Exercise 1.1: 128+64+32+16+8+4 = 252
Exercise 1.2: 257 = 256+1 = 100000001 binary
257/2 | = | 128 remains 1 | 1 | |
128/2 | = | 64 remains 0 | 0 | |
64/2 | = | 32 remains 0 | 0 | |
32/2 | = | 16 remains 0 | 0 | |
16/2 | = | 8 remains 0 | 0 | |
8/2 | = | 4 remains 0 | 0 | |
4/2 | = | 2 remains 0 | 0 | |
2/2 | = | 1 remains 0 | 0 | |
1/2 | = | 0 remains 1 | 1 |
19/2 | = | 9 remains 1 | 1 | |
9/2 | = | 4 remains 1 | 1 | |
4/2 | = | 2 remains 0 | 0 | |
2/2 | = | 1 remains 0 | 0 | |
1/2 | = | 0 remains 1 | 1 |
"19" decimal is "10011" binary
"10011" binary is: = 2^{0} + 2^{1} + 2^{4} = 1 + 2 + 16 = 19
101 | |||
+ | 1010 | ||
= | 1111 | = 2^{0} + 2^{1} + 2^{2} + 2^{3} | |
= 15 |
Exercise 1.5: 10000. No, it needs 5 bits.
127/2 | = | 63 remains 1 | 1 | |
63/2 | = | 31 remains 1 | 1 | |
31/2 | = | 15 remains 1 | 1 | |
15/2 | = | 7 remains 1 | 1 | |
7/2 | = | 3 remains 1 | 1 | |
3/2 | = | 1 remains 1 | 1 | |
1/2 | = | 0 remains 1 | 1 |
Exercise 1.9: "+128" is 10000000
one's complement of 10000000 is 01111111 adding 1 + 1 ------------ -128 = 10000000
Exercise 1.10: Largest number in 7 bits is 2^{8}-1, which is 127. Smallest number in one's complement is -(2^{8}-1), which is -127. The smallest two's complement is even 1 smaller: -(2^{8}-1 + 1), which is -128.
20/2 | is | 10 remainder 0 | 0 | |
10/2 | is | 5 remainder 0 | 0 | |
5/2 | is | 2 remainder 1 | 1 | |
2/2 | is | 1 remainder 0 | 0 | |
1/2 | is | 0 remainder 1 | 1 |
one's complement of 00010100 is 11101011 adding 1 + 1 ------------ -20 = 11101100
one's complement of 11101100 is 00010011 adding 1 + 1 ------------ 20 = 00010100
which is the same as before:
2^{2} + 2^{4} | = 4 + 16 | |
= 20 |
10111111 [ -65] 11111010 [ -6 ] + 11000001 [ -63] + 11111001 [ -7 ] ------------------ ------------------ = 00000000 V: 0 C: 1 = 11110011 V: 0 C: 1 x CORRECT ERROR x CORRECT ERROR 00010000 [ 16 ] 01111110 [ 126] + 01000000 [ 64 ] + 00101010 [ 42] ------------------ ------------------ = 01010000 V: 0 C: 0 = 10101000 V: 1 C: 0 x CORRECT ERROR CORRECT x ERROR
Exercise 1.13: No, because the result has a magnitude that is always less then 128, so that will always fit into seven bits.
largest number is | + (2^{15}-1) | = | 32767 |
smallest number is | - (2^{15}-1 +1) | = | -32768 |
Exercise 1.15: largest negative number is - (2^{23}-1 +1) = -8388608
"29" is in packed BCD "00101001"
"91" is in packed BCD "10010001"
Exercise 1.17: No, because the four bits "1010" of the first BCD digit is not valid; it is one of the six unused bit combinations.
