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

In reply to Re: floating point addition by davido
in thread floating point addition by kresike

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