course ???—?~??????
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19:08:13 `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 --> A^B=8 because there are 8 people who have dark hair and bright eyes AUB=35people n(A^B)=8 n(A U B)=27 because 12 people have dark hair and not bright eyes 8 have dark hair and bright eyes and 7 have bright eyes confidence assessment: 2
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19:08:28 The collection A ^ B consists of all the people with both dark hair and bright eyes, which corresponds to the overlap between the two circles (region I). There are 8 people in this overlap, so we say n(A ^ B) = 8. The collection A U B consists of all the people who have least one of the characteristics. This would include the 12 people with dark hair but not bright eyes, located in the first circle but outside the overlap (region II); plus the 7 people with bright eyes but not dark hair, located in the second circle but outside the overlap (region III); plus the 8 people with both characteristics, located in the overlap (region I). Thus we include the 12 + 8 + 7 = 27 people who might be located anywhere within the two circles.
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RESPONSE --> self critique assessment: 3
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19:12:52 `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 --> A' would have no dark haired people B' would not have bright eyes n(A')= 7 because 7 people have bright eyes n(B')=12 because 12 people have dark hair confidence assessment: 1
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19:14:15 Of the 35 people, those in A' are those outside of A. Since A consists of all the dark-haired people, A' consists of all the people lacking dark hair. This includes the 8 people outside of both circles (people having neither dark hair nor bright eyes, region IV) and the 7 people in the second circle but outside the overlap (people having bright eyes but not dark hair, region III). n(A ' ) is therefore 8 + 7 = 15. Since B consists of all the bright-eyed people, B' consists of all the people lacking bright eyes. This would include the 8 people outside both circles (region IV), all of whom lack both dark hair and bright eyes, and the 12 people in the first circle but outside the overlap (region II), who have dark hair but not bright eyes. n ( B ' ) is therefore 12 + 8 = 20.
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RESPONSE --> I forgot about the people outside the circles self critique assessment: 2
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19:19:13 `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 --> (AUB)' =no one because everyone had either dark hair or bright eyes or both (A^B)'= 19 because 12 have dark hair but not bright eyes and 7 have bright eyes but not dark hair confidence assessment: 1
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19:20:31 A U B consists of everyone having at least one of the characteristics (dark hair, bright eyes), and is represented by the numbers in the two circles (regions I, II, III). ( A U B ) ' consists of the people who do not have at least one of the characteristics, and is represented by the number outside both circles (region IV). This number is 8, representing the 8 people who have neither dark hair nor bright eyes. A ^ B stands for all the people with both of the two characteristics (represented by the overlap, region I), so ( A ^ B ) ' stands for all the people who do not have both of the two characteristics (represented by everything outside region I, or regions II, III and IV). [ Note that (A ^ B)' is not the same as the collection of people who have neither characteristic. Anyone who does not have both characteristics will be in ( A ^ B ) ' . ] ( A ^ B )' must include those who have neither characteristic, and also those who have only one of the characteristics. The 8 people outside both circles, the 12 people in the first circle but outside the overlap, and the 7 people in the second circle but outside the overlap all lack at least one characteristic to, so these 8 + 12 + 7 = 27 people make up( A ^ B ) '.
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RESPONSE --> self critique assessment: 0
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19:24:45 `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'UB'=8 A'^B'=19 confidence assessment: 1
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19:25:47 A ' U B ' consists of all the people who are in at least one of the sets A ' or B '. A ' consists of all the people who do not have dark hair, represented by every region of the diagram which does not include any of A. This will include the 7 people in B who are outside the overlapping region, and the 8 people who are outside of both A and B (regions III and IV. Since A consists of regions I and II, A' consists of regions III and IV). B ' consists of all the people who do not have bright eyes, represented by every region of the diagram which does not include any of B (regions II and IV). This will include the 12 people in A but outside the overlap, and the 8 people outside of both A and B. Thus A ' U B ' consists of everyone in at least one of A ' or B ', including the 7 people in B but outside the overlap (region III), the 12 people in A let outside the overlap (region II), and the 8 people outside of both A and B (region IV). These will be the people who lack at least one of the characteristics dark hair and/or bright eyes. Thus n(A' U B') = 7 + 12 + 8 = 27. Note that these are the same 27 people who are in ( A ^ B ) '. So at least in this case, ( A ^ B ) ' = A ' U B '. A ' ^ B ' consists of all the people in both A ' and B '. As before A ' includes the 7 people in B but not A (region III) as well as the 8 people outside both A and B (region IV), and B ' includes the 12 people in A but not B (region II) as well as the 8 people outside both A and B (region IV). The people in both A ' and B ' will be the 8 people outside both A and B, those who have neither dark hair nor bright eyes. We note that this is the same as the set ( A U B ) ', so at least for the present case we see that ( A ' ^ B ' = ( A U B ) '.
