sqrt(4) == 4^(0.5)

Added by: tomek, 2010 X 14, Last modified: 2010 X 14

The first problem is that 4^(0.5) gives not exactly decimal 2.
If you print the value as binary (the option is not available in the online calculator) but you can check it with this simple program:
#include <ttmath/ttmath.h>
#include <iostream>

typedef ttmath::Big<TTMATH_BITS(256), TTMATH_BITS(2048)> Float;

int main()
{
Float a, b;

a = 4;
b = "0.5";

a.Pow(b);

std::cout << "dec: " << a << std::endl;
std::cout << "bin: " << a.ToString(2) << std::endl;
}
You see that the binary result is:
10.00000000000000000000[..../a lot of zeroes/.....]001
At the end of the mantissa you have one bit set. This innacuracy comes from the nature of binary floating point numbers. 4^(0.5) is calculated by using a logarithm and exponent which are themselves calculated from some kind of series (a lot of mathematical operations: addind, dividing etc). So at the end you are only with an approximate result.

But why sqrt(4) is different? Because it returns exactly 2 (binary: 10). It is a feature implemented in sqrt function.
1. sqrt first calculates 4^(0.5)
2. because 4 is integer the sqrt function rounded the result to the nearest integer so:
(binary)10.000000000....1 becames: binary 10
3. the result binary 10 is squared: 10 * 10 giving: binary 100 which is exactly decimal 4, so sqrt as the result returns binary 10.

Now you can ask why 4^(0.5) is not printed as something like:
2.000000000000000000.....001
The simple answer is it is not so easy to implement. When printing such values in other words converting from binary to decimal we have to do some mathematical operations like calculating a logarithm, dividing etc. All those operations cause losing accuracy.

In the other hand it helps to print such values:
decimal: 0.1 which don't have finite binary representation.