Unless your interval is a fraction of a power of two or can be represented by a small number of them added together, your number can't be represented exactly in binary. Think about what you are asking your poor binary computer to do. .05 is 1/20. The first number that isn't bigger than your number is 1/32. Subtract that from 1/20. Now look for 1/2**n which is smaller than what's left. Continue until it comes out even or you run out of precision.

Decimal fractions are rarely representable exactly in computing memory.

To prevent errors from building up, round more often.

As for

3.65 + 0.05 = 3.69999999999999
I wouldn't say the error is in the first place. It's actually in the last place you showed. Any prior rounding would work. If the difference was significant in the first digit, you could round in the second place without getting the correct answer.

Decimals tie people in knots even when computers aren't involved. I remember a heated discussion in a math class I took, where the teacher was making no headway in convincing some student that .3 repeating is 1/3.

Phil

In reply to Re: adding numbers and floating point errors by philcrow
in thread adding numbers and floating point errors by smeenz

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