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09:42:13 Note that there are 4 questions in this assignment.
`q001. Let A stand for the collection of all whole numbers which have at least one even digit (e.g., 237, 864, 6, 3972 are in the collection, while 397, 135, 1, 9937 are not). Let A ' stand for the collection of all whole numbers which are not in the collection A. Let B stand for the collection { 3, 8, 35, 89, 104, 357, 4321 }. What numbers do B and A have in common? What numbers do B and A' have in common?......!!!!!!!!...................................
RESPONSE --> Collections A and B have four numbers in common; 8, 89, 104 and 4321.
Collections A ' and B have three numbers in common; 3, 35 and 357..................................................
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09:44:48 If we assume that dark brown, light brown or blonde, and bright red hair are mutually exclusive (i.e., someone can't be both one category and another, much less all three), then we have at least 8 + 2 + 9 = 19 people in the room, and it is not possible that we have exactly 17.
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RESPONSE --> If the three hair colors are exclusive, then there are 19 people in the classroom and thus there can't be only 17 since 8 brunettes + 2 redheads + 9 blondes = 19 people.
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09:46:07 `q003. I have in a room 6 people with dark hair and 10 people with blue eyes. There are only 14 people in the room. But 10 + 6 = 16, which is more than 14. How can this be?
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RESPONSE --> This is possible because the statement didn't say that a person with dark hair couldn't have blue eyes or vice versus. The two categories are not mutually exclusive thus it's possible for some of the six dark haired people to also be some of the ten blue eyed people.
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09:46:19 The key here is that there is nothing mutully exclusive about these categories-a person can have blue eyes as well as dark hair. So if there are 2 people in the room who have dark hair and blue eyes, which is certainly possible, then when we add 10 + 6 = 16 those two people would be counted twice, once among the 6 blue-eyed people and once among the 10 dark-haired people. So the 16 we get would be 2 too high. To get the correct number we would have to subtract the 2 people who were counted twice to get 16 - 2 = 14 people.
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RESPONSE --> Okay.
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09:47:43 `q004. In a set of 100 child's blocks 60 blocks are cubical and 40 blocks are cylindrical. 30 of the blocks are red and 20 of the red blocks are cubical. How many of the cylindrical blocks are red?
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RESPONSE --> If there are 30 red blocks and 20 of the 30 blocks are cubical that leaves 10 cylindrical red blocks.
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09:47:53 Of the 30 red blocks 20 are cubical, so the rest must be cylindrical. This leaves 10 red cylindrical blocks.
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RESPONSE --> Okay.
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݉U}C \ɪ͵ Student Name: assignment #002
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09:52:49 Note that there are 2 questions in this assignment.
`q001. We can represent the collection consisting of the letters a, b, c, d, e, f by a circle in which we write these letters. If we have another collection consisting of the letters a, c, f, g, k, we could represent it also by a circle containing these letters. If both collections are represented in the same diagram, then since the two collections have certain elements in common the two circles should overlap. Sketch a diagram with two overlapping circles. The two circles will create four regions (click below on 'Next Picture'). The first region is the region where the circles overlap. The second region is the one outside of both circles. The third region is the part of the first circle that doesn't include the overlap. The fourth region is the part of the second circle that doesn't include the overlap. Number these regions with the Roman numerals I (the overlap), II (first circle outside overlap), III (second circle outside overlap) and IV (outside both circles). Let the first circle contain the letters in the first collection and let the second circle contain the letters in the second collection, with the letters common to both circles represented in the overlapping region. Which letters, if any, go in region I, which in region II, which in region III and which in region IV?......!!!!!!!!...................................
RESPONSE --> In section 1 will be the letters a, c and f, as they are in both collections. In section 2 there will be b, d, and e. In section 3 there will be g and k. There will be nothing in section 4.
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09:53:09 The letters a, c and f go in the overlapping region, which we called Region I. The remaining letters in the first collection are b, d, and e, and they go in the part of the first circle that does not include the overlapping region, which we called Region II. The letters g and k go in the part of the second circle that does not include the overlapping region (Region III). There are no letters in Region IV.
