Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
References
Number Systems
COMP1021/9021 Lecture 12
We use a base 10 system for representing numbers and
arithmetic
Number Systems and Binary Arithmetic
The Sumerian culture used a base 60 number system
The Mayan culture developed a base 20 number system
Kai Engelhardt
Computers essentially use a base 2 (or binary) number system
Base 8 (octal) and base 16 (hexadecimal) number systems are
also commonly used in writing programs
CSE, UNSW (and NICTA)
In this lecture we will look at various number systems,
converting between them and also binary arithmetic
Revision: 1.2 of Date: 2006/01/13 06:25:53 UTC
almost identical to Maurice Pagnucco’s slides
2
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Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
Decimal Representation
References
Binary Representation
Can interpret decimal number 4705 as:
In a similar way, can interpret binary number 1011 as:
1 × 23 + 0 × 22 + 1 × 21 + 1 × 20
4 × 103 + 7 × 102 + 0 × 101 + 5 × 100
The base or radix is 2
Digits 0 and 1
The base or radix is 10
Digits 0 – 9
Place values:
···
···
Write number as 10112
(= 1110 )
Place values:
· · · 1000
···
103
Write number as 470510
100
102
10
101
1
100
8
23
4
22
2
21
1
20
Note use of subscript to denote base
3
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Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
Binary Representation
Binary Representation
Addition of 1-bit binary numbers can be achieved by half-adder
circuit.
Binary representation is convenient because
easily represented using two voltage levels: low and high
simple arithmetic tables that can be implemented using logic
gates
+
0
1
0
00
01
×
0
1
0
0
0
References
XOR
Sum
AND
Carry
1
01
10
1
0
1
General addition requires combination of half-adder circuits
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5
Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
Converting from base x to Decimal
Examples
Consider n-digit number in arbitrary base b: [d (n−1) . . . d (0) ]x
11012
Rewrite in polynomial form:
n−1
X
11012 = 1.23 + 1.22 + 0.21 + 1.20 = 1310
11012 = ((1.2 + 1).2 + 0).2 + 1 = 1310
(d (i) × bi )
110010112
110010112 =
1.27 + 1.26 + 0.25 + 0.24 + 1.23 + 0.22 + 1.21 + 1.20 = 20310
110010112 =
((((((1.2 + 1).2 + 0).2 + 0).2 + 1).2 + 0).2 + 1).2 + 1 = 20310
i=0
Or nested form:
310425
((. . . (d (n−1) b + d (n−1) )b + . . . + d (1) )b + d (0) )10
310425 = 3.54 + 1.53 + 0.52 + 4.51 + 2.50 = 202210
310425 = (((3.5 + 1).5 + 0).5 + 4).5 + 2 = 202210
Perform necessary arithmetic to achieve decimal conversion
7
References
8
Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
Converting to base x
Examples
Note that a base x number in nested form
((ax + b)x + c)x + d can be written:
abcdx
C
B
a
=
=
=
=
References
2510 ÷ 2
1210 ÷ 2
610 ÷ 2
310 ÷ 2
110 ÷ 2
Cx + d
Bx + c
ax + b
0x + a
=
=
=
=
=
12
6
3
1
0
r
r
r
r
r
↑
↑
↑
↑
↑
1
0
0
1
1
Write remainders from left to right reading from bottom to
top.
−→
2510 = 110012
Numbers a, b, c, d are all less than base x
Equations above imply that d , c, b, a are remainders when the
number abcdx is repeatedly divided by x
To convert an integer to base x
Check: 110012 = 1.24 + 1.23 + 0.22 + 0.21 + 1.20 = 2510
repeatedly divide quotient by x until quotient is 0
write the remainders in reverse
. . . or verify your conversion routines.
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Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
Examples
4873210 ÷ 5
9746 ÷ 5
1949 ÷ 5
389 ÷ 5
77 ÷ 5
15 ÷ 5
3÷5
=
=
=
=
=
=
=
9746
1949
389
77
15
3
0
References
Examples
r
r
r
r
r
r
r
2
1
4
4
2
0
3
819210 ÷ 8
1024 ÷ 8
128 ÷ 8
16 ÷ 8
2÷8
=
=
=
=
=
1024
128
16
2
0
r
r
r
r
r
0
0
0
0
2
819210 = 200008
4873210 = 30244125
200008 = 2.84 + 0.83 + 0.82 + 0.81 + 0.80 = 819210
Check:
3.56 + 0.55 + 2.54 + 4.53 + 4.52 + 1.51 + 2.50 = 4873210
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Number Systems
Binary Arithmetic
References
Number Systems
Decimal to Binary — With Division
Binary Arithmetic
References
Decimal to Binary — With Division
57310 ÷ 2
286 ÷ 2
143 ÷ 2
121 ÷ 2
60 ÷ 2
30 ÷ 2
15 ÷ 2
7÷2
3÷2
1÷2
Repeated division of quotient by 2 as in previous example.
