1. Algebra Problems… Solutions
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Algebra Problems…
Solutions
Set 17 part 1
By Herbert I. Gross and Richard A. Medeiros
© Herbert I. Gross

2. The set A has 8 members and However, the union of A and B
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Problem #1
The set A has 8 members and
the set B has 5 members,
However, the union of A and B
(A U B) has only 10 members.
How is this possible?
Answer: A ∩ B has 3 members.
© Herbert I. Gross 2

3. In terms of this exercise N(A) = 8,
Answer: A ∩ B has 3 members.
Solution for #1:
Preliminary Notation
We “invent” the notation N(A) to denote the number of members in the set A.
Thus, N(B) denotes the number of members in B, N(A U B) denotes the number of members in A U B and N(A ∩ B) denotes the number of members in A ∩ B.
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In terms of this exercise N(A) = 8,
N(B) = 5 and N(A U B) = 10.
© Herbert I. Gross

4. counted once because it was
Solution for # 1:
If c is a member of both A and B, it was counted twice in computing the number of members in A U B. That is, it was
counted once because it was
a member of A and once because it was also a member of B.
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In other words, all members in A ∩ B are counted twice.
Hence, we must subtract the number of members in A ∩ B from the sum of the number of members in A and the number of members in B
© Herbert I. Gross

5. N(A U B) = N(A) + N(B) – N(A ∩ B)
Solution for #1:
In more technical terms…
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N(A U B) = N(A) + N(B) – N(A ∩ B)
That is…
10 = – N(A ∩ B)
10 = N(A ∩ B)
-3 = -N(A ∩ B)
3 = N(A ∩ B)
© Herbert I. Gross

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Notes on #1
The formula…
N(A U B) = N(A) + N(B) – N(A ∩ B) plays an important role in problems that
involve being able to count accurately
(such as in determining the probability of a particular outcome occurring).
For example, in a standard
deck of playing cards there
are 13 spades and 12 face cards.
© Herbert I. Gross 6

7. Suppose you want to know the cards and 13 spades; and 13 + 12 = 25.
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Notes on #1
Suppose you want to know the
number of ways that in a single draw from the deck you can obtain either a spade or a face card. Clearly, there are 12 face
cards and 13 spades; and = 25.
= 25
© Herbert I. Gross 7

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Notes on #1
However, in arriving at 25, 3 cards were counted twice (namely the king, queen and
jack of spades), Hence, there are only 22 ways in which you can obtain your objective.
= 22
© Herbert I. Gross 8

9. two sets. In fact, while it might be N(A U B) = N(A) + N(B) – N(A ∩ B)
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Notes on #1
This exercise illustrates how one might tend to confuse adding with finding the number of members in the union of
two sets. In fact, while it might be
tempting to write…
N(A U B) = N(A) + N(B)
We see from the formula
N(A U B) = N(A) + N(B) – N(A ∩ B)
that this will be true only if N(A ∩ B) = 0 .
© Herbert I. Gross 9

10. A Geometric Interpretation
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A Geometric Interpretation
A geometric interpretation of the formula
N(A U B) = N(A) + N(B) – N(A ∩ B)
is known as a Venn diagram.
Let the set A be represented by
and let the set B be denoted by
© Herbert I. Gross 10

11. A Geometric Interpretation
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A Geometric Interpretation
A U B is represented by the total
area enclosed by the two rectangles,
and A ∩ B is represented by the area
of the region that is common to both rectangles (and is represented by ). 5 5
N(A) = 5 + 3 3 3 3
N(B) = 2 + 3 2 2
© Herbert I. Gross 11

12. A Geometric Interpretation
next 5 2 3
In summary, we see from the diagram that there are 5 members of A that do not belong to B; there are 2 members of B that don’t belong to A; there are 3 members that belong to both A and B; and a total of 10 (that is ) members that belong to either A or B.
© Herbert I. Gross 12

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Notes on #1
There is often a tendency to think of “Either..... or......” as meaning
“One or the other, but not both”.
However, the mathematical meaning is
“At least one”. Thus, with respect to
our earlier example of a spade or a face card”, if you were to pick the jack of spades you would still have won even though you picked both a spade and a face card.
© Herbert I. Gross 13

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Notes on #1
Venn diagrams work nicely for two or three sets. However, they do not work if there are more than three sets.
An alternative method involves using a table, in much the same way as we did when we wanted to record the outcomes of coins being flipped (Lesson 8).
© Herbert I. Gross 14

