sguinn6001emailvccsedu

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course 151

9/10 52.1,2.2

001. `Query 1

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Question: `qQuery 2.1.12 counting #'s 4 to 14

List the elements of the set.

 

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Your solution:

Since it says to list the set counting #’s 4 to 14 both of those numbers must be included in the set

And since it only asks for a list it doesn’t need to be in brackets

4,5,6,7,8,9,10,11,12,13,14

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Given Solution:

`a**A list of the elements would just be 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. **

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Self-critique (if necessary):

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Question: `qquery 2.1.24 listing for set of presidents between LBJ and Clinton

 

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Your solution:

{x|x is a president who served between LBJ and Bill Clinton}

{Lyndon Johnson, Richard Nixon, Gerald Ford, Jimmy Carter, Ronald Regan, George HW Bush, Bill Clinton}

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Given Solution:

`a** A listing would be {}{}{Lyndon Johnson, Richard Nixon, Gerald Ford, Jimmy Carter, Ronald Regan, George HW Bush, William Clinton}.{}{}Set-builder notation is {x|x is a president who served between Lyndon Johnson and William Clinton}

x is a variable and the condition 'x is a president who served between Lyndon Johnson and William Clinton' tells you what possible things the variable can be.

COMMON ERROR: It's incorrect to say {x | x is the set of presidents who served between Johnson and Clinton}.

x is a president, not a set of presidents. Should be {x|x is a president who served between Lyndon Johnson and William Clinton} **

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Self-critique (if necessary):

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Question: `q2.1.40 finite or infinite: set of rat #'s 0 to 1

 

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Your solution:

Infinite, because there is an infinite number of fractions between 0 and 1

{1/2, 1/3, ¼, 1/5, 1/6, ….} it can go on forever

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Given Solution:

`a** Rational numbers have form p/q, where p and q are integers. Numbers like 5/8, 57/31, -3/5, -57843/7843, etc.

The subset {1/2, 1/3, 1/4, 1/5, ... } is just by itself an infinite set of rational numbers between 0 and 1.

Then you have things like 348/937, and 39827389871 / 4982743789, and a whole infinite bunch of others. There are thus infinitely many rational numbers in any interval of the real line.

COMMON MISCONCEPTION: finite, because it doesn't go on forever

Rational numbers have form p/q, where p and q are integers. Numbers like 5/8, 57/31, -3/5, -57843/7843, etc. Not all of these lie between 0 and 1, of course. **

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Self-critique (if necessary):

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Question: `q2.1.48 n(A), A={x|x is a U.S. senator}

What is n(A) and why?

 

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Your solution:

n(A) stands for the cardinal number of the set

in this n(A)=100 because there are 100 senators in the us senate 2 from each state

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Given Solution:

`a** n(A) stands for the number of elements in the set--in this case for the number of senators. There are 100, 2 from each State. So n(A) = 100. **

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Self-critique (if necessary):

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Question: `qquery 2.1.54 {x|x is neagtive number}

 

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Your solution:

{x|x is a negative number} defined because anyone can tell whether or not a number is included in the set

{-1,-2,-3,-4,-5,-6, ….}

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Given Solution:

`a** This notation means all possible values of x such that x is a negative number.

The question is whether the set is well-defined or not.

It is in fact well-defined because there is a definite way to decide whether a given object is an element of the set, because there is a definite way to determine whether an object is a negative number or not.

ALTERNATIVE ANSWER: The set is well-defined because you have a criterion by which you can definitely decide whether something is or is not in the set. **

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Self-critique (if necessary):

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Question: `q2.1.74 (formerly 2.1.72) This was not assigned, but you should be able to answer based on your work on similar problems: It is or is it not true that 2 is not not subset of {7,6,5,4}?

 

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Your solution:

True because 2 is not a set it is just a number and cannot be a sub set of any set.

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2

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Given Solution:

`a** The statement is that 2 is not a subset. The statement is true because 2 isn't even a set, it's just a number. {2} is a set and could be a subset of something. 2 is just a number; it isn't a set so it can't be a subset of anything.

The usual answer is that 2 isn't a subset because 2 isn't in the set. However that's not the correct reason. The correct reason is that 2 isn't a set and a subset must be a set.

COMMON MISCONCEPTION: the statement says that 2 is not a subset, not that it is not an element of the set. So the reason it's not a subset is that 2 isn't a set at all, so it can't be a subset of anything. **

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Self-critique (if necessary):

I got the answer right but I wasn’t confident about it

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Question: `q2.1.86 (formerly 2.1.84). This was not assigned but you did answer several questions related to the sets C={4,10,12}, B={2,4,8,10}, and should be able to answer this.

Is it true or false that every element of C is also an element of B? Be sure to include your reasoning.

