TIL #053 – precision of Python floats

Precision of Python floats

Python floats are IEEE 754 double-precision binary floating-point numbers, commonly referred to as “doubles”, and take up 64 bits. Of those:

• 1 is for the sign of the number;
• 11 are for the exponent; and
• 52 are for the fraction.

We can verify experimentally that Python floats use 52 bits to store the fraction. The number 1 << 53 is an exact integer:

>>> n = 1 << 53
>>> n
9007199254740992

In binary, this number is a 1 followed by 53 0:

>>> bin(n)
'0b100000000000000000000000000000000000000000000000000000'
>>> bin(n)[2:]
'100000000000000000000000000000000000000000000000000000'
>>> bin(n)[2:].count("0")
53

Now, if you convert n to a float and add 1, nothing happens; whereas if you add 2, you get the correct value:

>>> float(n)
9007199254740992.0
>>> float(n) + 1
9007199254740992.0
>>> float(n) + 2
9007199254740994.0

Why?

Well, n + 1 in binary starts and ends with 1 and has 52 zeroes in the middle:

>>> bin(n + 1)
'0b100000000000000000000000000000000000000000000000000001'

Represented in scientific notation (in binary), this number would have 53 digits after the decimal point:

\[ 1.00000000000000000000000000000000000000000000000000001_2 \times 2^{53}\]

However, doubles only have 52 digits after the decimal point, so the final 1 is dropped and the number becomes

\[ 1.0000000000000000000000000000000000000000000000000000_2 \times 2^{53}\]

Which is exactly the same number.

However, if we add 2 instead of just 1, the final result is

\[ 1.00000000000000000000000000000000000000000000000000010_2 \times 2^{53}\]

which looks like

\[ 1.0000000000000000000000000000000000000000000000000001_2 \times 2^{53}\]

if we only use 52 digits after the decimal point, which is the same as the correct result.

The fact that n, n + 2, n + 4, ... give the correct results, and n + 1, n + 3, n + 5, ... don't, together with the fact that this phenomenon started at n = 1 << 53 and not n = 1 << 52 shows that Python floats use 52 bits to store the fraction of a number.

That's it for now! Stay tuned and I'll see you around!

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References

• “Double-precision floating-point format”, Wikipedia, https://en.wikipedia.org/wiki/Double-precision_floating-point_format [last visited 14-09-2022];