Why doesn't 0.9 recurring always equal 1

Question

Mathematically, 0.9 recurring can be shown to be equal to 1. This question however, is not about infinity, convergence, or the maths behind this.

The above assumption can be represented using doubles in C# with the following.

var oneOverNine = 1d / 9d;
var resultTimesNine = oneOverNine * 9d;

Using the code above, (resultTimesNine == 1d) evaluates to true.

When using decimals instead, the evaluation yields false, yet, my question is not about the disparate precision of double and decimal.

Since no type has infinite precision, how and why does double maintain such an equality where decimal does not? What is happening literally 'between the lines' of code above, with regards to the manner in which the oneOverNine variable is stored in memory?

Answer 1

It depends on the rounding used to get the closest representable value to 1/9. It could go either way. You can investigate the issue of representability at Rob Kennedy's useful page: http://pages.cs.wisc.edu/~rkennedy/exact-float

But don't think that somehow double is able to achieve exactness. It isn't. If you try with 2/9, 3/9 etc. you will find cases where the rounding goes the other way. The bottom line is that 1/9 is not exactly representable in binary floating point. And so rounding happens and your calculations are subject to rounding errors.