A-level Computing 2009/AQA/Problem Solving, Programming, Data Representation and Practical Exercise/Fundamentals of Data Representation/Binary number system





Before we jump into the world of number systems we'll need a point of reference, I recommend that you copy the following table that you can refer to throughout this chapter to check your answers.

Denary/Decimal
Denary is the number system that you have most probably grown up with. It is also another way of saying base 10. This means that there are 10 different numbers that you can use for each digit, namely: 0,1,2,3,4,5,6,7,8,9 Notice that if we wish to say 'ten', we use two of the numbers from the above digits, 1 and 0.

Using the above table we can see that each column has a different value assigned to it. And if we know the column values we can know the number, this will be very useful when we start looking at other base systems. Obviously, the number above is: five-thousands, nine-hundreds, seven-tens and three-units. 5*1000 + 9*100 + 7*10 + 3*1 = 597310

Binary
You should know denary pretty well by your age, but there are different base systems out there, and the most important one for computing is the binary base system. Binary is a base-2 number system, this means that there are two numbers that you can write for each digit: 0, 1 With these two numbers we should be able to write (or make an approximation) of all the numbers that we could write in denary.

Using the above table we can see that each column has a value assigned to it that is the power of two (the base number!), and if we take those values and the corresponding digits we can work out the value of the number: 1*64 + 1*32 + 1*8 + 1*2 = 106.

If you are asked to work out the value of a binary number, the best place to start is by labelling each column with its corresponding value and adding together all the columns that hold a 1. Let's take a look at another example:

000111112

So now all we need to do is to add the columns containing 1s together: 1*16 + 1*8 + 1*4 + 1*2 + 1*1 = 31

8+4 = 1210

64 + 16 + 8 + 1 = 8910

4 + 2 + 1 = 710

64 + 16 + 4 + 1 = 8510

Yes, take the first 0's column value and minus one = 128 - 1 = 127 = 64 + 32 + 16 + 8 + 4 + 2 + 1

000011112 = 16 - 1 = 15 = 8 + 4 + 2 + 1

000001112 = 8 - 1 = 7 = 4 + 2 + 1

Max and range
A common question that you'll need to know the answer to, and one that many people get wrong, is a question about the maximum denary value you can store in a set number of binary digits, or alternatively, the range of values that you can store in a set number of binary digits. Read carefully, these are not the same thing.

Consider the following example:

If I were to have 3 binary digits, the maximum value that I could store would be 1112, this equates to 4 + 2 + 1 = 710.

If I were to be asked, the range of numbers then we have 8 options: We could count these all out and write down: "There are 8 different values 3 binary digits can take". But this isn't very clever, what is you wanted to find out the range and maximum values for 34 bits, you can't be expected to write them all out. We are looking for a rule to save us the job and stop us making mistakes. Can you work out a rule in terms of $$n$$ for:
 * 1) 000
 * 2) 001
 * 3) 010
 * 4) 011
 * 5) 100
 * 6) 101
 * 7) 110
 * 8) 111

Maximum denary value of $$n$$ binary digits: Number of different values for $$n$$ binary digits:

Give both the maximum value and number of different values for the following n binary digits:

4

Maximum : $$2^n - 1 = 2^4 - 1 = 16 - 1 = 15$$

Range : $$2^n = 2^4 = 16$$

Maximum : $$2^n - 1 = 2^5 - 1 = 32 - 1 = 31$$

Range : $$2^n = 2^5 = 32$$

Maximum : $$2^n - 1 = 2^{8} - 1 = 256 - 1 = 255$$

Range : $$2^n = 2^{8} = 256$$

Maximum : $$2^n - 1 = 2^{10} - 1 = 1024 - 1 = 1023$$

Range : $$2^n = 2^{10} = 1024$$

highest address : $$2^n - 1 = 2^6 - 1 = 64 - 1 = 63$$

Different number of addresses : $$2^n = 2^6= 64$$

This is a very popular exam question!

Hexadecimal
You may notice from the table that one hexadecimal digit can represent exactly 4 binary bits. Hexadecimal is useful to us as a shorthand way of writing binary, and makes it easier to work with long binary numbers.

Hexadecimal is a base-16 number system which means we will have 16 different numbers to represent our digits. The only problem being that we run out of numbers after 9, and knowing that 10 is counted as two digits we need to use letters instead: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F We can do exactly the same thing as we did for denary and binary, and write out our table.

