Return built-in ints exactly from fractional

float() rounds integers beyond 2**53, so fractional(2**53 + 1) lost the low bit.

Author: itzzdev09

src/humanize/number.py

@@ -395,6 +395,11 @@ def fractional(value: NumberOrString) -> str:
     """
     import math
 
+    if isinstance(value, int) and not isinstance(value, bool):
+        # An integer has no fractional part, and float() would silently round
+        # one beyond 2**53 (e.g. 2**53 + 1), so return it exactly.
+        return str(value)
+
     try:
         number = float(value)
         if not math.isfinite(number):

tests/test_number.py

@@ -254,6 +254,8 @@ def test_apnumber(test_input: int | str, expected: str) -> None:
         (-(10**400), "-1" + "0" * 400),
         # Still genuinely infinite, because this is not an integer.
         ("1e400", "+Inf"),
+        (2**53 + 1, "9007199254740993"),
+        (-(2**53 + 1), "-9007199254740993"),
     ],
 )
 def test_fractional(test_input: float | str, expected: str) -> None: