What is the square root of #3# ?

2 Answers
Nov 3, 2016

Answer:

#sqrt3 = 1.732#

Explanation:

3 is not a perfect square, so does not have an exact square root.

#sqrt3# is an irrational number. The answer is an infinite, non-recurring decimal.

Using a calculator gives:

#sqrt3 = 1.7320508.....#

Give your answer rounded to a reasonable number of decimal places.

Mar 19, 2017

Answer:

The square root of #3# is an irrational number which we can approximate as:

#sqrt(3) ~~ 18817/10864 ~~ 1.7320508#

Explanation:

"The" square root of #3# is an irrational number. It is not expressible in the form #p/q# for integers #p, q# and its decimal expansion neither repeats nor terminates.

It can be expressed by a (non terminating) continued fraction:

#sqrt(3) = [1;bar(1,2)] = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+1/(1+...)))))))#

This positive square root is also known as the principal square root of #3#. The number #-sqrt(3)# is also a square root of #3#.

We can get rational approximations to #sqrt(3)# by truncating the continued fraction early.

For example:

#sqrt(3) ~~ [1;1,2,1] = 1+1/(1+1/(2+1/1)) = 7/4 = 1.75#

#sqrt(3) ~~ [1;1,2,1,2,1] = 1+1/(1+1/(2+1/(1+1/(2+1/1)))) = 26/15 = 1.7bar(3)#

#sqrt(3) ~~ [1;1,2,1,2,1,2,1] = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+1/1)))))) = 97/56 = 1.732bar(142857)#

#color(white)()#
Another method to find approximations to #sqrt(n)# begins with a rational approximation #p_0/q_0# then repeatedly applies the formulae:

#{ (p_(i+1) = p_i^2 + n q_i^2), (q_(i+1) = 2 p_i q_i) :}#

Let us start with #n=3#, #p_0=2#, #q_0=1#.

Then:

#{ (p_1 = p_0^2+n q_0^2 = 2^2+3*1^2 = 4+3=7), (q_1 = 2 p_0 q_0 = 2*2*1 = 4) :}#

#{ (p_2 = p_1^2 + n q_1^2 = 7^2+3*4^2 = 49+48 = 97), (q_2 = 2 p_1 q_1 = 2*7*4 = 56) :}#

#{ (p_3 = p_2^2 + n q_2^2 = 97^2+3*56^2 = 9409+9408 = 18817), (q_3 = 2 p_2 q_2 = 2*97*56 = 10864) :}#

Stopping here, we get:

#sqrt(3) ~~ 18817/10864 ~~ 1.7320508#

As you may notice, this method is exponentially faster than the continued fraction method, effectively doubling the number of terms each time.