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Defining R

  1. Sep 23, 2013 #1
    Hey,

    Ive a question in my textbook and I dont really know what to do!

    The question is:
    20130923_225545.jpg

    I know I need to prove that the relation is
    1)Symmetric
    2)Reflexive
    3)Transitive
    But how do I prove this
     
  2. jcsd
  3. Sep 23, 2013 #2
    Do you know what reflexivity, symmetry, and transitivity mean?
     
  4. Sep 23, 2013 #3
    Yes and this is my proof:

    Reflexive: x+x= 2.x and not 3.x so its not reflexive
    Symmetric: x+y = 3k so y+x = 3k
    Transitive: xRy and yRz
    x+y=3k for keZ
    x+y=3n for neZ
    So x=3k-y and y=3n-y so x+z=3k-y+3n-y= 3(k+n)-2y We see that its not transitive

    Because it hasnt al three properties it isnt an equivalence relationship
     
  5. Sep 23, 2013 #4
    I dont understand it really I read my textbook over and over but I still cant imagine how it works
     
  6. Sep 23, 2013 #5

    jbunniii

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    Even better would be to give a specific counterexample. Notice that if ##x = 3##, then ##x + x = 6 = 3\cdot 2## so the equation holds for some values of ##x##. Find one for which it does not hold.

    OK. Writing in more detail, you might say something like "if ##xRy## then ##x + y = 3k##. Since addition of integers is commutative, we have ##x + y = y + x##. Therefore ##y + x = 3k##, which means that ##yRx##."

    I think you mean ##y + z = 3n## for that last equation. What can you conclude about ##x + z##?
     
  7. Sep 23, 2013 #6

    Mark44

    Staff: Mentor

    From the definition you're given, xRy if and only if x + y = 3k, for some integer k. What this means is that for x and y to belong to the relation R, x and y have to add up to an integer multiple of 3. For example, 1R2 because 1 + 2 = 3 = 3 * 1.
    Some values of x belong to the relation, but others don't. As you note, x + x = 2x, so the only way a number x could belong to the relation is if x itself is a multiple of 3.
    0R0 = 0 + 0 = 3 * 0
    3R3 = 6 = 3 * 2
    but 1 doesn't work, nor does 2.
    I think you mean y + z = 3n for n ##\in## Z.
    If y is a multiple of 3, then we do have transitivity. For example, 0R3 because 0 + 3 = 3, and 3R6 because 3 + 6 = 9. From this, we see that 0R9 because 0 + 9 = 9 = 3*3.
     
  8. Sep 24, 2013 #7
    That x+z is not an integer?
     
  9. Sep 24, 2013 #8

    Mark44

    Staff: Mentor

    No. The context of what both jbunniii and I said was the part where you were investigating transitivity or the relation. You're given that xRy and yRz, so do x and z also belong to the relation?
     
  10. Sep 24, 2013 #9
    x,y,z are integers. xRy and yRz
    0R2 and 2R5
    0+2=2=3k and 2+5=7=3k we see that there doesnt exist an integer k such that 3k=2 and 3k=7 so
    xRz doesnt exists.
     
  11. Sep 24, 2013 #10

    Mark44

    Staff: Mentor

    But 0 and 2 don't belong to the relation R, so you can't say 0R2. And 2 and 5 don't belong to the relation either, so you can't say 2R5.

    Whatever numbers x, y, and z you work with, x and y have to belong to the relation, and y and z have to belong as well.
     
  12. Sep 24, 2013 #11
    but then its always transitive cause i cant find any false relations 3R6 and 6R9
    x+z=24=3*8
     
  13. Sep 24, 2013 #12

    Mark44

    Staff: Mentor

    That's not a good reason. Possibly you didn't look hard enough.
    How did you get 24? x = 3 and z = 9, so x + z = 12.

    What about 1R2 and 2R4?

    BTW, it's probably better to work with this algebraically rather than to try to guess combinations that work or don't work.

    For the record, you have already shown enough to convince me or your instructor that R as defined in this thread is NOT a relation. It is not in general reflexive (e.g. 1 and 1 aren't in the relation, nor are 2 and 2, but 0 and 0 are, as are 3 and 3).

    Also, the relation is transitive for some values of x, y, and z, but not for all such values. Most of what we've been doing here is trying to help you get a handle on what this is about.
     
  14. Sep 24, 2013 #13

    jbunniii

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    How did you get 24? If x = 3, y = 6, and z = 9, then x + z = 12. [EDIT: as Mark44 already pointed out before I submitted my reply.]

    Also, to prove transitivity, you must show that it is true in general, not for a specific example. Start by assuming that x + y and y + z are both multiples of 3. Then prove that this implies that x + z is also a multiple of 3. If you cannot show this in general, then try to find a counterexample.
     
  15. Sep 25, 2013 #14
    Ah yes I see it
    So if I have x+y 1R2 1+2=3=3.1
    and y+z 2R7 2+7=9=3.3
    then x+z 1R7 1+7=8=3.k we see that theres no integer k so that 3k=8 thus the relation is not transitive

    So if its wrong I only have to give a counterexample? And if its true you need to prove it
     
  16. Sep 25, 2013 #15

    HallsofIvy

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    Do you understand that this is saying that two numbers are "related" in this way if and only if their sum is a multiple of 3.
    Yes, to prove that a general statement is NOT true you only have to give one example in which is does not work- a "counter-example".

    2 is NOT related to itself in this way because 2+ 2= 4 is not a multiple of 3. Therefore, it is not "reflexive".

    It is obviously symmetric.

    4+ 2= 6= 2(3) so 4 and 2 are related. 2+ 1= 3 so 2 and 1 are related. But 4+ 1= 5 which is NOT a multiple of 3 so this is NOT transitive.
     
  17. Sep 25, 2013 #16
    Yes thank you very much for the help!
     
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