How do you find the square root of 15?

1 Answer
Jun 26, 2016

Answer:

#sqrt(15)# is not simplifiable.

We can find rational approximations #31/8#, #244/63#

Explanation:

#15=3xx5# has no square factors, so #sqrt(15)# cannot be simplified.

It is not expressible as a rational number. It is an irrational number a little less than #4#.

Since #15 = 4^2-1# is of the form #n^2-1#, #sqrt(15)# has a fairly simple continued fraction expansion:

#sqrt(15) = [3;bar(1,6)] = 3+1/(1+1/(6+1/(1+1/(6+1/(1+1/(6+1/(1+...)))))))#

We can truncate this continued fraction expansion early to get rational approximations to #sqrt(15)#.

For example:

#sqrt(15) ~~ [3;1,6,1] = 3+1/(1+1/(6+1/1)) = 3+1/(1+1/7) = 3+7/8 = 31/8 = 3.875#

#sqrt(15) ~~ [3;1,6,1,6,1] = 3+1/(1+1/(6+1/(1+1/(6+1/1)))) = 3+1/(1+1/(6+1/(1+1/7)))#

#= 3+1/(1+1/(6+7/8)) = 3+1/(1+8/55) = 3+55/63 = 244/63 = 3.bar(873015)#

Actually:

#sqrt(15) ~~ 3.87298334620741688517#