course Mth 151
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11:08:03 `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 --> confidence assessment:
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cKҮьҖף assignment #003 003. Intersection, Union, Complement, de Morgans Laws Liberal Arts Mathematics I 09-16-2007
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11:16:36 `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 --> In the union of A^B, it would include all elements that are common to both A & B. U= 35 A= region II = 12 B= region IV =7 Region III = the intersection of A & B = 8 n(A U B) = the set of all elements that abelong to either of the sets. A = region II = 12 B = region IV = 7 12 + 7 + 8 (intersection) = 27 people confidence assessment: 2
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11:16:51 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 --> I understand and got it correct. self critique assessment: 2
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11:22:57 `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 --> The people in A' do not have dark hair. The people in B' do not have bright eyes. A' are the elements not in A = regions 1 & 4 B' are the elements not in B = regions 1 & 2 A' = 7 people B' = 12 people confidence assessment: 1
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11:24:56 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 understand. I forgot to add the 8 people outside of both circles. self critique assessment: 2
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11:29:36 `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 --> (A U B)' would be people with neither characteristic of dark hair or bright eyes. This would = 8 people in region I. (A^B) would be people with either dark hair, bright eyes, or neither characteristic. This would = 12 + 7 + 8= 27 people. confidence assessment: 2
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11:30:46 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 --> I understand. I got it correct. self critique assessment: 2
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11:46:49 `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 --> The union of A' B' would be regions 1, 2, 4. This would equal 8 + 12 + 7 = 27 people. These people would have at least one characteristic or neither of the characteristics. A'^B' would be the people who don't have both characteristics. This could result in them having one of the characteristics or neither of the characteristics. This would equal 12 + 7 + 8= 27 people. confidence assessment: 1
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11:48:08 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 --> I understand and I got it correct but I understand it more clearly because your explanation goes into more detail that what I had pictured. self critique assessment: 2
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12:01:53 `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)' = people who don't have any of the characteristics = 8 people A' U B' = people who don't have dark hair and who don't have bright eyes = 8 people.
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12:02:13 ( 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 --> self critique assessment:
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12:02:19 ( 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 --> self critique assessment:
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12:02:50 ( 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 understand but I figured it out differently. Is my response correct? self critique assessment: 1
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ˮcVl٣، assignment #004 004. Subsets; One-to-One Correspondences. Liberal Arts Mathematics I 09-16-2007 ~ԷTӯּx~үé assignment #004 004. Subsets; One-to-One Correspondences. Liberal Arts Mathematics I 09-16-2007
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14:55:25 `qNote that there are 4 questions in this assignment. `q001. From the collection of letters a, b, c, d, e, how many smaller collections having at least one element may be formed?
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RESPONSE --> 1 element = 5 2 elements =10 3 elements = 10 4 elements = 5 5 + 10 + 10 + 5 = 30 collections confidence assessment: 2
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14:55:43 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 --> I understand. self critique assessment: 2
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{wbEJwWx߂[v؝ assignment #005 005. Infinite Sets Liberal Arts Mathematics I 09-16-2007
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18:12:17 `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, 5 <-> 9....} confidence assessment: 2
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18:12:38 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
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RESPONSE --> I understand and got it correct. self critique assessment: 2
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18:15:01 `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 --> In the second set of numbers, we are using the odd numbers. The next pair would be 4 <-> 7, 5 <-> 9... confidence assessment: 2
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18:15:26 We might say something like 'choose the next odd number'. That wouldn't be too bad. Even clearer would be to note that the numbers 1, 3, 5, ... are each 1 less than the 'double-the-counting-number' numbers 2, 4, 6. So our rule could be the 'double-the-first-number-and-subtract-1' rule. If we double each of the numbers 1, 2, 3, ... and subtract 1, we get 1, 3, 5, ... .
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RESPONSE --> I understand. self critique assessment: 2
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18:19:06 `q003. The 'double-the-number' rule for the correspondence [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ] could be made even clearer. First we we let n stand for the nth number of the set {1, 2, 3, ... }, like 10 stands for the 10th number, 187 stands for the 187th number, so whatever it is and long as n is a counting number, n stands for the nth counting number. Then we note that the correspondence always associates n with 2n, so the correspondence could be written0 [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... , n <--> 2n, ... ]. This tells us that whatever counting number n we choose, we will associate it with its double 2n. Since we know that any even number is a double of the counting number, of the form 2n, this rule also tells us what each even number is associated with. So we can argue very specifically that this rule is indeed a 1-to-1 correspondence. In terms of n, how would we write the rule for the correspondence [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ]?
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RESPONSE --> {1 <-> 1, 2 <-> 3, 3 <-> 5, ...n <-> 2 + n,...} confidence assessment: 2
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18:19:52 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 --> i understand but would you have to use the ""double rule"" or could you add 2 like I did? self critique assessment: 2
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18:22:28 `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.
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RESPONSE --> {1 <-> 5, 2 <-> 10, 3 <-> 15,...n <-> n x 5,...} confidence assessment: 2
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18:22:40 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 --> I understand. self critique assessment: 2
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18:24:33 `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, 4 <-> 22,...n <-> n+5...} confidence assessment: 2
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18:25:40 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. self critique assessment: 2
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18:31:37 `q006. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 3, 10, 17, ... }.
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RESPONSE --> {1 <-> 3, 2 <-> 10, 3 <-> 17, 4 <-> 24,...n <-> 7n - 4...} confidence assessment: 2
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18:31:59 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 --> I understand and, it took me a little while, but I figured it out correctly. self critique assessment: 2
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18:34:15 `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 --> {1/2<-> 1, 1/3 <-> 2, 1/4 <-> 3, 1/5 <-> 4, 1/6 <-> 5...} confidence assessment: 2
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18:34:47 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 --> I understand. self critique assessment: 2
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18:35:58 `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 --> I don't know. confidence assessment: 0
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18:38:07 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 --> I am not sure how the sequence follows when it changes from 1/3, 2/3 to 1/4, 2/4. Why don't you continue to 3/4, etc. ? self critique assessment: 2
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