1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook