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Algebra Problems…Solutions
Algebra Problems…Solutions
© 2007 Herbert I. Gross
Set 17 part 1By Herbert I. Gross and Richard A. Medeiros
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Problem #1
© 2007 Herbert I. Gross
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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.
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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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© 2007 Herbert I. Gross
In terms of this exercise N(A) = 8, N(B) = 5 and N(A U B) = 10.
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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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© 2007 Herbert I. Gross
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In other words, all members in A ∩ B are counted twice.
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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
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Solution for #1:In more technical terms…
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© 2007 Herbert I. Gross
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N(A U B) = N(A) + N(B) – N(A ∩ B)
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That is…
10 = 8 + 5 – N(A ∩ B)
10 = 13 + -N(A ∩ B)
-3 = -N(A ∩ B)
3 = N(A ∩ B)
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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 thatinvolve being able to count accurately
(such as in determining the probability of a particular outcome occurring).
© 2007 Herbert I. Gross
For example, in a standard deck of playing cards there are 13 spades and 12 face
cards.
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Notes on #1
Suppose you want to know thenumber of ways that in a single draw from the deck you can obtain either a spade or a
face card. Clearly, there are 12 facecards and 13 spades; and 13 + 12 = 25.
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© 2007 Herbert I. Gross
= 25
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Notes on #1
© 2007 Herbert I. Gross
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.
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= 22
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Notes on #1
© 2007 Herbert I. Gross
This exercise illustrates how one might tend to confuse adding with finding the
number of members in the union oftwo sets. In fact, while it might be
tempting to write…
N(A U B) = N(A) + N(B)
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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.
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A Geometric Interpretation
© 2007 Herbert I. Gross
A geometric interpretation of the formula N(A U B) = N(A) + N(B) – N(A ∩ B)
is known as a Venn diagram.
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Let the set A be represented by
and let the set B be denoted by .
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A Geometric Interpretation
© 2007 Herbert I. Gross
A U B is represented by the totalarea enclosed by the two rectangles,
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and A ∩ B is represented by the area of the region that is common to both
rectangles (and is represented by ).
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2
5N(A) = 5 + 3
N(B) = 2 + 3
5
2
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333
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next A Geometric Interpretation
© 2007 Herbert I. Gross
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 5 + 3 + 2) members that belong to
either A or B.
5
23
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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”.
© 2007 Herbert I. Gross
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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.
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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.
© 2007 Herbert I. Gross
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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).
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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…
© 2007 Herbert I. Gross
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A B A U B A ∩ B0 0 0 00 1 1 01 0 1 01 1 1 1
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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.
© 2007 Herbert I. Gross
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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.
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Notes on #1
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 thescope of our course.
© 2007 Herbert I. Gross
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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.
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Problem #2a
© 2007 Herbert I. Gross
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Let A = {(x,y,z): x + y + z = 9}.
Is (1,2,3) a member of A?
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Answer: No
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Answer: NoSolution for 2a:To belong to A, (x,y,z) must possess the property that…
x + y + z = 9
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© 2007 Herbert I. Gross
1 + 2 + 3 = 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…
The fact that 1 + 2 + 3 = 9 is a false statement means that (x,y,z} doesn’t
pass the test for membership in the set 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.
© 2007 Herbert I. Gross
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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.
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Notes on #2a
There are a number of ways to rewrite the equation x + y + z = 9.
© 2007 Herbert I. Gross
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.
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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…
© 2007 Herbert I. Gross
Hence, (5,3,1) is a member of A.
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z = 9 – 5 – 3 = 1
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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…
© 2007 Herbert I. Gross
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.
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y = 9 – z – x
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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…
© 2007 Herbert I. Gross
Hence, (4,3,2) is a member of A.
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y = 9 – 2 – 4 = 3
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Problem #2b
© 2007 Herbert I. Gross
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Let A = {(x,y,z): x + y + z = 9}.
Is (1,2,6) a member of A?
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Answer: Yes
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Answer: YesSolution for #2b:To belong to A, (x,y,z) must possess the property that…
x + y + z = 9
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© 2007 Herbert I. Gross
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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…1 + 2 + 6 = 9
The fact that 1 + 2 + 3 = 6 is a true statement means that (1,2,3} satisfies the
test for membership in the set A
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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.
