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Re: floating point addition

by davido (Cardinal)
on Jan 12, 2015 at 16:55 UTC ( [id://1112964]=note: print w/replies, xml ) Need Help??


in reply to floating point addition

If the floating point number cannot be represented precisely as n/(2^m) then it must be represented imprecisely in a binary form, and when that happens, rounding errors occur. In base ten, the formula is k/((2^n)*(5^m)). We're just more accustomed to the mental leap that 0.33 in base ten usually means 1/3rd, even though this is an imprecise representation of a non-terminating expansion. When we see the same thing in numbers that have lost precision due to being stored in base-2, we panic. ;)

A few years ago I put together a more thorough explanation: Re: shocking imprecision


Dave

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