So now all we need to do is to add the columns containing values together, but remember that A = 10, B = 11, C = 12, D = 13, E = 14, F = 15. 3*4096 + 4*256 + (A)10*16 + (F)15*1 = 1348716

You might be wondering why we would want to use hexadecimal when we have binary and denary, and when computer store and calculate everything in binary. The answer is that it is entirely for human ease. Consider the following example:

All the numbers are the same and the easiest version to remember/understand for humans is the base-16. Hexadecimal is used in computers for representing numbers for human consumption, having uses for things such as memory addresses and error codes. ''NOTE: Hexadecimal is used as it is shorthand for binary and easier for people to remember. It DOES NOT take up less space in computer memory, only on paper or in your head!'' Computers still have to store everything as binary whatever it appears as on the screen.

16 1  A  1 16 * 10 + 1 * 1 = 16110

16 1  F  F 16 * 15 + 1 * 15 = 25510

16 1  0  D 16 * 0 + 1 * 13 = 1310

16 1  3  7 16 * 3 + 1 * 7 = 5510

Converting Between Bases
The sum that you saw previously to convert from hex to denary seemed a little cumbersome and in the exam you wouldn't want to make any errors, we therefore have to find an easier way to make the conversion.

Since 4 binary bits are represented by one hexadecimal digit, it is simple to convert between the two. You can group binary bits into groups of 4, starting from the right, and adding extra 0's to the left if required, and then convert each group to their hexadecimal equivalent. For example, the binary number 0110110011110101 can be written like this:

0110 1100 1111 0101

and then by using the table above, you can convert each group of 4 bits into hexadecimal: 0110 1100 1111 0101  6    C    F    5

So the binary number 0110110011110101 is 6CF5 in hexadecimal. We can check this by converting both to denary. First we'll convert the binary number, since you already know how to do this:

By multiplying the columns and then adding the results, the answer is 27893.

Notice that the column headings are all 2 raised to a power, $$1 = 2^0$$, $$2 = 2^1$$, $$4 = 2^2$$, $$8 = 2^3$$, and so on. To convert from hexadecimal to denary, we must use column headings that are powers with the base 16, like this:

$$5 \times 1 = 5$$

$$15 \times 16 = 240$$ (You should memorize the values A-F)

$$12 \times 256 = 3072$$

$$6 \times 4096 = 24576$$

Totalling them all up gives us 27893, showing that 0110110011110101 is equal to 6CF5.

To convert from denary to hexadecimal, it is recommended to just convert the number to binary first, and then use the simple method above to convert from binary to hexadecimal.

In summary, to convert from one number to another we can use the following rule: Hexadecimal <-> Binary <-> Denary

1   2  (Hex) 0001 0010 (Binary) 128 64 32 16 8  4  2  1   0  0  0  1  0  0  1  0 = 16+2 = 18 (decimal)

A   5  (Hex) 1010 0101 (Binary) 128 64 32 16 8  4  2  1   1  0  1  0  0  1  0  1  = 128+32+4+1 = 165 (decimal)

7   F  (Hex) 0111 1111 (Binary) 128 64 32 16 8  4  2  1   0  1  1  1  1  1  1  1  = 64+32+8+4+2+1 = 127 (decimal)

1   0  (Hex) 0001 0000 (Binary) 128 64 32 16 8  4  2  1   0  0  0  1  0  0  0  0  = 16(decimal)

1010 1101 (Binary) A   D  (Hex)

0011 0111 (Binary) 3   7  (Hex)

1010 1111 (Binary) A   F  (Hex)

1110 1010 0001 (Binary) E   A    1  (Hex)

128 64 32 16 8  4  2  1   0  1  0  1  0  1  1  1  = 64+16+4+2+1 = 87(decimal)

0101 0111 (Binary) 5   7  (Hex)

128 64 32 16 8  4  2  1   0  0  0  0  1  1  0  0  = 8+4 = 12(decimal)

0000 1100 (Binary) 0   C  (Hex)

128 64 32 16 8  4  2  1   0  1  1  1  0  1  0  1  = 64+32+16+4+1 = 117(decimal)

0111 0101 (Binary) 7   5  (Hex)