Click below on 'Next Picture' for a picture.......!!!!!!!!...................................
RESPONSE --> Okay.
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09:57:51 `q002. Suppose that we have a total of 35 people in a room. Of these, 20 have dark hair and 15 have bright eyes. There are 8 people with dark hair and bright eyes.
Draw two circles, one representing the dark-haired people and the other representing the bright-eyed people. Represent the dark-haired people without bright eyes by writing this number in the part of the first circle that doesn't include the overlap (region II). Represent the number of bright-eyed people without dark hair by writing this number in the part of the second circle that doesn't include the overlap (region III). Write the appropriate number in the overlap (region I). How many people are included in the first circle, and how many in the second? How many people are included in both circles? How many of the 35 people are not included in either circle?......!!!!!!!!...................................
RESPONSE --> Of the 35 people, 20 have dark hair, 15 have bright eyes and 8 have both. Thus, the first circle, showing all of the dark haired people without bright eyes would show 12 people, as 20 dark haired people - 8 with bright eyes = 12 dark haired people without bright eyes. Likewise, 15 bright eyed people - 8 with dark hair = 7 bright eyed people without dark hair.
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09:58:36 Of the 20 dark-haired people in the preceding example, 8 also have bright eyes. This leaves 12 dark-haired people for that part of the circle that doesn't include the overlap (region I).
The 8 having both dark hair and bright eyes will occupy the overlap (region I). Of the 15 people with bright eyes, 8 also have dark hair so the other 7 do not have dark hair, and this number will be represented by the part of the second circle that doesn't include the overlap (region III). We have accounted for 12 + 8 + 7 = 27 people. This leaves 35-27 = 8 people who are not included in either of the circles. The number 8 can be written outside the two circles (region IV) to indicate the 8 people who have neither dark hair nor bright eyes (click below on 'Next Picture').......!!!!!!!!...................................
RESPONSE --> I didn't include the 8 people for section 4.
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QXLw Student Name: assignment #003
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10:06:44 `q001. Note that there are 5 questions in this assignment.
Again we have a total of 35 people in a room. Of these, 20 have dark hair and 15 have bright eyes. There are 8 people with dark hair and bright eyes. Let A stand for the collection of people who have dark hair and B for the collection who have bright eyes. The Intersection of these two collections is denoted A ^ B, and stands for the collection of all people who have both dark hair and bright eyes. The Union of these two collections is denoted A U B, and stands for the collection of all people who have at least one of these characteristics. In terms of the diagram you made for the preceding problem, describe the collection A ^ B and the collection A U B. Give the number of people in each of these collections (these numbers are designated by the notation n ( A ^ B) and n(A U B) ). Refer to the diagrams you have made.......!!!!!!!!...................................
RESPONSE --> For A^B, the number of people with both dark hair and bright eyes is 8. Thus A^B = 8.
For A U B, the number of people with either dark hair, bright eyes or both is 35. Thus A U B = 35..................................................
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10:07:29 The collection A ^ B consists of all the people with both dark hair and bright eyes, which corresponds to the overlap between the two circles (region I). There are 8 people in this overlap, so we say n(A ^ B) = 8.
The collection A U B consists of all the people who have least one of the characteristics. This would include the 12 people with dark hair but not bright eyes, located in the first circle but outside the overlap (region II); plus the 7 people with bright eyes but not dark hair, located in the second circle but outside the overlap (region III); plus the 8 people with both characteristics, located in the overlap (region I). Thus we include the 12 + 8 + 7 = 27 people who might be located anywhere within the two circles.......!!!!!!!!...................................
RESPONSE --> I forgot about that other 8 people with neither characteristic.
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10:12:27 `q002. Continuing the preceding example, we let A' stand for the people who are not in the collection A, and we let B' stand for the people who are not in the collection B.
What are the characteristics of the people in A', and what characterizes people in B' ? What are n(A ') and n(B '), the numbers of people in A' and B' ?......!!!!!!!!...................................
RESPONSE --> The characteristics of the people in A' are the people without dark hair. n(A') is 15 people. The characteristics of people in B' are the people without bright eyes. Thus n(B') is 15 people.
