course Mth 151 I did well with my textbook work but the q_a_ assignmnets, I had some trouble with. I think I understand most of it now.
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22:30:00 `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,...]
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22:30:09 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
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RESPONSE --> ok
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22:31:49 `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 --> there are the same number of elements to each 1-1 corr. .
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22:32:15 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 --> ok, I didn't understand the question.
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22:34:42 `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
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22:34:50 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 --> ok
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22:35:40 `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.
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RESPONSE --> n <--> -5
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22:35:59 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 --> ok
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22:36:49 `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }.
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RESPONSE --> n <--> n+5
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22:37:31 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 --> ok
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22:38:31 `q006. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 3, 10, 17, ... }.
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RESPONSE --> n<-- > =7
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22:39:36 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 --> ok
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22:42:25 `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
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22:43:07 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 --> ok
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22:45:54 `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 --> ? would you add the two sets of fractions to make the union and then add enough counting numbers to make a one to one corr, then work the prob.?
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22:46:19 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 --> ok I thougt so.
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Cğа畇рҎ Student Name: assignment #006
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22:48:10 `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
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22:48:16 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 --> ok
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22:51:51 `q002. Find the likely next two elements of the sequence 1, 2, 4, 8, 15, 26, ... .
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RESPONSE --> 42 , 64
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22:53:07 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 --> ok
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22:54:17 `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
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22:54:22 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 --> ok
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22:59:52 `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 --> 4, a constant number. the next number is 162.
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23:00:22 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 --> ok
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23:02:00 `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 --> 55, 89
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23:02:29 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 --> ok, i forgot the 3rd element.
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ߝֻאt assignment #005 űpZ Liberal Arts Mathematics I 02-05-2006
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21:42:38 Query 2.5.12 n({9, 12, 15, ..., 36})
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RESPONSE --> 2.5 # 12 n= 10 ( 9,12,15,18,21,24,27,30,33,36 )
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21:43:04 ** There are 10 numbers in the set: 9, 12, 15, 18, 21, 24, 27, 30, 33, 36 **
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RESPONSE --> ok
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21:45:08 Query 2.5.18 n({x | x is an even integer }
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RESPONSE --> 2.5 # 18 (2,4,6,8,.... ) n = aleph-null
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21:45:22 ** {x | x is an even integer } indicates the set of ALL possible values of the variable x which are even integers. Anything that satisfies the description is in the set. This is therefore the set of even integers, which is infinite. Since this set can be put into 1-1 correspondence with the counting numbers its cardinality is aleph-null. **
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RESPONSE --> ok
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21:46:47 Query 2.5.18 how many diff corresp between {stallone, bogart, diCaprio} and {dawson, rocky, blaine}?
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RESPONSE --> Six?
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21:48:11 ** Listing them in order, according to the order of listing in the set. We have: [ {S,D},{B,R},{Dic.,BL}] , [{S,bl},{B,D},{Dic.,R}], [{S,R},{B,Bl},{dic.,D}] [ {S,D},{B,DL},{Dic.,R}], [{S,bl},{B,R},{Dic.,D}], [{S,R},{B,D},{dic.,B1}] for a total of six. Reasoning it out, there are three choices for the character paired with Stallone, which leaves two for the character to pair with Bogart, leaving only one choice for the character to pair with diCaprio. **
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RESPONSE --> ok. I thought so. I compared this with the problem in my book that I had worked out and got six diff corr, but I wasn't for sure if I done it right.
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21:51:23 2.5.36 1-1 corresp between counting #'s and {-17, -22, -27, ...}
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RESPONSE --> {-17, -22, -27, -32,...} ,n + 5,... { 1 , 2, 3, 4, ...} ,n, ...
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21:52:15 **You have to describe the 1-1 correspondence, including the rule for the nth number. A complete description might be 1 <-> -17, 2 <-> -22, 3 <-> -27, ..., n <-> -12 + 5 * n. You have to give a rule for the description. n <-> -12 - 5 * n is the rule. Note that we jump by -5 each time, hence the -5n. To get -17 when n=1, we need to start with -12. THE REASONING PROCESS TO GET THE FORMULA: The numbers in the first set decrease by 5 each time so you need -5n. The n=1 number must be -17. -5 * 1 = -5. You need to subtract 12 from -5 to get -17. So the formula is -5 n - 12. **
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RESPONSE --> ok I wasn't for sure how to show what I had on my paper on the computer, but I got the answer right.
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21:54:12 2.5.42 show two vert lines, diff lengths have same # of points
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RESPONSE --> they are both infinte, but can have more spaced out points
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21:54:50 ** This is a pretty tough question. One way of describing the correspondence (you will probably need to do the construction to understand): Sketch a straight line from the top of the blue line at the right to the top of the blue line at the left, extending this line until it meets the dotted line. Call this meeting point P. Then for any point on the shorter blue line we can draw a straight line from P to that point and extend it to a point of the longer blue line, and in our 1-1 correspondence we match the point on the shorter line with the point on the longer. From any point on the longer blue line we can draw a straight line to P; the point on the longer line will be associated with the point we meet on the shorter. We match these two points. If the two points on the long line are different, the straight lines will be different so the points on the shorter line will be different. Thus each point on the longer line is matched with just one point of the shorter line. We can in fact do this for any point of either line. So any point of either line can be matched with any point of the other, and if the points are different on one line they are different on the other. We therefore have defined a one-to-one correspondence. **
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RESPONSE --> Ok I think I get it now.
