course Mth 151 sorry this is late!! ٚ{ޟyaassignment #003
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21:55:32 `q001. Note that there are 5 questions in this assignment. Again we have a total of 35 people in a room. Of these, 20 have dark hair and 15 have bright eyes. There are 8 people with dark hair and bright eyes. Let A stand for the collection of people who have dark hair and B for the collection who have bright eyes. The Intersection of these two collections is denoted A ^ B, and stands for the collection of all people who have both dark hair and bright eyes. The Union of these two collections is denoted A U B, and stands for the collection of all people who have at least one of these characteristics. In terms of the diagram you made for the preceding problem, describe the collection A ^ B and the collection A U B. Give the number of people in each of these collections (these numbers are designated by the notation n ( A ^ B) and n(A U B) ). Refer to the diagrams you have made.
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RESPONSE --> 1. A ^ B : Have both characteristics, 8 people 2. A U B : Have all the people who have at least one characteristic. 35 people 3. n( A ^ B ) = 8 4. n( A U B ) = 35 confidence assessment: 2
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22:09:06 `q002. Continuing the preceding example, we let A' stand for the people who are not in the collection A, and we let B' stand for the people who are not in the collection B. What are the characteristics of the people in A', and what characterizes people in B' ? What are n(A ') and n(B '), the numbers of people in A' and B' ?
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RESPONSE --> there are 12 in A so there is 15 in n(A') there is 7 in B so there is 20 in n(B') confidence assessment: 1
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22:18:29 `q003. ( A U B ) ' stands for the everyone outside A U B, and ( A ^ B ) ' stands for everyone outside A ^ B. What characterizes the people in each of these collections, and how many people are there in each?
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RESPONSE --> there are 0 people in ( A U B)' there are 8 peolpe in ( A ^ B )' confidence assessment: 1
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22:34:57 `q004. How many people are in A ' U B ', and how could those people be characterized? Answer the same for A ' ^ B '.
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RESPONSE --> A' U B' = 8 A' ^ B' = 35 confidence assessment: 1
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23:07:49 `q005. Succinctly describe the relationships between ( A U B ) ', A ' U B ', (A ^ B) ' and A ' ^ B '.
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RESPONSE --> (A U B)' stands for every thing out side of A U B, the center between A and B. 8 is the ans. A' U B' consist of everyone in at least on A' or B'. the 7 peolpe in B, but not out side the overlap, and 8 people outside of A and B. (A ^ B)' stands for everything out side of A^B 12+7+8=27 A' ^ B' stands for every thing out side of A and B confidence assessment: 1
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You then start the process over with the next question, clicking on the Next Question/Solution button at top left, etc. &#
Zq矔ҸeyϨ{ assignment #004 004. Subsets; One-to-One Correspondences. Liberal Arts Mathematics I 10-01-2007 ܼ⯮M~ assignment #008 008. Arithmetic Sequences Liberal Arts Mathematics I 10-01-2007 sN¬SH`ힸ assignment #004 004. Subsets; One-to-One Correspondences. Liberal Arts Mathematics I 10-09-2007......!!!!!!!!...................................
23:20:07 We will list the original collection by placing its elements between braces: { a, b, c, d, e }. The collection {a, b, c, d} is a smaller collection obtained by eliminating e from the original collection. Similarly we can eliminate d or c or b or a to get the 4-element collections {a, b, c, e}, {a, b, d, e}, { a, c, d, e} and {b, c, d, e}. Alternatively we could simply include either a or b or c or d or e in a 1-element collection, obtaining {a}, {b}, {c}, {d} and {e}. It should be clear that these are the only ways to form collections of 1 or 4 elements. To form a collection of 2 elements we could include a and one other element, obtaining { a, b}, { a, c }, { a, d } and { a, e }. Or we could include b and one other element (excluding a, since we already have the collection { a, b } which is identical to the collection { b, a } since it has exactly the same elements). We obtain { b, c }, { b, d } and { b, e }. {}Or we could include c and one other element (other than a or b, since these have already been listed) to obtain { c, d } and { c, e }. Finally we could include d and the only other element left, e, to get { d, e}. This gives us a complete listing of the 10 sets we can form with 2 of the original elements. This leaves us the 3-element sets, which can be formed by excluding the 2-element sets. Working in reverse order, we can exclude { d, e } to get { a, b, c }, or { c, e } to get { a, b, d }, etc.. The remaining sets we get in this fashion are { a, b, e}, { a, c, d }, { a, c, e}, { a, d, e}, { b, c, d}, {b, c, e}, {b, d, e}, {c, d, e}. We thus have 10 three-element sets. The total number of smaller sets containing at least one element is therefore 5 + 5 + 10 + 10 = 30.
