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Math Help - Comparing sets

  1. #1
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    Comparing sets

    Hey everyone,

    Thanks for taking the time to look at my thread.

    I have this question, and a particular part of it is confusing me, I was hoping someone could help me reach the right place.

    The question is:

    Let A = {(x,y) is element of ZxZ : | x-y | <= 1},
    B = {0,1,2,3,4,5}
    C = {3,4,5}

    Determine |A Intersection (B x C) |

    Now I know what im supposed to do. Determine the cardinality of the resultant set of A Intersection (B x C). B and C are right there, however
    acquiring the members of A is confusing me. It appears to me that the set in A would have elements consisting of ordered pairs, because of (x,y), but this doesn't seem right to me. Z x Z means the domain of Integers by cartesian product right? so like... (0,0), (0,1), (0,2)... then (1,0), (1,1)...
    and so on? I believe the set A is supposed to have singular elements in it, as opposed to pairs, but how do I get this set. I would appreciate any help =] and I apologise for the vagueness of the question. I haven't done maths for a few years and this is my first experience with this branch.

    Cheers =]
    Last edited by yoonsi; August 1st 2009 at 10:43 PM. Reason: Fixed a typo
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by yoonsi View Post
    Hey everyone,

    Thanks for taking the time to look at my thread.

    I have this question, and a particular part of it is confusing me, I was hoping someone could help me reach the right place.

    The question is:

    Let A = {(x,y) is element of ZxZ : | x-y | <= 1},
    B = {0,1,2,3,4,5}
    C = {3,4,5}

    Determine |A Intersection (B x C) |

    Now I know what im supposed to do. Determine the cardinality of the resultant set of A Intersection (B x C). B and C are right there, however
    acquiring the members of A is confusing me. It appears to me that the set in A would have elements consisting of ordered pairs, because of (x,y), but this doesn't seem right to me. Z x Z means the domain of Integers by cartesian product right? so like... (0,0), (0,1), (0,2)... then (1,0), (1,1)...
    and so on? I believe the set A is supposed to have singular elements in it, as opposed to pairs, but how do I get this set. I would appreciate any help =] and I apologise for the vagueness of the question. I haven't done maths for a few years and this is my first experience with this branch.

    Cheers =]
    No, actually, the set A ist clearly a set of pairs. To elaborate a bit: The notation A = \{(x,y) \in \mathbb{Z}\times \mathbb{Z} : | x-y | \leq 1\} means that A is the set of all pairs (x,y) from \mathbb{Z}\times \mathbb{Z} that satisfy the additional condition that |x-y|\leq 1, i.e. that their coordinates differ by at most \pm 1. Thus the pairs (0,0), (1,0), (0,1), (-1,0),(7,8), (-8,-7) are all in A, to give a few examples.
    I think the easiest way to determine the cardinality of the the intersection of A with B\times C is to check all the pairs in B\times C, whether they are in \mathbb{Z}\times\mathbb{Z} (which is always true) and satisfy |x-y|\leq 1 (which may or may not be true): those pairs belong to A as well and are therefore elements of the intersection.
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  3. #3
    Senior Member Danneedshelp's Avatar
    Joined
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    B\times{C}= \{(0,3),(0,4),(0,5),(1,3),(1,4),<br />
(1,5),(2,3), (2,4),(2,5),(3,3),(3,4),<br />
(3,5),(4,3),(4,4),<br />
(4,5),(5,3),(5,4),(5,5)\}

    Let, A = \{(x,y)\in{Z\times{Z}}  : | x-y |\leq{ 1}\}\

    So, A\cap{(B\times{C})}=\{(2,3),(3,3),(3,4),(4,3),(4,4  ),(4,5),(5,4),(5,5)\}
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  4. #4
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    Thanks!

    Thank you both very much, I appreciate that you took the time. Looking at it now, it seems very obvious now, and I feel kind of silly.
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  5. #5
    Newbie
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    Quote Originally Posted by yoonsi View Post
    Hey everyone,

    Thanks for taking the time to look at my thread.

    I have this question, and a particular part of it is confusing me, I was hoping someone could help me reach the right place.

    The question is:

    Let A = {(x,y) is element of ZxZ : | x-y | <= 1},
    B = {0,1,2,3,4,5}
    C = {3,4,5}

    Determine |A Intersection (B x C) |

    Now I know what im supposed to do. Determine the cardinality of the resultant set of A Intersection (B x C). B and C are right there, however
    acquiring the members of A is confusing me. It appears to me that the set in A would have elements consisting of ordered pairs, because of (x,y), but this doesn't seem right to me. Z x Z means the domain of Integers by cartesian product right? so like... (0,0), (0,1), (0,2)... then (1,0), (1,1)...
    and so on? I believe the set A is supposed to have singular elements in it, as opposed to pairs, but how do I get this set. I would appreciate any help =] and I apologise for the vagueness of the question. I haven't done maths for a few years and this is my first experience with this branch.

    Cheers =]
    Hey yoonsi...

    A is a set of ordered pairs from the integers just as you supposed. The caveat is that the difference of the two numbers must be <=1 which makes life a little easier.

    Since we are going to look at the intersection of A with BxC, and both B and C have integers > 0 and < 5 let's restrict the pairs in A to between 0 and 5.

    Thus we have ordered pairs from {0,1,2,3,4,5} which have a difference of <=1.
    {(0,0),(0,1),
    (1,0),(1,1),(1,2),
    (2,1),(2,2),(2,3),
    (3,2),(3,3),(3,4),
    (4,3),(4,4),(4,5),
    (5,4),(5,5)}

    Now you can determine BxC to get another set of ordered pairs, then find the intersection.

    Hope that helps.
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