ALGEBRA - solved problems
  Sets
1.     If S is the set { 1, 2, 3 } then power set of S,
Solution:        P (S) = { {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}.
2.    Assuming the set of natural numbers N is the universal set given are sets, 
                 A = { x N x < 5 }B = { x N 3 x < 8 } and  C = { x N x = 2n 1, n N }.
   Find:    a)  A B,    b)  A U B,    c)  A B,    d)  B A   and   e)  C'.
Solution:  As  A = { x N x < 5 } = { 1, 2, 3, 4 }, 
                      B = { x N 3 x < 8 } = { 3, 4, 5, 6, 7 }
               and  C = { x N x = 2n 1, n N } = { 1, 3, 5, 7, . . . }
         then   a)  A B = { 3, 4 },    b)  A U B = { 1, 2, 3, 4, 5, 6, 7 },  
                  c)  A B = { 1, 2 },    d)  B A = { 5, 6, 7 } 
         and    e)  C' = { x N x = 2n } = { 2, 4, 6, 8, . . . }. 
3.    If given set  {1, 2, 3, 4, 5, 6, 7}  then one of its possible partitions is
Solution:        { {1, 3, 7}, {2}, {4, 5}, {6} }.
  Relations
4.    Given is set S = {1, 3, 5, 7, 9} and relations on S:
                   R1 = {(1, 3), (3, 5), (5, 1), (3, 7), (7, 9), (9, 1)},
                   R2 = {(1, 1), (3, 3), (5, 5), (7, 7), (9, 1)},
                   R3 = {(1, 5), (3, 7), (7, 1), (9, 5), (1, 1)},
                   R4 = {S S}.
Which of these relations is function from S to S ?
Solution:   A binary relation R, a subset of the Cartesian product A B, is said to be a function from A to B if for each x A there is exactly one y B, such that the pair (x, y) is in subset R. The set A is called the domain of the function, and the set B is called the codomain of the function.
     Thus,       R2 = {(1, 1), (3, 3), (5, 5), (7, 7), (9, 1)} is the function from S to S.
5.    Supplement the relation  R = {(1, 1), (2, 2), (3, 2), (4, 1)}, defined on the set  S = {1, 2, 3, 4}, with 
minimal number of elements of the product set  S S  such that the relation becomes symmetric.
Solution:    The given relation should be supplemented with pairs 
(1, 4)  and  (2, 3).
The relation   R = {(1, 1), (1, 4), (2, 2), (2, 3), (3, 2), (4, 1)}   is symmetric (in
relation to the main diagonal).

  Binary numbers
6.    Convert given binary number to its decimal equivalent by writing it in a place-value notation:
Solution:       1 0 1 1 1 0 1 = 1 260 25 + 1 24 + 1 23 + 1 220 21 + 1 20 =
                                        =    64   +     0    +   16    +    8    +    4    +     0    +     1    = 93.
7.    Convert given decimal number to its binary equivalent:
Solution:      (113)10  =>   ( 1 1 1 1 0 0 0 1 )2
To convert a decimal number to its binary equivalent
divide given decimal and each successive quotient by
2 noting remainders from right to left, i.e., form the
lowest place value to the higher. The remainders can
only be 0 and 1 since divisions are by 2. The division 
ends by the quotient zero
113   2 =  1 1 1 1 0 0 0 1
56 <=====
28
14
7
3
1
0
Binary operations
8.    Add given binary numbers:
Solution:   The binary system works under the same principles as the decimal system as show basic rules for binary addition.
(1) 0 + 0 = 0
(2) 0 + 1 = 1
(3) 1 + 0 = 1
(4) 1 + 1 = 1 0 (with a carry of 1)
(5) 1 + 1 + 1 = 1 1 (with a carry of 1)
or
(1) (2) and (3)    (4)    (5)
           1
0        1      1     1
+ 0    +  0  +  1 +  1
0        1    10   11
When adding two multiple digits numbers a carry has to be added to the next higher place value digit.
1 1    carry
1 1 0 1 0 =>    26
+ 1 1 1 0 0 => + 28
1 1 0 1 1 0   sum    54
The method of subtraction by adding complement of the subtrahend applied to decimals
9.    Calculate the difference D = A − B, where A = 423 minuend (M) and B = 79 subtrahend (S).
Solution:  423 79 = 423 + (100 79 100) = 423 + 21100 = 344, by using this method we can avoid 
subtraction of larger from smaller digits thus, when calculating the complement of the subtrahend (CS) we
write 100 79 = 99 + 1 79 = 21.
The method of subtraction by adding complement of the subtrahend applied to binary numbers
10.    Calculate the difference D = A − B, where  A = 101000(40) and  B = 10111(23).
Solution:   The calculation of the complement, 
1000000 10111 = 111111 + 1 10111 = 101000 + 1 = 101001 (CS)
 then,   101000 10111 = 101000 + (1000000 101111000000) = 101000 + 101001 1000000 = 10001
Proof: the binary complement of 10111(23) is 101001(41) since 10111(23) + 101001(41) = 1000000(64) = 26.
Computers form the complement in two steps, first invert all bits of the number by changing all of the ones to zeroes and all of the zeroes to ones (which is called one's complement). Then, add 1 to the result, thus forming, so called the two's complement of the given number.
Since computers use fixed-length fields, the complement from above example is shown in 8 bits:
0 0 0 1 0 1 1 1 (23) 
1 1 1 0 1 0 0 0 1's complement
+ 1

1 1 1 0 1 0 0 1 2's complement
0 0 0 1 0 1 1 1   (23)
+ 1 1 1 0 1 0 0 1  (133) CS

1 0 0 0 0 0 0 0 0 = 256 = 28
Then by adding the minuend and the complement of the subtrahend obtained is the difference. 
Notice that the leading 1 (overflow), that occurs in the highest-order bit of the result, equals the corresponding place value 28 which must be subtracted from the result as the above algorithm shows.  
0 0 1 0 1 0 0 0  Minuend
+ 1 1 1 0 1 0 0 1  CS

1 0 0 0 1 0 0 0 1  Result
Binary multiplication is the same as repeated binary addition, and binary division is the repeated process of subtraction, just as in decimal division.

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