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Homework Help: Sets and Relations - quick one

  1. Oct 28, 2005 #1
    What are the conditons on A, B and C for (AuB)nC = Au(BnC) ?

    Is it that AnBnC ?

    Can someone explain if they are different and why? :confused:

    Now If A = {irrationals}, B= {integers} and C={reals} does the equality from above hold in this case?

    I answered yes the equality holds as AcBcC.

    Any suggestions?
     
    Last edited: Oct 28, 2005
  2. jcsd
  3. Oct 28, 2005 #2
    The irrationals are not a subset of the integers. :wink:
     
  4. Oct 28, 2005 #3
    so for the first part of the question

    What are the conditons on A, B and C for (AuB)nC = Au(BnC) ?

    Is it that AnBnC ? This is correct

    And the second part we get AcC only right?
     
  5. Oct 28, 2005 #4
    [itex]A\cap B\cap C[/itex] is not a complete sentence. Did you mean the condition for the equality to be true is that [itex]A\subseteq B \subseteq C[/itex] ?
    An easy way to check is to look at the condition of elements of each set. For example, if [itex]S = ((A\cup B)\cap C)[/itex], then S = {x: ([itex]x\in A[/itex] and [itex]x\in C[/itex]) or ([itex]x\in B[/itex] and [itex]x\in C[/itex])} which can be written more concisely (or draw a Venn diagram). Do the same for the other side.
    For the second part, yes, [itex]A\subset C[/itex], but you didn't answer the question posed. :smile:
     
  6. Oct 28, 2005 #5

    HallsofIvy

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    AnBnC isn't even a "condition"! It's a set and you have to say something about it to get a "condition". (AuB)nC consists of all things that are in either A or B and in C. Au(BnC) consists of all things that are in A or in both B and C. Drawing Venn diagrams might help. Suppose "z" is in A but not in C. then it would certainly be in Au(BnC) since it is in A. But since it is not in C it is not[\b] in (AuB)nC. So one possible condition is that A is a subset of C. Is that sufficient?
     
  7. Oct 28, 2005 #6
    I did mean AcBcC actually not AnBnC sorry.

    But actually the conditions for (AuB)nC = Au(BnC) in real numbers not our examble. Is that AcC and BcC right?

    And in our example which is A= irrationals B= integers and C= reals is satisfies as AcC and BcC right?
     
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