EBK ENHANCED DISCOVERING COMPUTERS & MI
1st Edition
ISBN: 9780100606920
Author: Vermaat
Publisher: YUZU
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Chapter 12, Problem 38SG
Explanation of Solution
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2. Signed Integers
Unsigned binary numbers work for natural numbers, but many calculations use negative
numbers as well. To deal with this, a number of different methods have been used to represent
signed numbers, but we will focus on two's complement, as it is the standard solution for
representing signed integers.
2.1 Two's complement
• Most significant bit has a negative value, all others are positive. So, the value of an n-digit
-2
two's complement number can be written as: Σ2 2¹ di 2n-1 dn
• Otherwise exactly the same as unsigned integers.
i=0
-
• A neat trick for flipping the sign of a two's complement number: flip all the bits (0 becomes 1,
or 1 becomes 0) and then add 1 to the least significant bit.
• Addition is exactly the same as with an unsigned number.
2.2 Exercises
For questions 1-3, answer each one for the case of a two's complement number and an
unsigned number, indicating if it cannot be answered with a specific representation.
1. (15 pts) What is the largest integer…
can u solve this question
1. Unsigned Integers
If we have an n-digit unsigned numeral dn-1d n-2...do in radix (or base) r, then the value of that
numeral is
n−1
r² di
Σi=0
which is basically saying that instead of a 10's or 100's place we have an r's or
r²'s place. For binary, decimal, and hex r equals 2, 10, and 16, respectively.
Just a reminder that in order to write down a large number, we typically use the IEC or SI
prefixing system:
IEC: Ki = 210, Mi = 220, Gi = 230, Ti = 240, Pi = 250, Ei = 260, Zi = 270, Yi = 280;
SI: K=103, M = 106, G = 109, T = 10¹², P = 1015, E = 10¹8, Z = 1021, Y = 1024.
1.1 Conversions
a. (15 pts) Write the following using IEC prefixes: 213, 223, 251, 272, 226, 244
21323 Ki8 Ki
223 23 Mi 8 Mi
b. (15 pts) Write the following using SI prefixes: 107, 10¹7, 10¹¹, 1022, 1026, 1015
107 10¹ M = 10 M
=
1017102 P = 100 P
c. (10 pts) Write the following with powers of 10: 7 K, 100 E, 21 G
7 K = 7*10³
Chapter 12 Solutions
EBK ENHANCED DISCOVERING COMPUTERS & MI
Ch. 12 - Prob. 1SGCh. 12 - Prob. 2SGCh. 12 - Prob. 3SGCh. 12 - Prob. 4SGCh. 12 - Prob. 5SGCh. 12 - Prob. 6SGCh. 12 - Prob. 7SGCh. 12 - Prob. 8SGCh. 12 - Prob. 9SGCh. 12 - Prob. 10SG
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