square root of square of x equals x

Question

Given a double precision floating-point (non-negative) number x, does square root of its square always equals to itself?

In other words, is there any loss of precision if doing the following:

x = <non-negative double>
y = x^2
z = sqrt(y)

so that:

x == z

I'm not interested in the case when the square becomes infinity or zero, just numbers that fit the double-precision.

Answer 1

#include <stdio.h>
#include <math.h>

int main(void) {
  double x = 1.0000000000000001E-160;
  double square = x*x;
  double root = sqrt(square);
  if (root != x) {
    printf("%.20g\n", x);
    printf("%.20g\n", root);
  }
}

outputs

1.0000000000000001466e-160
9.9999443357584897793e-161

What's happening here is that x is large enough that its square is non-zero, but small enough that its square is only representable as a denormalised number, which reduces the available precision.

I get the impression that @MarkDickinson's comment on @LưuVĩnhPhúc's answer is largely correct though. If both x and x*x are positive normalised numbers, then I'm not able to find examples where x != sqrt(x*x) even with a quick brute force (over a few small ranges), though this should not be considered proof.