What is the square root of 5?

2 Answers
Mar 14, 2018

The square root of #5# can't be simplified father than it already is, so here is #sqrt5# to ten decimal places:

#sqrt5~~2.2360679775...#

Mar 14, 2018

Answer:

#sqrt(5) = 2+1/(4+1/(4+1/(4+1/(4+1/(4+...))))) ~~ 2889/1292 ~~ 2.236068# is an irrational number.

Explanation:

All positive numbers normally have two square roots, a positive one and a negative of the same size. We denote the positive (a.k.a. principal) square root of #n# by #sqrt(n)#.

A square root of a number #n# is a number #x# such that #x^2 = n#. So if #x^2 = n# then also #(-x)^2 = n#.

However, popular usage is that "the square root" refers to the positive one.

Suppose we have a positive number #x# which satisfies:

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

Then multiplying both sides by #(2+x)# we get:

#x^2+2x = 2x+5#

Then subtracting #2x# from both sides we get:

#x^2=5#

So we have found:

#sqrt(5) = 2+1/(2+sqrt(5))#

#color(white)(sqrt(5)) = 2+1/(4+1/(4+1/(4+1/(4+1/(4+...)))))#

SInce this continued fraction does not terminate, we can tell that #sqrt(5)# cannot be represented as a terminating fraction - i.e. a rational number. So #sqrt(5)# is an irrational number a little smaller than #2 1/4 = 9/4#. For better rational approximations you can terminate the continued fraction after more terms.

For example:

#sqrt(5) ~~ 2+1/(4+1/4) = 2+4/17 = 38/17 ~~ 2.235#

Unpacking these continued fractions can be a little tedious, so I generally prefer to use a different method, namely the limiting ratio of an integer sequence defined recursively.

Define a sequence by:

#{ (a_0 = 0), (a_1 = 1), (a_(n+2) = 4a_(n+1)+a_n) :}#

The first few terms are:

#0, 1, 4, 17, 72, 305, 1292, 5473#

The ratio between terms will tend to #2+sqrt(5)#.

So we find:

#sqrt(5) ~~ 5473/1292 - 2 = 2889/1292 ~~ 2.236068#