by Wong Poh Ling B031210033
Showing posts with label Arithmetic for Computer Science. Show all posts
Showing posts with label Arithmetic for Computer Science. Show all posts
Sunday, 21 October 2012
Wednesday, 17 October 2012
Binary Subtraction
0 –
0 = 0
1 –
0 = 1
0 –
1 = 1
Digit “0” will transform to digit “1”,while 1 is
borrow from the left side.
One’s Complement Addition and Subtraction
N-bit represent integers in the range –(2N-1-1)
to +(2N-1-1) or call modulo (2N-1-1)
Example of Number Presentation :
The table show that all possible value is in 4-bit system:
No.
|
Positive binary (+)
|
Negative binary(-)
|
0
|
0000
|
1111
|
1
|
0001
|
1110
|
2
|
0010
|
1101
|
3
|
0011
|
1100
|
4
|
0100
|
1011
|
5
|
0101
|
1010
|
6
|
0110
|
1001
|
7
|
0111
|
1000
|
8
|
1111
|
0000
|
Two’s Complement Addition and Subtraction
N-bit be defined as complement with respect to 2N Equivalent taking one’s complement and then plus “1” N-bit can represent in the range –(2N-1) to
+(2N-1-1)
Example :
Signal overflow :
Different for unsigned and two’s complement
representation. Not necessarily occur when it carry out of sign, but occur
when the result is of the opposite sign. Two example of 8-bit for present the result of two’s
complement addition.
-lee Man
Binary Addition
the Hexadecimal System
Decimal:
The hexadecimal can be
converted to decimal by using the 16n system…
Eg:
C616
Hexadecimal C 6
Decimal * 12×161 6×160 =
192 + 6 = 19810
* (C is equal to 12 in the Number System
Conversion of hexadecimal)
Eg 2 :
FFD4.F16
Hexadecimal F F D 4 . F
Decimal *15×163 15×162 **13×161 4×160 . 15×16-1
=61440 + 3840 +
208 + 4+0.9375
=65492.937510
(* 15 is equal to
the F in the Number System Conversion of
hexadecimal…
**13 is equal to the D in the Number System
Conversion of hexadecimal…)
Binary:
The conversion of
hexadecimal to binary
by 3 methods:
·
According
to the Number System Conversion between hexadecimal and binary
·
By using
the 2n system
Eg :
2C16
Method 1: Converting
to decimal and then to binary
Hexadecimal 2 C
Decimal 2×161 12×160
= 32+ 12 =4410
Then convert the
decimal into binary by using division…
2
|
44
|
2
|
22 0 |
2
|
11 0
|
2
|
5
1
|
2
|
2
1
|
1
0
|
4410 =
1011002
And also can written
as …
2C16 =
1011002
Method 2 : According
to the Number System Conversion between hexadecimal and binary
Hexadecimal 2 C
Binary 0010 1100
·
2C16 = 10
11002
Method 3: By using the
2n system
---when converting the binary into hexadecimal ,we
need to group the binary into 4…and the 4 of grouped binary are indicated by
the 2n system…
For example, the most left of the grouped binary indicate the 23,next
is 22 , followed by 21, and lastly is 20
So,
2C16 = 0010 11002
(00102 = (
0×23)+(0×22)+(1×21)+(0×20)=0+0+2+0=216;
11002 =(1×23) + (1×22) + (0×21) + (0×20) = 8+4+0+0=
12=C16)
Eg 2:
C6.F16
We convert according
the Number System Conversion between the hexadecimal and binary…
Hexadecimal C 6 . F
Binary 1100 0110 1111
So,
C6.F16
= 1100 0110.11112
Octal:
There are few ways to
convert the hexadecimal to octal:
·
Convert
the hexadecimal into decimal and then to octal
·
Convert
the hexadecimal into binary and then to octal
Eg :
4A16
Method 1: Convert the
hexadecimal into decimal and then to octal
Hexadecimal 4 A
Decimal 4×161 10×160
= 64 + 10 =7410
Convert to octal by
dividing the decimal with 8…
8
|
 74 |
8
|
9 2
|
1 1
|
4A16 = 1128
Method 2 : Convert the
hexadecimal into binary and then to octal
Hexadecimal 4 A
Binary * 0100 **1010
* ((0×23) + (1×22) + (0×21) + (0×20) =416)
**((1×23) + (0×22) + (1×21) + (0×20) = 10 =A16)
4A16 = 100
10102
Convert to octal by
grouping the binary into 3…
100 10102 = 001
| 001 | 0102
Then , according to
the Number System Conversion between Octal and binary…
Octal
|
Binary
|
0
|
000
|
1
|
001
|
2
|
010
|
3
|
011
|
4
|
100
|
5
|
101
|
6
|
110
|
7
|
111
|
10010102 =
1128
Or …
By using the 2n
system…
100 10102
= 001 |
001 | 0102
=((0×22)+(0×21)+(1×20)) |((0×22)+(0×21)+(1×20)) | ((0×22)+(1×21)+(0×20))
=1 | 1
| 28
Therefore ,
4A16 = 1128
Eg 2 :
6A.B16
Converting to binary …
Hexadecimal 6 A . B
Binary 0110 1010 .
