为什么 IEEE 浮点精度中 2x - x == x？

我希望这仅在尾数的最后一位为时才成立0。否则，为了减去它们（因为它们的指数相差 1），x首先会损失一些精度，结果最终要么向上舍入，要么向下舍入。

但一个快速实验表明，对于任何随机数（包括带有 aa 尾随位的数） ，它似乎总是成立（假设x和是有限的） 。2x1

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")

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