#$&* course Mth 151 005. Infinite Sets
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Given Solution: This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ]. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Self-critique Rating: ********************************************* Question: `q002. Writing [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ] for the correspondence between { 1, 2, 3, ... } and { 1, 3, 5, ... } isn't bad, but the pattern here might be a bit less clear to the reader than the correspondence [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ] given for { 1, 2, 3, ... } and { 2, 4, 6, ... }. That is because in the latter case it is clear that we are simply doubling the numbers in the first set to get the numbers in the second. It might not be quite as clear exactly what the rule is in the correspondence [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ], except that we know we are pairing the numbers in the two sets in order. Without explicitly stating the rule in a form as clear as the doubling rule, we can't be quite as sure that our rule really works. How might we state the rule for the correspondence [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ] as clearly as the 'double-the-first-number' rule for [ 1 <--> 2, 2 <--> 4, 3 <--> 6, ... ]? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Either subtract 1 or add 1 to the even numbers. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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, ... . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `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, ... ]? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2n-1 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rule for this correspondence is 'double and subtract 1', so n would be associated with 2n - 1. The correspondence would thus be [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... , n <--> 2n-1, ... ]. Note how this gives a definite formula for the rule, removing all ambiguity. No doubt is left as to how to figure which number goes with which. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Multiply each element in the first set by 5. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5n+2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: First we note that the numbers in the second set are going up by 5 each time. This indicates that we will probably somehow have to use 5n in our formula. Just plain 5n gives us 5, 10, 15, ... . It's easy to see that these numbers are each 2 less than the numbers 7, 12, 17, ... . So if we add 2 to 5n we get the numbers we want. Thus the rule is n <--> 5n+2, or in a bit more detail [ 1 <--> 7, 2 <--> 12, 3 <--> 17, ..., n <--> 5n+2, ... ]. Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5n+2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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, ... ]. Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 151 006. Sequences and Patterns
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Given Solution: The difference between 1 and 2 is 1; between 2 and 4 is 2; between 4 and 7 is 3; between 7 and 11 is 4. So we expect that the next difference will be 5, which will make the next element 11 + 5 = 16. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Find the likely next two elements of the sequence 1, 2, 4, 8, 15, 26, ... . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Taking the numbers from the previous problem and adding them to this set of numbers confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 + 22 = 64 as the next element. So the original sequence would continue as 1, 2, 4, 8, 15, 26, 42, 68, ... . Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Each number is being doubled. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: One obvious pattern for this sequence is that each number is doubled to get the next. If this pattern continues then the sequence would continue by doubling 8 to get 16. The sequence would therefore be 1, 2, 4, 8, 16, ... . Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. There are two important types of patterns for sequences, one being the pattern defined by the differences between the numbers of the sequence, the other being the pattern defined by the ratios of the numbers of the sequence. In the preceding sequence 1, 2, 4, 8, 16, ..., the ratios were 2/1 = 2; 4/2 = 2; 8/4 = 2; 16/8 = 2. The sequence of ratios for 1, 2, 4, 8, 16, ..., is thus 2, 2, 2, 2, a constant sequence. Find the sequence of ratios for the sequence 32, 48, 72, 108, ... , and use your result to estimate the next number and sequence. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Dividing each number by the number before it, the results come to 1.5. Making the next number after 108 (108 * 1.5) being 162. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Find the sequence of ratios for the sequence 1, 2, 3, 5, 8, 13, 21... , and estimate the next element of the sequence. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Keep dividing the number’s one after another creating a different number to multiply by. Making the next element 34. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The ratios are 2/1 = 2; 3/2 = 1.5; 5/3 = 1.66...; 8/5 = 1.60; 13/8 = 1.625; 21/13 = 1.615. The sequence of ratios is 2, 1.5, 1.66..., 1.625, 1.615, ... . We see that each number in the sequence lies between the two numbers that precede it -- 1.66... lies between 2 and 1.5; 1.60 lies between 1.5 and 1.66...; 1.625 lies between 1.66... and 1.60; 1.615 lies between 1.60 and 1.625. We also see that the numbers in the sequence alternate between being greater than the preceding number and less than the preceding number, so that the intervals between the numbers get smaller and smaller. So we expect that the next number in the sequence of ratios will be between 1.615 and 1.625, and if we pay careful attention to the pattern we expect the next number to be closer to 1.615 than to 1.625. We might therefore estimate that the next ratio would be about 1.618. We would therefore get 1.618 * 21 = 33.98 for the next number in the original sequence. However, since the numbers in the sequence are all whole numbers, we round our estimate up to 34. Our conjecture is that the sequence continues with 1, 2, 3, 5, 8, 13, 21, 34, ... . Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `q006. Without using ratios, can you find a pattern to the sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, ..., and continue the sequence for three more numbers? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The numbers before the next element are being added together. confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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, ... . ********************************************* Question: `q007. What is the pattern of the sequence ... 4 7 11 18 ... and what are the next three numbers in this sequence? What positive numbers could precede the 4, how did you figure them out and how do you know you got them all? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The next 3 numbers in the sequence are 29, 47, and 76. The numbers are being added with the number before it. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 151 007. Triangular, Square, Pentagonal Numbers
