23 text assignment

course Mth 151

Pnnܐ~ϫ{olfyassignment #003

003. `Query 3

College Algebra

09-10-2008

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19:43:38

Query 2.3.15 (Y ^ Z')U X, univ={a,..g}, X={a,c,e,g}, Y = {a,b,c}, Z = {b, ..., f}

What is the set (Y ^ Z')U X?

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RESPONSE -->

Y= [a,b,c] and the compliment of Z gives us Z'= [a,g]. They share only [a]. When you union [a] with X, that gives us the set

[a,c,e,g]

confidence assessment: 2

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19:43:54

**Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'.

So (Y ^ Z') U X = {a, c, e, g}, the set of all elements which lie in at least one of the sets (Y ^ Z') U X. **

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RESPONSE -->

OK

self critique assessment: 3

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19:46:04

Give the intersection of the two sets Y and Z'

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RESPONSE -->

The set Y has these elements [a,b,c], and the set Z', which is a compliment of everything NOT in set Z is [a,g]. The intersection is what the two sets have in common, which is the element a. So [a] is the intersection of the two sets.

confidence assessment: 2

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19:46:12

**Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'.**

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RESPONSE -->

OK

self critique assessment: 3

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19:48:05

Query 2.3.30 describe in words (A ^ B' ) U (B ^ A')

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RESPONSE -->

The intersection of A and the compliment of B unioned with the intersection of B and the compliment of A'.

confidence assessment: 2

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19:51:06

** a description, not using a lot of set-theoretic terms, of (A ^ B' ) U (B ^ A') would be, all the elements that are in A and not in B, or that are not in A and are in B

Or you might want to say something like 'elements which are in A but not B OR which are in B but not A'.

STUDENT SOLUTION WITH INSTRUCTOR COMMENT:everything that is in set A and not in set B or everything that is in set B and is not in set A.

INSTRUCTOR COMMENT: I'd avoid the use of 'everything' unless the word is necessary to the description. Otherwise it's likely to be misleading. **

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RESPONSE -->

I did not delve into the question as deeply as I could have. I still have problems translating the problems into words. Or maybe I have a good grasp on it and just solved it in an alternate way.

self critique assessment: 2

A verbal description of a set should never contain set-theoretic words like union, intersection, complement, etc.. The description should be in everyday language.

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19:52:30

2.3.51 always or not always true: n(A U B) = n(A)+n(B)

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RESPONSE -->

This is always true. The cardinal numbers of a and b will always been combined if a union is made between the two, it's simple addition.

confidence assessment: 2

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19:53:28

** This conclusion is contradicted by many examples, including the one of the dark-haired and bright-eyed people in the q_a_.

Basically n(A U B) isn't equal to n(A) + n(B) if there are some elements which are in both sets--i.e., in the intersection.

}

MORE DETAIL: The statement can be either true or false, depending on the sets A and B; it is not always true.

The statement n(A U B) = n(A)+n(B) means that the number of elements in A U B is equal to the sum of the number of elements in A and the number of elements in B.

The statement would be true for A = { c, f } and B = { a, g, h} because A U B would be { a, c, f, g, h} so n(A U B) = 5, and n(A) + n(B) = 2 + 3 = 5.

The statement would not be true for A = { c, f, g } and B = { a, g, h} because A U B would be the same as before so n(AUB) = 5, while n(A) + n(B) = 3 + 3 = 6.

The precise condition for which the statement is true is that A and B have nothing in common. In that case n(A U B) = n(A) + n(B). A more precise mathematical way to state this is to say that n(A U B) = n(A) + n(B) if and only if the intersection A ^ B of the two sets is empty. **

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RESPONSE -->

I will remember this, and have saved these as notes. I will now remember that that can be either true or false, but never always true, depending on the elements.

self critique assessment: 2

Remember also the qa questions related to hair and eye color.

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19:58:44

Query 2.3.60 X = {1,3,5}, Y = {1,2,3}. Find (X ^ Y)' and X' U Y'.

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RESPONSE -->

U= [1,2,3,4,5}

(x ^ y)' are elements that aren't shared between x and y.. So that leaves us with [2,4,5].

The union of elements not in X and elements not in Y gives us {2,4,5] also.

confidence assessment: 3

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19:59:07

** X ^ Y = {1,3} so (X ^ Y) ' = {1,3}' = {2, 4, 5}.

(X ' U Y ' ) = {2, 4} U {4, 5} = {2, 4, 5}

The two resulting sets are equal so a reasonable conjecture would be that (X ^ Y)' = X' U Y'. **

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RESPONSE -->

OK

self critique assessment: 3

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20:03:49

2.3.72 A = {3,6,9,12}, B = {6,8}.

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RESPONSE -->

A X B gives us the subset and ordered pairs of

[(3,6), (3,8), (6,6),(6,8),(9,6),(9,8),(12,6),(12,8). That gives us the Cartesian product of 8, which is ab=ba.

