What is #sqrt(2)# ?

The positive square root of #2#, #sqrt(2)# is not a rational number. That is, it cannot be represented in the form #p/q# with integers #p, q# with #q != 0#.

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Here's a sketch of a proof by contradiction...

Suppose #sqrt(2) = p/q# for some positive integers #p, q#.

Note that #p > q# and without loss of generality, we may suppose that #p, q# is the smallest such pair of integers.

Then:

#2 = (p/q)^2 = p^2/q^2#

So:

#p^2 = 2q^2#

So #p^2# is divisible by #2# and since #2# is prime, #p# must also be divisible by #2#.

So:

#p = 2m#

for some positive integer #m# with #m < p#.

So we have:

#2q^2 = p^2 = (2m)^2 = 2*2m^2#

Dividing both ends by #2m^2# we find:

#(q/m)^2 = q^2/m^2 = 2#

So #sqrt(2) = q/m# and #p > q > m#

That is #q, m# is a smaller pair of positive integers whose quotient is #sqrt(2)#, contradicting our choice of #p, q#.

Hence there is no such pair of integers, and #sqrt(2)# is irrational.

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Approximations

In common with the square root of any positive integer, #sqrt(2)# does have a representation as a recurring continued fraction. In this case a particularly simple one:

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

We can truncate this continued fraction to get rational approximations to #sqrt(2)#.

For example:

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

That's not stunning good, but we can take more terms to find:

#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 the approximation corresponding to the dimensions of a sheet of A4 paper which is #297"mm" xx 210"mm"# (just multiply #99# and #70# by #3"mm"#).

See https://socratic.org/s/aHnJqjrg for more details.

Using a calculator or computer you can find closer approximations such as:

#sqrt(2) ~~ 1.414213562373#