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Given Solution: The original triangle had the three points A, B and C. When you extended the two sides you marked the new endpoints, then you marked the point in the middle of the third side. So you've got 6 points marked. The construction of these numbers is shown in the figure below. We begin with a single dot: We label this point A and construct a triangle containing this point as a vertex. We place similar dots at the vertices of this triangle. We now 'scale up' the triangle by doubling the lengths of its sides: We divide this triangle into triangles of the original size, and place dots at each of these vertices. The first figure has a single 'dot', the second has 3 'dots', and the third has 6 'dots'. Note the similarity with the figures below. The first depicts the pattern illustrated in this question. The second illustrates the pattern extended one steps: The third depicts the pattern as it would appear if extended 12 steps beyond the original triangle: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `q002. Extend the two sides that meet at A by distances equal to the original lengths AC and AB and mark the endpoints of the newly extended segments. Each of the newly extended sides will have 4 marked points. Now connect the new endpoints to form a new right triangle. Mark points along the new side at the same intervals that occur on the other two sides. How many marked points are on your new triangle, and how many in the whole figure? Your solution: 4 and 10 confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You added the two new endpoints when you extended the sides. You then should have marked two new points on the new third side, so that each side contains 4 points including its endpoints. Your figure will now contain 10 marked points. The construction is shown below. First we extend the two sides by a length equal to that of the original triangle: Next we join the 'free' endpoints of those new sides to form a triangle. Now we place points along the new side and join them to complete the 'small' triangles within our new figure: We have added four new dots. The figure below depicts only the 'dots', without the triangles: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Continue the process for another step-extend each side by a distance equal to the original point-to-point distance. How many points do you have in the new triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 15 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You will add an endpoint to each newly extended side, so each of the new sides will contain 5 points. You will then have to add 3 equally spaced points to the new side, giving you a total of 13 points on the new triangle. In addition there are two marked points inside the triangle, for a total of 15 points. Click on 'Next Picture' to see the construction. The line segments along two sides of the triangle have again been extended and points marked at the ends of these segments. The new endpoints have been connected to form the third side of a larger triangle, and equally spaced points have been constructed along that side. `routine triangle4 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Continue the process for one more step. How many points do you have in the new triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 21 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You will add an endpoint to each newly extended side, so each of the new sides will contain 6 points. You will then have to add 4 equally spaced points to the new side, giving you a total of 15 points on the new triangle. There are also 5 marked points inside the triangle for a total of 21 marked points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. The sequence of marked points is 3, 6, 10, 15, 21. What do expect will be the next number in this sequence? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 28. The number started being added with 3+3, but then went up 1 leading up to 21+7=28 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The differences between these numbers are 3, 4, 5, 6. The next difference, according to this pattern, should be 7, which would make the next number 28. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. How can you tell, in terms of the process you used to construct these triangles, that the next number should be 7 greater? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: When you extend the triangle again, you will add two new endpoints and each side will now have 7 points. The 7 points on the new triangle will be all of the new points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. How do you know this sequence will continue in this manner? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Each time you extend the triangle, each side increases by 1. All the new marked points are on the new side, so the total number of marked points will increase by 1 more than with the previous extension. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. How do you know this sequence will continue in this manner? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Each time you extend the triangle, each side increases by 1. All the new marked points are on the new side, so the total number of marked points will increase by 1 more than with the previous extension. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #*&!