How do you find the square root of 203?

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Answer 1 · 2016-08-31T19:08:04

Use a Newton Raphson method to find:

#sqrt(203) ~~ 6497/456 ~~ 14.247807#

#203 = 7 * 29# has no square factors. So its square root cannot be simplified.

It is an irrational number between #14# and #15#, since:

#14^2 = 196 < 203 < 225 = 15^2#

As such, it cannot be represented in the form #p/q# for integers #p, q#.

We can find rational approximations using a Newton Raphson method:

To approximate the square root of #n#, start with an approximation #a_0# then iterate using the formula:

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

I prefer to re-formulate this slightly using separate integers #p_i, q_i# where #p_i/q_i = a_i# and formulae:

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

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

If the resulting pair of integers have a common factor, then divide by that before the next iteration.

So for our example, let #n=203#, #p_0 = 29# and #q_0 = 2# (i.e. #a_0 = 14.5#) ...

#{ (p_0 = 29), (q_0 = 2) :}#

#{ (p_1 = p_0^2+ n q_0^2 = 29^2 + 203*2^2 = 841+812 = 1653), (q_1 = 2 p_0 q_0 = 2*29*2 = 116) :}#

Note that both #p_1# and #q_1# are divisible by #29#, so divide both by that:

#{ (p_(1a) = 1653/29 = 57), (q_(1a) = 116/29 = 4) :}#

Next iteration:

#{ (p_2 = p_(1a)^2 + n q_(1a)^2 = 57^2+203*4^2 = 3249+3248 = 6497), (q_2 = 2 p_(1a) q_(1a) = 2*57*4 = 456) :}#

If we stop here, we get:

#sqrt(203) ~~ 6497/456 ~~ 14.247807#

Each iteration roughly doubles the number of significant figures in the approximation.