Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 17 part 1...

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Algebra Problems…Solutions

Set 17 part 1By Herbert I. Gross and Richard A. Medeiros

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.

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.

In terms of this exercise N(A) = 8, N(B) = 5 and N(A U B) = 10.

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.

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

Solution for #1:In more technical terms…

N(A U B) = N(A) + N(B) – N(A ∩ B)

That is…

10 = 8 + 5 – N(A ∩ B)

10 = 13 + -N(A ∩ B)

-3 = -N(A ∩ B)

3 = N(A ∩ B)

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).

For example, in a standard deck of playing cards there are 13 spades and 12 face

cards.

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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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.

Notes on #1

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)

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.

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 .

A Geometric Interpretation

A U B is represented by the totalarea 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 ).

5N(A) = 5 + 3

N(B) = 2 + 3

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next A Geometric Interpretation

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.

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.

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).

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 ∩ B0 0 0 00 1 1 01 0 1 01 1 1 1

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.

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.

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.

Problem #2a

Let A = {(x,y,z): x + y + z = 9}.

Is (1,2,3) a member of A?

Answer: No

Answer: NoSolution for 2a:To belong to A, (x,y,z) must possess the property that…

x + y + z = 9

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.

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.

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.

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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…

Hence, (5,3,1) is a member of A.

z = 9 – 5 – 3 = 1

Notes on #2a

In a similar way, we might have chosen to rewrite the equation x + y + z = 9

in the equivalent form…

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.

y = 9 – z – x

Notes on #2a

For example, if we let z = 2 and x = 4, then equation y = 9 – z – x tells us that…

Hence, (4,3,2) is a member of A.

y = 9 – 2 – 4 = 3

Problem #2b

Let A = {(x,y,z): x + y + z = 9}.

Is (1,2,6) a member of A?

Answer: Yes

Answer: YesSolution for #2b:To belong to A, (x,y,z) must possess the property that…

x + y + z = 9

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

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.

1 + 2 + z = 9

1 + 2 + 6 = 9

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 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…

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

Problem #3

How are sets A and B related if A = {1,2,3}

andB = {x:(x – 1)(x – 2)(x – 3) = 0}?

Answer: They are equal.

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.

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.

Solution for #3:

Namely for a number x to belong to B it must satisfy the equation…

(x – 1)(x – 2)(x – 3) = 0

Notice that the equation… (x – 1)(x – 2)(x – 3) = 0

tells us that every member of A is also a member of B.

Solution for #3:

For example, since the product of any number and 0 is 0, we see that…

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

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.

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.

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.

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}.

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.

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).

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.

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).

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.

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.

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.

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.

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.

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}

The key point is that to be member of A, a positive integer has to be divisible by 3.

Solution for #4:In a similar way, we may represent B in the form…

B = {2n:n is a positive integer}

The key point is that to be member of B, a positive integer has to be divisible by 2.

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.

Hence, A ∩ B = {6n:n is a positive integer}; or in roster format…

A ∩ B = {6,12,18,..., 6n, ...}

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.

Notes on #4

For example, 4 × 6 is divisible by both 4 and 6.

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.

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.

Notes on #4

So in terms of a diagram…

2 4 = 2 × 2

6 = 3 × 2

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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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)}

Solution for #5:In set builder notation we know that…

A ∩ B = {(x,y):y = 4x + 7and y = 2x + 13}.

The only way in which y can be equal to both 4x + 7 and 2x + 13 is if…

4x + 7 = 2x + 13

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

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.

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.

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

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.

Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 17 part 1...

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