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10:12:42 Of the 35 people, those in A' are those outside of A. Since A consists of all the dark-haired people, A' consists of all the people lacking dark hair. This includes the 8 people outside of both circles (people having neither dark hair nor bright eyes, region IV) and the 7 people in the second circle but outside the overlap (people having bright eyes but not dark hair, region III). n(A ' ) is therefore 8 + 7 = 15.
Since B consists of all the bright-eyed people, B' consists of all the people lacking bright eyes. This would include the 8 people outside both circles (region IV), all of whom lack both dark hair and bright eyes, and the 12 people in the first circle but outside the overlap (region II), who have dark hair but not bright eyes. n ( B ' ) is therefore 12 + 8 = 20.......!!!!!!!!...................................
RESPONSE --> Okay.
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10:16:54 `q003. ( A U B ) ' stands for the everyone outside A U B, and ( A ^ B ) ' stands for everyone outside A ^ B. What characterizes the people in each of these collections, and how many people are there in each?
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RESPONSE --> A U B stands for all the people with either dark hair or bright eyes, so (A U B)' stands for everyone with neither dark hair or bright eyes. Thus with 35 people in the group, 20 of which have dark hair and 15 of which have bright eyes, 8 of which have both, that leaves 8 people with neither characteristic. (A^B) stands for everyone with both characteristics. (A^B)' stands for people without both characteristics. Thus (A^B)' = 27.
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10:17:09 A U B consists of everyone having at least one of the characteristics (dark hair, bright eyes), and is represented by the numbers in the two circles (regions I, II, III). ( A U B ) ' consists of the people who do not have at least one of the characteristics, and is represented by the number outside both circles (region IV). This number is 8, representing the 8 people who have neither dark hair nor bright eyes.
A ^ B stands for all the people with both of the two characteristics (represented by the overlap, region I), so ( A ^ B ) ' stands for all the people who do not have both of the two characteristics (represented by everything outside region I, or regions II, III and IV). [ Note that (A ^ B)' is not the same as the collection of people who have neither characteristic. Anyone who does not have both characteristics will be in ( A ^ B ) ' . ] ( A ^ B )' must include those who have neither characteristic, and also those who have only one of the characteristics. The 8 people outside both circles, the 12 people in the first circle but outside the overlap, and the 7 people in the second circle but outside the overlap all lack at least one characteristic to, so these 8 + 12 + 7 = 27 people make up( A ^ B ) '.......!!!!!!!!...................................
RESPONSE --> Okay.
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10:21:05 `q004. How many people are in A ' U B ', and how could those people be characterized? Answer the same for A ' ^ B '.
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RESPONSE --> A' U B' is the group of people that either don't have dark hair or don't have bright eyes. Thus of the 35 people, we must take away everyone with dark hair or bright eyes. That leaves us with 8.
A' ^ B' is the group of people that don't have dark hair and/or don't have bright eyes. That number is also 8..................................................
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10:21:22 A ' U B ' consists of all the people who are in at least one of the sets A ' or B '.
A ' consists of all the people who do not have dark hair, represented by every region of the diagram which does not include any of A. This will include the 7 people in B who are outside the overlapping region, and the 8 people who are outside of both A and B (regions III and IV. Since A consists of regions I and II, A' consists of regions III and IV). B ' consists of all the people who do not have bright eyes, represented by every region of the diagram which does not include any of B (regions II and IV). This will include the 12 people in A but outside the overlap, and the 8 people outside of both A and B. Thus A ' U B ' consists of everyone in at least one of A ' or B ', including the 7 people in B but outside the overlap (region III), the 12 people in A let outside the overlap (region II), and the 8 people outside of both A and B (region IV). These will be the people who lack at least one of the characteristics dark hair and/or bright eyes. Thus n(A' U B') = 7 + 12 + 8 = 27. Note that these are the same 27 people who are in ( A ^ B ) '. So at least in this case, ( A ^ B ) ' = A ' U B '. A ' ^ B ' consists of all the people in both A ' and B '. As before A ' includes the 7 people in B but not A (region III) as well as the 8 people outside both A and B (region IV), and B ' includes the 12 people in A but not B (region II) as well as the 8 people outside both A and B (region IV). The people in both A ' and B ' will be the 8 people outside both A and B, those who have neither dark hair nor bright eyes. We note that this is the same as the set ( A U B ) ', so at least for the present case we see that ( A ' ^ B ' = ( A U B ) '.......!!!!!!!!...................................
