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Is this relation equivalence relation ?

  1. May 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Relation is x^y = y^x........x and y belong to integers


    2. Relevant equations



    3. The attempt at a solution
    Well i have already proven that they are reflexive and symmetric. I have doubt with transitive
    I did the follwoing way x^y = y^x.......(1) and y^z = z^y.........(2)
    from(1) x^z = y^(zx/y)
    from(2) z^x = y^(zx/y)
    Therefore x^z = z^x proving it is transitive

    But the test paper solution does it this way
    from(1) y = x^(y/x) and from(2) y = z^(y/z)
    Now x^(y/x) = z^(y/z) ..... so its not transitive
    Which is correct? Did i do anything wrong?
     
  2. jcsd
  3. May 3, 2012 #2
    How did it follow from x^(y/x) = z^(y/z) that x and z are not equivalent?
    Raise both sides to the power of (xz/y) , we will get x and z as equivalent.
    I think both methods prove transitivity. I can't see how the test paper solution is giving different answer after it has shown this :- x^(y/x) = z^(y/z) .

    Note :- we don't have to worry about y being zero here. As 0 won't be equivalent to any other integer. As for any non-zero integer x ,
    0^x(=0) won't be same as x^0(=1) .
     
  4. May 3, 2012 #3
    So it is an equivalent relation right?
    When the book gives the wrong solution i get real angry - more when it wastes so much time
     
  5. May 3, 2012 #4
    Anyways can you find three numbers which are equivalent according to the above relation. I am unable to.
    I could think of pairs of numbers which are equivalent. For ex:- 2^4 = 4^2 .
    I am unable to find 3 such numbers which are equivalent.:cry:
     
  6. May 3, 2012 #5
    Ah , as it happens other than the equivalence in the obvious case where an integer is equivalent to itself i.e x~x , there are only two pairs of integers which are equivalent , i.e (-2)~(-4) and (2)~(4) .

    Thus there is no equivalence class containing 3 members or more for this equivalence relation. Either we have equivalence class of one member each i.e ...,{-3},{-1},{0},{1},{3}... or the two equivalence classes both of which contain two members i.e {2,4} and {-2,-4} .

    This maybe a reason why your solution book gave the wrong answer for transitivity. Because there is no equivalence class of the form {x,y,z} where x,y,z are integers.

    But there is another very common equivalence relation for which such a thing happens. I.e the equality relation.
    I.e x~y if x=y. For this relation , all the equivalence classes contain one member each. :smile:
     
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