A-level Computing/AQA/Paper 2/Fundamentals of data representation/Twos complement

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Nearly all computers work purely in binary. That means that they only use ones and zeros, and there's no - or + symbol that the computer can use. The computer must represent negative numbers in a different way.

We can represent a negative number in binary by making the most significant bit (MSB) a sign bit, which will tell us whether the number is positive or negative. The column headings for an 8 bit number will look like this:

-128	64	32	16	8	4	2	1
MSB							LSB
1	0	1	1	1	1	0	1

Here, the most significant bit is negative, and the other bits are positive. You start with -128, and add the other bits as normal. The example above is -67 in denary because: (-128 + 32 + 16 + 8 + 4 + 1 = -67)

-1 in binary is 11111111.

Note that you only use the most significant bit as a sign bit if the number is specified as signed. If the number is unsigned, then the msb is positive regardless of whether it is a one or not.

Template:CPTExample If the MSB is 0 then the number is positive, if 1 then the number is negative.

0000 0101 (positive)
1111 1011 (negative)

Template:Robox/Close

Template:ExampleRobox Let's say you want to convert -35 into Binary Twos Complement. First, find the binary equivalent of 35 (the positive version)

32  16   8   4   2   1 
 1   0   0   0   1   1

Now add an extra bit before the MSB, make it a zero, which gives you:

64 32  16   8   4   2   1 
 0  1   0   0   0   1   1

Now 'flip' all the bits: if it's a 0, make it a 1; if it's a 1, make it a 0:

64 32  16   8   4   2   1 
 1  0   1   1   1   0   0

This new bit represents -64 (minus 64). Now add 1:

64 32  16   8   4   2   1 
 1  0   1   1   1   0   0
                      + 1
 1  0   1   1   1   0   1

If we perform a quick binary -> denary conversion, we have: -64 + 16 + 8 + 4 + 1 = -64 + 29 = -35 Template:Robox/Close

Converting Negative Numbers

To find out the value of a twos complement number we must first make note of its sign bit (the most significant, left most bit), if the bit is a zero we work out the number as usual, if it's a one we are dealing with a negative number and need to find out its value.

Template:CPTExample To find the value of the negative number we must find and keep the right most 1 and all bits to its right, and then flip everything to its left. Here is an example:

1111 1011 note the number is negative

1111 1011 find the right most one

1111 1011 
0000 0101 flip all the bits to its left

We can now work out the value of this new number which is:

128  64  32  16   8   4   2   1 
  0   0   0   0   0   1   0   1
                      4   +   1 = −5   (remember the sign you worked out earlier!)

Template:Robox/Close

Template:ExampleRobox To find the value of the negative number we must take the MSB and apply a negative value to it. Then we can add all the heading values together

1111 1011 note the number is negative
-128  64  32  16   8   4   2   1 
   1   1   1   1   1   0   1   1
-128 +64 +32 +16  +8      +2  +1 = -5

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How about a more complex example? Template:ExampleRobox

1111 1100 note the number is negative 

1111 1100 find the right most one

1111 1100
0000 0100 flip all the bits to its left

128  64  32  16   8   4   2   1 
  0   0   0   0   0   1   0   0
                      4         = −4   (remember the sign you worked out earlier!)

Template:Robox/Close

Template:ExampleRobox To find the value of the negative number we must take the MSB and apply a negative value to it. Then we can add all the heading values together

1111 1100 note the number is negative

-128  64  32  16   8   4   2   1 
   1   1   1   1   1   1   0   0
-128 +64 +32 +16  +8  +4          = -4

Template:Robox/Close

So we know how to work out the value of a negative number that has been given to us. How do we go about working out the negative version of a positive number? Like this, that's how...

Template:CPTExample Take the binary version of the positive number

0000 0101 (5)
0000 0101 find the right most one

0000 0101 
1111 1011 flip all the bits to its left

So now we can see the difference between a positive and a negative number

0000 0101 (5)
1111 1011 (−5)

Template:Robox/Close

Template:CPTExample Take the binary version of the positive number

starting with -128, we know the MSB is worth -128. We need to work back from this:

-128  64  32  16   8   4   2   1 
   1   1   1   1   1   0   1   0   
-128 +64 +32 +16  +8      +1      = -5

0000 0101 (5)
1111 1011 (−5)

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Range of two's complement values

If the msb is being used to provide negative values it follows that the maximum possible value will be limited. The number of possible values remains the same and the range of these numbers will include negative values as well as the positive values.

