Let's say we consider IEEE 754 double precision floating-point numbers, and we use RNTE - Round To Nearest, Ties to Even - rounding. I know that the RNTE rounding works this way: given two consecutive floating-point numbers, a and b, and a real number x,

  • if a <= x < (a+b)/2, x gets translated to a
  • if (a+b)/2 < x <= b, x gets translated to b
  • if x = (a+b)/2, x gets translated to the number with an even significand between a and b

For example, I want to know which are the real numbers that are "mapped" to 1, where 1 is a floating-point number. How should I proceed? I thought that I must consider both the cases where 1 is a and where 1 is b in the above definition, and so consider both the previous and the next floating-point number. Is it correct? Is there a simpler way?

  • $\begingroup$ Stricto sensu, there is no answer to this question. Because computers cannot represent real numbers, so "x gets translated to ..." is meaningless, no such translation process exists. $\endgroup$ Aug 8 at 8:40
  • $\begingroup$ Also note that the floating-point numbers do not follow a regular arithmetic progression, so one can question the "nearest neighbor" rule. $\endgroup$ Aug 8 at 8:42

3 Answers 3


Today, my professor explained how to sove this problem. Since I think that it might be useful to other users, I'll post here the answer.

Given a floating-point number $f$ in a particular encoding (for example IEEE 754 single/double precision), we want to know which subset of real numbers corresponds to $f$, that is to say which real numbers gets rounded to f when converted to the same encoding. We're using RNTE (Round To Nearest, Ties to Even) rounding.

We represent the encoding as a quadruple $\mathbb{F}(\beta,t,M_1,M_2)$, where $\beta$ is the base, $t$ is the number of significand digits after the point and $M_1$,$M_2$ are the lower and upper bounds for the exponent range, respectively.

For example, let's choose to work with IEEE 754 double precision numbers, which we can denote as $\mathbb{F}(2,52,-1023,1024)$.

Let $f = 17 = (10001)_{2} = 1.\overbrace{00010...0}^\text{52 digits} \times 2^{4} \in \mathbb{F}$.

We have to determine the predecessor and the successor of $f$, both of which are in $\mathbb{F}$. Let's call them $a$ and $b$, respectively. In order to do this, we have to find the distance between $a$ and $f$ and between $f$ and $b$. We know that $17 \in [16, 32] = [2^{4},2^{5}]$, so we can use the formula $y-x=\beta^{p-t}$ to get those distances, where $x,y \in [\beta^{p},\beta^{p+1}]$, $\beta=2$, $p=4$, $p+1=5$ and $t=52$.

So, $a = 17 - 2^{4-52} = 17 - 2^{-48}$,

and $b = 17 + 2^{4-52} = 17 + 2^{-48}$

Now we can apply the RNTE rounding, which states that a real number $x$ is rounded to its nearest number in $\mathbb{F}$ and, if there are two candidates, it chooses the one with an even mantissa (or significand). The latter case means that $x$ is the midpoint between two consecutive numbers in $\mathbb{F}$.

The midpoint between $a$ and $17$ is $ x_1 = 17 - \frac{1}{2} 2^{-48} = 17 - 2^{-49}$,

the midpoint between $17$ and $b$ is $ x_2 = 17 + \frac{1}{2} 2^{-48} = 17 + 2^{-49}$

Each real number between $x_1$ and $x_2$ gets rounded to $17$, including $x_1$ and $x_2$, because 17 has an even mantissa in our representation. So the answer is $\{x \in \mathbb{R} \mid 17 - 2^{-49} \leq x \leq 17 + 2^{-49}\}$.

  • $\begingroup$ Just saw this. Looked remarkably like my answer except your professor missed the case where f is a power of two. $\endgroup$
    – gnasher729
    Apr 9 at 22:48

The powers of two are more interesting than most other numbers.

Assume you are using double precision IEEE 754 format floating point numbers. Let $u = 2^{-52}$, Then the next floating point number larger than 1 is $1+u$, and the next floating point number smaller than 1 is $1 - u/2$.

A real number x is rounded to 1 if $1-u/4 ≤ x ≤ 1+u/2$. For any floating-point number f with $1 < f < 2$, x is rounded to f if $f - u/2 ≤ x ≤ f + u/2$ if the last bit of f is even, and if $f - u/2 < x < f + u/2$ if the last bit of f is odd.

Numbers x with $f - u/2 < x < f + u/2$ are always rounded to f, because f is the nearest floating point number. For the numbers $x = f ± u/2$, $f$ and $f ± u$ are equally far away, so the round-to-even rule decides.

  • $\begingroup$ Thank you for the answer. Could you please tell me what is the general rule being applied here? How can I solve this problem for any given floating-point number? One more thing: I read a solved excercise where $f = 1+2^{-52}$, and the answer is $1+2^{-53} < x < 1+2^{-52}+2^{-53}$, I don't understand how this is achieved. $\endgroup$ Nov 13, 2020 at 0:58
  • $\begingroup$ Since f = 1 + 2^-52, that's exactly the f - u/2 < x < f + u/2 from my answer. $\endgroup$
    – gnasher729
    Nov 15, 2020 at 12:24

Short answer:

If the mantissa is not all ones, the next floating-point number is $1$ ulp more.

  • E.g. $1.1000_b\times 2^7\to1.1001_b\times 2^7$ ($192_d\to200_d$), the difference is $2^{7-4}=8$.

If the mantissa is all ones, the next floating-point number is all zeroes with the next exponent, and the difference is again 1 ulp.

  • E.g. $1.1111_b\times 2^7\to1.0000_b\times 2^8$ ($248_d\to256_d$), the difference is $2^{7+1}-(2^{7+1}-2^{7-4})=8$.

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