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RESPONSE --> 4 elements b,c,d,e a,c,d,e a,b,d,e a,b,c,e 4ways 3 element a,b,e, a,c,d a,c,e a,d,e b,c,e b,c,d b,d,e c,d,e 8 ways 2 elements a,b a,c a,d a,e 10 ways b,c b,d b,e c,d e,c e,d 1 element a b c d e 5ways self critique assessment: 2
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23:26:59 `q003. There are four possible one-to-one correspondences between the sets {a, b, c} and {1, 2, 3} which were not described in the preceding exercise. One of them would be [ a <--> 3, b <--> 2, c <--> 1 ]. What are the other three possible one-to-one correspondences?
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RESPONSE --> it fail b/c 3 is used with both a, c. and it fail b/c 1 is not included confidence assessment: 2
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23:35:44 If we designate the correspondence [ a <--> 1, b <--> 2, c <--> 3 ] as the '123' correspondence, [a <--> 2, b <--> 1, c <--> 3 ] as the '213' correspondence and [a <--> 3, b <--> 2, c <--> 1 ] as the '321' correspondence, in each case listing the numbers associated with a, b, c in that order, we see that the remaining three correspondences could be designated 132, 231 and 312. These correspondences could of course be written out as [ a <--> 1, b <--> 3, c <--> 2 ], [ a <--> 2, b <--> 3, c <--> 1 ] and [ a <--> 3, b <--> 1, c <--> 2 ]. Note that 123, 132, 213, 231, 312, 321 represent the six ways of rearranging the digits 1, 2, 3 into a 3-digit number, listed in increasing order.
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RESPONSE --> 123 [ a <-> 1, b<-> 2, c <-> 3] 132 [ a <-> 1, b <->3, c <-> 2] 213 [ a <-> 2, b <-> 1, c <-> 3] 231 [ a <-> 2, b <-> 3, c<-> 1] 312 [ a <-> 3, b <->1, c <->2] 321 [ a <-> 3, b<-> 2, c<-> 1] self critique assessment: 2
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23:37:23 `q004. Explain why it is not possible to put the sets { a, b, c} and {1, 2, 3, 4} into a one-to-one correspondence.
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RESPONSE --> It not poss. b/c ther is only 3 element in the 1st set (a,b,c) and there is four in the 2nd set ( 1,2,3,4) confidence assessment: 3
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̋|HQxr assignment #005 005. Infinite Sets Liberal Arts Mathematics I 10-09-2007
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23:42:53 `q001. Note that there are 8 questions in this assignment. The set { 1, 2, 3, ... } consists of the numbers 1, 2, 3, etc.. The etc. has no end. This set consists of the familiar counting numbers, which most of us have long known to be unending. This is one example of an infinite set. Another is the set of even positive numbers { 2, 4, 6, ... }. This set is also infinite. There is an obvious one-to-one correspondence between these sets. This correspondence could be written as [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ], where the ... indicates as before that the pattern should be clear and that it continues forever. Give a one-to-one correspondence between the sets { 1, 2, 3, ... } and the set { 1, 3, 5, ... } of odd numbers.