Repeat until quotient is 0. Output remainders in reverse
sequence to the one in which they were generated: LIFO!
=
=
=
=
=
=
=
=
=
=
286
143
121
60
30
15
7
3
1
0
r
r
r
r
r
r
r
r
r
r
1
0
1
1
0
0
1
1
1
1
57310 = 11110011012
Check:
1.29 +1.28 +1.26 +1.26 +0.25 +0.24 +1.23 +1.22 +0.21 +1.20 =
57310
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Number Systems
Binary Arithmetic
References
Number Systems
Decimal to Binary — Without Division
Decimal to Binary — Without Division
Convert 17410 to binary:
174 − 128 = 46 −→ 1 in 27 place
64 > 46
−→ 0 in 26 place
46 − 32 = 14
−→ 1 in 25 place
16 > 14
−→ 0 in 24 place
14 − 8 = 6
−→ 1 in 23 place
6−4=2
−→ 1 in 22 place
2−2=0
−→ 1 in 21 place
1>0
−→ 0 in 20 place
= 101011102
Find largest power of 2 smaller than current number. Subtract
this from the number and repeat on the difference
Usually not practical for larger numbers
15
Binary Arithmetic
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References
Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
Octal Representation
Binary to Octal
08
0002
Can interpret octal number 70538 as:
7 × 83 + 0 × 82 + 5 × 81 + 3 × 80
The base or radix is 8
Digits 0, 1, 2, 3, 4, 5, 6, 7
18
0012
28
0102
38
0112
48
1002
58
1012
68
1102
78
1112
Idea: Collect bits into groups of three starting from right to left
“pad” out left-hand side with 0’s if necessary
Place values:
· · · 512
···
83
Write number as 70538
(= 362710 )
64
82
8
81
Convert each group of three bits into its equivalent octal
representation (given in table above)
1
80
Example, 10111110001010012
011
3
001
1
17
111
7
000
0
101
5
0012
18
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Number Systems
Binary Arithmetic
References
Number Systems
Octal to Binary
Binary Arithmetic
References
Hexadecimal Representation
Can interpret hexadecimal number 3AF1 as:
3 × 163 + 10 × 162 + 15 × 161 + 1 × 160
Reverse the previous process
Convert each octal digit into equivalent 3-bit binary
representation
The base or radix is 16
Digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F
Example, 270158
Place values:
2
010
7
111
0
000
1
001
· · · 4096
···
163
Write number as 3AF116
(= 1508910 )
58
1012
19
256
162
16
161
1
160
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Number Systems
Binary Arithmetic
References
Number Systems
Binary to Hexadecimal
016
00002
816
10002
116
00012
916
10012
216
00102
A16
10102
316
00112
B16
10112
416
01002
C16
11002
516
01012
D16
11012
Binary Arithmetic
References
Binary to Hexadecimal
616
01102
E16
11102
716
01112
F16
11112
10111110001010012
Idea: Collect bits into groups of four starting from right to left
Convert each group of four bits into its equivalent hexadecimal
representation (given in table above)
Example, 10111110001010012
21
1011
B
1110
E
0010
2
10012
916
0010
2
1111
F
0101
5
1100
C16
101111010111002
“pad” out left-hand side with 0’s if necessary
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Number Systems
Binary Arithmetic
References
Number Systems
Hexadecimal to Binary
Binary Arithmetic
carry
Convert each hexadecimal digit into equivalent 4-bit binary
representation
1
1
2097
+
5
Example, AD516
A
1010
D
1101
References
Decimal Arithmetic — Addition
Reverse the previous process
2102
5
01012
Important principle of “sum” and “carry”.
