What is the square root of 230?

1 Answer
Jun 27, 2018

Answer:

#sqrt(230) = 15+1/(6+1/(30+1/(6+1/(30+1/(6+1/(30+...))))))#

Explanation:

Suppose #x > 0# satisfies:

#x = 15+1/(6+1/(15+x))#

Simplifying the right hand side we find:

#x = 15+1/(6+1/(15+x))#

#color(white)(x) = 15+(15+x)/(96+6x)#

#color(white)(x) = (1380+91x)/(91+6x)#

Multiplying both ends by #(91+6x)# this becomes:

#6x^2+91x = 1380+91x#

Subtracting #91x# from both sides, this becomes:

#6x^2 = 1380#

Dividing both sides by #6#, we find:

#x^2 = 230#

So #x = sqrt(230)# and:

#sqrt(230) = 15+1/(6+1/(15+sqrt(230)))#

#color(white)(sqrt(230)) = 15+1/(6+1/(30+1/(6+1/(30+1/(6+1/(30+...))))))#

Since this continued fraction does not terminate, we can conclude that #sqrt(230)# is not expressible as a terminating fraction. In other words, it is irrational.

Terminating before the first occurrence of #30# we get an efficient first rational approximation:

#sqrt(230) ~~ 15+1/6 = 91/6#

with the property that:

#91^2 = 8281 = 8280+1 = 230 * 6^2 + 1#

For greater accuracy you can terminate the continued fraction later or use this rational approximation #91/6# as the initial approximation in a Babylonian method.