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RESPONSE --> [ 1<-> 1, 2<->3, 3 <-> 5,.......] confidence assessment: 2
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23:45:38 `q002. Writing [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ] for the correspondence between { 1, 2, 3, ... } and { 1, 3, 5, ... } isn't bad, but the pattern here might be a bit less clear to the reader than the correspondence [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ] given for { 1, 2, 3, ... } and { 2, 4, 6, ... }. That is because in the latter case it is clear that we are simply doubling the numbers in the first set to get the numbers in the second. It might not be quite as clear exactly what the rule is in the correspondence [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ], except that we know we are pairing the numbers in the two sets in order. Without explicitly stating the rule in a form as clear as the doubling rule, we can't be quite as sure that our rule really works. How might we state the rule for the correspondence [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ] as clearly as the 'double-the-first-number' rule for [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ]?
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RESPONSE --> yes we can, the rule does work b/c your follow the numbers in the set. confidence assessment: 2
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23:53:07 The rule for this correspondence is 'double and subtract 1', so n would be associated with 2n - 1. The correspondence would thus be [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... , n <--> 2n-1, ... ]. Note how this gives a definite formula for the rule, removing all ambiguity. No doubt is left as to how to figure which number goes with which.
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RESPONSE --> 2n 2n-1= self critique assessment: 1
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23:54:06 It should be clear that each element of the second set is 5 times as great as the corresponding element the first set. The rule would therefore be n <--> 5n.
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RESPONSE --> n <-> 5n self critique assessment: 3
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23:57:08 First we note that the numbers in the second set are going up by 5 each time. This indicates that we will probably somehow have to use 5n in our formula. Just plain 5n gives us 5, 10, 15, ... . It's easy to see that these numbers are each 2 less than the numbers 7, 12, 17, ... . So if we add 2 to 5n we get the numbers we want. Thus the rule is n <--> 5n+2, or in a bit more detail [ 1 <--> 7, 2 <--> 12, 3 <--> 17, ..., n <--> 5n+2, ... ].
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RESPONSE --> n <-> 5n+2 5n defines the set 5,10,15... +2 defines the 7,12,17.... self critique assessment: 3
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23:58:00 `q006. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 3, 10, 17, ... }.
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RESPONSE --> 2n+7 confidence assessment: 2
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00:02:44 `q007. Surprisingly, the numbers { 1, 2, 3, ... } can be put into correspondence with the set of all integer-divided-by-integer fractions (the set of all such fractions is called the set of Rational Numbers). This set would include the fractions 1/2, 1/3, 1/4, ..., as well as fractions like 38237 / 819872 and 232987 / 3. It is a bit surprising that this set could be in 1-1 correspondence with the counting numbers, because just the fractions 1/2, 1/3, 1/4, ... can be put into one-to-one correspondence with the set {1, 2, 3, ... }, and these fractions are less than a drop in the bucket among all possible fractions. These fractions all have numerator 1. The set will also contain the fractions of numerator 2: 2/1, 2/2, 2/3, 2/4, ... . And the fractions with numerator 3: 3/1, 3/2, 3/3, 3/4, ... . We could go on, but the idea should be clear. It certainly seems like there should be more fractions than counting numbers. But it isn't so, as you will see in the lectures and the text. Give a one-to-one correspondence between just the fractions 1/2, 1/3, 1/4, ... and the counting numbers {1, 2, 3, ... }.
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RESPONSE --> [1 <-> 1/2, 2 <-> 1/3, 3 <-> 1/4, n<-> 1/n] confidence assessment: 2
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00:05:40 We could alternate between the two sets, using odd numbers with fractions whose numerator is 1 and even numbers with fractions whose numerator is 2. The correspondence would be [ 1 <--> 1/2, 2 <--> 2/2, 3 <--> 1/3, 4 <--> 2/3, 5 <--> 1/4, 6 <--> 2/4, ... ]. It would be a little bit tricky, but not all that difficult, to write this rule in terms of n. However, we won't go into that here.
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RESPONSE --> [ 1<-> 1/2, 2 <-> 2/2, 3 <-> 1/3, 4<-> 2/3, 5 <->1/4, 6<-> 2/4,....] self critique assessment: 2
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