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Notes on #1
For example, we may use 1 to indicate that a member belongs to a set and 0 to indicate that it doesn’t. So, for example, with respect to two sets (which we will denote by A and B) the table might look like… A B
A U B
A ∩ B 1 1 1
© Herbert I. Gross 15

16. This is where Boolean algebra is used.
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Notes on #1
However, just as in the case with flipping coins, as the number of sets increases the number of rows in our chart become rather unmanageable.
This is where Boolean algebra is used.
That is just as there are “rules of the game” in traditional algebra, there are rules in the “game” of Boolean algebra. Some of the rules are the same as they are in traditional algebra, but some are different as well.
© Herbert I. Gross 16

17. The study of Boolean Algebra is very valuable but is beyond the
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Notes on #1
For example, the union of sets is commutative. That is, A U B = B U A.
However, whereas a + a = 2a, A U A = A.
An in depth study of unions and intersections is included in the course known as Boolean Algebra.
The study of Boolean Algebra is very valuable but is beyond the
scope of our course.
© Herbert I. Gross 17

18. Answer: No Problem #2a Let A = {(x,y,z): x + y + z = 9}.
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Problem #2a
Let A = {(x,y,z): x + y + z = 9}.
Is (1,2,3) a member of A?
Answer: No
© Herbert I. Gross 18

19. pass the test for membership in the set A.
Answer: No
Solution for 2a:
To belong to A, (x,y,z) must possess the property that…
x + y + z = 9
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If we replace x by 1, y by 2 and z by 3 in the equation x + y + z = 9, we obtain the false statement…
= 9
The fact that = 9 is a false statement means that (x,y,z} doesn’t
pass the test for membership in the set A.
© Herbert I. Gross

20. However, we are given the test for membership to belong to A.
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Notes on #2a
As defined above, the set A is described implicitly. That is, we are not told specifically what the members of A are.
However, we are given the test for membership to belong to A.
This is the value of the set-builder notation. Namely it gives an objective criterion by which to determine whether a member
belongs to it.
© Herbert I. Gross 20

21. There are a number of ways to rewrite the equation x + y + z = 9.
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Notes on #2a
There are a number of ways to rewrite the equation x + y + z = 9.
For example, the equation x + y + z = 9 is equivalent to the equation z = 9 – x – y.
From the equation z = 9 – x – y, we see that if we choose the values of x and
y in a completely arbitrary manner, the value of z is uniquely determined.
© Herbert I. Gross 21

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Notes on #2a
For example, if we let x = 5 and y = 3, then equation z = 9 – x – y tells us that…
z = 9 – 5 – 3 = 1
Hence, (5,3,1) is a member of A.
© Herbert I. Gross 22

23. in the equivalent form…
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Notes on #2a
In a similar way, we might have chosen to rewrite the equation x + y + z = 9
in the equivalent form…
y = 9 – z – x
In this case, we could choose x and z at random, and then the value of y would be uniquely determined by the equation
y = 9 – z – x.
© Herbert I. Gross 23

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Notes on #2a
For example, if we let z = 2 and x = 4, then equation y = 9 – z – x tells us that…
y = 9 – 2 – 4 = 3
Hence, (4,3,2) is a member of A.
© Herbert I. Gross 24

25. Answer: Yes Problem #2b Let A = {(x,y,z): x + y + z = 9}.
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Problem #2b
Let A = {(x,y,z): x + y + z = 9}.
Is (1,2,6) a member of A?
Answer: Yes
© Herbert I. Gross 25

26. To belong to A, (x,y,z) must possess the property that… x + y + z = 9
Answer: Yes
Solution for #2b:
To belong to A, (x,y,z) must possess the property that…
x + y + z = 9
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If we replace x by 1, y by 2 and z by 6 in the equation above we obtain the
true statement…
= 9
The fact that = 6 is a true statement means that (1,2,3} satisfies the test for membership in the set A
© Herbert I. Gross

27. x + y + z = 9 tells us that z must equal 6.
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Notes on #2b
Notice that in both parts of this exercise the value of x was 1 and the value of y was 2. However (1,2,3) didn’t belong to set A, but (1,2,6) does. The reason for this is that once we know that x = 1 and y = 2, the equation
x + y + z = 9 tells us that z must equal 6.
z = 9
= 9
© Herbert I. Gross

