course mth 151
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fࣅF~Wy Student Name: assignment #005
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10:14:39 `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...
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10:14:46 This correspondence can be written [ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ].
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RESPONSE --> ok
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10:16:52 `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 --> i'm really not sure, when you do the correspondence thats the way the odds fall into place
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10:17:08 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 --> ok i understand
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10:18:59 `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<>n+1
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10:19:12 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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10:20:07 `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.
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RESPONSE --> n<>5n
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10:20:11 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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10:21:44 `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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10:21:52 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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10:22:46 `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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10:22:50 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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10:24:35 `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 --> 2<>1/2 3<>1/3 4<>1/4
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10:24:44 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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10:27:15 `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 --> 1<>1/2 2<>1/3 3<>1/4 and the 1<>2/2 2<>2/3 3<>2/4
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10:27:20 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
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q^~氋 assignment #005 Fᚭy|Cב Liberal Arts Mathematics I 06-07-2006
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09:59:45 Query 2.5.12 n({9, 12, 15, ..., 36})
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RESPONSE --> {(36-9)/3}+1 = 10
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10:00:00 ** 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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10:01:04 Query 2.5.18 n({x | x is an even integer }
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RESPONSE --> infinite
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10:01:26 ** {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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10:04:21 Query 2.5.18 how many diff corresp between {stallone, bogart, diCaprio} and {dawson, rocky, blaine}?
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RESPONSE --> stallone <> rocky, bogart <> blaine, di Caprio <> dawson
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10:05:10 ** 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 --> i thought we were suppose to show just the one correspondence. I understand how you got six.
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10:10:29 2.5.36 1-1 corresp between counting #'s and {-17, -22, -27, ...}
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RESPONSE --> -17<>1 -22<>2 -27<>3 -32<>4...-n-5<>n
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10:10:58 **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
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10:12:16 2.5.42 show two vert lines, diff lengths have same # of points
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RESPONSE --> 1<>1 2<>2 3<>3 4<>4 etc
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10:13:02 ** 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 --> that was tough question, i was confused about what it really wanted
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