How do you find the square root of 7?

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Sep 1, 2016

Answer:

#sqrt(7) ~~ 2.645751311#

Explanation:

Since #7# is a prime number, it has no square factors and its square root cannot be simplified.

It is an irrational number, so cannot be exactly represented by #p/q# for any integers #p, q#.

We can however find good rational approximations to #sqrt(7)#.

First note that:

#8^2 = 64 = 63+1 = 7*3^2 + 1#

This is in Pell's equation form:

#p^2 = n q^2 + 1#

with #n = 7#, #p = 8# and #q = 3#.

This means that #8/3# is an economical approximation for #sqrt(7)# and it also means that we can use #8/3# to derive the continued fraction expansion of #sqrt(7)#:

#8/3 = 2 + 1/(1+1/(1+1/1))#

and hence we can deduce:

#sqrt(7) = [2;bar(1,1,1,4)] = 2 + 1/(1+1/(1+1/(1+1/(4+1/(1+1/(1+1/(1+1/(4+...))))))))#

The next economical approximation is given by truncating the continued fraction expansion just before the next #4#, i.e.

#sqrt(7) ~~ [2;1,1,1,4,1,1,1] = 2 + 1/(1+1/(1+1/(1+1/(4+1/(1+1/(1+1/1)))))) = 127/48 = 2.6458bar(3)#

This is also a solution of Pell's equation for #7#, since we find:

#127^2 = 16129 = 16128+1 = 7*48^2+1#

If you want more accuracy, truncate just before the next #4# or the one after.

By expanding the repeating part of the continued fraction for #sqrt(7)# we can derive a generalised continued fraction:

#sqrt(7) = 21/8+(7/64)/(21/4+(7/64)/(21/4+(7/64)/(21/4+(7/64)/(21/4+...))))#

Using a calculator, we find:

#sqrt(7) ~~ 2.645751311#

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