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course Mth 151

Submitting assignment2.5 QA

005. Infinite Sets*********************************************

Question: `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.

SOLUTION:[ 1 <--> 1, 2 <--> 3, 3 <--> 5, ... ]

confidence rating #$&*: Very confident

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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, ... ]?

SOLUTION: Add together the current and previous number in the first set to arrive at the number in the second set.

confidence rating #$&*: Very confident

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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 written [ 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, ... ]?

SOLUTION: (n*2)-1

confidence rating #$&*: Somewhat confident

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Given Solution: 2n-1

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Self-critique (if necessary): My formula was less algebraic than the givven solution

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Question: `q004. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 5, 10, 15, ... }.

SOLUTION: n<-->5n where n is the number in the first set

confidence rating #$&*: Very confident

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Question: `q005. Write an unambiguous rule involving n for the correspondence between { 1, 2, 3, ... } and { 7, 12, 17, ... }.

SOLUTION:n<-->5n+2

confidence rating #$&*: Very confident

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NOTE: There are only 5 questions here "

Self-critique (if necessary):

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Self-critique rating:

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&#Your work looks very good. Let me know if you have any questions. &#