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: Laws of set algebra

  1. Jan 2, 2007 #1

    Use the laws of set algebra to show that for the sets X,Y,Z:
    a) X + X’Y + ZX’ = X + Y + Z
    b) XY + X’Y’ + X’Y + XY’ = U
    c) Z’(X + Y’)(X’ +Z) = (X + Y’ + Z)’
    d) XYZ + X’ + Y’ + Z’ = U


    I can prove these using venn diagrams but what are the laws of set algebra?

    also U means the union of X, Y and Z...no?
  2. jcsd
  3. Jan 2, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I would imagine things like the distributive laws, the associative laws, DeMorgan's laws, et cetera.
  4. Jan 2, 2007 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    U generally denotes the universal set; a set large enough to contain all sets under consideration (for the given question). For example, Ac=U\A.

    (Another way to think of it, is as the "rectangle" in which the sets are drawn when constructing a Venn diagram) In this instance, it is enough to take U to be the union of X, Y and Z.
    Last edited: Jan 2, 2007
  5. Jan 2, 2007 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    No, you can't prove them using venn diagrams. You can justify them, but I doubt anyone would accept Venn diagrams as a 'proper' proof. It is easy to turn Venn diagram arguments into proper arguments though.
  6. Jan 2, 2007 #5
    'It is easy to turn Venn diagram arguments into proper arguments though.'
    yeah?...how? could you give me an example?
  7. Jan 3, 2007 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Look at the diagram. A set equality A=B is always best shown (if no clever methods are at hand) by demonstrating that anything in A is in B, and vice versa. If you look at the venn diagram for a), say, you can read off how to do this because you can see where elements in subsets of the LHS can be found in subset of the RHS.

    Of course, here you just use the laws set algebras some of which are (and these are comutative operations)

    X'=U\X (set difference), or U=X+X'

    In particular Y=XY+X'Y
  8. Jan 3, 2007 #7


    User Avatar
    Science Advisor

    Obviously it does not here because then b and d would make no sense! U here means the "universal set". "X+ Y" is the union of the two sets X and Y and XY means the intersection.
  9. Jan 3, 2007 #8
    using the laws posted, this what i did:

    a) X + X'Y + ZX'
    = X +Y - XY + Z -XZ
    =X + Y + X'Y - Y + Z + X'Z - Z
    =X + X'Y + X'Z
    =X + Y + XY + XZ + Z
    = X + Y+ Z (RHS)
    somehow i feel that i did a) wrong

    b) XY + X'Y' + X'Y + XY'
    =Y - X'Y + X'Y + X + YX
    =Y + X + YX

    c) Z'(X+Y)(X'+Z)
    =Z'XX' + XZZ' + X'YZ' + YZZ'
    =0 + 0 + X'YZ' + 0

    d) don't know where to start...whatever i do i get stuck

    is a, b and c correct?
  10. Jan 3, 2007 #9
    ( i should start a new thread for this but i hate seeing my name a million times)


    if f: R -> A A={x|x belongs to R and -1<x<1}

    x^2=y^2 is not injective...is it?

    because x= + or - y

    (the question is longer...if i turn out to be right i'll post it up properly and ask for help)
  11. Jan 3, 2007 #10

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    And what relation does f have to any of x or y in this question? y^2=x^2 does not define y as a function of x for precisely the reason given in your post, so this can't have any relation to f, can it?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook