#$&* course Mth 151 1/17/2012 9:51PM 002. `Query 2
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Given Solution: `a** It's true because all elements of C are in the universal set, and because there are elements of U that aren't in C. You have to have both conditions, since a proper subset cannot be identical to other set. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 2.2.30 phi s D YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The statement is true because an empty set can be in any subset. confidence rating #$&*: 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q2.2.33 D not s B Is the statement true or false and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The statement is true because there are elements in D that are not in B, so D is not a subset of B. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** D is a subset of B if every element of D is an element of B-i.e., if D doesn't contain anything that B doesn't also contain. The statement says that D is not a subset of B. This will be so if D contains at least one element that B doesn't. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q2.2.36 there are exactly 31 subsets of B YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the number of subsets you do 2^n, n is the number of elements. There are 5 elements in B, so its 2^5 which equals 32, not 31, so the statement is false. 2^5=32 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 2.2.40 there are exactly 127 proper subsets of U Is the statement true or false and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find proper subsets you do the same thing to find subsets then subtract 1. There are 7 elements in U, so it would be 2^7-1 which equals 127, so the statement would be true 2^7-1=127 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The set is not a proper subset of itself, and the set itself is contained in the 2^n = 2^7 = 128 subsets of this 7-element set. This leaves 128-1 = 127 proper subsets. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 2.2.48 U={1,2,...,10}, complement of {2,5,7,9,10} What is the complement of the given set? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The complement is all the elements not in the given set but were in U, so the complement would be {1, 3, 4, 6, 8} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery 2.2.63 in how many ways can 3 of the five people A, B, C, D, E gather in a suite? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using trial and error I listed all the possible combinations and ended up with 10. {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} confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The answer here would consist of a list of all 3-element subsets: {a,b,c}, {a,b,d}, {a,b,e}, {a,c,d} etc. There are ten such subsets. Using a,b,c,d,e to stand for the names, we can list them in alphabetical order: {a,b,c), {a,b,d}, {a,b,e}, {a,c,d}, {a,c,e}, {a,d,e|, {b,c,d}, {b,c,e}, {b,d,e}, {c, d, e}** " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qquery 2.2.63 in how many ways can 3 of the five people A, B, C, D, E gather in a suite? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using trial and error I listed all the possible combinations and ended up with 10. {a, b, c} {a, b, d} {a, b, e} {a, c, d} {a, c, e} {a, d, e} {b, c, d} {b, c, e} {b, d, e} {c, d, e} confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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}** " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!