c.
x (20p+x) <= c
Which is the formula used by the wikipedia page to find the next digit. Does that help?


Gold Member
There is an explanation in George Crystal's(sp) Algebra(2 volumes), mine is in a box somewhere, but if you can find a copy it's in there.

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