为何在IEEE浮点精度下2x - x恒等于x？

我原本以为只有当浮点数的尾数（mantissa）最后一位为0时，等式2x - x == x才成立。否则，因为2x和x的指数相差1，相减时x会丢失一位精度，结果会被向上或向下舍入。

但实际实验显示，只要x和2x都是有限值，不管随机数的尾数最后一位是不是1，这个等式始终成立。

import random
import struct
from collections import Counter


def float_to_bits(f: float) -> int:
    """
    Convert a double-precision floating-point number to a 64-bit integer.
    """
    # Pack the float into 8 bytes, then unpack as an unsigned 64-bit integer
    return struct.unpack(">Q", struct.pack(">d", f))[0]


def check_floating_point_precision(num_trials: int) -> float:
    true_count = 0
    false_count = 0
    bit_counts = Counter()

    for _ in range(num_trials):
        x = random.uniform(0, 1)
        if 2 * x - x == x:
            true_count += 1
        else:
            false_count += 1

        bits = float_to_bits(x)

        # Extract the last three bits of the mantissa
        last_three_bits = bits & 0b111
        bit_counts[last_three_bits] += 1

    return (bit_counts, true_count / num_trials)


num_trials = 1_000_000
(bit_counts, proportion_true) = check_floating_point_precision(num_trials)

print(f"The proportion of times 2x - x == x holds true: {proportion_true:.6f}")
print("Distribution of last three bits (mod 8):")
for bits_value in range(8):
    print(f"{bits_value:03b}: {bit_counts[bits_value]} occurrences")

运行结果：

The proportion of times 2x - x == x holds true: 1.000000
Distribution of last three bits (mod 8):
000: 312738 occurrences
001: 62542 occurrences
010: 125035 occurrences
011: 62219 occurrences
100: 187848 occurrences
101: 62054 occurrences
110: 125129 occurrences
111: 62435 occurrences

内容的提问来源于stack exchange，提问作者dspyz