How do you find the exact square root of 39?

1 Answer
Oct 6, 2015

Answer:

#39 = 3 * 13# has no square factors, so its square root cannot be simplified. It is an irrational number, so cannot be represented by a fraction or by a terminating or repeating decimal expansion.

Explanation:

If you would like to prove that #sqrt(39)# is irrational, you can prove it in a similar way to #sqrt(15)# as shown here:

How do you prove that square root 15 is irrational?

You can find a succession of rational approximations for #sqrt(39)# using a Newton Raphson type method.

Typically you would start with an approximation #a_0# and iterate using a formula like:

#a_(i+1) = (a_i^2 + n) / (2a_i)#

where #n = 39# is the number you are trying to approximate the square root of.

I prefer to split #a_i# into a fraction #p_i/q_i# and iterate using the formulas:

#p_(i+1) = p_i^2 + n q_i^2#

#q_(i+1) = 2 p_i q_i#

If the resulting #p_(i+1)# and #q_(i+1)# have a common factor, then divide both by that factor before the next iteration...

So in our case, let #n = 39#, #p_0 = 6# and #q_0 = 1#. We use an initial approximation #sqrt(39) ~~ 6/1# since #6^2 = 36#.

Then:

#p_1 = p_0^2 + n q_0^2 = 6^2 + 39*1^2 = 36+39 = 75#

#q_1 = 2 p_0 q_0 = 2 * 6 * 1 = 12#

Now both of these are divisible by #3#, so divide both by #3# to get:

#p_(1a) = 25#

#q_(1a) = 4#

If we stopped here we would get #sqrt(39) ~~ 25/4 = 6.25#

Next iteration:

#p_2 = p_(1a)^2 + n q_(1a)^2 = 25^2 + 39*4^2 = 625 + 624 = 1249#

#q_2 = 2 p_(1a) q_(1a) = 2 * 25 * 4 = 200#

If we stopped here we would get #sqrt(39) ~~ 1249/200 = 6.245#

Next iteration:

#p_3 = p_2^2 + n q_2^2 = 1249^2 + 39*200^2 = 1560001 + 1560000 = 3120001#

#q_3 = 2 * 1249 * 200 = 499600#

If we stop here, we get the approximation:

#sqrt(39) ~~ 3120001 / 499600 ~~ 6.2449979983987#

Actually:

#sqrt(39) ~~ 6.24499799839839820584#