What is the square root of 89?

The square root of ##89## is a number which when squared gives ##89##.

##sqrt(89) ~~ 9.434##

Since ##89## is prime, ##sqrt(89)## cannot be simplified.

You can approximate it using a Newton Raphson method.

I like to reformulate it a little as follows:

Let ##n = 89## be the number you want the square root of.

Choose ##p_0 = 19##, ##q_0 = 2## so that ##p_0/q_0## is a reasonable rational approximation. I chose these particular values since ##89## is about halfway between ##9^2 = 81## and ##10^2 = 100##.

Iterate using the formulas:

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

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

This will give a better rational approximation.

So:

##p_1 = p_0^2 + n q_0^2 = 19^2 + 89 * 2^2 = 361+356 = 717##

##q_1 = 2 p_0 q_0 = 2 * 19 * 2 = 76##

So if we stopped here, we would get an approximation:

##sqrt(89) ~~ 717/76 ~~ 9.434##

Let's go one more step:

##p_2 = p_1^2 + n q_1^2 = 717^2 + 89 * 76^2 = 514089 + 514064 = 1028153##

##q_2 = 2 p_1 q_1 = 2 * 717 * 76 = 108984##

So we get an approximation:

##sqrt(89) ~~ 1028153/108984 ~~ 9.43398113##

This Newton Raphson method converges fast.

##color(white)()## Actually, a rather good simple approximation for ##sqrt(89)## is ##500/53##, since ##500^2 = 250000## and ##89 * 53^2 = 250001##

##sqrt(89) ~~ 500/53 ~~ 9.43396##

If we apply one iteration step to this, we get a better approximation:

##sqrt(89) ~~ 500001 / 53000 ~~ 9.4339811321##

##color(white)()##Footnote

All square roots of positive integers have repeating continued fraction expansions, which you can also use to give rational approximations.

However, in the case of ##sqrt(89)## the continued fraction expansion is a little messy so not so nice to work with:

##sqrt(89) = [9; bar(2, 3, 3, 2, 18)] = 9+1/(2+1/(3+1/(3+1/(2+1/(18+1/(2+1/(3+...)))))))##

The approximation ##500/53## above is ##[9; 2, 3, 3, 2]##