28. set A tells us that since (1,2,6) is a member
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Notes on #2b
Because addition is associative and commutative, the test for membership in
set A tells us that since (1,2,6) is a member
of set A then so also are (1,6,2), (2,1,6), (2,6,1), (6,1,2) and (6,2,1). That is…
= 9
= 9
= 9
= 9
= 9
= 9
© Herbert I. Gross

29. How are sets A and B related if
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Problem #3
How are sets A and B related if
A = {1,2,3}
and
B = {x:(x – 1)(x – 2)(x – 3) = 0}?
Answer: They are equal.
© Herbert I. Gross 29

30. B are, but we are told the test for membership in B.
Answer: They are equal.
Solution for #3:
The set A is described by the roster method. That is, its members are explicitly listed. That is, for a number to belong to A,
it must be either 1, 2 or 3.
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On the other hand the set builder notation is used to describe set B. That is, we are not told explicitly what the members of
B are, but we are told the test for membership in B.
© Herbert I. Gross

31. Namely for a number x to belong to B it must satisfy the equation…
Solution for #3:
Namely for a number x to belong to B it must satisfy the equation…
(x – 1)(x – 2)(x – 3) = 0
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Notice that the equation…
(x – 1)(x – 2)(x – 3) = 0
tells us that every member of A is
also a member of B.
© Herbert I. Gross

32. For example, since the product of any number and 0 is 0, we see that…
Solution for #3:
For example, since the product of any number and 0 is 0, we see that…
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If x = 1, x – 1 = 0; and if x – 1 = 0,
(x – 1)(x – 2)(x – 3) = 0
If x = 2, x – 2 = 0; and if x – 2 = 0,
(x – 1)(x – 2)(x – 3) = 0
If x = 3, x – 3 = 0; and if x – 3 = 0,
(x – 1)(x – 2)(x – 3) = 0
© Herbert I. Gross

33. Thus, with respect to the equation (x – 1)(x – 2)(x – 3) = 0,
Solution for #3:
On the other hand, the only way a product can equal 0 is if at least one of its factors is equal to 0.
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Thus, with respect to the equation
(x – 1)(x – 2)(x – 3) = 0,
the only way the product can equal 0 is
if either (x – 1) = 0, (x – 2) = 0 or (x – 3) = 0; that is only if x = 1, x = 2, or x = 3.
Hence, in the roster method format
B = {1,2,3}. Thus the sets A and B are, in effect, two different names for same set.
© Herbert I. Gross

34. an equation from the set builder notation to the roster notation.
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Notes on #3
In terms of implicit versus explicit, we may say that part of the mission of algebra is to help us convert the solution set of
an equation from the set builder notation to the roster notation.
For example, in terms of this exercise, if we define the set B by {x:(x – 1)(x – 2)(x – 3) = 0} then algebra is the process of showing that B = {1,2,3}.
© Herbert I. Gross

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Notes on #3
In general we’re more comfortable seeing the solution set in its roster format. However the set-builder notation tells us the test for membership.
For example, in a previous problem, we
were looking at the set…
A = {(x,y,z): x + y + z = 9}.
In this case, it is impossible to list all the members of A because there are infinitely many such members
(one for each choice of x and y).
© Herbert I. Gross

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Notes on #3
However, as we saw in the solution of the exercise, we could use the test for membership to determine whether a
given “triplet” of numbers (x,y,z) belonged to the set A.
As another example, it is known that there are infinitely many prime numbers
(recall that a prime number is any whole
number greater than 1 that is divisible only by 1 and itself).
© Herbert I. Gross

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Notes on #3
However, there is no known formula for listing them all. But, given any whole number greater than 1 we can test
to see if it has any divisors other than 1 and itself.
For example, given the number 1,234,576,926 we see at once that it
is even and hence divisible by 2.
Therefore, it doesn’t belong to the set of prime numbers.
© Herbert I. Gross

38. roster representation. For example, when we see {1,2,3} we
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Final Note on #3
As a final note, observe that in a very important way the set builder notation tells us things we might not observe from the
roster representation.
For example, when we see {1,2,3} we
do not know the sense in which these three numbers were chosen for membership. For example, we might have chosen them because they were the first three positive integers. However, when we see
{x:(x – 1)(x – 2)(x – 3) = 0} we know immediately what the test for membership was.
© Herbert I. Gross

39. Answer: It is the set of all multiples of 6.
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Problem #4
Let A be the set of positive integers that are multiples of 3 and let B be the set of positive integers that are multiples of 2.
Describe the set A ∩ B.
Answer: It is the set of all multiples of 6.
© Herbert I. Gross 39

