Eu esperaria que isso só acontecesse quando o último bit da mantissa fosse 0. Caso contrário, para subtraí-los (já que seus expoentes diferem em 1), xperderia um pouco de precisão primeiro e o resultado acabaria sendo arredondado para cima ou para baixo.
Mas um experimento rápido mostra que isso parece sempre valer (assumindo que xe 2xsão finitos) para qualquer número aleatório (incluindo aqueles com um 1bit à direita).
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