Correct IFDRational.__float__() return value

[nyxst4ck]

Changes proposed

IFDRational.__float__ returns the wrong value for a non-integral numerator.

IFDRational delegates __eq__, __int__, __round__, __repr__ and all arithmetic to self._val (the normalized Fraction), but never defined __float__. So float() falls back to numbers.Rational.__float__, which does int(self.numerator) / int(self.denominator). For a non-integral numerator the stored _numerator/_denominator are un-normalized, so the result disagrees with every other accessor:

>>> from PIL.TiffImagePlugin import IFDRational
>>> r = IFDRational(1.5, 3)   # value 1.5 / 3 == 0.5
>>> r == 0.5
True
>>> int(r), repr(r)
(0, '0.5')
>>> float(r)
0.3333333333333333          # int(1.5) / int(3) == 1/3, should be 0.5

This is reachable via the public IFDRational(value, denominator) constructor.

Fix

Delegate __float__ to self._val, the same way __int__, __round__, __repr__ and __eq__ are delegated, so float() is consistent with the normalized fraction.

Tests

Added test_float to Tests/test_tiff_ifdrational.py: it asserts float(IFDRational(1.5, 3)) == 0.5, and that the integral and 0/0 (nan) cases stay correct. It fails on main (0.333… != 0.5) and passes with the change; Tests/test_tiff_ifdrational.py is green (6 passed). ruff and black clean.

[radarhere]

Suggested change

	# normalized fraction. For a non-integral numerator (1.5 / 3 == 0.5) the
	# inherited Rational.__float__ used int(numerator) / int(denominator) and
	# returned 1/3 instead.
	# normalized fraction

I don't think we need to document prior incorrect behaviour.

[radarhere]

Suggested change

	# Integral and 0/0 (nan) cases stay correct.
	# Integer numerator and 0/0 (nan) cases