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RESPONSE --> self critique assessment:
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19:28:14 `q005. Succinctly describe the relationships between ( A U B ) ', A ' U B ', (A ^ B) ' and A ' ^ B '.
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RESPONSE --> (AUB)'=A'UB' (A^B)'=A'^B' confidence assessment: 2
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19:29:56 ( A U B ) ' = A ' ^ B ' and ( A ^ B ) ' = A ' U B '. The collection outside of the union A U B is the intersection A ' ^ B ', and the collection outside the intersection A ^ B is the union A ' U B '. The ' operation changes union to intersection and intersection to union.
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RESPONSE --> I do not understand any of this. I know that I am supposed to pinpoint what I dont understand but I dont know how to say it. I just dont get any of it self critique assessment: 2
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?{??????`???? assignment #005 005. Infinite Sets Liberal Arts Mathematics I 09-15-2008
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20:12:16 `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<->5,4<->7....} confidence assessment: 3
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20:12:23 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
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RESPONSE --> self critique assessment: 3
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20:20:08 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 --> {1<->5, 2<->10, 3<->15,....n<->5n,...} self critique assessment: 3
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20:25:43 `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }.
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RESPONSE --> {1<->7, 2<->12, 3<->17,...,n<->5n+2...} confidence assessment: 3
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20:26:17 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 --> I understand this pretty well. self critique assessment: 3
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20:48:00 `q006. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 3, 10, 17, ... }.
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RESPONSE --> Im not sure what the formula would be but I know that you have to add 7 to the next answer (1<->3,2<->10,3<->17, 4<->24...} Also there is a difference of 6 if you use the first set 1+2=3, 2+8=10, 3+14=17 you add 6 more than you previously added confidence assessment: 0
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20:48:24 The numbers in the second set are going up by 7 each time, so we will probably 7n in our formula. Just plain 7n gives us 7, 14, 21, ... . These numbers are each 4 greater than the numbers 3, 10, 17, ... . So if we subtract 4 from 7n we get the numbers we want. Thus the rule is n <--> 7n-4, or [ 1 <--> 3, 2 <--> 10, 3 <--> 17, ..., n <--> 7n-4, ... ].
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RESPONSE --> self critique assessment: 3
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20:57:31 `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, 4<->1/5... n<->1/n+1} confidence assessment: 3
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20:57:51 The correspondence would be [ 1 <--> 1/2, 2 <--> 1/3, 3 <--> 1/4, ... ]. The denominator of the fraction is always 1 greater than the counting number. So if the counting number is n, the denominator the corresponding fraction is n + 1. We would therefore write the correspondence as n <--> 1 / (n+1), or in a bit more detail [ 1 <--> 1/2, 2 <--> 1/3, 3 <--> 1/4, ... , n <--> 1/(n+1), ... ].
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RESPONSE --> self critique assessment: 3
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21:06:49 `q008. How might we write a one-to-one correspondence between the set {1, 2, 3, ... } of counting numbers and the set union { 1/2, 1/3, 1/4, ... } U { 2/2, 2/3, 2/4, ... } of fractions with numerator 1 or 2?
