Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are floating-point numbers fundamentally different from the sets of rational or real numbers in any way? Are there any good mathematical treatments of floating-point numbers? Are they even interesting to study mathematically?

share|cite|improve this question
up vote 6 down vote accepted

They are fundamentally different in that (given any particular floating-point representation) there are only finitely many of them. And the basic arithmetic operations on floating-point numbers do not satisfy nearly as many nice laws as arithmetic on true reals or rationals do -- for example, $(a+b)+c = a+(b+c)$ is not true for floating point; neither is $(a+b)-b = a$.

There's some mathematical reasoning about floating-point arithmetic going on in numerical analysis, in particular with respect to how much error relative to true arithmetic on reals various formulas can introduce. That often (but not always) has a somewhat conservative character, overestimating the errors rather than going to the trouble of modeling them exactly.

Also, some people have developed complete machine-readable formalizations of floating-point arithmetic, in the context of using them to create computer-verified proofs that specific microprocessor designs implement it correctly.

share|cite|improve this answer
Maybe not quite floating-point numbers anymore, but what if you you use a finite number of bits to represent the significand but allow for an arbitrarily big exponent? – Aqwis Apr 23 '12 at 23:45
If you use a finite number of bit, you have the same problem as above: $a=1$, $b=1*2^{64}$. $a+b=1*2^{64}$ (or use higher exponent if you have higher precision number). $(a+b)-b=0$, but $1\neq0$... – carlop Apr 24 '12 at 0:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.