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Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 17 part 1 By Herbert I. Gross and Richard A. Medeiros next

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Problem #1 © 2007 Herbert I. Gross next 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. next

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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. next © 2007 Herbert I. Gross In terms of this exercise N(A) = 8, N(B) = 5 and N(A U B) = 10. next

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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. next © 2007 Herbert I. Gross next 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

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Solution for #1: In more technical terms… next © 2007 Herbert I. Gross next N(A U B) = N(A) + N(B) – N(A B) next That is… 10 = – N(A B) 10 = N(A B) - 3 = - N(A B) 3 = N(A B)

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next 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). © 2007 Herbert I. Gross For example, in a standard deck of playing cards there are 13 spades and 12 face cards. next

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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. next © 2007 Herbert I. Gross = 25

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next 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. next = 22 next

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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 of two sets. In fact, while it might be tempting to write… N(A U B) = N(A) + N(B) next 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. next 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 total area enclosed by the two rectangles, next and A B is represented by the area of the region that is common to both rectangles (and is represented by ). 2 5 N(A) = N(B) =

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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 dont 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

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next 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 next 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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next 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 next 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 doesnt. 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 next ABA U BA B

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next 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 next 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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next 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 the scope of our course. © 2007 Herbert I. Gross next 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 next Let A = {(x,y,z): x + y + z = 9}. Is (1,2,3) a member of A? next Answer: No

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Solution for 2a: To belong to A, (x,y,z) must possess the property that… x + y + z = 9 next © 2007 Herbert I. Gross = 9 next 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 = 9 is a false statement means that (x,y,z} doesnt pass the test for membership in the set A.

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next 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 next 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. next

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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. next z = 9 – 5 – 3 = 1

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next 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. next y = 9 – z – x next

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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. next y = 9 – 2 – 4 = 3

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Problem #2b © 2007 Herbert I. Gross next Let A = {(x,y,z): x + y + z = 9}. Is (1,2,6) a member of A? next Answer: Yes

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Solution for #2b: To belong to A, (x,y,z) must possess the property that… x + y + z = 9 next © 2007 Herbert I. Gross next 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 next

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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) didnt 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. next © 2007 Herbert I. Gross next z = = 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 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… next © 2007 Herbert I. Gross next = 9 Notes on #2b = = = = = 9

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Problem #3 © 2007 Herbert I. Gross next How are sets A and B related if A = {1,2,3} and B = {x:(x – 1)(x – 2)(x – 3) = 0}? next Answer: They are equal.

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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. next © 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. next

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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 next © 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. next

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Solution for #3: For example, since the product of any number and 0 is 0, we see that… next © 2007 Herbert I. Gross If x = 1, x – 1 = 0; and if x – 1 = 0, (x – 1)(x – 2)(x – 3) = 0 next 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. next © 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. next 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 of an equation from the set builder notation to the roster notation. © 2007 Herbert I. Gross next 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 were more comfortable seeing the solution set in its roster format. However the set-builder notation tells us the test for membership. next © 2007 Herbert I. Gross next Notes on #3 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).

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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. next © 2007 Herbert I. Gross next Notes on #3 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).

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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. next © 2007 Herbert I. Gross next Notes on #3 For example, given the number 1,234,576,926 we see at once that it is even and hence divisible by 2. Therefore, it doesnt 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 the roster representation. next © 2007 Herbert I. Gross next Final Note on #3 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.

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Problem #4 © 2007 Herbert I. Gross next 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. next Answer: It is the set of all multiples of 6.

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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} next © 2007 Herbert I. Gross The key point is that to be member of A, a positive integer has to be divisible by 3. next

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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} next © 2007 Herbert I. Gross The key point is that to be member of B, a positive integer has to be divisible by 2. next

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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. next © 2007 Herbert I. Gross Hence, A B = {6n:n is a positive integer}; or in roster format… A B = {6,12,18,..., 6n,...} next

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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. © 2007 Herbert I. Gross next 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. next © 2007 Herbert I. Gross next 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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next © 2007 Herbert I. Gross next Notes on #4 So in terms of a diagram… = 2 × 2 6 = 3 × The diagram shows that to be divisible by both 4 and 6 it has to be a multiple of 2 × 2 × 3 next 4 6

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Problem #5 © 2007 Herbert I. Gross next Let A = {(x,y):y = 4x + 7} and let B = {(x,y):y = 2x + 13}. Describe the set A B. next Answer: A B = {(3,19)}

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Solution for #5: In set builder notation we know that… A B = {(x,y):y = 4x + 7and y = 2x + 13}. next © 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 next

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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 next © 2007 Herbert I. Gross next 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. next

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© 2007 Herbert I. Gross next For example, suppose we wanted to see whether (5,27) is a member of AB. Notes on #5 Even if we didnt 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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next © 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 next

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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. next

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