RESPONSE --> Okay.
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10:25:06 `q005. Succinctly describe the relationships between ( A U B ) ', A ' U B ', (A ^ B) ' and A ' ^ B '.
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RESPONSE --> (A U B) stands for people having either dark hair or bright eyes or both. A' U B' stands for people having neither characteristic. (A^B) stands for people having both dark hair and bright eyes while A' ^B' stands for people having one or the other characteristic but not both.
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10:25:16 ( A U B ) ' = A ' ^ B ' and ( A ^ B ) ' = A ' U B '. The collection outside of the union A U B is the intersection A ' ^ B ', and the collection outside the intersection A ^ B is the union A ' U B '. The ' operation changes union to intersection and intersection to union.
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RESPONSE --> Okay
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Tɵgo쯖Ffŏ Student Name: assignment #004
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10:31:38 Note that there are 4 questions in this assignment.
`q001. From the collection of letters a, b, c, d, e, how many smaller collections having at least one element may be formed?......!!!!!!!!...................................
RESPONSE --> There can be twelve smaller collections that I can count.
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10:32:04 We will list the original collection by placing its elements between braces: { a, b, c, d, e }.
The collection {a, b, c, d} is a smaller collection obtained by eliminating e from the original collection. Similarly we can eliminate d or c or b or a to get the 4-element collections {a, b, c, e}, {a, b, d, e}, { a, c, d, e} and {b, c, d, e}. Alternatively we could simply include either a or b or c or d or e in a 1-element collection, obtaining {a}, {b}, {c}, {d} and {e}. It should be clear that these are the only ways to form collections of 1 or 4 elements. To form a collection of 2 elements we could include a and one other element, obtaining { a, b}, { a, c }, { a, d } and { a, e }. Or we could include b and one other element (excluding a, since we already have the collection { a, b } which is identical to the collection { b, a } since it has exactly the same elements). We obtain { b, c }, { b, d } and { b, e }. {}Or we could include c and one other element (other than a or b, since these have already been listed) to obtain { c, d } and { c, e }. Finally we could include d and the only other element left, e, to get { d, e}. This gives us a complete listing of the 10 sets we can form with 2 of the original elements. This leaves us the 3-element sets, which can be formed by excluding the 2-element sets. Working in reverse order, we can exclude { d, e } to get { a, b, c }, or { c, e } to get { a, b, d }, etc.. The remaining sets we get in this fashion are { a, b, e}, { a, c, d }, { a, c, e}, { a, d, e}, { b, c, d}, {b, c, e}, {b, d, e}, {c, d, e}. We thus have 10 three-element sets. The total number of smaller sets containing at least one element is therefore 5 + 5 + 10 + 10 = 30.......!!!!!!!!...................................
RESPONSE --> I didn't think to go backwards.
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10:33:05 `q002. A one-to-one correspondence between two sets is a rule that associates each element of the each with exactly one element of the other. A natural one-to-one correspondence between the sets { a, b, c } and { 1, 2, 3 } would be to associate a with 1, b with 2, c with 3. This correspondence might be represented as [ a <--> 1, b <--> 2, c <--> 3 ].
This isn't the only possible one-to-one correspondence between these sets. Another might be [ a <--> 2, b <--> 1, c <--> 3 ]. In each case, every element of each set is associated with exactly one element of the other. Another correspondence between the sets might be [ a <--> 3, b<-->2, c<-->3 ]. This correspondence is not one-to-one. In what way does it fail to be a one-to-one correspondence (remember that a one-to-one correspondence is one in which every element of each set is associated with exactly one element of the other).......!!!!!!!!...................................
