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| Boolean
Algebra |
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Binary
subtraction |
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The
method of subtraction by adding the complement of the subtrahend
applied to decimals |
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The
method of subtraction by adding the complement of the subtrahend
applied to binary numbers |
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| Binary
subtraction |
| Computers
perform binary subtraction by adding the minuend (M) and the
complement of the subtrahend (the
number to be subtracted). |
| The
sum of the complement and the given number equals the
corresponding place value (the particular power of the base of a
counting system). |
| The
place values in the binary system are the powers of the base b
= 2, just as the place values in the decimal system are the
powers of ten. |
| Thus,
for example the decimal complement of 79 is 21
since 79 + 21 = 100 = 102. |
| The
binary complement of 79 is 110001(49) since 1001111(79)
+ 110001(49) = 10000000(128) = 27. |
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| The
method of subtraction by adding the complement of the subtrahend
applied to decimals: |
| Example: |
Calculate the
difference D = A − B,
where A = 423 minuend (M) and B = 79 subtrahend
(S). |
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| Solution:
423 − 79 =
423 + (100 − 79 −
100) = 423 + 21− 100 = 344,
by using this method we can avoid |
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subtraction of larger
from smaller digits thus, when calculating the complement of the
subtrahend (CS) we write 100 − 79 =
99 + 1 − 79 = 21. |
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| The
method of subtraction by adding the complement of the subtrahend
applied to binary numbers: |
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Calculate the
difference D = A − B,
where A = 101000(40) and B = 10111(23). |
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| The
calculation of the complement, 1000000 −
10111 = 111111 + 1 − 10111 =
101000 + 1 = 101001 (CS) |
| Solution:
101000 −
10111 = 101000
+ (1000000 −10111−1000000) =
101000 + 101001 −
1000000 = 10001 |
| Proof:
the binary complement of 10111(23) is 101001(41)
since 10111(23) + 101001(41) =
1000000(64) = 26. |
| Computers
form the complement in two steps, first invert all bits of the
number by changing all of the ones to zeroes and all of the
zeroes to ones (which is called one's complement). Then, add 1
to the result, thus forming, so called the two's complement of
the given number. |
| Since
computers use fixed-length fields, the complement from above
example is shown in 8 bits: |
|
| 0 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
(23) |
| 1 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
1's
complement |
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+ |
1 |
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| 1 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
2's
complement |
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0 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
(23) |
| + |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
(133)
CS |
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| 1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
=
256 = 28
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| Then
by adding the minuend and the complement of the subtrahend
obtained is the difference. |
| Notice that the
leading 1 (overflow), that occurs in the
highest-order bit of the result, equals the
corresponding place value 28
which
must be subtracted from the result as the above
algorithm shows. |
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0 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
Minuend |
| + |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
CS |
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| 1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
Result |
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| Binary
multiplication is the same as repeated binary addition, and
binary division is the repeated process of subtraction, just as
in decimal division. |
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| Beginning
Algebra Contents |
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