How do you find the square root of 2?

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Answer 1 · 2016-08-30T21:33:04

Use a continued fraction to find rational approximations.

#sqrt(2)# is an irrational number, not expressible in the form #p/q# for integers #p, q#.

We can find rational approximations in several ways. Here I show a method called continued fractions...

Consider the number #t = sqrt(2)+1#

Then:

#t = sqrt(2)+1#

#= 2 + (sqrt(2)-1)#

#= 2 + ((sqrt(2)-1)(sqrt(2)+1))/(sqrt(2)+1)#

#= 2 + (2-1)/(sqrt(2)+1)#

#= 2 + 1/(sqrt(2)+1)#

#= 2 + 1/t#

Given that:

#t = 2 + 1/t#

notice that we can substitute this expression for #t# on the right hand side to find:

#t = 2 + 1/(2+1/t)#

and again:

#t = 2 + 1/(2+1/(2+1/t))#

In fact:

#t = 2 + 1/(2+1/(2+1/(2+1/(2+1/(2+...)))))#

Now remember #t = sqrt(2) + 1#, so we have:

#sqrt(2) = 1 + 1/(2+1/(2+1/(2+1/(2+1/(2+...)))))#

This is called a continued fraction.

There is a shorter notation for a continued fraction using square brackets. Using this notation we can write:

#sqrt(2) = [1;2,2,2,2,2,...] = [1;bar(2)]#

To find a rational approximation for #sqrt(2)# we can truncate this continued fraction early.

For example:

#sqrt(2) ~~ [1;2,2,2] = 1+1/(2+1/(2+1/2)) = 1+1/(2+2/5) = 1+5/12 = 17/12 ~~ 1.41bar(6)#

For more accuracy, truncate a little later:

#sqrt(2) ~~ [1;2,2,2,2,2] = 1+1/(2+1/(2+1/(2+1/(2+1/2)))) = 99/70 = 1.4bar(142857)#

This is actually the same accuracy as an approximation to #sqrt(2)# as the ratio of the sides of a sheet of A4 (#297"mm" xx 210"mm"#).

In fact #sqrt(2)# is closer to #1.41421356237#