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© 2007 Herbert I. Gross
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1 + 2 + z = 9
1 + 2 + 6 = 9
Notes on #2b
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Because addition is associative and commutative, the test for membership in
set A tells us that since (1,2,6) is a memberof set A then so also are (1,6,2), (2,1,6),
(2,6,1), (6,1,2) and (6,2,1). That is…
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© 2007 Herbert I. Gross
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1 + 2 + 6 = 9
Notes on #2b
1 + 6 + 2 = 9
2 + 1 + 6 = 9
2 + 6 + 1 = 9
6 + 1 + 2 = 9
6 + 2 + 1 = 9
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Problem #3
© 2007 Herbert I. Gross
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How are sets A and B related if A = {1,2,3}
andB = {x:(x – 1)(x – 2)(x – 3) = 0}?
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Answer: They are equal.
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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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© 2007 Herbert I. Gross
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.
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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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© 2007 Herbert I. Gross
Notice that the equation… (x – 1)(x – 2)(x – 3) = 0
tells us that every member of A is also a member of B.
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Solution for #3:
For example, since the product of any number and 0 is 0, we see that…
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© 2007 Herbert I. Gross
If x = 1, x – 1 = 0; and if x – 1 = 0, (x – 1)(x – 2)(x – 3) = 0
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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
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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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© 2007 Herbert I. Gross
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.
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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.
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In terms of implicit versus explicit, we may say that part of the mission of algebra is to
help us convert the solution set ofan equation from the set builder notation
to the roster notation.
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© 2007 Herbert I. Gross
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Notes on #3
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}.
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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.
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© 2007 Herbert I. Gross
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Notes on #3
For example, in a previous problem, wewere 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).
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However, as we saw in the solution of the exercise, we could use the test for
membership to determine whether agiven “triplet” of numbers (x,y,z) belonged
to the set A.
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© 2007 Herbert I. Gross
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Notes on #3
As another example, it is known that there are infinitely many prime numbers
(recall that a prime number is any wholenumber greater than 1 that is divisible only
by 1 and itself).
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However, there is no known formula for listing them all. But, given any whole
number greater than 1 we can testto see if it has any divisors other than 1
and itself.
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© 2007 Herbert I. Gross
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Notes on #3
For example, given the number 1,234,576,926 we see at once that itis even and hence divisible by 2.
Therefore, it doesn’t belong to the set of prime numbers.
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As a final note, observe that in a very important way the set builder notation tells us
things we might not observe from theroster representation.
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© 2007 Herbert I. Gross
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Final Note on #3
For example, when we see {1,2,3} wedo 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.
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Problem #4
© 2007 Herbert I. Gross
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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.
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Answer: It is the set of all multiples of 6.
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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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© 2007 Herbert I. Gross
The key point is that to be member of A, a positive integer has to be divisible by 3.
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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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© 2007 Herbert I. Gross
The key point is that to be member of B, a positive integer has to be divisible by 2.
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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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© 2007 Herbert I. Gross
Hence, A ∩ B = {6n:n is a positive integer}; or in roster format…
A ∩ B = {6,12,18,..., 6n, ...}
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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.
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© 2007 Herbert I. Gross
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Notes on #4
For example, 4 × 6 is divisible by both 4 and 6.
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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.
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© 2007 Herbert I. Gross
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Notes on #4
The reason for this is that 4 and 6 share the factor 2 in common.
That is 4 = 2 × 2 and 6 = 2 × 3.
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© 2007 Herbert I. Gross
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Notes on #4
So in terms of a diagram…
3
2 4 = 2 × 2
6 = 3 × 2
2
3222
The diagram shows that to be divisible by both 4 and 6 it has to be a multiple of
2 × 2 × 3
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4
6
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Problem #5
© 2007 Herbert I. Gross
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Let A = {(x,y):y = 4x + 7} and let B = {(x,y):y = 2x + 13}.
Describe the set A ∩ B.
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Answer: A ∩ B = {(3,19)}
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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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© 2007 Herbert I. Gross
The only way in which y can be equal to both 4x + 7 and 2x + 13 is if…
4x + 7 = 2x + 13
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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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© 2007 Herbert I. Gross
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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.
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© 2007 Herbert I. Gross
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For example, suppose we wanted to see whether (5,27) is a member of A∩B.
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.
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© 2007 Herbert I. Gross
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
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© 2007 Herbert I. Gross
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.
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.
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