Assignment 5

course Mth 151

?Vx?||???????assignment #005?w?????}??yO?????Liberal Arts Mathematics I

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

06-28-2006

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16:27:52

Query 2.5.12 n({9, 12, 15, ..., 36})

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RESPONSE -->

There is a difference of 3 inbetween each number, so there would be a total of 10 elements.

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16:27:58

** 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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16:29:08

Query 2.5.18 n({x | x is an even integer }

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RESPONSE -->

aleph-null - If you start with an even integer and continue on then you will have an infinite number.

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16:29:30

** {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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16:30:58

Query 2.5.18 how many diff corresp between {stallone, bogart, diCaprio} and {dawson, rocky, blaine}?

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RESPONSE -->

6 - If you leave the 1st set as {S, B, D} and you change the other sets, it will be 6 different sets.

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16:31:44

** 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 got the same answer, but I didn't work it out that way.

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16:34:32

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, ...}

-17 with 1

-22 with 2

-27 with 3

-32 with 4

n-5 with n

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16:38:24

**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 -->

I forgot to figure the difference between the 2nd set (1) and the 1st set (-17). I was only thinking about the difference in the 1st set (the negative integer set).

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16:38:58

2.5.42 show two vert lines, diff lengths have same # of points

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I have no clue about this one.

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16:41:27

** 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 -->

I am glad you said it was tough. If this one is on the test, I'll just skip it.

It won't be on the test.

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Good work. Let me know if you have questions.