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Misc improvements to the float tutorial #102052

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Merged
rhettinger merged 10 commits into from
Feb 19, 2023
Merged

Misc improvements to the float tutorial #102052

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1e2ec34
Example is more intelligible without the leading zeros
rhettinger Feb 19, 2023
082d7a8
Leading zeros are easier to read and match the surrounding text.
rhettinger Feb 19, 2023
2ff520d
Compare sum() to straight addition and to math.fsum()
rhettinger Feb 19, 2023
22e1a96
Show math.isclose() for comparing inexact values
rhettinger Feb 19, 2023
cb1f78a
Add doctests throughout
rhettinger Feb 19, 2023
e9e959a
Add a more useful example of round() for comparisons
rhettinger Feb 19, 2023
8f02dc5
Shorter example
rhettinger Feb 19, 2023
d8bcc4c
Fix doctests: Missing import and counterfactual example.
rhettinger Feb 19, 2023
cd8abc5
Compare sum() and fsum() with straight addition
rhettinger Feb 19, 2023
d6a6b6b
Add section author
rhettinger Feb 19, 2023
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Leading zeros are easier to read and match the surrounding text.
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@rhettinger
rhettinger committed Feb 19, 2023
commit 082d7a84335cd635bb6a8ac218f9f3c413399dfa
6 changes: 3 additions & 3 deletions Doc/tutorial/floatingpoint.rst
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Expand Up @@ -108,21 +108,21 @@ simply rounding the *display* of the true machine value.
One illusion may beget another. For example, since 0.1 is not exactly 1/10,
summing three values of 0.1 may not yield exactly 0.3, either::

>>> .1 + .1 + .1 == .3
>>> 0.1 + 0.1 + 0.1 == 0.3
False

Also, since the 0.1 cannot get any closer to the exact value of 1/10 and
0.3 cannot get any closer to the exact value of 3/10, then pre-rounding with
:func:`round` function cannot help::

>>> round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1)
>>> round(0.1, 1) + round(0.1, 1) + round(0.1, 1) == round(0.3, 1)
False

Though the numbers cannot be made closer to their intended exact values,
the :func:`round` function can be useful for post-rounding so that results
with inexact values become comparable to one another::

>>> round(.1 + .1 + .1, 10) == round(.3, 10)
>>> round(0.1 + 0.1 + 0.1, 10) == round(0.3, 10)
True

Binary floating-point arithmetic holds many surprises like this. The problem
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