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IEϥ٭ɢ assignment #006 űpZ Liberal Arts Mathematics I 02-05-2006
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21:56:17 Query 1.1.4 first 3 children male; conclusion next male. Inductive or deductive?
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RESPONSE --> inductive
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21:56:24 ** The argument is inductive, because it attempts to argue from a pattern. **
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RESPONSE --> ok
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21:58:38 Query 1.1.8 all men mortal, Socrates a man, therefore Socrates mortal.
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RESPONSE --> deductive, because the conclusion that socrates is mortal follows the two premises( all men mortal and socrates a man).
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21:58:45 ** this argument is deductive--the conclusions follow inescapably from the premises. 'all men' is general; 'Socrates' is specific. This goes general to specific and is therefore deductive. COMMON ERROR: because it is based on a fact, or concrete evidence. Fact isn't the key; the key is logical inevitability. The argument could be 'all men are idiots, Socrates is an man, therefore Socrates is an idiot'. The argument is every bit as logical as before. The only test for correctness of an argument is that the conclusions follow from the premises. It's irrelevant to the logic whether the premises are in fact true. **
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RESPONSE --> ok
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22:00:20 Query 1.1.20 1 / 3, 3 / 5, 5/7, ... Probable next element.
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RESPONSE --> 11/13 the pattern is odd numbers starting with 3.
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22:00:57 **The numbers 1, 3, 5, 7 are odd numbers. We note that the numerators consist of the odd numbers, each in its turn. The denominator for any given fraction is the next odd number after the numerator. Since the last member listed is 5/7, with numerator 5, the next member will have numerator 7; its denominator will be the next odd number 9, and the fraction will be 7/9. There are other ways of seeing the pattern. We could see that we use every odd number in its turn, and that the numerator of one member is the denominator of the preceding member. Alternatively we might simply note that the numerator and denominator of the next member are always 2 greater than the numerator and denominator of the present member. **
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RESPONSE --> ok
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22:05:31 Query 1.1.23 1, 8, 27, 64, ... Probable next element.
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RESPONSE --> 216
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22:07:33 ** This is the sequence of cubes. 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125. The next element is 6^3 = 216. Successive differences also work: 1 8 27 64 125 .. 216 7 19 37 61 .. 91 12 18 24 .. 30 6 6 .. 6 **
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RESPONSE --> ok
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22:08:50 Query 1.1.36 11 * 11 = 121, 111 * 111 = 12321 1111 * 1111 = 1234321; next equation, verify.
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RESPONSE --> 11111 * 11111 = 123454321
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22:08:57 ** We easily verify that 11111*11111=123,454,321 **
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RESPONSE --> ok
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22:10:01 Do you think this sequence would continue in this manner forever? Why or why not?
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RESPONSE --> yes , because the numbers will keep getting larger with the same nunbers in diff orders
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22:10:18 ** You could think forward to the next few products: What happens after you get 12345678987654321? Is there any reason to expect that the sequence could continue in the same manner? The middle three digits in this example are 8, 9 and 8. The logical next step would have 9, 10, 9, but now you would have 9109 in the middle and the symmetry of the number would be destroyed. There is every reason to expect that the pattern would also be destroyed. **
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RESPONSE --> ok
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22:11:19 Query 1.1.46 1 + 2 + 3 + ... + 2000 by Gauss' method
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RESPONSE --> 1,000 * 2001 = 2,001,000
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22:11:26 ** Pair up the first and last, second and second to last, etc.. You'll thus pair up 1 and 2000, 2 and 1999, 3 and 1998, etc.. Each pair of numbers totals 2001. Since there are 2000 numbers there are 1000 pairs. So the sum is 2001 * 1000 = 2,001,000 **
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RESPONSE --> ok
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22:15:33 Query 1.1.57 142857 * 1, 2, 3, 4, 5, 6. What happens with 7? Give your solution to the problem as stated in the text.
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RESPONSE --> The answers are 142,857 , 285,714 , 428,571 , 571,428 , 714,285 , 857,142 . With 7, 142,857*7= 999,999 1-6 kept the same numbers but in a different order, then when * by 7 all the numbers were the same.
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22:15:44 ** Multiplying we get 142857*1=142857 142857*2= 285714 142857*3= 428571 142857*4=571428 142857*5= 714285 142857*6=857142. Each of these results contains the same set of digits {1, 2, 4, 5, 7, 8} as the number 1428785. The digits just occur in different order in each product. We might expect that this pattern continues if we multiply by 7, but 142875*7=999999, which breaks the pattern. **
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RESPONSE --> ok
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22:17:12 What does this problem show you about the nature of inductive reasoning?
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RESPONSE --> When you think that you know the answer, by looking at the pattern, the pattern may be broken somewhere down the line.We assume that we know the pattern.
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22:17:19 ** Inductive reasoning would have led us to expect that the pattern continues for multiplication by 7. Inductive reasoning is often correct it is not reliable. Apparent patterns can be broken. **
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RESPONSE --> ok
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