How do you find a one-decimal place approximation for #sqrt 50#?

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Answer 1 · 2015-10-21T23:51:19

#sqrt(50) ~~ 7 + 1/14 ~~ 7.1#

Since #50 = 7^2 + 1# is of the form #n^2 + 1#, its square root is a very simple continued fraction:

#sqrt(50) = [7;bar(14)] = 7+1/(14+1/(14+1/(14+...)))#

A rough approximation can be made by truncating early:

#sqrt(50) ~~ [7;14] = 7+1/14 = 99/14 = 7.0dot(7)1428dot(5)#

If we want a better one, just include more terms:

#sqrt(50) ~~ [7;14;14] = 7+1/(14+1/14) = 7+14/197 = 1393/197 ~~ 7.071066#

Actually #sqrt(50) ~~ 7.07106781186547524400#