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References
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Addend
Augend
Sum
Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic — Addition
Binary Arithmetic
References
Binary Arithmetic — Addition
Similar idea: “sum” and “carry”
1010 + 1210 = 2210
Four cases to consider
Addend
Augend
Sum
Carry
0
0
—
0
0
0
1
—
1
0
1
0
—
1
0
1
1
—
0
1
1
1
1
0
0
1
1
1
0
1
0
0
0
+
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Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
References
Multiplying Binary Numbers by 2
Dividing Binary Numbers by 2
Consider decimal multiplication: 10010 × 1010 = 100010 ;
shift digits one place to the left introducing 0 to the right
Consider decimal division: 10010 ÷ 1010 = 1010 ;
shift digits one place to the right removing rightmost digit
Similar idea applies to binary numbers
Similar idea applies to binary numbers
Binary multiplication: 1002 × 210 = 10002 (i.e.,
410 × 210 = 810 );
shift digits one place to the left introducing 0 to the right
Binary division: 1002 ÷ 210 = 102 (i.e., 410 ÷ 210 = 210 );
shift digits one place to the right removing rightmost bit
C’s bitwise shift: >> shift right
C’s bitwise shift: << shift left
int x = 12;
x = x >> 1;
printf("%d\n", x);
int x = 12;
x = x << 1;
printf("%d\n", x);
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Number Systems
Binary Arithmetic
References
Number Systems
Bitwise Shift
Binary Arithmetic
References
Negative Binary Numbers
Several possible representations
Signed magnitude
One’s complement
Two’s complement (used in C for integers)
Example: Bitwise shift by two places (i.e., ÷4 and ×4)
0 0 0 0 1 1 0 0 ( = 12)
0 0 0 0 0 0 1 1 >> 2
(= 3)
0 0 1 1 0 0 0 0 << 2
(= 48)
What happens when we bitwise shift signed quantities?
Most significant bit (MSB; left-most bit) determines whether
number is positive or negative
Most significant bit (MSB)
1
29
1
Least significant bit (LSB)
0
1
0
0
1
0
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Number Systems
Binary Arithmetic
References
Number Systems
Signed Magnitude
Binary Arithmetic
References
Signed Magnitude
Consider 4-bit binary
00002
00012
00102
00112
01002
01012
01102
01112
Utilise MSB as a “sign” bit: 0 = positive; 1 = negative
Remaining bits express number in binary as usual
Example
5310 = 001101012
−5310 = 101101012
How many numbers can be represented with this scheme?
010
110
210
310
410
510
610
710
10002
10012
10102
10112
11002
11012
11102
11112
010
−110
−210
−310
−410
−510
−610
−710
Can express −(2n−1 − 1). .2n−1 − 1, i.e., 2n − 1 numbers.
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Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
One’s Complement
References
One’s Complement
Consider 4-bit binary
00002
00012
00102
00112
01002
01012
01102
01112
Invert 0’s and 1’s
MSB indicates sign: 0 = positive; 1 = negative
Example
5310 = 001101012
−5310 = 110010102
How many numbers can be represented with this scheme?
010
110
210
310
410
510
610
710
−710
−610
−510
−410
−310
−210
−110
010
10002
10012
10102
10112
11002
11012
11102
11112
Can express −(2n−1 − 1) . . 2n−1 − 1, i.e., 2n − 1 numbers.
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Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
Two’s Complement
Two’s Complement
Need more efficient representation
Positive numbers as usual
Negative numbers:
−532
Compute one’s complement representation
Add 1
MSB indicates sign: 0 = positive; 1 = negative
Example
5310 = 001101012
−5310 = 110010112
N = −d
0
1
0
1
1
0
1
0
0
1
1
0
0
1
1
1
0
0
1
0
1
0
1
0
1
1
0
1
0
0
1
0
1
0
1
1
1
0
1
0
0
0
1
0
1
1
one’s complement
+1
two’s complement
0
1
1
0
one’s complement
+1
two’s complement
−482
Theorem
The representation of a numer N as n-bit two’s complement is a
vector (d (i) )0≤i<n of binary digits satisfying:
(n−1)
References
n−1
×2
+
n−2
X
(d
(i)
i
×2 )
i=0
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Number Systems
Binary Arithmetic
References
Number Systems
Two’s Complement
Consider 4-bit binary
00002
00012
00102
00112
01002
01012
01102
01112
010
110
210
310
410
510
610
710
10002
10012
10102
10112
11002
11012
11102
11112
Binary Arithmetic
References
A Shortcut to Two’s Complements
Scan the original binary from right to left, leaving all least
significant zeros and the first one unchanged and then
complement the remaining digits to the left of the first
one.
[Bri03]
−810
−710
−610
−510
−410
−310
−210
−110
Example
1001011001100000
⇒ 1001011001100000
⇒ 0110100110100000
Can express −(2n−1 ). .2n−1 − 1, i.e., 2n numbers.
Exercise: verify the table above
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Number Systems
Binary Arithmetic
References
Number Systems
Two’s Complement Advantages
Binary Arithmetic
Two’s Complement Advantages cont.
Extending, say, from int to long merely requires to pad with the
MSB. This obviously works for non-negative numbers. For negative
numbers observe that, when extending from n bit two’s
complement to n + k bits, we replace the summand −(2n−1 ) by
More efficient than signed magnitude and one’s complement
Zero represented once only
− (2n+k−1 ) +
Used in C to represent negative integers
Can be used to implement subtraction via addition
n+k−2
X
(2i )
i=n−1
= −2 × (2n+k−2 ) + 2n+k−2 +
Define subtraction x − y to be x + (−y )
That is, convert −y to two’s complement form and add it to x
as in unsigned binary. This works neither with signed
magnitudes nor ones’ complements.
n+k−3
X
i=n−1
= −2n+k−2 +
n+k−3
X
i=n−1
..