RESPONSE --> Because the set uses 3 twice and does not use 1 at all.
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10:33:13 [ a <--> 3, b<-->2, c<-->3 ] fails to be a one-to-one correspondence for two reasons. In the first place, 3 is associated with a and with c, and every element of each set is to be associated with exactly one element of the other. 3 is associated with two elements of the other set.
It also fails because the element 1 of the second set is not associated with anything in the first set.......!!!!!!!!...................................
RESPONSE --> Okay.
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10:38:12 `q003. There are four possible one-to-one correspondences between the sets {a, b, c} and {1, 2, 3} which were not described in the preceding exercise. One of them would be [ a <--> 3, b <--> 2, c <--> 1 ]. What are the other three possible one-to-one correspondences?
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RESPONSE --> [ a <----> 2 , b <----> 1 , c<----> 3 [ a <----> 1 , b <----> 2 , c<----> 3 [ a <----> 3 , b <----> 3 , c<----> 2
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10:38:55 If we designate the correspondence [ a <--> 1, b <--> 2, c <--> 3 ] as the '123' correspondence, [a <--> 2, b <--> 1, c <--> 3 ] as the '213' correspondence and [a <--> 3, b <--> 2, c <--> 1 ] as the '321' correspondence, in each case listing the numbers associated with a, b, c in that order, we see that the remaining three correspondences could be designated 132, 231 and 312. These correspondences could of course be written out as [ a <--> 1, b <--> 3, c <--> 2 ], [ a <--> 2, b <--> 3, c <--> 1 ] and [ a <--> 3, b <--> 1, c <--> 2 ].
Note that 123, 132, 213, 231, 312, 321 represent the six ways of rearranging the digits 1, 2, 3 into a 3-digit number, listed in increasing order.......!!!!!!!!...................................
RESPONSE --> Okay
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10:39:25 `q004. Explain why it is not possible to put the sets { a, b, c} and {1, 2, 3, 4} into a one-to-one correspondence.
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RESPONSE --> Because there are only 3 letters and 4 numbers, thus one number will be left out.
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10:39:31 One set has 3 elements and the other has 4 elements. A 1-to-1 correspondence has to match each element of each set with exactly one element of the other. It would not be possible to find four different elements of the first set to match with the four elements of the second.
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RESPONSE --> Okay.
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nQdkԷϕz} Student Name: assignment #005
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10:41:34 `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.......!!!!!!!!...................................
RESPONSE --> [ 1 <----> 1, 2 <----> 3, 3<----> 5 ....]
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10:41:43 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
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RESPONSE --> Okay.
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10:45:34 `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, ... ]?......!!!!!!!!...................................
RESPONSE --> The rule of correspondance can be stated like this: the numbers in each set will correspond to the number in the same position in the second set.
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10:45:51 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 --> Okay.
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10:47:37 `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, ... ]?......!!!!!!!!...................................
RESPONSE --> 2n - 1
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10:47:47 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.......!!!!!!!!...................................
RESPONSE --> Okay.
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10:48:17 `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.
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RESPONSE --> 5n.
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10:48:27 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 --> Okay.
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10:57:32 `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }.
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RESPONSE --> I can recognize the pattern but I can't figure out a rule for it. The rule I would suggest is n <---> n + 5, but that doesn't account for how 1 corresponds to 7.
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10:58:05 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, ... ].......!!!!!!!!...................................
RESPONSE --> Okay, I understand now.
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11:03:40 `q006. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 3, 10, 17, ... }.
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RESPONSE --> n <---> 7n - 4
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11:03:47 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, ... ].......!!!!!!!!...................................
RESPONSE --> Okay.
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11:06:23 `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, ... }.......!!!!!!!!...................................
RESPONSE --> Set 1: 1/1, 1/2, 1/3 ... Set 2: 2/1, 2/2, 2/3 ... Set 3: 3/1, 3/2, 3/3 ...
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11:06:48 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), ... ].......!!!!!!!!...................................
RESPONSE --> Okay.
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11:10:11 `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
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11:10:28 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.......!!!!!!!!...................................
RESPONSE --> Okay.
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