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

I'm trying to make floating point number systems a bit more intuitive for myself. There are a few things I am confused about, and I think the best way to clear up my confusions would be for someone to guide me through one question: How many numbers are there in a floating point number system given the base, precision, max exponent, and min exponent (B, P, emax, emin)?

Wikipedia provides this formula. If someone could explain each term's significance that would be extremely helpful! The +1's are the most confusing...

From Wikipedia:

$$2(B − 1)(B^{P − 1})(\text{emax} − \text{emin} + 1) + 1$$


share|cite|improve this question
If you have the number of bits in representation then its quite simple. You can arrive at the number directly. – BiGYaN Jan 31 '12 at 10:25

First we consider the significand. The significand is a string of $P$ digits in the base $B$, but except for the representation of the number $0$ it is normalized so that the first digit is not $0$. Since there are $B$ base-$B$ digits, $0,1,\dots,B-1$, there are $B-1$ possible first digits, $1,\dots,B-1$. Each of the remaining $P-1$ slots in the significand can be filled with any of the $B$ digits, so these $P-1$ slots can be filled in $B^{P-1}$ different ways. There are therefore altogether $(B-1)B^{P-1}$ ways to fill all $P$ slots of the significand.

Example: If $B=10$ and $P=3$, you’re just counting the possible $3$-digit numbers in base $10$, i.e., the integers from $100$ through $999$: there are $9$ choices for the first digit, since $0$ is not permitted, and $10$ for each of the other two digits, for a total of $9\cdot 10^2=900$ choices. As a check on this, observe that there are certainly $999$ integers between $1$ and $999$ inclusive, and we’re throwing out the $99$ integers from $1$ through $99$, so we must be left with $999-99=900$ integers altogether.

The significand is signed, so each of these $(B-1)B^{P-1}$ different values can appear either positively or negatively; this doubles the actual number of choices for the significand, making it $2(B-1)B^{P-1}$.

Now we consider the exponent. It can take any integer value between $\text{emin}$ and $\text{emax}$ inclusive. Let $a=\text{emin}-1$; then $\text{emin}=a+1$, $\text{emin}+1=a+2$, $\text{emin}+2=a+3$, and in general $\text{emin}+k=a+(k+1)$ for any $k$. Let $d=\text{emax}-\text{emin}$; then $\text{emax}=\text{emin}+d$, so $\text{emax}=a+(d+1)$. The integers from $\text{emin}$ through $\text{emax}$ are therefore $$a+1,a+2,\dots,a+d,a+(d+1)\;,\tag{1}$$ and it’s clear from looking at the second term of each of these sums that there are $d+1$ integers in the list $(1)$. But $d+1=\text{emax}-\text{emin}+1$, so there are $\text{emax}-\text{emin}+1$ possible values of the exponent.

Example: If $\text{emax}=8$ and $\text{emin}=-7$, $a=-7-1=-8$, and $d=8-(-7)=15$, the possible exponents are the integers from $a+1=-7$ through $a+(d+1)=-8+17=8$, and there are $d+1=16$ of them. As a check, there are clearly $7$ possible negative exponents, $8$ possible positive exponents, and $0$, for a total of $7+8+1=16$ possible exponents.

Each of the $2(B-1)B^{P-1}$ possible values of the signed significand can be combined with any of the $\text{emax}-\text{emin}+1$ possible values of the exponent, so there are altogether $$2(B-1)B^{P-1}(\text{emax}-\text{emin}+1)\tag{2}$$ possible combinations, each representing a different number. However, $(2)$ counts only the non-zero numbers that can be represented; the system can also represent the number $0$, so the grand total is $$2(B-1)B^{P-1}(\text{emax}-\text{emin}+1)+1$$ representable numbers.

share|cite|improve this answer

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.