Range of $n$ binary digits using two's complement representation:

Rule:

Minimum value = $- 2^{(n - 1)}$

For example, for 3 digits: $- 2^{(n - 1)} = - 2^{(3 - 1)} = - 4$

Rule:

Maximum value = $2^{(n - 1)} - 1$

For example, for 3 digits: $- 2^{(n - 1)} - 1 = - 2^{(3 - 1)} - 1 = 3$

Binary Subtraction

Template:ExampleRobox When it comes to subtracting one number from another in binary things can get very messy.

X (82 denary) 0101 0010
Y (78 denary) 0100 1110 −

An easier way to subtract Y from X is to add the negative value of Y to the value of X

X−Y = X+(−Y)

To do this we first need to find the negative value of Y (78 denary)

0100 1110 find the right most one

0100 1110 
1011 0010 flip all the bits to its left

Now try the sum again

   0101 0010     X( 82 denary) 
   1011 0010 +   Y(−78 denary)
   0000 0100
(¹)¹¹¹   ¹       the one carried over the bit 9 is ignored

Which comes out as:

128  64  32  16   8   4   2   1 
  0   0   0   0   0   1   0   0
                      4         = 4 = 82-78

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Template:CPTExercise Template:CPTQuestionTab Find the answers to the following sums in binary, show your working

  0110 1100 (108)
- 0000 0111 (7)

Template:CPTQuestionTabEnd Template:CPTAnswerTab Convert the 0000 0111 into a negative number 1111 1001 = -7 Add both numbers together:

   0110 1100
 + 1111 1001
   0110 0101 = 101
(¹)¹¹¹¹         the one carried over the bit 9 is ignored

Template:CPTAnswerTabEnd Template:CPTQuestionTab

  0001 1111 (31)
- 0001 0011 (19)

Template:CPTQuestionTabEnd Template:CPTAnswerTab Convert the 0001 0011 into a negative number 1110 1101 = -19 Add both numbers together:

   0001 1111
 + 1110 1101
   0000 1100 = 12
(¹)¹¹¹¹ ¹¹¹    the one carried over the bit 9 is ignored

Template:CPTAnswerTabEnd Template:CPTQuestionTab

  0111 0111 (119)
- 0101 1011 (91)

Template:CPTQuestionTabEnd Template:CPTAnswerTab Convert the 0101 1011 into a negative number 1010 0101 = -91 Add both numbers together:

   0111 0111
 + 1010 0101
   0001 1100 = 28
(¹)¹¹   ¹¹¹    the one carried over the bit 9 is ignored

Template:CPTAnswerTabEnd Template:CPTQuestionTab

23 (hex)  - 1F (hex)

Template:CPTQuestionTabEnd Template:CPTAnswerTab Convert the HEX values to binary
0010 0011 = 23 HEX or 35 denary
0001 1111 = 1F HEX or 31 denary
Now let's find the negative value of 1F
1110 0001 = -31
Add both numbers together:

   0010 0011
 + 1110 0001
   0000 0100 = 4
(¹)¹¹¹   ¹¹    the one carried over the bit 9 is ignored

Template:CPTAnswerTabEnd Template:CPTQuestionTab

  0001 0010 (10)
- 1110 0001 (-31)

Template:CPTQuestionTabEnd Template:CPTAnswerTab They have tried to trick you. What is a negative number minus a negative number? X - (-Y) = X + Y
Let's start by finding the value of the bottom number: 1110 0001 -> 0001 1111 = 31
And by working this out we have the positive value (0001 1111) Add both numbers together:

   0001 0010 (10)
 + 0001 1111 (31)
   0011 0001 = 49
(¹)  ¹¹ ¹¹     the one carried over the bit 9 is ignored

Template:CPTAnswerTabEnd Template:Robox/Close Template:BookCat

A-level Computing/AQA/Paper 2/Fundamentals of data representation/Twos complement

Converting Negative Numbers

Range of two's complement values

Binary Subtraction

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