BCD: 0000 0101 0001 0010 0011 0001 0010 0011 23123 in binary: 0101 1010 0101 0011 -23123 in two's complement: 1010 0101 1010 1101
23123/2 | is | 11561 remains 1 | 1 | |
11561/2 | is | 5780 remains 1 | 1 | |
5780/2 | is | 2890 remains 0 | 0 | |
2890/2 | is | 1445 remains 0 | 0 | |
1445/2 | is | 722 remains 1 | 1 | |
722/2 | is | 361 remains 0 | 0 | |
361/2 | is | 180 remains 1 | 1 | |
180/2 | is | 90 remains 0 | 0 | |
90/2 | is | 45 remains 0 | 0 | |
45/2 | is | 22 remains 1 | 1 | |
22/2 | is | 11 remains 0 | 0 | |
11/2 | is | 5 remains 1 | 1 | |
5/2 | is | 2 remains 1 | 1 | |
2/2 | is | 1 remains 0 | 0 | |
1/2 | is | 0 remains 1 | 1 |
one's complement of 23123 is 1010 0101 1010 1100 adding 1 + 1 ---------------------- -12123 = 1010 0101 1010 1101
"222" is 0000 0011 0000 0010 0010 0010
"111" is 0000 0011 0000 0001 0001 0001
222 × 111 is 24642
"24642" is 0000 0101 0000 0010 0100 0110 0100 0010
Exercise 1.20: BCD needs 32 bits, two's complement needs 16 bits.
BCD "9999" is 0000 0100 0000 1001 1001 1001 1001 1001 two's complement "9999" is 0010 0111 0000 1111
9999/2 | is | 4999 remains 1 | 1 | |
4999/2 | is | 2499 remains 1 | 1 | |
2499/2 | is | 1249 remains 1 | 1 | |
1249/2 | is | 624 remains 1 | 1 | |
624/2 | is | 312 remains 0 | 0 | |
312/2 | is | 156 remains 0 | 0 | |
156/2 | is | 78 remains 0 | 0 | |
78/2 | is | 39 remains 0 | 0 | |
39/2 | is | 19 remains 1 | 1 | |
19/2 | is | 9 remains 1 | 1 | |
9/2 | is | 4 remains 1 | 1 | |
4/2 | is | 2 remains 0 | 0 | |
2/2 | is | 1 remains 0 | 0 | |
1/2 | is | 0 remains 1 | 1 |
Exercise 1.21: 7 decimal digits.
largest positive number is | 2^{23}-1: | 8,388,607 |
largest negative number is | -2^{23}: | -8,388,608 |
"0" | in even parity | is "0 0110000" |
"1" | ,, ,, ,, ,, | is "1 0110001" |
"2" | ,, ,, ,, ,, | is "1 0110010" |
"3" | ,, ,, ,, ,, | is "0 0110011" |
"4" | ,, ,, ,, ,, | is "1 0110100" |
"5" | ,, ,, ,, ,, | is "0 0110101" |
"6" | ,, ,, ,, ,, | is "0 0110110" |
"7" | ,, ,, ,, ,, | is "1 0110111" |
"8" | ,, ,, ,, ,, | is "1 0111000" |
"9" | ,, ,, ,, ,, | is "0 0111001" |
"A" | in even parity | is "0 1000001" |
"B" | ,, ,, ,, ,, | is "0 1000010" |
"C" | ,, ,, ,, ,, | is "1 1000011" |
"D" | ,, ,, ,, ,, | is "0 1000100" |
"E" | ,, ,, ,, ,, | is "1 1000101" |
"F" | ,, ,, ,, ,, | is "1 1000110" |
"A" | in non-parity ASCII | is "0 1000001" |
"?" | ,, ,, ,, ,, | is "0 0111111" |
"3" | ,, ,, ,, ,, | is "0 0110011" |
"b" | ,, ,, ,, ,, | is "0 1100010" |
Exercise 1.25: "1010 1010" binary is hexadecimal "AA".
Exercise 1.26: "FA" hexadecimal is binary "1111 1010"
Exercise 1.27: "01 000 001" binary is octal "101".
Exercise 1.28: Addition and subtraction is simple.
1024 is 2^{10}, which is | in direct binary |
in signed binary | 00000100 00000000 |
in two's complement |
Exercise 1.30: The V-bit is the overflow bit, which shows if a signed number has changed sign after an operation unintentionally. In practice, the overflow should only be acted upon in two's complement representation, and can be ignored in direct (unsigned) binary representation.
+16 in two's complement is 00010000
+17 in two's complement is 00010001
+18 in two's complement is 00010010
-16 in one's complement is 11101111
-16 in two's complement is 11110000
-17 in one's complement is 11101110
-17 in two's complement is 11101111
-18 in one's complement is 11101101
-18 in two's complement is 11101110
Exercise 1.32: 4D 45 53 53 41 47 45
M | 4D |
E | 45 |
S | 53 |
S | 53 |
A | 41 |
G | 47 |
E | 45 |