course Mth 151 This is my assignment queries for 2.3, 2.4, 2.5 for your review. ³‡×ÅÓƇˆ}¶åäVЙh€~î¢assignment #003
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12:53:43 `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 --> If n(A) represents the people with dark hair and n(B) represents the people with bright eyes then n(A)=20 - 8 which is the intersection of people that have both which would be n(A)=12 and n(B)=15-8 which would be n(B)= 7. The formula I used to find the union of A and B is n(A U B)=n(A) + n(B) - n(A ^ B) which brought me to the answer of n(A U B) =20 + 15 - 8= 35- 8 = 27 for the union. confidence assessment: 2
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12:55:06 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 used a different formula than this but I do remember working this wa in the homework assignments. self critique assessment: 2
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13:05:27 `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 characteristics of the people in A' are the people that do not have dark hair. The people in B' are the people that do not have bright eyes. Therefore n(A') would be everyone that is not in n(A) and n(A^B) which would be 7 which is n(B) and then 8 which is the number of people that does not fall into the categories of dark hair and bright eyes. So n(A') =7+8=15 and then n(B')= 12+8=20. confidence assessment: 2
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13:06:05 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 --> This is the correct answer that I had. self critique assessment: 3
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13:17:00 `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)' stands for everyone outside of A U B so (A U B)' = 20+ 15 - 8 = 27 people that do not have both dark hair and bright eyes. (A^B)' stands for everyone outside A ^ B so (A ^ B)' = 12 + 7 + 8= 27 so the total number of people 35 - 27 which is the union leaves 8 people that does not have darka hair and / or bright eyes. confidence assessment: 2
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13:22:04 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 did get 27 for the answer of (A ^ B)'. self critique assessment: 3
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13:44:41 `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 --> To me both of these answers would be 8 due to the fact that they are the people that do not have at least one or both of the characteristics of the other 27 people. confidence assessment: 1
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13:50:39 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 --> When I figured this way before I got the same answer as the 27 but because it was what I answered in the previous section I thought that it could not be the right answer. I understand how you arrive at this answer. self critique assessment: 2
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13:58:41 `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)' is the elements that are not in the union of A and B. A' U B' would be the union of the elements that are not in A and not in B. (A ^ B)' would be the the elements that are not in the intersection of A and B. A' ^ B' would be the set of elements that are not in A with the intersection of the elements that are not in B. confidence assessment: 3
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14:01:11 ( 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 how this works. self critique assessment: 2
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Ûó´¹ÃÛ‘yhÞ͹’¯^÷£}Ššr¦ÝÅò assignment #004 004. Subsets; One-to-One Correspondences. Liberal Arts Mathematics I 09-01-2008
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20:47:57 `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 --> From the collection of letters a,b,c,d,e there are at least 25 collections with at least one element. (a,a) (a,b) (a,c) (a,d) (a,e) and so forth with each letter of the collection. confidence assessment: 3
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20:51:10 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 now that I read this. I was pairing them in lorders of 2 elements instead of taking them by 5,4,3,2,1 element sets. self critique assessment: 2
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20:53:55 `q002. A one-to-one correspondence between two sets is a rule that associates each element of the each with exactly one element of the other. A natural one-to-one correspondence between the sets { a, b, c } and { 1, 2, 3 } would be to associate a with 1, b with 2, c with 3. This correspondence might be represented as [ a <--> 1, b <--> 2, c <--> 3 ]. This isn't the only possible one-to-one correspondence between these sets. Another might be [ a <--> 2, b <--> 1, c <--> 3 ]. In each case, every element of each set is associated with exactly one element of the other. Another correspondence between the sets might be [ a <--> 3, b<-->2, c<-->3 ]. This correspondence is not one-to-one. In what way does it fail to be a one-to-one correspondence (remember that a one-to-one correspondence is one in which every element of each set is associated with exactly one element of the other).
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RESPONSE --> The way this fails is that a and c are both paired with the same number 3. If one of these had been paired with the number 1 and then other with the number 3 and b still paired with 2 then it would have been a one to one correspondence. confidence assessment: 3
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20:54:17 [ a <--> 3, b<-->2, c<-->3 ] fails to be a one-to-one correspondence for two reasons. In the first place, 3 is associated with a and with c, and every element of each set is to be associated with exactly one element of the other. 3 is associated with two elements of the other set. It also fails because the element 1 of the second set is not associated with anything in the first set.
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RESPONSE --> self critique assessment: 3
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21:00:02 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 --> The answer is [a<->2, b<->3,c<->1] , [a<->1,b<->2,c<->3], [a<->2,b<->1,c<->3]. This is the answer that I had worked but for some reason I skipped the place to enter it in. This appearas to be correct with what you have listed. self critique assessment: 3
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21:01:28 `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 is not possible becasue one to one corresponds means that eack element would have to have a letter matched with a number and you actually would have 1 number left that would not be matched. confidence assessment: 3
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21:01:46 One set has 3 elements and the other has 4 elements. A 1-to-1 correspondence has to match each element of each set with exactly one element of the other. It would not be possible to find four different elements of the first set to match with the four elements of the second.
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RESPONSE --> Correct self critique assessment: 3
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`‹±¿¼¼ÀžªãÖŒ›àIêƇ¹Õ§~kÅñùú²«Å assignment #005 005. Infinite Sets Liberal Arts Mathematics I 09-01-2008
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21:33:47 `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 --> The one- to-one - correspondence would be [1<->1, 2<->3, 3<->5]. confidence assessment: 2
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21:34:01 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
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RESPONSE --> Correct self critique assessment: 3
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21:38:57 `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 --> I'm not sure exactly how to write the rule but I do know that we take 1+1 to give us 2 and then 2+1 to give us 3 , 3+2 to give us 5, 4+3 to give us 7 and so on. confidence assessment: 2
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21:39:44 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 --> This makes sense to say it that way as opposed to my way. self critique assessment: 2
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21:41:34 `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 --> We could write the rule for the correspondence as n<->2n-1. confidence assessment: 2
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21:41:56 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 --> This is correct. self critique assessment: 3
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21:43:30 `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.
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RESPONSE --> The rule would be n<->5n confidence assessment: 2
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21:43:43 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 --> Correct self critique assessment: 3
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21:50:18 `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }.
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RESPONSE --> The closest thing I have would be n<->5n-3 confidence assessment: 2
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21:51:07 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 knew that I could not get the first number I was close but not correct. self critique assessment: 2
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21:52:51 `q006. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 3, 10, 17, ... }.
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RESPONSE --> This rule would be n<->7n-4. confidence assessment: 2
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21:53:06 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 --> Correct self critique assessment: 3
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22:02:32 `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 --> n<->1/n+1 which would be 1<->1/2, 2<->1/3, 3<->1/4 confidence assessment: 1
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22:03:29 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 --> This is correct in what I was thinking but I did not place the n+1 in ( ) self critique assessment: 3
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22:05:56 `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'm not sure how to write this but it must have somethiing to do with the diagram in the text on pg 89 figure 22. confidence assessment: 0
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22:07:48 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 never really thought about putting the correspondence with odd and even numbers but I will keep this in mind for the future. self critique assessment: 2
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