1011
6A.B16 =
1101010.10112
Converting into octal…
1101010.10112
= 001 |101 |010. |101 | 100
(According to the
Number System Conversion between octal and binary)
= 1 |
5 | 2 .
| 5 |48
Therefore,
6A.B16 =
152.548
by Chong Cai Ning B031210080 shinbachi123@hotmail.com
Tuesday, 16 October 2012
the Octal System
For this section, let’s take (65)8 for
example.
Conversion of Octal to Decimal
Octal Number
|
6
|
5
|
Significance of each bit
|
81
|
80
|
Weighted value
|
6(8) = 40
|
5(1) = 5
|
Solved Multiplication
|
48
|
5
|
Therefore = (65)8 = 48 + 5 = 53
Conversion of Octal fraction to decimal fraction
Take (0.5)8 for an example.
Octal number
|
5
|
Significance of bit
|
8-1
|
Weighted value
|
5(8-1)
|
Solved multiplication
|
0.625
|
Therefore, (0.5)8 = 0.625, and (65.5)8
= 53.625
Conversion of Octal to Binary
1.
Convert each octal digit to its 3-bit equivalent.
2. Combine the numbers
(65)8
Octal number
|
6
|
5
|
Binary number
|
110
|
101
|
Combine them: 1101012
8-bit binary number = 001101012
Conversion of octal fraction to binary fraction
0.58
Octal number
|
5
|
Binary number
|
101
|
Therefore (0.5)8 = 0.101 and (65.5)8
= (110101.101)2
the Binary System
Conversion of
binary to ….
Decimal:
The steps to convert the binary numbers to the decimal:
- · Use the bit of binary number one-by-one (starting with the most significant (leftmost) bit. Begin with the value 0
- · Then , repeatedly double the prior value
- · Add the next bit to produce the next value
- · The next value is taken down as the prior value and multiply the 2 and also add the next bit.
- Eg: 102
- (0×2)+1=1( 1 = this value is then taken down to multiply 2 and adding the second bit,0 )
- (1×2)+0=2
- The operation is repeated until the last bit
Eg:
Prior value
|
×2and +
|
Next bit
|
Next value
|
0
|
1
|
1
|
|
1
|
0
|
2
|
|
2
|
1
|
5
|
|
5
|
1
|
11
|
|
11
|
0
|
22
|
|
22
|
1
|
45
|
|
45
|
0
|
90
|
10110102 =
90 10
Step 2:
Binary 1 0
1 1 0 1
0
Decimal 1×26 0×25
1×24 1×23
0×22 1×21 0×20
= 90
Eg 2:
Step 1:
11111012
Prior value
|
×2 and+
|
Next bit
|
Next value
|
0
|
1
|
1
|
|
1
|
1
|
3
|
|
3
|
1
|
7
|
|
7
|
1
|
15
|
|
15
|
1
|
31
|
|
31
|
0
|
62
|
|
62
|
1
|
125
|
11111012 = 125 10
Step 2:
Binary 1 1 1 1 1 0 1
Decimal 1×26
1×25 1×24 1×23 1×22 0×21 1×20
=125
:The conversion of binary fraction to decimal fraction=
Eg :
0.110102
Weight
|
...20
|
2-1
|
2-2
|
2-3
|
2-4
|
2-5
|
|
Value represented
|
… 1
|
0.5
|
0.25
|
0.125
|
0.0625
|
0.03125
|
|
Binary
|
…0
|
1
|
1
|
0
|
1
|
0
|
Converting the binary
fraction to decimal fraction:
Binary 1 1 0 1 0
Decimal 1×2-1 1×2-2 0×2-3 1×2-4 0×2-5
=0.5 + 0.25 +
0.0625
=0.8125 10
:: therefore , 0.110102
= 0.8125 10
Eg:
0.112
Weight
|
…20
|
2-1
|
2-2
|
2-3
|
Value represented
|
…1
|
0.5
|
0.25
|
0.125
|
Binary
|
…0
|
1
|
1
|
0
|
Converting binary
fraction to decimal fraction=
Binary 1 1
Decimal 1×2-1 1×2-2 =
0.5 + 0.25 = 0.75 10
:: therefore, 0.112
= 0.7510
Hexadecimal:
The following below is the Number System Conversion between hexadecimal and binary:
Hexadecimal
|
Binary
|
0
|
|
0001
|
|
2
|
0010
|
3
|
0011
|
4
|
0100
|
5
|
0101
|
6
|
0110
|
7
|
0111
|
8
|
1000
|
9
|
1001
|
A
|
1010
|