 

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Your solution:

False because 12 is in C but not in B

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Given Solution:

`a** Specifically it's false because the element 12 is in C but not in B. **

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

002. `Query 2

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Question: `q2.2.24 U={a,b,...,g}, A={a,e}, B={a,b,e,f,g}, C={b,f,g}, D={d,e} Is C ps U?

Is the statement true or false and why?

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Your solution:

True because there is at least one element in U that is not contained in C

But all of C is contained in U

confidence rating #$&*

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Given Solution:

`a** It's true because all elements of C are in the universal set, and because there are elements of U that aren't in C. You have to have both conditions, since a proper subset cannot be identical to other set. **

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Self-critique (if necessary):

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Question: `qQuery 2.2.30 phi s D

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Your solution:

The empty set is a subset of all sets

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2

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Given Solution:

`a** Note that You should be responding to problem 2.2.30 from the homework you worked out on paper. The shorthand notation is for my reference and won't always make sense to you. For clarification, though, the symbol for the empty set is the Greek letter phi.

One set is a subset of another if every element of that set is in the other. To show that a set isn't a subset of another you have to show something in that set that isn't in the other.

There's nothing in the empty set so you can never do this--you can never show that it has something the other set doesn't. So you can never say that the empty set isn't a subset of another set.

Thus the empty set is a subset of any given set, and in particular it's a subset of D.

ALTERNATIVE ANSWER: As the text tells you, the empty set is a subset of every set.

ANOTHER ALTERNATIVE

Every element of the empty set is in D because there is no element in the empty set available to lie outside of D.

ONE MORE ALTERNATIVE: The empty set is a subset of every set. Any element in an empty set is in any set, since there's nothing in the empty set to contradict that statement. **

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Self-critique (if necessary):

I like the last alternate answer better than the book answer

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Question: `q2.2.33 D not s B

Is the statement true or false and why?

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Your solution:

True because B does not include all the elements included in D

confidence rating #$&*

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Given Solution:

`a** D is a subset of B if every element of D is an element of B-i.e., if D doesn't contain anything that B doesn't also contain.

The statement says that D is not a subset of B. This will be so if D contains at least one element that B doesn't. **

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Self-critique (if necessary):

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Question: `q2.2.36 there are exactly 31 subsets of B

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Your solution:

False there are 32 subsets of B, because to find the number of subsets you take 2^n where n equals the number of elements in a given set B has 5 elements which equals 2^5 which is

2*2*2*2*2=32 subsets

confidence rating #$&*

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Given Solution:

`a** If a set has n elements then is has 2^n subsets, all but one of which are proper subsets. B has 5 elements so it has 2^5 = 32 subsets. So the statement is false.

There are exactly 31 proper subsets of B, but there are 32 subsets of B. **

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Self-critique (if necessary):

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Question: `qQuery 2.2.40 there are exactly 127 proper subsets of U

Is the statement true or false and why?

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Your solution:

To find proper subsets the equation is 2^n-1

True because U contains 7 elements which is 2^7=2*2*2*2*2*2*2=128 subsets

But because U is not a proper subset itself you have to subtract 1 which is 128-1=127

confidence rating #$&*

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Given Solution:

`a** The set is not a proper subset of itself, and the set itself is contained in the 2^n = 2^7 = 128 subsets of this 7-element set. This leaves 128-1 = 127 proper subsets. **

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Self-critique (if necessary):

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Question: `qQuery 2.2.48 U={1,2,...,10}, complement of {2,5,7,9,10}

What is the complement of the given set?

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Your solution:

{1,3,4,6,8} is the complement because it is not contained in the second set but are contained in the U set

confidence rating #$&*

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Given Solution:

`a** the complement is {1,3,4,6,8}, the set of all elements in U that aren't in the given set. **

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Self-critique (if necessary):

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Question: `qquery 2.2.63 in how many ways can 3 of the five people A, B, C, D, E gather in a suite?

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Your solution:

{A,B,C,D,E}

{A,B,C}, {A,B,D},{A,B,E},{A,C,D},{A,C,E},{A,D,E},{B,C,D},{B,C,E}{B,D,E}{C,D,E}

There are 10 different subsets of 3 elements of the set

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Given Solution:

`a** The answer here would consist of a list of all 3-element subsets: {a,b,c}, {a,b,d}, {a,b,e}, {a,c,d} etc. There are ten such subsets.

Using a,b,c,d,e to stand for the names, we can list them in alphabetical order:

{a,b,c), {a,b,d}, {a,b,e}, {a,c,d}, {a,c,e}, {a,d,e|, {b,c,d}, {b,c,e}, {b,d,e}, {c, d, e}**

"

Very nicely done.

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