x的平方的平方根等于x [英] square root of square of x equals x

问题描述

给出一个双精度浮点(非负)数字 x ，它的平方根是否总是等于它的平方根?

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

换句话说，如果执行以下操作，是否会损失精度:

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

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

这样:

x == z

我对平方变为无穷大或为零的情况不感兴趣，只是适合双精度的数字.

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

推荐答案

#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);
  }
}

输出

1.0000000000000001466e-160
9.9999443357584897793e-161

这里发生的是 x 足够大以至于其平方非零，但又足够小以至于其平方只能表示为非正规数，从而降低了可用精度.

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.

我觉得@MarkDickinson对@LưuVĩnhPhúc的回答的评论在很大程度上是正确的.如果 x 和 x * x 均为正归一化数字，那么我将无法找到 x！= sqrt(x * x)即使使用快速蛮力(在很小的范围内)，也不应将其视为证明.

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.

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