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RESPONSE --> {1<->1/2, 2<->1/3, 3<->1/4, ... n<->1/n+1}U {2/2, 2/3, 2/4,...} confidence assessment: 1
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21:07:22 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 --> self critique assessment: 3
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???—?~?????? assignment #003 003. Intersection, Union, Complement, de Morgans Laws Liberal Arts Mathematics I 09-15-2008
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19:08:13 `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 --> A^B=8 because there are 8 people who have dark hair and bright eyes AUB=35people n(A^B)=8 n(A U B)=27 because 12 people have dark hair and not bright eyes 8 have dark hair and bright eyes and 7 have bright eyes confidence assessment: 2
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19:08:28 The collection A ^ B consists of all the people with both dark hair and bright eyes, which corresponds to the overlap between the two circles (region I). There are 8 people in this overlap, so we say n(A ^ B) = 8. The collection A U B consists of all the people who have least one of the characteristics. This would include the 12 people with dark hair but not bright eyes, located in the first circle but outside the overlap (region II); plus the 7 people with bright eyes but not dark hair, located in the second circle but outside the overlap (region III); plus the 8 people with both characteristics, located in the overlap (region I). Thus we include the 12 + 8 + 7 = 27 people who might be located anywhere within the two circles.
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RESPONSE --> self critique assessment: 3
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19:12:52 `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 --> A' would have no dark haired people B' would not have bright eyes n(A')= 7 because 7 people have bright eyes n(B')=12 because 12 people have dark hair confidence assessment: 1
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19:14:15 Of the 35 people, those in A' are those outside of A. Since A consists of all the dark-haired people, A' consists of all the people lacking dark hair. This includes the 8 people outside of both circles (people having neither dark hair nor bright eyes, region IV) and the 7 people in the second circle but outside the overlap (people having bright eyes but not dark hair, region III). n(A ' ) is therefore 8 + 7 = 15. Since B consists of all the bright-eyed people, B' consists of all the people lacking bright eyes. This would include the 8 people outside both circles (region IV), all of whom lack both dark hair and bright eyes, and the 12 people in the first circle but outside the overlap (region II), who have dark hair but not bright eyes. n ( B ' ) is therefore 12 + 8 = 20.
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RESPONSE --> I forgot about the people outside the circles self critique assessment: 2
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19:19:13 `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 --> (AUB)' =no one because everyone had either dark hair or bright eyes or both (A^B)'= 19 because 12 have dark hair but not bright eyes and 7 have bright eyes but not dark hair confidence assessment: 1
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19:20:31 A U B consists of everyone having at least one of the characteristics (dark hair, bright eyes), and is represented by the numbers in the two circles (regions I, II, III). ( A U B ) ' consists of the people who do not have at least one of the characteristics, and is represented by the number outside both circles (region IV). This number is 8, representing the 8 people who have neither dark hair nor bright eyes. A ^ B stands for all the people with both of the two characteristics (represented by the overlap, region I), so ( A ^ B ) ' stands for all the people who do not have both of the two characteristics (represented by everything outside region I, or regions II, III and IV). [ Note that (A ^ B)' is not the same as the collection of people who have neither characteristic. Anyone who does not have both characteristics will be in ( A ^ B ) ' . ] ( A ^ B )' must include those who have neither characteristic, and also those who have only one of the characteristics. The 8 people outside both circles, the 12 people in the first circle but outside the overlap, and the 7 people in the second circle but outside the overlap all lack at least one characteristic to, so these 8 + 12 + 7 = 27 people make up( A ^ B ) '.
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RESPONSE --> self critique assessment: 0
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19:24:45 `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'UB'=8 A'^B'=19 confidence assessment: 1
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19:25:47 A ' U B ' consists of all the people who are in at least one of the sets A ' or B '. A ' consists of all the people who do not have dark hair, represented by every region of the diagram which does not include any of A. This will include the 7 people in B who are outside the overlapping region, and the 8 people who are outside of both A and B (regions III and IV. Since A consists of regions I and II, A' consists of regions III and IV). B ' consists of all the people who do not have bright eyes, represented by every region of the diagram which does not include any of B (regions II and IV). This will include the 12 people in A but outside the overlap, and the 8 people outside of both A and B. Thus A ' U B ' consists of everyone in at least one of A ' or B ', including the 7 people in B but outside the overlap (region III), the 12 people in A let outside the overlap (region II), and the 8 people outside of both A and B (region IV). These will be the people who lack at least one of the characteristics dark hair and/or bright eyes. Thus n(A' U B') = 7 + 12 + 8 = 27. Note that these are the same 27 people who are in ( A ^ B ) '. So at least in this case, ( A ^ B ) ' = A ' U B '. A ' ^ B ' consists of all the people in both A ' and B '. As before A ' includes the 7 people in B but not A (region III) as well as the 8 people outside both A and B (region IV), and B ' includes the 12 people in A but not B (region II) as well as the 8 people outside both A and B (region IV). The people in both A ' and B ' will be the 8 people outside both A and B, those who have neither dark hair nor bright eyes. We note that this is the same as the set ( A U B ) ', so at least for the present case we see that ( A ' ^ B ' = ( A U B ) '.
