Results 1 to 2 of 2

Math Help - Relations help..

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    35

    Relations help..

    I am new to discrete math, and am having trouble with a couple problems..

    What is the inverse of the following relations:
    Rsub1 = {(a, b) | a < b}
    Rsub2 = {(a, b) | a divides b}

    Also, i need help with composition.. :
    What is Rsub3 of Rsub4 if
    Rsub3 = {(a, b) | a > b}
    Rsub4 = {(a, b) | a >= b}
    Last edited by Manizzle; January 26th 2009 at 05:25 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Inverse Relations

    Hello Manizzle
    Quote Originally Posted by Manizzle View Post
    I am new to discrete math, and am having trouble with a couple problems..

    What is the inverse of the following relations:
    Rsub1 = {(a, b) | a < b}
    Rsub2 = {(a, b) | a divides b}

    Also, i need help with composition.. :
    What is Rsub3 of Rsub4 if
    Rsub3 = {(a, b) | a > b}
    Rsub4 = {(a, b) | a >= b}
    When a relation is defined on a set, it forms a set of ordered pairs, like (a, b). (An ordered pair is just a pair of 'things' - often, but not always, numbers - in a certain order.) The 'things', a and b, that make up the ordered pair, are related in the way that is described in the definition of the relation.

    So, for instance, if on the set \{0, 1, 2, 3\} the relation R_1 = \{(a, b) | a < b\} is defined, then R_1 will form a set of ordered pairs (a, b) in which the first number, a, is less than the second number, b. So the set that's formed is:

    \{(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)\}

    (Obviously, if R_1 is defined on a different set of numbers - all the integers, for instance - then you'll get a different, but similar, set of ordered pairs.)

    Now the inverse relationship simply reverses the order of the numbers in the ordered pair. So if R_1 is

    \{(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)\}, then

    R_1^{-1} = \{(1, 0), (2, 0), (3, 0), (2, 1), (3, 1), (3, 2)\}.

    So your job in finding R_1^{-1} is to define this new set of ordered pairs using a rule, so that it looks like:

    R_1^{-1} = \{(a, b) | a and b are related by such-and-such a rule \}

    Clearly, then, if R_1 = \{(a, b) | a < b\}, then R_1^{-1} = \{(a, b) | a > b\}

    For R_2, where the rule is " a divides b" typical ordered pairs might be (depending on which numbers are in the original set)

    (2, 10), (3, 15), (6, 6), ...

    because in each ordered pair, the first number divides into the second without leaving a remainder.

    When we reverse these to form some typical ordered pairs in R_2^{-1} we get:

    (10, 2), (15, 3), (6, 6), ...

    So, can you see how we could describe the inverse rule? (The word 'multiple' comes to mind!)

    Yes, it's R_2^{-1} = \{(a, b) | a is a multiple of b\}

    Combining relations means first forming one set of ordered pairs using the first relation (which, confusingly is written down second, because we don't work from left to right!), and then using the second relation to form a second set of ordered pairs from the first.

    So, for instance, we might start with a, and form the ordered pair (a, b) using the first relation R_4. Then use the b from this ordered pair with the relation R_3 to form another ordered pair (b, c).

    Finally, if we're going to say what's happened when these relations are combined, we will need to describe the relation that gives all the ordered pairs like (a, c)

    In your question, then,

    R_3 = \{(a, b) | a > b\} and

     R_4 = \{(a, b) | a >= b\}

    and we want to know what R_3 \circ R_4 is. So we do R_4 first, getting ordered pairs like:

    (3, 1), (3, 2), (3, 3), ... where the second number in the pair is never greater than the first.

    Then we take the second number from each ordered pair, and with it form a new ordered pair using R_3. For instance, we could get:

    (1, 0), (2, 1), (3, 2), ... where the second number in the pair is now definitely smaller than the first.

    Looking at the overall effect, using these examples, we get the ordered pairs:

    (3, 0), (3, 1), (3, 2), ...

    Notice that we can't get (3, 3) now, because R_3 insists that the first number is strictly greater than the second. So the combined relation is simply:

    \{(a, b) | a > b\}

    In other words, it's the same as R_3.

    Sorry if that was a bit long, but I hope you understand it a bit better now.

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Relations and Functions - Inverse Relations Question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 13th 2011, 01:20 PM
  2. Replies: 1
    Last Post: September 19th 2011, 02:09 PM
  3. [SOLVED] Relations on A
    Posted in the Discrete Math Forum
    Replies: 10
    Last Post: November 21st 2010, 12:25 PM
  4. Relations in a set
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 5th 2010, 11:03 PM
  5. relations help (3)
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: April 18th 2010, 05:49 AM

Search Tags


/mathhelpforum @mathhelpforum