.
= −(2n−1 )
39
References
40
(2i )
(2i )
Number Systems
Binary Arithmetic
References
Number Systems
Binary Subtraction via Addition and Two’s
Complement
1
1
0
0
1
1
0
0
0
0
0
1
1
0
0
1
1
0
0
0
0
0
1
1
1
0
1
0
0
0
1
0
1
1
0
1
Recall that in two’s complement we can represent numbers in
the range −(2n−1 ) to (2n−1 − 1)
If we consider two positive binary numbers x and y where
x + y > 2n−1 − 1 we can see that this sum will result in a
negative number (the MSB is 1)
positive overflow
If we consider two negative binary numbers −x and −y where
x + y > 2n−1 we can see that this sum will result in a positive
number (the MSB is 0)
negative overflow
Can use XOR gate in hardware to determine overflow condition
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Number Systems
Binary Arithmetic
References
Number Systems
Positive Overflow in Two’s Complement
Addition of 4-bit positive
0
0
0
Binary Arithmetic
References
Negative Overflow in Two’s Complement
numbers without overflow
x x x
x x x +
x x x
Addition of 4-bit negative numbers
1 x x
1 x x
1 1 x x
Carry into MSB must have been 0; carry out of MSB is 0
Addition of 4-bit positive
0
0
1
without overflow
x
x +
x
Carry into MSB must have been 1; carry out of MSB is 1
numbers with overflow
x x x
x x x +
x x x
Addition of 4-bit negative numbers
1 x x
1 x x
1 0 x x
Carry into MSB must have been 1; carry out of MSB is 0
with overflow
x
x +
x
Carry into MSB must have been 0; carry out of MSB is 1
carry in 6= carry out means overflow has occurred
carry in 6= carry out means overflow has occurred
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Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
References
Converting Fractions to Base x
Fractions
In the same way that we use a decimal point (.) to represent
fractional quantities for decimal numbers, we use the radix
point to represent fractional quantities in any base
Convert integer component as usual
To convert fractional component use “separate and multiply”
technique
Consider fractional component of abcd .pqrsx
For example, abcd .pqrsx (note: base x) represents
ax 3 + bx 2 + cx 1 + dx 0 + px −1 + qx −2 + rx −3 + sx −4
px −1 + qx −2 + rx −3 + sx −4
Multiplying by x gives: p + qx −1 + rx −2 + sx −3
Multiplying remaining fractional component by x gives:
q + rx −1 + sx −2
Numbers to the left of radix point represent integer component
Numbers to the right of radix point represent fractional
component
Repeat until fractional part exhausted or you have sufficient
digits (Note: process is not guaranteed to terminate)
Polynomial above can be evaluated to determine equivalent
decimal representation.
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Number Systems
Binary Arithmetic
References
Number Systems
Example
Binary Arithmetic
References
Example
Convert 23.312510 to base 8
Convert 23.312510 to base 2
Integer component
Integer component
2310 ÷ 2
11 ÷ 2
5÷2
2÷2
1÷2
2310 ÷ 8 = 2 r 7
2÷8
= 0 r 2
Fractional component
.312510 × 8 = 2.5
.510 × 8
= 4.0
Therefore, 23.312510 = 27.248
47
References
Overflow in Two’s Complement
Consider 8-bit binary representation
3510 − 1510 = 2010
0 0
1 1
0 0
−5310 − 4810 = −10110
1 1
1 1
1 0
Binary Arithmetic
48
=
=
=
=
=
11
5
2
1
0
r
r
r
r
r
1
1
1
0
1
Number Systems
Binary Arithmetic
References
Number Systems
Binary Arithmetic
Example continued
Floating Point Numbers
Beyond the scope of this course
Broadly speaking, floating point numbers are expressed as
Fractional component
.312510 × 2 =
.625 × 2
=
.25 × 2
=
.5 × 2
=
Therefore, 23.312510 = 10111.01018
±d0 .d1 d2 · · · dp−1 × β exp
0.625
1.25
0.5
1.0
where
d0 .d1 d2 · · · dp−1 is the significand (or mantissa)
β is the base (so the di are expressed in this base)
exp is the exponent
Idea is that of scientific notation
International Standard IEEE754 (1985)
β = 2; p = 24 (single precision); p = 53 (double precision)
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Number Systems
Binary Arithmetic
References
References
Robert Britton.
MIPS Assembly Language Programming.
Prentice Hall, 2003.
Sample chapters available at ftp://ftp.prenhall.com/pub/
esm/sample_chapters/cs/britton/index.html.
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References