How do you find the simplest radical form of 433?

1 Answer
Dec 16, 2017

Answer:

Hmmm...

Explanation:

If you mean the simplest form of #sqrt(433)# then it is #sqrt(433)# since #433# is a prime number.

If you mean the simplest expression involving a radical with value #433#, then you might choose #433 sqrt(1)# or possibly #sqrt(187489)# (since #433*433 = 187489#).

Approximations

Since #433# is a prime number, its square root cannot be simplified. In addition, it is not close to a square number or half way between two square numbers. So it is a little fiddly to get a rapidly convergent continued fraction expressing it.

Note that:

#20^2 = 400 < 433 < 441 = 21^2#

So a reasonable approximation is somewhere between #20# and #21#. Linearly interpolating, we find it is about #20+33/41 ~~ 20.8 = 104/5#

Then:

#433 - (104/5)^2 = 9/25#

Now:

#sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+...)))#

So:

#sqrt(433) = 104/5+(9/25)/(208/5+(9/25)/(208/5+(9/25)/(208/5+...)))#

#color(white)(sqrt(104)) = 1/5(104+9/(208+9/(208+9/(208+...))))#

Hence, if we define a sequence recursively by:

#{ (a_0 = 0), (a_1 = 1), (a_(n+2) = 208a_(n+1)+9a_n) :}#

then the ratio between successive terms rapidly converges to #5sqrt(433)+104#

The first few terms are:

#0, 1, 208, 43273, 9002656#

So:

#5sqrt(433) ~~ 9002656/43273-104 = 4502264/43273#

So:

#sqrt(433) ~~ 4502264/(5 * 43273) = 4502264/216365 ~~ 20.808652046#