$100th$ digit after decimal point in $\sqrt 2$

Question

I want to find the $100th$ digit after decimal in $\sqrt 2$.

$\sqrt 2 = 1.41421356237...$

for example $7th$ digit after decimal is $5$. Is there any general method for such problems? I thought it might use the representation of real numbers as a cauchy sequence, but was not able to proceed further.

Thanks in advance!

Answer 1

Yes, there is a general method for this problem; basically, there is a way to modify the "long division" algorithm to give you square roots instead of quotients. A detailed explanation and worked example can be found here:

https://xlinux.nist.gov/dads/HTML/squareRoot.html

An alternative technique: suppose we want the first few digits of the square root of two. Take the integer 2000000000000. (Note that there must be an even number of zeros.) There is some integer n such that when squared, it is less than 2000000000000, and (n+1) squared is greater. You can find that integer via binary search in about 40 guesses; the digits that n and n+1 have in common are definitely the digits of the square root of two. Want more digits? Add more zeros. Each additional pair of zeros will increase the binary search length by a small amount.

Answer 2

Every CAS can do this

$$1.4142135623730950488016887242096980785696718\\75376948073176679737990732478462107038850387534327641573$$

Answer 3

This could be of help to you in your pursuit. Good luck!