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RESPONSE --> self critique assessment:
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19:28:14 `q005. Succinctly describe the relationships between ( A U B ) ', A ' U B ', (A ^ B) ' and A ' ^ B '.
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RESPONSE --> (AUB)'=A'UB' (A^B)'=A'^B' confidence assessment: 2
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19:29:56 ( A U B ) ' = A ' ^ B ' and ( A ^ B ) ' = A ' U B '. The collection outside of the union A U B is the intersection A ' ^ B ', and the collection outside the intersection A ^ B is the union A ' U B '. The ' operation changes union to intersection and intersection to union.
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RESPONSE --> I do not understand any of this. I know that I am supposed to pinpoint what I dont understand but I dont know how to say it. I just dont get any of it self critique assessment: 2
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?{??????`???? assignment #005 005. Infinite Sets Liberal Arts Mathematics I 09-15-2008
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20:12:16 `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<->5,4<->7....} confidence assessment: 3
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20:12:23 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
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RESPONSE --> self critique assessment: 3
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20:20:08 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 --> {1<->5, 2<->10, 3<->15,....n<->5n,...} self critique assessment: 3
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20:25:43 `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }.
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RESPONSE --> {1<->7, 2<->12, 3<->17,...,n<->5n+2...} confidence assessment: 3
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20:26:17 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 --> I understand this pretty well. self critique assessment: 3
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20:48:00 `q006. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 3, 10, 17, ... }.
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RESPONSE --> Im not sure what the formula would be but I know that you have to add 7 to the next answer (1<->3,2<->10,3<->17, 4<->24...} Also there is a difference of 6 if you use the first set 1+2=3, 2+8=10, 3+14=17 you add 6 more than you previously added confidence assessment: 0
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20:48:24 The numbers in the second set are going up by 7 each time, so we will probably 7n in our formula. Just plain 7n gives us 7, 14, 21, ... . These numbers are each 4 greater than the numbers 3, 10, 17, ... . So if we subtract 4 from 7n we get the numbers we want. Thus the rule is n <--> 7n-4, or [ 1 <--> 3, 2 <--> 10, 3 <--> 17, ..., n <--> 7n-4, ... ].
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RESPONSE --> self critique assessment: 3
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20:57:31 `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, 4<->1/5... n<->1/n+1} confidence assessment: 3
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20:57:51 The correspondence would be [ 1 <--> 1/2, 2 <--> 1/3, 3 <--> 1/4, ... ]. The denominator of the fraction is always 1 greater than the counting number. So if the counting number is n, the denominator the corresponding fraction is n + 1. We would therefore write the correspondence as n <--> 1 / (n+1), or in a bit more detail [ 1 <--> 1/2, 2 <--> 1/3, 3 <--> 1/4, ... , n <--> 1/(n+1), ... ].
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RESPONSE --> self critique assessment: 3
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21:06:49 `q008. How might we write a one-to-one correspondence between the set {1, 2, 3, ... } of counting numbers and the set union { 1/2, 1/3, 1/4, ... } U { 2/2, 2/3, 2/4, ... } of fractions with numerator 1 or 2?
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RESPONSE --> {1<->1/2, 2<->1/3, 3<->1/4, ... n<->1/n+1}U {2/2, 2/3, 2/4,...} confidence assessment: 1
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21:07:22 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 --> self critique assessment: 3
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