40. Answer: It is the set of all multiples of 6. Solution for #4:
Using the roster method we see that
A = {3,6,9,12,15,..., 3n,...}.
Because the set A has infinitely many members, we might prefer to write it in the set builder notation; namely…
A = {3n:n is a positive integer}
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The key point is that to be member of A,
a positive integer has to be divisible by 3.
© Herbert I. Gross

41. In a similar way, we may represent B in the form…
Solution for #4:
In a similar way, we may represent B in the form…
B = {2n:n is a positive integer}
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The key point is that to be member of B,
a positive integer has to be divisible by 2.
© Herbert I. Gross

42. Thus, to belong to both A and B (that is, to A ∩ B) the positive
Solution for #4:
Thus, to belong to both A and B
(that is, to A ∩ B) the positive
integer has to be divisible by both 2 and 3. Any number that is divisible by both 2 and 3 is also divisible by 6; and conversely,
any number divisible by 6 is also divisible by both 2 and 3.
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Hence, A ∩ B = {6n:n is a positive integer}; or in roster format…
A ∩ B = {6,12,18,..., 6n, ...}
© Herbert I. Gross

43. For example, 4 × 6 is divisible
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Notes on #4
In our solution to this exercise, we used the fact that 6 was the least common multiple of 2 and 3. It is always true that the product of two numbers is divisible by each of the two numbers.
For example, 4 × 6 is divisible
by both 4 and 6.
© Herbert I. Gross

44. The reason for this is that 4 and 6 share the factor 2 in common.
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Notes on #4
However, in the case of 4 and 6, 4 × 6 is not the least common multiple of 4 and 6. Rather 12 is the least common multiple of 4 and 6.
The reason for this is that 4 and 6 share the factor 2 in common.
That is 4 = 2 × 2 and 6 = 2 × 3.
© Herbert I. Gross

45. So in terms of a diagram…
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Notes on #4
So in terms of a diagram… 4 2 2
4 = 2 × 2 2 2 2
6 = 3 × 2 3 3 6
The diagram shows that to be divisible by both 4 and 6 it has to be a multiple of
2 × 2 × 3
© Herbert I. Gross

46. Answer: A ∩ B = {(3,19)} Problem #5 Let A = {(x,y):y = 4x + 7} and
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Problem #5
Let A = {(x,y):y = 4x + 7} and
let B = {(x,y):y = 2x + 13}.
Describe the set A ∩ B.
Answer: A ∩ B = {(3,19)}
© Herbert I. Gross 46

47. In set builder notation we know that…
Answer: A ∩ B = {(3,19)}
Solution for #5:
In set builder notation we know that…
A ∩ B = {(x,y):y = 4x + 7and y = 2x + 13}.
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The only way in which y can be equal to both 4x + 7 and 2x + 13 is if…
4x + 7 = 2x + 13
© Herbert I. Gross

48. To solve the equation 4x + 7 = 2x + 13 for
Solution for #5:
To solve the equation 4x + 7 = 2x + 13 for
x, we may first subtract 2x from both sides of the equation to obtain…
2x + 7 = 13
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We then subtract 7 from both sides of
the equation 2x + 7 = 13 to obtain…
2x = 6
…and from the equation 2x = 6, it follows that x = 3.
That is, if x = 3 then 4x + 7 = 2x + 13 = 19.
© Herbert I. Gross

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Notes on #5
Even if we didn’t know how to use algebra to convert from the set builder notation to the roster method, we could still use the test for membership.
For example, suppose we wanted to see whether (5,27) is a member of A∩B.
© Herbert I. Gross

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Notes on #5
If we replace x by 5 and y by 27 in the equation y = 4x + 7, we obtain the true statement…
27 = 4(5) + 7.
However, if we replace x by 5 and y by 27 in the equation y = 2x + 13, we obtain the false statement…
27 = 2(5) + 13
© Herbert I. Gross

51. to B. Hence, (5, 27) is not a member of the set A ∩ B.
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Notes on #5
In other words (5,27) belongs to A but not
to B. Hence, (5, 27) is not a member of the set A ∩ B.
The beauty of the algebraic solution is that it eliminates the need to test other pairs of numbers. Namely, the algebraic solution told us that (x,y) was a member of the set A ∩ B if and only if x = 3 and y = 19.
© Herbert I. Gross