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 Î N |  x < 5 },  B = { x Î N |  3 ≤ x < 8 } and  C = { x Î N |  x = 2n − 1, n Î N }.
   Find:    a)  A ∩ B,    b)  A U B,    c)  A − B,    d)  B − A   and   e)  C'.
Solution:  As  A = { x Î N |  x < 5 } = { 1, 2, 3, 4 }, 
                      B = { x Î N |  3 ≤ x < 8 } = { 3, 4, 5, 6, 7 }
               and  C = { x Î N |  x = 2n − 1, n Î 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 Î 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 Î A there is exactly one y Î 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 · 26 +  0 · 25 + 1 · 24 + 1 · 23 + 1 · 22 +  0 · 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 + 21− 100 = 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 −10111−1000000) = 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.
  Natural numbers and integers, rules and properties
Operations on integers
11.    Add integers with same signs.
Solution:   To add integers having the same sign, keep the same sign and add the absolute value of each number.
12.    Add integers with different signs.
Solution:   To add integers with different signs, keep the sign of the number with larger absolute value and subtract smaller absolute value from the larger.
13.    Subtract given integers:
Solution:   Subtract an integer by adding its opposite,
                          a)   5 -  7 =  5 + (- 7) = - 2,             b)   - 1 -  8 =  - 1 - ( + 8) = - 1+ (- 8) = - 9, 
                          c)   4 - ( - 3) =  4 + ( + 3) = 7,          d)   - 5 -  (- 9) =  - 5 + (+ 9) = - 5+ 9 = 4.
14.    The use of parentheses: 
Solution:   Evaluate given expressions,
                          a)   (- 5 + 3) + 6 = - 5 + 3 + 6 = 4,             b)   - 7 + (- 3 + 8) = - 7 - 3 + 8 = - 2, 
                          c)   6 - ( 5 - 12)  = 6 -  5 + 12 = 13,            d)   - (- 2 + 7) + 3 = 2 - 7 + 3 = - 2.
Multiplication and division of integers
15.    Multiply given integers: 
Solution:            a)   7 · 0 = 0,                       b)   (- 1) · 18 = - 18, 
                          c)   (- 3) · (- 4) = 12,             d)   3 · (- 4) = -12.
16.    Multiply given integers: 
Solution:            a)   (- 3 + 5) · 6 = (- 3) · 6 + 5 · 6 = - 18 + 30 = 12,
                          b)   ( 7 -  4) · (- 3) = 7 · (- 3) - 4 · (- 3) = - 21+ 12 = - 9,
                          c)   (- 5 ) · 3 + 2 · 3 =  (- 5 + 2) · 3 = (- 3) · 3 = - 9,
                          d)   9 · (- 6) - 2 · (- 6) = (9 - 2) · (- 6) = 7 · (- 6) = - 42.
17.    Solve:
Solution:          a)   (- 7 + 4) · (- 2) = (- 7) · (- 2) + 4 · (- 2) = 14 -  8 = 6       or        (- 3) · (- 2) = 6,
                         b)   - 9 · a + 5 · a = (- 9 + 5) · a = - 4a,        c)   3 · (1 - a) + 5 = 3 - 3 · a + 5 = 8 - 3a.
18.    Divide given integers: 
Solution:           a)   0 ¸ 5 = 0,                                   b)   - 3 ¸ 1 =  - 3, 
                          c)   - 28 ¸ (- 4) = 28 ¸ 4 = 7,            d)  45 ¸ (- 15) = - 45 ¸ 15 = - 3.
If both the dividend and divisor signs are the same the quotient will be positive, if they are different, the quotient will be negative.
19.    Divide given integers: 
Solution:            a)   (36 ¸ 6) ¸2 = 6 ¸ 2 = 3,      while      b)   36 ¸(6 ¸ 2)= 36 ¸ 3 = 12, 
                          c)   18 ¸ 6 ¸ 3 = 3 ¸ 3 = 1,       while      d)  18 ¸(6 ¸ 3)= 18 ¸ 2 = 9.
Order of operations - parenthesis as grouping symbols
20.    Solve:
Solution:          a)   - 8 -  3 · (- 4) + 15 ¸ (- 5) = - 8 - (- 12) + (- 3) = - 8 + 12 - 3 = 1.
When expressions have more than one operation, we have to follow rules for the order of operations.
First do all multiplication and division operations working from left to right. Next do all addition and subtraction.
                        b)   - 3 + 2 · [- 2 - 3 · (- 4 ¸ 2 + 1)] = - 3 + 2 · [- 2 - 3 · (- 2 + 1)]
                                   = - 3 + 2 · [- 2 - 3 · (- 1)] = - 3 + 2 · [- 2 + 3] = - 3 + 2 = - 1.
We often use grouping symbols, like parentheses, to help us organize complicated expressions into simpler ones. 
Do operations in parentheses and other grouping symbols first. If there are grouping symbols  within other grouping symbols do the innermost first.
The integer prime-factorization
21.    Find the prime factors of  350.
Solution:    The sequence  of prime numbers begins  2, 3, 5, 7, 11, 13, 17, 19, 23, ...
Every positive integer greater than 1 can be factored as a product of prime numbers.
Given a number n divide by the first prime from the list of primes, if it does not divide cleanly, divide n by the next prime, and so on.
Thus, the prime factors of  350 = 2 · 5 · 5 · 7.
