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!

Transitive math help

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data
    R is the relation of the set R; xRy if and only if (x + y)2 = 1.
    Is it transitive.

    2. Relevant equations



    3. The attempt at a solution
    Every description of transitive involves 3 elements. Here, I just have x and y. It is defined as xRy so there is a relation but is it transitive all by itself with just those two elements?
     
  2. jcsd
  3. Feb 18, 2014 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Write down what xRy means, what yRz means and what xRz means using your given equation. Then decide whether you can prove the third from the first two or find a counterexample.
     
  4. Feb 18, 2014 #3
    This confuses me because there is no z. What kind of relationship can y or x have with something that doesn't exist? I can't describe what it means for yRz at all because of this.

    I can say that xRy stems from the fact that x and y are a pair that satisfies the given equation. R contains the pair (x,y). Since there is no z so: (x,y)##\bar{R}##z. (Can't find a way to strikeout R.. so \bar = strikeout)

    But like I said, does it matter? If I just have the two elements is that enough to say that it is transitive?
     
    Last edited: Feb 18, 2014
  5. Feb 18, 2014 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, you aren't understanding this at all. R is a relation between numbers. Any two numbers may or may not be related. Is 1R3? To answer you have to ask whether (1+3)^2=1. It doesn't make any sense to talk about whether (x,y)Rz because (x,y) is not a number. If w and z are numbers it makes sense to say wRz if ##(w+z)^2 = 1##. It will be true for some choices of w and z and not for others.

    So I ask you again: What are you given and what are you trying to prove?
     
  6. Feb 18, 2014 #5
    I shouldnt have said (x,y) but rather x or y. I was trying to save myself time.

    Are x,y,z suppose to be elements in the same set then? Ive been thinking of them as in different real sets.
     
  7. Feb 18, 2014 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    What is the definition of transitivity? Does it ever involve R where there's an instance with more than two elements?
     
  8. Feb 18, 2014 #7
    I thought I already defined it. If I did so incorrectly please tell me how/why! I'm not sure I understand your question but R is only a reltionship between 2 elements as far as ive seen. But my question is still are they all in the same set? In the end I think it probably doesnt matter. z could be in there but it just isnt used. My gut tells me it is transitive but I have no basis for the assumption.

    Sorry grammar, on phone.
     
  9. Feb 18, 2014 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    No, you haven't defined what it means for R to be transitive, and it seems pretty clear from your questions that you don't know the definition. Even if you do know it, just humor us and tell us what the precise definition of transitivity is.
     
  10. Feb 18, 2014 #9
    I'm sorry, I thought I already had but couldn't look because I couldn't scroll up on my phone. As my book explains it:if xRy and yRz then xRz. This makes perfect sense to me. But in my case, what is z? That is the part I don't get.
     
  11. Feb 18, 2014 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I'm not sure I understand your confusion here. Suppose two elements x and y are related if x=y. Transitivity would mean if a=b and b=c, then a=c. Your question is asking "what is c?"
     
  12. Feb 18, 2014 #11
    Yes, because z is not in my equation anywhere. Do I just say that z exists but x and y are not related? Do I even need to mention z?
     
  13. Feb 18, 2014 #12

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    What equation are you referring to? There are no equations here.
     
  14. Feb 18, 2014 #13
    The ##(x+y)^2=1## at the top
     
  15. Feb 18, 2014 #14

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You do realize that x and y are dummy variables, right? Reread what LCKurtz said in post #4.
     
  16. Feb 18, 2014 #15
    okay so given that y=1-x from the equation. If x=2 then y=-1 then 2R-1. Then if z=1-y then -1R2. So x=z. Am I on the right track now?
     
  17. Feb 18, 2014 #16

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Not really. You shouldn't be solving for y or z. You need to prove the statement for any x, y, and z that satisfy the given conditions — or find a counterexample to show R isn't transitive.

    xRy means that (x+y)2=1.
    yRz means that (y+z)2=1.

    That's all you know about x, y, and z. Given that, you want to show that (x+z)2=1 from which you can conclude that xRz. Then you will have shown that xRy and yRz implies xRz, i.e. that R is transitive.

    Or you might decide R is not transitive. To show that, you want to find a counterexample. Find specific values of x, y, and z such that x and y are related and y and z are related, but x and z are not related.
     
  18. Feb 19, 2014 #17

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, that isn't given. If [itex](x+ y)^2= 1[/itex] then either [itex]x+ y= 1[/itex], whence [itex]y= 1- x[/itex] or [itex]x+ y= -1[/itex], whence [itex]y= -1- x[/itex].

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted