course Mth 151 ??????????€??assignment #005005. Infinite Sets
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14:12:03 `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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14:21:59 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
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RESPONSE --> I didn't put the brackets and I extended it one more in case you wanted me to predict the next element. self critique assessment: 2
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14:30:05 `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 sets {1,2,3,...} and {1,3,5...} the correspondence is 1 to 1, 2 to 3, and 3 to 5. confidence assessment: 2
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14:34:08 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 didn't understand what was being asked. Now I understand that what was asked is how does each of the numbers in the set relate to each other or how are they were derived in sequence. self critique assessment: 2
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14:40:38 `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 --> I am not sure how to word it. confidence assessment: 0
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14:41:43 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. I know it in my head how to do it but to express it I go blank. self critique assessment: 2
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15:01:53 `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.
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RESPONSE --> this set corresponds to numbers in blocks of 5 confidence assessment: 3
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15:02:59 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 am forgetting to using the n such as n<->5n. self critique assessment: 2
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15:05:46 `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }.
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RESPONSE --> this is blocks of 5 also so it would be n<->5n confidence assessment: 3
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15:06:44 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 see I need to add two to the 5n self critique assessment: 2
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15:08:43 `q006. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 3, 10, 17, ... }.
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RESPONSE --> this one is in blocks of 7 so it would be n<->3n+7 confidence assessment: 2
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15:12:58 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 used three instead because the first number was three. I think I am also writing my formula down wrong. self critique assessment: 2
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15:16:37 `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, and 1/4<->3] confidence assessment: 3
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15:17:55 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 --> so the single number always goes first whether or not the set is listed first? self critique assessment: 2
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15:43:15 `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 --> n<->n+1 for the first 3 and n<->2n+1 for the other 3 confidence assessment: 2
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15:44:59 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 can't tell on these questions if you want me to write out the correspondence like here or in terms of n. Here is what I was thinking but put it in terms of n. self critique assessment: 2
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16:22:10 `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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16:22:46 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 --> I didn't write it out what I counted in my head. self critique assessment: 2
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16:29:21 `q002. Find the likely next two elements of the sequence 1, 2, 4, 8, 15, 26, ... .
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RESPONSE --> I figure out the sequence until I get up to the 15 but that is as far as I can go. confidence assessment: 0
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16:40:10 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 --> I could get that the difference was 1,2,4,7,11 but I thought I was wrong because it doesn't make sense like the first problem. also the answer is the differences not adding on the original problem can you understand what I am saying because it is difficult for me to express. I don't understand what is in common to make the next element in 1,2,4,7,11 to be 16 and then 22. How is this posible? What am I missing?
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16:42:10 `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 --> Wouldn't this be 1+1=2, 2+2=4, 4+4=8 and 8+8=16 making the next element 16. confidence assessment: 3
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16:42:35 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 --> I understand this one it makes sense. self critique assessment: 3
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17:50:34 `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 --> This is what I can tell there is a difference of 16 between 32 and 48, a difference of 24 between 48 and 72, a difference of 36 between 72 and 108 which if you divide them all by 4 the sequence would be 4,6,9,13 from here there is a difference of 2 between 4 and 6, a difference of 3 between and 6 and 9, and a difference of 4 between 9 and 13 so going back to my sequence 4,6,9,13 4*4 =16, 6*4=24,9*4=36, and 13*4=52. Which then brings me to my original sequence of 32,48,72,108,... 16*2=32 ,24*2=48, 36*2= 72, 52*2=104 which blows my theory. Anyway I think the next one would be 160 however my theory is not right. confidence assessment: 1
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17:53:40 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 --> I was sort of close considering my theory was way off. I never would have thought it would have been a decimal number. self critique assessment: 2
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17:56:46 `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 --> Well here I go again. 1+2=3, 2+3=5, 3+5=8, 5+8=13, 8+13=21and 13+21=34 therefore the next element would be 34. confidence assessment: 3
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17:59:03 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 --> Why is this answer so complicated to get too? Isn't much easier to add the numbers like I did to get the answer? self critique assessment: 2
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18:13:50 `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 --> in the sequence 1,1,2,3,5,8,13,21,34,... if you add1+1=2,2+3=5,5+8=13,13+21=34 then 21+34=55 so the next element would be 55 confidence assessment: 3
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19:07:41 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 --> I forgot the other two 21+34=55,34+55=89 and 55+89=144 so the other two would be 89 and 144. I wrote this before I looked at the answer. self critique assessment: 2
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