350 | 2
175 | 5
35 | 5
7 | 7
  1 | 
The greatest common divisor (GCD)
22.    Find the greatest common divisor of  72 and 90.
Solution:    The greatest common divisor of two integers is the largest integer that divides both numbers.
The greatest common divisor can be computed by determining the prime factors of the two numbers and comparing the factors,
72 = 2 · 2 · 2 · 3 · 3,      90 = 2 · 3 · 3 · 5,    =>     GCD(72, 90) = 2 · 3 · 3 = 18.
The least common multiple (LCM)
23.     Find the least common multiple or LCM of,  24, 54 and 60.
Solution:    A common multiple is a number that is a multiple of two or more numbers. 
The least common multiple of two integers a and b is the smallest positive integer that is a multiple of both a and b. To find the least common multiple of two numbers:
 - first list the prime factors of each number, then
- multiply each factor the greatest number of times it occurs in either number.
If the same factor occurs more than once in both numbers, multiply the factor the greatest number of times it occurs.
Thus, the prime factors:    24 = 2 · 2 · 2 · 3,    54 = 2 · 3 · 3 · 3,    60 = 2 · 2 · 3 · 5,
=>       LCM(24, 54, 60) = 2 · 2 · 2 · 3 · 3 · 3 · 5 = 1080.
  Operations on real numbers
24.    Add or subtract given real numbers:
Solution:       a)   −7 − 4 = − 7 + (− 4) = − 11,                b )   + 3 − 9 = 3 + (− 9) = − 6,
                     c)     5 − (− 6) = 5 + (+ 6) = 11,                 d )    −3 − (− 9) =  −3 + 9 = 6,
                     e)     0.5 − 1.5 = 0.5 + (− 1.5) = − 1,           f )    −1.5 − 0.25 = −1.5 + (− 0.25) = −1.75.
25.    Multiply or divide given real numbers:
Solution:       a)   −7 · (− 4) = 28,            b)   3 · (− 9) = − 27,             c)   − 6 · 0.1 = − 0.6,
                     d)   6 ¸ (− 0.1) = 6 · (− 10)  = − 60,        e)   − 8 ¸ (− 0.125) = − 8 · (− 1000/125) = 64
Decimal representation of rational numbers or fractions
26.    Given fractions write as decimal numbers:
Solution:
27.    Given fractions write as decimal numbers:
Solution:
28.    Given decimal write as fraction:
Solution:
Terminating decimals
29.    Given fractions write as terminating decimals:
Solution:
The irreducible fractions, i.e., the vulgar fractions in lowest terms whose the only prime factors in a denominator are 2 and/or 5 can be converted to terminating decimals.
That is, the terminating decimals represent rational numbers whose fractions in the lowest terms are of the form a/(2n · 5m).
Recurring decimals
30.    Given fractions write as purely recurring decimals:
Solution:
The irreducible fractions, i.e., the vulgar fractions in lowest terms whose prime factors in the denominator are other than 2 or 5, that is, the prime numbers from the sequence (3, 7, 11, 13, 17, 19, ...) convert to the purely recurring decimals, i.e., the decimals which start their recurring cycle immediately after the decimal point.
Mixed recurring decimals
31.    Given fractions write as mixed recurring decimals:
Solution:
The irreducible fractions, i.e., the vulgar fractions in lowest terms whose denominator is a product of 2's and/or 5's besides the prime numbers from the sequence (3, 7, 11, 13, 17, 19, ...) convert to the mixed recurring decimals, i.e., the decimals that have some extra digits before the repeating sequence of digits.
The repeating sequence may consist of just one digit or of any finite number of digits. The number of digits in the repeating pattern is called the period. All recurring decimals are infinite decimals.
All fractions can be written either as terminating decimals or as recurring/repeating decimals.
32.    Given terminating decimals write as fractions:
Solution:
33.    Convert purely recurring decimals to fractions:
Solution:
When converting the purely recurring decimal less than one to fraction, write the group of repeating digits to the numerator, and to the denominator of the equivalent fraction write as much 9’s as is the number of digits in the repeating pattern.
34.    Convert mixed recurring decimals to fractions:
Solution:
When converting the mixed recurring decimal less than one to fraction, write the difference between the number formed by the entire sequence of digits, including the digits of the recurring part, and the number formed only by the digits of the non-recurring pattern to its numerator.
To the denominator of the equivalent fraction write as much 9’s as is the number of digits in the repeating pattern and add as much 0’s as is the number of digits in the non-recurring pattern.
35.    Given decimals convert to scientific notation:
Solution:
a)   302 567 908 = 3.02567908 · 108      b)   0.000040635 = 4.0635 · 10−5.
36.    Convert from scientific to decimal notation:
Solution:
a)   2.09085 · 107 = 20908500               b)   7.81 · 10−5 = 0.0000781 
  Order of operations
37.    Solve given expression:
Solution:
- 8 -  3 · (- 4) + 15 ¸ (- 5) = - 8 - (- 12) + (- 3) = - 8 + 12 - 3 = 1.
38.    Solve given expression:
Solution:
- 3 + 2 · [- 2 - 3 · (- 4 ¸ 2 + 1)] = - 3 + 2 · [- 2 - 3 · (- 2 + 1)] = - 3 + 2 · [- 2 - 3 · (- 1)]  
                                                                         = - 3 + 2 · [- 2 + 3] = - 3 + 2 = - 1.
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