Computer Arithmetic - 國立臺灣師範大學

133 Pages·2012·0.91 MB·English

Computer Arithmetic
1
Numerical Analysis
NTNU
Tsung-Min Hwang
September 14, 2003
Department of Mathematics – NTNU Tsung-Min Hwang September 14, 2003
Computer Arithmetic
2
1 Floating-Point Number and Roundoff Error
Department of Mathematics – NTNU Tsung-Min Hwang September 14, 2003
Normalized scientific notation for the decimal number system of x:
(cid:15)
n
x = r 10 ;
(cid:6) (cid:2)
where
1
r < 1;
10 (cid:20)
and n is an integer (positive, negative, or zero).
– r is called the mantissa and n is the exponent.
– The leading digit in the fraction is not zero.
– For example,
42:965 = 0:42965 102;
(cid:2)
0:00234 = 0:234 10(cid:0)2:
(cid:0) (cid:0) (cid:2)
Computer Arithmetic
2
1 Floating-Point Number and Roundoff Error
Normalized scientific notation for the decimal number system of x:
(cid:15)
n
x = r 10 ;
(cid:6) (cid:2)
where
1
r < 1;
10 (cid:20)
and n is an integer (positive, negative, or zero).
Department of Mathematics – NTNU Tsung-Min Hwang September 14, 2003
– r is called the mantissa and n is the exponent.
– The leading digit in the fraction is not zero.
– For example,
42:965 = 0:42965 102;
(cid:2)
0:00234 = 0:234 10(cid:0)2:
(cid:0) (cid:0) (cid:2)
Computer Arithmetic
2
1 Floating-Point Number and Roundoff Error
Normalized scientific notation for the decimal number system of x:
(cid:15)
n
x = r 10 ;
(cid:6) (cid:2)
where
1
r < 1;
10 (cid:20)
and n is an integer (positive, negative, or zero).
– r is called the mantissa and n is the exponent.
– The leading digit in the fraction is not zero.
Department of Mathematics – NTNU Tsung-Min Hwang September 14, 2003
– For example,
42:965 = 0:42965 102;
(cid:2)
0:00234 = 0:234 10(cid:0)2:
(cid:0) (cid:0) (cid:2)
Computer Arithmetic
2
1 Floating-Point Number and Roundoff Error
Normalized scientific notation for the decimal number system of x:
(cid:15)
n
x = r 10 ;
(cid:6) (cid:2)
where
1
r < 1;
10 (cid:20)
and n is an integer (positive, negative, or zero).
– r is called the mantissa and n is the exponent.
– The leading digit in the fraction is not zero.
– For example,
42:965 = 0:42965 102;
(cid:2)
0:00234 = 0:234 10(cid:0)2:
(cid:0) (cid:0) (cid:2)
Department of Mathematics – NTNU Tsung-Min Hwang September 14, 2003
Computer Arithmetic
3
Department of Mathematics – NTNU Tsung-Min Hwang September 14, 2003
Scientific notation for the binary number system of x:
(cid:15)
m
x = q 2
(cid:6) (cid:2)
with
1
q < 1;
2 (cid:20)
and some integer m. For example,
(1001:1101) = 1 23 + 1 20 + 1 2(cid:0)1 + 1 2(cid:0)2 + 1 2(cid:0)4
2
(cid:2) (cid:2) (cid:2) (cid:2) (cid:2)
= 0:10011101 24
(cid:2)
= (9:8125)
10
Computer Arithmetic
3
Scientific notation for the binary number system of x:
(cid:15)
m
x = q 2
(cid:6) (cid:2)
with
1
q < 1;
2 (cid:20)
and some integer m.
Department of Mathematics – NTNU Tsung-Min Hwang September 14, 2003
For example,
(1001:1101) = 1 23 + 1 20 + 1 2(cid:0)1 + 1 2(cid:0)2 + 1 2(cid:0)4
2
(cid:2) (cid:2) (cid:2) (cid:2) (cid:2)
= 0:10011101 24
(cid:2)
= (9:8125)
10
Computer Arithmetic
3
Scientific notation for the binary number system of x:
(cid:15)
m
x = q 2
(cid:6) (cid:2)
with
1
q < 1;
2 (cid:20)
and some integer m. For example,
(1001:1101) = 1 23 + 1 20 + 1 2(cid:0)1 + 1 2(cid:0)2 + 1 2(cid:0)4
2
(cid:2) (cid:2) (cid:2) (cid:2) (cid:2)
= 0:10011101 24
(cid:2)
= (9:8125)
10
Department of Mathematics – NTNU Tsung-Min Hwang September 14, 2003
Computer Arithmetic
4
Department of Mathematics – NTNU Tsung-Min Hwang September 14, 2003
2
Example 1.1 What is the binary representation of ?
3
2
Solution: To determine the binary representation for , we write
3
2
= (0:a a a : : :) :
1 2 3 2
3
Multiply by 2 to obtain
4
= (a :a a : : :) :
1 2 3 2
3
Therefore, we get a = 1 by taking the integer part of both sides. Subtracting 1, we have
1
1
= (0:a a a : : :) :
2 3 4 2
3
Repeating the previous step, we arrive at
2
= (0:101010 : : :) :
2
3
Computer Arithmetic
4
2
Example 1.1 What is the binary representation of ?
3
Department of Mathematics – NTNU Tsung-Min Hwang September 14, 2003
2
Solution: To determine the binary representation for , we write
3
2
= (0:a a a : : :) :
1 2 3 2
3
Multiply by 2 to obtain
4
= (a :a a : : :) :
1 2 3 2
3
Therefore, we get a = 1 by taking the integer part of both sides. Subtracting 1, we have
1
1
= (0:a a a : : :) :
2 3 4 2
3
Repeating the previous step, we arrive at
2
= (0:101010 : : :) :
2
3