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Twos complement representation

  • Ones Complement

The ones complement of a number is represented by flipping the number's bits one at a time. For example, the value 01001001 becomes 10110110.

Suppose you are working with B-bit numbers. Then the ones complement for the number N is 2 B size 12{2 rSup { size 8{B} } } {} -1 -N

  • Twos complement

The twos complement of a number N in a B-bit system is 2 B size 12{2 rSup { size 8{B} } } {} -N.

There is a simple way to calculate a twos complement value: invert the number's bits and then add 1.

For a concrete example, consider the value 17 which is 00010001 in binary. Inverting gives 11101110. Adding one makes this 11101111.

  • Twos Complement Representation

Like sign magnitude, twos complement representation uses the most significant bit as a sign bit, making it easy to test whether an integer is positive or negative. It differs from the use of the sign-magnitude representation in the way that the other bits are interpreted.

Consider an n-bit integer. A, in twos complement representation. If A is positive, then the sign bit an-1 is zero. The remaining, bits represent the magnitude of the number in the same fashion as for sign magnitude:

A = i = 0 n 2 a i 2 i size 12{A= Sum cSub { size 8{i=0} } cSup { size 8{n - 2} } {a rSub { size 8{i} } 2 rSup { size 8{i} } } } {} If a n 1 = 0 size 12{a rSub { size 8{n - 1} } =0} {}

The number zero is identified as positive and therefore has a 0 sign bit and a mag­nitude of all 0s. We can see that the range of positive integers that may he repre­sented is from 0 (all of the magnitude bits are 0) through- 1 (all of the magnitude bits are 1). Any larger number would require more bits.

For a negative number A (A<0), the sign bit,is one. The remain­ing n-1 bits can take on any one of a n 1 2 n 1 size 12{a rSub { size 8{n - 1} } 2 rSup { size 8{n - 1} } } {} values. Therefore, the range of negative integers that can be represented is from -1 to - 2 n 1 size 12{2 rSup { size 8{n - 1} } } {}

This is the convention used in twos complement representation, yielding the following expression for negative and positive numbers:

A = 2 n 1 a n 1 + i = 0 n 2 a i 2 i size 12{A= - 2 rSup { size 8{n - 1} } a rSub { size 8{n - 1} } + Sum cSub { size 8{i=0} } cSup { size 8{n - 2} } {a rSub { size 8{i} } 2 rSup { size 8{i} } } } {}

The range of A is from - 2 n 1 size 12{2 rSup { size 8{n - 1} } } {} to 2 n 1 size 12{2 rSup { size 8{n - 1} } } {} -1.

Example: Using 8 bit to represent

+50= 0011 0010

-70=1011 1010

Converting between different bit lengths

The rule for twos complement integers is to move the sign hit to the new leftmost position and fill in with copies of the sign bit. For positive numbers, fill in with zeros, and for negative numbers, till in with ones.

For example:

+18 = 00010010

+18 = 00000000 00010010

-18 = 10010010

-18 = 11111111 10010010

3. integer arithmetic

3.1 negation

In sign-magnitude representation, the rule for forming the negation of an integer is simple: Invert the sign bit.

In twos complement representation, the negation of an integer can he formed with the following rules:

1. Take the Boolean complement of each bit of the integer (including the sign bit). That is. set each 1 to 0 and each 0 to 1.

2. Treating the result as an unsigned binary integer, add 1.

This two-step process is referred to as the twos complement operation, or the taking of the twos complement of an integer

Example:

Negation Special Case 1:

0 = 00000000

Bitwise not 11111111

Add 1 to LSB +1

Result 1 00000000

Overflow is ignored, so: - 0 = 0

Negation Special Case 2:

-128 = 10000000

bitwise not 01111111

Add 1 to LSB +1

Result 10000000

Questions & Answers

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Lukman Reply
The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
SABAL Reply
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
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Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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4
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x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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(61/11,41/11,−4/11)
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x+2y-z=7
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-1
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Source:  OpenStax, Computer architecture. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10761/1.1
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