Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the
difference in floating point calculations between different language implementations becomes significant.
For example, a difference between '''32''' bit and '''64''' bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
;Task:
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
'''100000000000000.01''' may be approximately equal to '''100000000000000.011''',
even though '''100.01''' is not approximately equal to '''100.011'''.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
:# 100000000000000.01, 100000000000000.011 (note: should return ''true'')
:# 100.01, 100.011 (note: should return ''false'')
:# 10000000000000.001 / 10000.0, 1000000000.0000001000
:# 0.001, 0.0010000001
:# 0.000000000000000000000101, 0.0
:# sqrt(2) * sqrt(2), 2.0
:# -sqrt(2) * sqrt(2), -2.0
:# 3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
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