B X A gives us the set and ordered pairs of [(6.3).(6.6).(6,9),(6,12),(8,3),(8,6),(8,9),(8,12)] which gives us a Cartesian product of 8 also.

confidence assessment: 3

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20:05:21

** (A X B) = {(3,6),(3,8),(6,6),(6,8),(9,6),(9,8),(12,6), (12,8)}

(B X A) = (6,3),(6,6),(6,9),(6,12),(8,3),(8,6),(8,9),(8,12)}

How is n(A x B) related to n(A) and n(B)?

n(S) stands for the number of elements in the set S, i.e., its cardinality.

n(A x B) = n(A) * n(B) **

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RESPONSE -->

I understand that the formula for finding the cartesian product is the cardinal number of set A multiplied by the cardinal product of B, which gives the big cardinal number found in the basic formula of the cartesian product.

self critique assessment: 2

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20:09:28

2.3.84 Shade A U B

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RESPONSE -->

I separated the diagram into regions.

A= 1

B=2

^=3

U=4

I took what was in A which is (1,3) and what was in circle B which is (3,2) and made a union of those elements which gave me [1,3,2} which is all the elements in A and B, including the overlap. So I shaded regions 1,2 and 3, leaving U unshaded.

confidence assessment: 3

Your meaning is clear and in terms of the regions you define your solution is correct.

However note that region 4 is not U. U is the universal set and consists of all of A, all of B, and everything outside these two sets. The regions you intend would be denoted

A B (your region 1)

B A (your region 2)

A ^ B (your region 3) and

U (A U B) (region 4). Your region 4 could also be denoted (A U B) .

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20:09:35

** everything in A and everything in B would be shaded. The rest of the universal set (the region outside A and B but still in the rectangle) wouldn't be. **

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RESPONSE -->

ok

self critique assessment: 3

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20:19:39

Query 2.3.100 Shade (A' ^ B) ^ C

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RESPONSE -->

I drew a Venn Diagram with 3 circles and a universal set on the outside. I classified the following as such

A= (1,2,4,5)

B= (2,3,5,6)

C= (4,5,6,7)

U= (1,2,3,4,5,6,7)

I took what wasn't in A which gave me

(3,6,7)

and found what they shared with what was in B and got

(3,6)

I then found what the intersect was between that set and all of circle B, and the only thing in common was

(6).

So I found the corresponding region 6, which is the overlap shared with B, in other words I shaded the bottom part of circle B, and the upper right part of circle C

Shaded= Region 6.

confidence assessment: 3

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20:20:38

** you would have to shade every region that lies outside of A and also inside B and also inside C. This would be the single region in the overlap of B and C but not including any part of A. Another way to put it: the region common to B and C, but not including any of A **

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RESPONSE -->

I am happy that I was able to find the correct region. I will simplify my answers more in future problems to save redundancy or confusion.

self critique assessment: 2

There was nothing at all wrong with your solution, except that there is also an eighth region, consisting of U (A U B U C), also denoted (A U B U C) .

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20:27:04

Describe the shading of the set (A ^ B)' U C.

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RESPONSE -->

(A ^ B)' is everything in A, B, and C excluding the middle overlaps. When you make a union of that with C, you shade all regions except that shared by A and B, but not C. In general, all is shared except for that which A and B share together exclusively.

confidence assessment: 3

Very good, succinct and accurate description.

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20:27:30

** All of C would be shaded because we have a union with C, which will include all of C.

Every region outside A ^ B would also be shaded. A ^ B is the 'overlap' region where A and B meet, and only this 'overlap' would not be part of (A ^ B) '. The 'large' parts of A and B, as well as everything outside of A and B, would therefore be shaded.

Combining this with the shading of C the only the part of the diagram not shaded would be that part of the 'overlap' of A and B which is not part of C. **

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RESPONSE -->

OK

self critique assessment: 3

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20:34:42

2.3.114 Largest area of A shaded (sets A,B,C). Write a description using A, B, C, subset, union, intersection symbols, ', - for the shaded region.

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RESPONSE -->

(B U C) - A

confidence assessment: 2

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20:34:56

** Student Answer and Instructor Response:

(B'^C')^A

Instructor Response:

Good. Another alternative would be A - (B U C ), and others are mentioned below.

COMMON ERROR: A ^ (B' U C')

INSTRUCTOR COMMENT: This is close but A ^ (B' U C') would contain all of B ^ C, including a part that's not shaded. A ^ (B U C)' would be one correct answer. **

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RESPONSE -->

OK

self critique assessment: 3

Note that (B U C) - A is different from A - (B U C) .

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