How do you simplify #sqrt41#?

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Answer 1 · 2017-06-11T15:17:34

#sqrt(41) ~~ 6.4031242374# is an irrational number which cannot be simplified.

#41# is a prime number, so has no square factors.

As a result its square root cannot be simplified. It is an irrational number.

We find:

#6^2 = 36 < 41 < 49 = 7^2#

So:

#6 < sqrt(41) < 7#

To get a good approximation for #sqrt(41)# note that:

#64^2 = 4096#

So:

#sqrt(41) ~~ sqrt(40.96) = 6.4 = 32/5#

In general:

#sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+...)))#

Putting #a=32/5# and #b=1/25#, we find:

#sqrt(41) = 32/5+(1/25)/(64/5+(1/25)/(64/5+(1/25)/(64/5+...)))#

We can get rational approximations for #sqrt(41)# by truncating this continued fraction.

For example:

#sqrt(41) ~~ 32/5+(1/25)/(64/5) = 2049/320#

#sqrt(41) ~~ 32/5+(1/25)/(64/5+(1/25)/(64/5)) = 131168/20485 ~~ 6.4031242374#