course Mth 151 y~ڪNb男passignment #005
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16:03:11 `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 corrrespondence would be [1<->1, 2<->3, 3<->5, .....]. Eventhough the 3 is listed twice the corresponding position isn't the same (one is in the first group and the other is in the second group) confidence assessment: 3
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16:05:14 `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 --> The one to one correspondence can not be listed twice as the first correspondence but can be listed as the first correspondence in one group and the second correspondence in the second. confidence assessment: 2
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16:06:44 `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 --> You could say to mutiply the first number by 2 to obtain the second number. confidence assessment: 3
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16:12:51 `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.
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RESPONSE --> You could mutiple the first number by 5 to obtain the second number. [1 <-> 5, 2 <-> 10, 3 <-> 15, ..... n <-> n(5).....] confidence assessment: 2
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16:49:42 `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, n<->n+7.....] confidence assessment: 1
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16:54:15 `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, 1/3<->2/1, 1/4<->3/1) confidence assessment: 2
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17:02:35 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'm sorry but I could not figure this one out. I tried and tried but couldn't figure it out. self critique assessment:
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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