Fractional powers of negative numbers

[jamshark70]

@Spacechild1

You can raise a negative number to some fractional powers and get a real number answer, but only if the denominator of the fraction (in its lowest terms) is odd.

That’s the theory. In practice, such an exponent can’t be expressed accurately with floating point numbers, so it’s not really possible - unless the language has first class support for rational numbers.

That’s an excellent point. In base-two floating-point, the denominator must be even.

Consider decimals. If you have a finite number of digits after the decimal point, then the denominator may consist of any combination of prime factors of the base. For decimal, those prime factors are 2 and 5: 0.333 = 333/1000 = 333 / (2*2*2*5*5*5) but there is no finite-precision decimal where the denominator includes a factor of 3 or 7 or 11.

In binary, the base has only one prime factor: 2. And of course floating point can’t represent infinite precision (double-precision allows 53 bits in the mantissa).

a = 1/3;

a.high32Bits.asBinaryString(32)
a.low32Bits.asBinaryString(32)

-> 00111111110101010101010101010101
-> 01010101010101010101010101010101

=

sign bit = 0
exponent = 01111111101
mantissa = implicit 1. then 0101010101010101010101010101010101010101010101010101

So the denominator of this fraction 1.0101010101… will be a very large power of two, but it is a power of two, hence even, hence -1.pow(1/3) cannot be computed using a system of finite-precision binary fractions.

TL;DR -1.pow(0.33333333333333) = -1.pow(33333333333333 / 100000000000000) which is likewise an even denominator.

TL;DR It’s hard to argue with the IEEE; you’ll probably lose

hjh