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 August 4th, 2010, 06:15 AM #1 Newbie   Joined: Aug 2010 Posts: 7 Thanks: 0 Cardinality, Sets, in one triple Question. Hey guys, this is my first post, (Hi) was just wondering if i could get your help. I'm studying for my repeats and you guys can save me. If X = {1,2,3,4}, Y = {2,4,6} what is the cardinality of the following sets? (i) A = {x|x mod 2 = 0 and 0 <=x<=20} (ii) B = X * X * Y (iii) C = {(x,y)|x ¬'NOT EQUALS' SIGN HERE¬ y and x,y E (as in Element of) X I am sorry for not being able to type the actual symbols i see in part (iii) of this question. please explain your train of thought in solving this. I am trying hard to understand the right way to approach this question quickly, Thank you for your time guys...
 August 4th, 2010, 06:22 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Cardinality, Sets, in one triple Question. Welcome! I assume you have trouble calculating the sets in question, because if you could figure that out you could, presumably, count the elements. But what is the trouble? They seem straightforward. Can you explain where you're stuck? Is there a symbol you don't know? For the second, I assume the * is the Cartesian product, that is, you're looking for triples (x1, x2, y) where x and x2 are in X and y is in Y. If there are |X| choices for the first element, |X| choices for the second element, and |Y| choices for the third element, there are |X| * |X| * |Y| choices overall.
 August 4th, 2010, 06:45 AM #3 Newbie   Joined: Aug 2010 Posts: 7 Thanks: 0 Re: Cardinality, Sets, in one triple Question. Thanks CRG you are plain awesome!!! I have just truly learnt the meaning of Cartesian, and having an example in front of me has certainly made it less intimidating for me to read through (love it when math comes down to real-world examples in my head) which i will! I've tried them out, with the following answers:- (i) A = {2,4,6,8,10,12,14,16,18,20}, Therefore A has a cardinality of 10 (elements). (ii) B = Cartesian Product of 'X times X times Y' or better yet '4 by 4 by 3' elements each to give a total of 48 in cardinality?! (iii) C = UNSOLVED!!! I wanna make sure you agree with me having the right answers since you're the pro I really wanna know what the '|' symbol stands for or means, as in 'x|x'. Hard to specifically search for in a book.
August 4th, 2010, 07:52 AM   #4
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Re: Cardinality, Sets, in one triple Question.

Quote:
 Originally Posted by Aeonitis I really wanna know what the '|' symbol stands for or means, as in 'x|x'. Hard to specifically search for in a book.
In this case, | is "such that". $C= \{(x,y)|x \neq y \wedge x,y \in X\}$ is "C equals the set containing pairs (x, y) such that x is not equal to y and x and y are both in X". You can see that the elements in this set are just those of the Carteasian product X x X that are not of the form (x, x). How many are in the product? How many have both members equal? Subtract and you have your answer.

 August 4th, 2010, 08:32 AM #5 Newbie   Joined: Aug 2010 Posts: 7 Thanks: 0 Re: Cardinality, Sets, in one triple Question. You rock CRG!!! Thank you so much, i hope to bump into you a lot in the future hee hee I will post the full answer for future questioneers (i) A = {2,4,6,8,10,12,14,16,18,20}, Therefore A has a cardinality of 10 (elements). (ii) B = Cartesian Product of 'X times X times Y' or better yet '4 by 4 by 3' elements each to give a total of 48 in cardinality?! (iii) C = {(x,y)|x ? y and x,y ? X} pairs x,y {such as (1,1),(1,2),etc...} drawn from set X with a cardinality of '4 by 4 = 16' as in the question "x,y ? X". Due to the statement 'x ? y' pairs can't come in equals, discarding the following four sets (1,1),(2,2),(3,3),(4,4). The end product is 16-4 giving a cardinality of 12 for set 'C'.

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