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Number Systems and Logic
Prepared by
Dr P Marais
(Modified by D Burford)
Counting….
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, …
 What comes next?
 Why do we have ten digits?
Number Representations
 We are used to base/radix 10 (decimal)
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
10, 11, 12, 13, 14, 15, 16, 17, 18, 19
10, 20, 30, …, 100
10, …, 100, …, 1000, …, 10 000
Decimal
 Decimal Example: 1041.20310 :
1
0
4
1
103 102 101 100
.
2
0
3
10-1 10-2 10-3
 Represents:
1*1000 + 0*100 + 4*10 + 1*1 + 2/10 + 0/100 + 3/1000
Number Representations
 General radix r number, Nr
dp dp-1 … d2 d1 d0 .
rp rp-1
…
r2 r1 r0
d–1 d–2 … d–q
r-1 r-2
…
r-q
 Represents:
dprp + dp-1rp-1+ …+ d2 r2 + d1r1 + d0 r0 + d–1r--1+ d–2r--2+…d–qr-q
Binary
 Computer uses presence/absence of voltage
 Digits 0 and 1(off/on).
 i.e. base 2
 Bit = Binary digit
Binary
 Binary Example: 1001.1012 :
1
0
0
1
23
22
21
20
.
1
0
1
2-1
2-2
2-3
 Represents:
1*8 + 0*4 + 0*2 + 1*1 + 1/2 + 0/4 + 1/8 = 9.62510
Binary
 Integers (whole numbers) stored in n-bit words
 n-bit binary number has range: 0 to (2n-1)
 Largest 8-bit number:
=
1111 11112
= 1 0000 00002 - 12
= 28
-1
= 25610 - 110
= 25510
Binary to Decimal Conversion
1. quot = number; i = 0;
2. repeat until quot = 0:
1. quot = quot/2;
2. digit i = remainder;
3. i++;

gives digits from least to
most significant bit
Example:
33/2 = 16
16/2 = 8
8/2 = 4
4/2 = 2
2/2 = 1
1/2 = 0
rem 1
rem 0
rem 0
rem 0
rem 0
rem 1
=> 3310 = 1000012
(lsb)
(msb)
Converting fractional numbers



Convert int and frac. parts
separately
i = 0;
repeat until N = 1.0 or i = n:
 N = FracPart(N); N *= 2;
 digit i = IntPart(N); i++
Example:
0.8125*2 = 1.625
0.625 *2 = 1.250
0.25 *2 = 0.500
0.5 *2 = 1.000
int = 1
int = 1
int = 0
int = 1
=> 0.812510 = 0.11012
Caution: Many numbers cannot be represented accurately:
0.310 = 0.0[1001]...2 (bracket repeats, limited by bit size)
(msb)
(lsb)
Binary Addition
 Adding binary numbers:
1+0=0+1=1
0+0=0
1 + 1 = 0 carry 1
 Add 10910to 13610:
0110 1101
+1000 1000
____________
Binary Addition
 Adding binary numbers:
1 + 0 = 0 + 1 = 1;
0 + 0 = 0;
1 + 1 = 0 carry 1
 Add 10910to 13610:
0110 1101
+1000 1000
1111 0101
Check it!
Binary Addition: Overflow
 Possibility of overflow
 Add 210 to 25410 :
1111 1110
+
0000 0010
[1] 0000 0000 = 25610
 We only have 8 bits to store answer...so it's zero!
Binary Addition: Overflow
 Overflow: what happens?
 Program can generate an “exception” to let us know
 Usually many bits used (e.g. Intel word is 32-bits) so
occurrence is reduced
Signed Numbers
 So far, only have positive numbers
 What about negative numbers?
 What about subtraction?
Signed Numbers
 Can use left-most bit to encode sign
 0 -> pos, 1 -> neg
+510
0 00001012
-510
1 00001012
Signed Numbers
 For 8-bits, range is:
-(27-1)
1 11111112
-12710
+(27-1)
0 11111112
+12710
 Problems:
– Two zeros!!
– Addition not straight forward
– Must check sign-bit before adding (bad for hardware implementors)
 This is non-sensical and wasteful: can use extra bit pattern
One’s Complement
 Try “one's complement”:
– negative numbers obtained by flipping bits
– positive numbers unchanged
– Example:
+510
0000 01012
-510
1111 10102
 Left-most bit still indicates sign
Subtraction with One’s complement
 Now easy to subtract: complement number and add:
5-6
= 0000 0101 + complement (0000 0110)
=
0000 0101
+ 1111 1001
= 1111 1110
= complement (0000 0001) = (-1)
Subtraction with One’s complement
 Caution: If overflow,
– if numbers have different signs,
 carry is added into lsb
– if numbers have same sign,
 actual overflow has occured
Subtraction with One’s complement
 Evaluate 10–7 using 8-bit one’s complement
arithmetic:
Subtraction with One’s complement
 Evaluate 10–7 using 8-bit one’s complement
arithmetic:
10 - 7
= 0000 1010 + complement (0000 0111)
=
0000 1010
+ 1111 1000
= 0000 0010 carry 1
= 0000 0010 + 0000 0001 = 00000011 = 310
Two’s Complement
 Still have two zeros  two’s complement
– Complement then add 1
– Our number range now asymmetric: -27 to 27-1
– Used extra zero bit pattern
Two’s Complement
 Now when we add, discard carry
10 - 7
= 00001010 + 2complement (00000111)
=
00001010
+ 11111001
= 00000011 carry 1 (discard)
= 3
 Same overflow test can be used (same/different signs)
Octal and Hexadecimal
 Base 8 (octal) and base 16 (Hexadecimal) are
sometimes used (powers of 2)
 Octal uses 8 digits (0-7)
 Hex uses 16 “digits”:
0, 1, 2, 3, 4, 5, 6, 7 ,8, 9, A, B, C, D, E, F
Octal and Hexadecimal
 Each octal digit represents 3-bits
 Each hex digit represents 4-bits
 Examples:
1310 = 11012
= (001)(101)2
= (1101)2
= 158
= D16
Octal and Hexadecimal
 Conversion from decimal as for bin:
– divide/multiply by 8 or 16 instead
– May be easier to convert to binary first
 Binary to octal or hexadecimal
– group bits into 3 (octal) or 4 (hex) from LS bit
– pad with leading zeros if required
Octal and Hexadecimal: Example
 0100 0110 1101 01112
Octal and Hexadecimal: Example
 0100 0110 1101 01112
= (000) (100) (011) (011) (010) (111)
= 433278
= (0100) (0110) (1101) (0111)
= 46D716
 Note padding at front of number
Octal and Hexadecimal
 To convert from hex/octal to binary: reverse procedure
FF16 = (1111)(1111)2
3778 = (011)(111)(111)2
 NOTE: for fractional conversion, move from left to right
and pad at right end:
0.110011010112
0.112 = 0.(110)2
= 0. (110) (011) (010) (110)
= 0.63268
= 0.68
 Convert fractional/integer part separately