B
|
1011
|
C
|
1100
|
D
|
1101
|
E
|
1110
|
F
|
1111
|
To convert binary to hexadecimal:
·
- Grouped the binary into 4
For
example : 1100 00112 = 1100 | 00112
- · If the binary not enough to group into 4, then insert an extra 0 bit on the left side of binary (called padding )
For
example : 11000112 = 0110 |
00112 (0 is the extra 0 bits inserted
)
- · After grouping the binary , according the Number System Conversion ,convert the binary into hexadecimal
Eg :
1100 00112
First step :Grouping the binary into 4
1100 00112 = 1100
| 00112
Second step : According to the Number System Conversion…
1100 00112 = C316
Eg 2:
1011002
First step : Grouping the binary into 4
Since 1011002
is not enough to divide into 4
each ,therefore insert two extra
0 bits on the
left side of binary.
1011002 = 0010 |
11002Second step :According to the Number System Conversion…
1011002 =3C16
Eg 3:
1011.112
First step : Grouping the binary
into 4
Since 0.112
is a decimal point therefore insert 2 0 bits on the right side of the
decimal
point…
1011 .112 = 1011. |11002
Second step : According to the Number System Conversion between
hexadecimal and
the
binary…
1011.112 =B. C16
Octal:
It is easy to convert the binary into octal numeral system.
The steps to convert the binary into octal numeral system are:
- · Grouping the binary into 3 each
For
example: 1100112 = 110
| 0112
- · Likes the hexadecimal , if there isn't enough to group the binary into 3,then insert an extra 0 bit on the left side of the binary . This process is called padding.
For
example: 100112 = 010 | 0112
( 0 is the extra 0 bit
inserted)
- · Then according to the table as below:
Number
System conversion between octal and binary:
Octal
|
Binary
|
0
|
000
|
1
|
001
|
2
|
010
|
3
|
011
|
4
|
100
|
5
|
101
|
6
|
110
|
7
|
111
|
Eg:
110
0112
1.
Grouping
the binary into 3:
110 0112 =
110 | 0112
2.
According
to the table above:
110
= 6 ; 011
= 3;
So, 110
0112 = 638
Apart of that , there is also a way to convert binary to octal :
Eg 2:
1.
Grouping
the binary into 3 :
110 0112 =
110 | 0112
2.
Convert
the binary by using 2n system:
110 = (1×22) + (1×21) + (0 ×20) | 011= (0×22) + (1×21) + (1×20)
=4+2+0 | = 0 + 2 + 1
=6 | =3
Therefore,
110 0112 = 638
Eg 3:
10 0112
1.
Grouping
the binary into 3… but this binary is not enough to group into 3 so we have to
insert an extra 0 bit at the left side of this binary…
10 0112 = 010 | 0112 ( 0 is the extra 0 bit inserted )
2.
Then,
convert the binary into octal either according to the Number System Conversion
between octal and binary or by using the 2n system…
Method 1: According to Number
System Conversion between octal and binary
010 = 2 | 011 =3
Therefore,
Method 2 :By using 2n
system |
010 = (0×22) + ( 1×21) + (0×20) | 011= ( 0×22) + (1×21) + (1×20)
=2
| =2+1
=3
10 0112 = 238
Eg 4:
Eg 4:
100011.112
1. Grouping the binary into
3…
100011.112 = 100
| 011. |1102
2.By using 2n
system
100=
(1×22) + (0×21) + (0×20) | 011=(0×22) +(1×21) +(1×20) .| 110= (1×22)+(1×21)+(0×20)
= 4 +0 +0 | =0 + 2 +1 | =4 + 2 +0
= 4 | = 3 | =6
100011.112 = 43
.6 8
-Cai Ning
-Cai Ning
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