I want to convert $\frac{2}{7}$ to a binary number in a $32$ bit computer. That is, $1$ bit is assigned to the sign of the number, $8$ bits are assigned to the exponent, and $23$ bits are assigned to the mantissa.

So $x = \pm q \times 2^{m}$ where $\frac{1}{2} \leq q < 1$ (if $x \neq 0$) and $m = e - 127$ is an integer. Suppose the leading binary digit $1$ is shifted just to the left of the binary point. In this case, the representation would be $q = (1.f)_{2}$ and $1 \leq q < 2$. So in effect, the machine has a $24$-bit mantissa.

The binary representation of $\frac{2}{7}$ is $\left ( 0.010 \overline{010} \right )_{2}$. In normalized notation, this is $ \left ( 0.10\overline{010} \right )_{2} \times 2^{-1}$.

I want to write out fully what this number would like in the $32$ bit computer. So, I should write out $24$ bits for the mantissa.

$$x = \left ( 0.\underbrace{10010010010010010010010}_{23 \text{ bits}}\underbrace{\_}_{24\text{'th bit}} \right )_{2} \times 2^{-1}$$

For the $24th$ bit, do I put a $0$? There is not enough room for the entire $3$-period of $\overline{010}$ so what do I do?

  • 3
    $\begingroup$ This is a limitation of floating point variables - there is no "recurring" notation. Also, please note that the normalised mantissa always has a 1 in the most significant place (except for value 0) and in IEEE representations that bit is not stored. $\endgroup$ – Weather Vane Sep 5 '17 at 16:55
  • $\begingroup$ It depends on the standard of calculation, but ideally you'd round to the nearest 24th digit. Since the "full" digits continue: $01001\dots$, the rounding digit would be $1$, not $0$. $\endgroup$ – Thomas Andrews Sep 5 '17 at 16:55
  • $\begingroup$ If you have access to a computer and a programming language, you should be able to access the memory and actually see how it is stored. $\endgroup$ – steven gregory Sep 5 '17 at 16:59
  • $\begingroup$ @WeatherVane Yes, the reason why my work in my post has a $24$-bit mantissa is exactly the reason you stated - that $1$ is not stored. This question stems from an exercise "find the roundoff error" which means check if rounding or chopping the number gives a better approximation to the original binary number. So maybe this unavoidable misrepresentation is intentional by the exercise! $\endgroup$ – Kazi Sep 5 '17 at 17:04
  • $\begingroup$ Also note that IEEE defines four possible rounding modes so this passes the buck to the programmer. $\endgroup$ – Weather Vane Sep 5 '17 at 17:07

There's no particular reason why your three-period has to be written $010.$ You have multiple choices, depending on where you choose to start looking for a repeating block. In particular, $$ 0.010\overline{010}_2 = 0.01\overline{001}_2 = 0.0\overline{100}_2. $$

And of course even if you did end up with three available bits at the end of the computer word, allowing you to write one copy of your three-period there, you would still have only an approximation, because all the other three-periods (there are infinitely many) don't fit in that space.

So you have to find out how the computer is set up to do rounding of floating-point numbers in this case. Using the default rule for IEEE-754 binary (round to nearest, ties to even--see the other answer), you can start to figure out which way to round (up to $1$ or down to $0$) by looking at the value of the binary digits that don't fit in the IEEE format. The least significant bit in the single-precision representation of $\frac27$ has place value $2^{-25},$ and the bits to the right of it have value $$ 0.10\overline{010} \times 2^{-25}. $$ Since this is greater than $\frac12 \times 2^{-25},$ you round up. (If it were less, you would round down, and if it were exactly equal you would look at the digit with place value $2^{-25}$ to figure out which way the "round to even" rule goes.)

By the way, notice that the IEEE-754 single-precision representation of $\frac27,$ as demonstrated in the other answer, has exponent bits $01111101,$ implying that $$m = e - 127 = 01111101_2 - 127 = -2,$$ not $-1$ as you seem to be assuming when you write, "$x = \pm q \times 2^{m}$ where $\frac{1}{2} \leq q < 1$ (if $x \neq 0$) and $m = e - 127$ is an integer."

  • $\begingroup$ David, thank you for answer. I had been using an exponent of $-1$ because I thought it would make the binary arithmetic easier, but I think I should have just done $-2$. I also find that $x_-$ is a better approximation to $x$ on the 32 bit computer which means NOT using the rounded number and instead using the chopped number. $\endgroup$ – Kazi Sep 6 '17 at 15:37
  • $\begingroup$ Just out of curiosity: how did you decide $x_-$ is a better approximation? $\endgroup$ – David K Sep 6 '17 at 17:51
  • $\begingroup$ David, note the problem scope is for a 32 bit computer. To check if the chopped or rounded representation of $x$ is a better approximation, one must compare the absolute errors of $x_-$ and $x_+$. That is, compare the values of $|x - x_{-}|$ and $|x_+ - x|$. Check the amount of error for $x_-$ and $x_+$ then pick the term which is the least amount of it. That term will be the better approximation of $x$. $\endgroup$ – Kazi Sep 7 '17 at 14:39
  • $\begingroup$ @Kazi I didn't mean in general how do you decide, I meant exactly how in this specific case did you decide $x_-$ was better. For $\frac27$ in 32-bit IEEE-754 binary format, $|x_+ - x|=\frac 37\times 2^{-25}$ but $|x - x_-|=\frac 47 \times 2^{-25},$ so by that criterion $x_+$ is the better approximation. $\endgroup$ – David K Sep 7 '17 at 17:34

C program:

#include <stdio.h>
int main(int a){
	float f = 2.0/7;
	unsigned int i = *(unsigned int*)&f;

Try it online!

This gives 00111110100100100100100100100101:


So, the mantissa is 1.00100100100100100100101 (first 1 implicit).

$$\cdots010010\underline0\color{grey}{10010\cdots} \mapsto \cdots010010\underline1$$

Rounding to the nearest representable number.

The official standard is behind paywall, so I can only quote a secondary source:

IEEE Standard 754 Floating Point Numbers, Steve Hollasch, 2015 Dec 2:

Algebraic operations covered by IEEE 754, namely + , - , · , / , √ and Binary <-> Decimal Conversion with rare exceptions, must be Correctly Rounded to the precision of the operation’s destination unless the programmer has specified a rounding other than the default. If it does not Overflow, a correctly rounded operation’s error cannot exceed half the gap between adjacent floating-point numbers astride the operation’s ideal ( unrounded ) result.

(Emphasis mine.)

A tertiary source:

IEEE 754, Wikipedia:

The standard defines five rounding rules. [...]

Round to nearest, ties to even – rounds to the nearest value; if the number falls midway it is rounded to the nearest value with an even (zero) least significant bit; this is the default for binary floating-point and the recommended default for decimal.

(Emphasis mine.)

  • $\begingroup$ Your online code should be double f = 2.0/7; $\endgroup$ – John Alexiou Sep 5 '17 at 17:47
  • $\begingroup$ @ja72 OP is asking 32 bit, i.e. float. $\endgroup$ – Kenny Lau Sep 5 '17 at 17:52
  • $\begingroup$ Right. Carry on. $\endgroup$ – John Alexiou Sep 5 '17 at 18:15

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