course Mth 151 I'm sorry these are late, I just forgot to send them in! S훼K^ˡǚassignment #005
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13:52:32 `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: 3
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13:52:39 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
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RESPONSE --> self critique assessment: 3
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13:54:00 `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 --> Pair each whole number with and even number. confidence assessment: 3
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13:54:20 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 --> self critique assessment: 3
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13:55:51 `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 --> n+2 confidence assessment: 2
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13:56:05 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 --> self critique assessment: 3
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13:56:38 `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.
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RESPONSE --> 5n confidence assessment: 3
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13:56:45 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 --> self critique assessment: 3
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13:58:32 `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }.
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RESPONSE --> confidence assessment: 0
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13:58:53 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 --> self critique assessment: 3
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14:00:25 `q006. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 3, 10, 17, ... }.
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RESPONSE --> 7n confidence assessment: 2
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14:00:44 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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14:02:52 `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<-> 1/4, 3 <-> 1/3 ] confidence assessment: 2
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14:03:16 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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14:05:32 `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 ] [ 1 <-> 2/2, 2 <-> 2/3, 3<-> 2/4] confidence assessment: 2
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14:06:01 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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ၲD][ǘ| assignment #006 006. Sequences and Patterns Liberal Arts Mathematics I 02-03-2008
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14:07:48 `q001. Note that there are 6 questions in this assignment. Find the likely next element of the sequence 1, 2, 4, 7, 11, ... .
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RESPONSE --> 16 confidence assessment: 3
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14:07:56 The difference between 1 and 2 is 1; between 2 and 4 is 2; between 4 and 7 is 3; between 7 and 11 is 4. So we expect that the next difference will be 5, which will make the next element 11 + 5 = 16.
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RESPONSE --> self critique assessment: 3
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14:10:00 `q002. Find the likely next two elements of the sequence 1, 2, 4, 8, 15, 26, ... .
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RESPONSE --> confidence assessment: 0
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14:12:05 The difference between 1 and 2 is 1; the difference between 2 and 4 is 2, the difference between 4 and 8 is 4; the difference between 8 and 15 is 7; the difference between 15 and 26 is 11. The differences form the sequence 1, 2, 4, 7, 11, ... . As seen in the preceding problem the differences of this sequence are 1, 2, 3, 4, ... . We would expect the next two differences of this last sequence to be 5 and 6, which would extend the sequence 1, 2, 4, 7, 11, ... to 1, 2, 4, 7, 11, 16, 22, ... . If this is the continuation of the sequence of differences for the original sequence 1, 2, 4, 8, 15, 26, ... then the next two differences of this sequence would be 16 , giving us 26 + 16 = 42 as the next element, and 22, giving us 42 + 26 = 68 as the next element. So the original sequence would continue as 1, 2, 4, 8, 15, 26, 42, 68, ... .
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RESPONSE --> self critique assessment: 3
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14:12:46 `q003. What would be the likely next element in the sequence 1, 2, 4, 8, ... . It is understood that while this sequence starts off the same as that in the preceding exercise, it is not the same. The next element is not 15, and the pattern of the sequence is different than the pattern of the preceding.
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RESPONSE --> 16 confidence assessment: 3
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14:12:53 One obvious pattern for this sequence is that each number is doubled to get the next. If this pattern continues then the sequence would continue by doubling 8 to get 16. The sequence would therefore be 1, 2, 4, 8, 16, ... .
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RESPONSE --> self critique assessment: 3
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14:19:29 `q004. There are two important types of patterns for sequences, one being the pattern defined by the differences between the numbers of the sequence, the other being the pattern defined by the ratios of the numbers of the sequence. In the preceding sequence 1, 2, 4, 8, 16, ..., the ratios were 2/1 = 2; 4/2 = 2; 8/4 = 2; 16/8 = 2. The sequence of ratios for 1, 2, 4, 8, 16, ..., is thus 2, 2, 2, 2, a constant sequence. Find the sequence of ratios for the sequence 32, 48, 72, 108, ... , and use your result to estimate the next number and sequence.
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RESPONSE --> confidence assessment: 0
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14:19:44 The ratios are 48/32 = 1.5; 72 / 48 = 1.5; 108/72 = 1.5, so the sequence of ratios is 1.5, 1.5, 1.5, 1.5, ... . The next number the sequence should probably therefore be 108 * 1.5 = 162.
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RESPONSE --> self critique assessment: 0
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14:22:03 `q005. Find the sequence of ratios for the sequence 1, 2, 3, 5, 8, 13, 21... , and estimate the next element of the sequence.
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RESPONSE --> 34 confidence assessment: 3
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14:22:23 The ratios are 2/1 = 2; 3/2 = 1.5; 5/3 = 1.66...; 8/5 = 1.60; 13/8 = 1.625; 21/13 = 1.615. The sequence of ratios is 2, 1.5, 1.66..., 1.625, 1.615, ... . We see that each number in the sequence lies between the two numbers that precede it -- 1.66... lies between 2 and 1.5; 1.60 lies between 1.5 and 1.66...; 1.625 lies between 1.66... and 1.60; 1.615 lies between 1.60 and 1.625. We also see that the numbers in the sequence alternate between being greater than the preceding number and less than the preceding number, so that the intervals between the numbers get smaller and smaller. So we expect that the next number in the sequence of ratios will be between 1.615 and 1.625, and if we pay careful attention to the pattern we expect the next number to be closer to 1.615 than to 1.625. We might therefore estimate that the next ratio would be about 1.618. We would therefore get 1.618 * 21 = 33.98 for the next number in the original sequence. However, since the numbers in the sequence are all whole numbers, we round our estimate up to 34. Our conjecture is that the sequence continues with 1, 2, 3, 5, 8, 13, 21, 34, ... .
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RESPONSE --> self critique assessment: 3
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14:23:55 `q006. Without using ratios, can you find a pattern to the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, ..., and continue the sequence for three more numbers?
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RESPONSE --> you add the previous 2 to get the 3 number 55, 89, 144 confidence assessment: 3
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14:24:04 The pattern is that each element from the third on is the sum of the two elements that precede it. That is, 1+1=2, 2+1=3; 3+2=5; 5+3=8; 8+5=13; 13+8=21; 21+13=34; . The next three elements would therefore e 34+21=55; 55+34=89; 89+55=144. . The sequence is seen to be 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... .
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RESPONSE --> self critique assessment: 3
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