Fundamentals of Data Representation: Binary fractions

PAPER 2 - ⇑ Fundamentals of data representation ⇑
← Twos complement	Binary fractions	Rounding errors →

Binary Fractions - using Fixed Point notation

So far we have only looked at whole numbers (integers), we need to understand how computers represent fractions.

From the Specification : Introduction to Principles of Computation

Know how numbers with a fractional part can be represented in:

fixed point form in binary in a given number of bits
floating point form in binary in a given number of bits.

Be able to convert for each representation from:

decimal to binary of a given number of bits
binary to decimal of a given number of bits.

You should have learned at Primary School how a decimal fraction works:

10	1		${\frac {1}{10}}$	${\frac {1}{100}}$
$10^{1}$	$10^{0}$		$10^{-1}$	$10^{-2}$
1	2	.	7	5

As you can see, the column headings have been extended to $10^{-1}={\frac {1}{10}}$ and $10^{-2}={\frac {1}{100}}$ . We can do the same thing in binary with the column headings $2^{-1}={\frac {1}{2}}$ , $2^{-2}={\frac {1}{4}}$ , and so on. The number 12.75 in 8 bit binary with 4 bits after the binary point is therefore 8 + 4 + 0.5 + 0.25:

8	4	2	1		${\frac {1}{2}}$	${\frac {1}{4}}$	${\frac {1}{8}}$	${\frac {1}{16}}$
$2^{3}$	$2^{2}$	$2^{1}$	$2^{0}$		$2^{-1}$	$2^{-2}$	$2^{-3}$	$2^{-4}$
1	1	0	0	.	1	1	0	0

Notice that for the same number of bits after the point, the binary fraction provides less accuracy. It can only take 4 different values, whereas the decimal number can have 100 different values with two digits. You'll see in a moment how this can cause trouble.

Example: converting decimal to binary decimal using fixed point notation

We are going to convert the number 6.125 into a binary fraction by using the grid below

8	4	2	1		${\frac {1}{2}}$	${\frac {1}{4}}$	${\frac {1}{8}}$	${\frac {1}{16}}$
0	1	1	0	.	0	0	1	0

This seems simple enough as 6.125 = 4 + 2 + 0.125, but what about this more interesting number: 6.4

8	4	2	1		${\frac {1}{2}}$	${\frac {1}{4}}$	${\frac {1}{8}}$	${\frac {1}{16}}$
0	1	1	0	.	0	1	1	0

But this doesn't look right?! This number isn't correct as it only reaches 4 + 2 + 0.25 + 0.125 = 6.375, we need more bits for the binary fraction places. However, a computer might restrict you to the number of bits you can use, so we'll use the number closest to the one we were aiming for. You could feel a bit annoyed at this, but don't worry, you make this compromise every time you try to represent ${\frac {1}{3}}$ with the decimal factions, 0.33333333.

So you might ask how a computer does complicated mathematics if it struggles so hard with fractions. The answers we have looked at so far have only used one byte, computers can use far more space than this. They can also manipulate the number of bits they have been given in two ways:

increase the number of bits to increase range of number
increase number of bits after the decimal point to increase accuracy

In practice they will also use clever techniques such as floating point numbers (see below).

Exercise: converting from denary to binary fractions

Now try some questions yourself and see how you get on. Remember, where there aren't enough bits for the decimal place, write down the number closest to your target number. In each case use 8 bits for the binary with four bits after the decimal point:

7.5

Answer:

0111.1000

4.5625

Answer:

0100.1001

1.6

Answer:

0001.1010 (this is the closest we are going to get)

3.3333333

Answer:

0011.0101 (this is the closest we are going to get)

Try and convert these binary fractions into denary:

0111.0100

Answer:

7.25

1011.1001

Answer:

11.5625 (notice that we treated this as a positive number, some of you might already know about twos complement, if you haven't heard about it before, don't worry, we'll get there very soon)
To interpret it as a two's complement number let's flip it: 0100.0111 = 4 + 0.25 + 0.125 + 0.0625 = -4.4375

If I want to increase the range of numbers stored in a fixed point binary number, what should I do?

Answer:

Increase the number of bits before the decimal point

If I want to increase the accuracy of numbers stored in a fixed point binary number, what should I do?

Answer:

Increase the number of bits after the decimal point