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Proving a function is bijective

  1. Dec 13, 2012 #1
    1. The problem statement, all variables and given/known data

    The relation R, on the set ℝxℝ, is defined by (x,y)R(a,b) if and only if x+b=a+y for any real number x,y,a, and b.

    Let G denote the set of equivalence classes of the relation R.

    Question: Let f: ℝ -> G be defined by f(a)=[(0,a)]. Prove that f is a bijection.


    3. The attempt at a solution

    I think I have to come up with an equation, for f(a)=[(0,a)]. I also think I could prove this using induction.

    I am very confused and need a leg up on how to start! Any help would be greatly appreciated.
     
  2. jcsd
  3. Dec 13, 2012 #2

    Dick

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    I think you need to understand what the question is actually asking first. [(0,a)] is the equivalence class consisting of all (x,y) such that (x,y)R(0,a). What does that set look like if you graph it in the xy plane?
     
  4. Dec 13, 2012 #3
    Do you mean the graph of the set (0,a)? The x-values are stationary at 0 and the y-values are varying.

    I know that I need to prove that the function is both onto and 1:1. Ergh, am I really confused
     
  5. Dec 13, 2012 #4

    Dick

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    I still don't think you really understand what they are asking. No. (0,a) is just a point. I mean the graph of the equivalence class [(0,a)] with a fixed. The brackets mean 'equivalence class'. (0,a)R(x,y) means 0+y=x+a, right?
     
  6. Dec 13, 2012 #5
    Yes.. so the graph is y=x+a then?
     
  7. Dec 13, 2012 #6

    Dick

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    Right. So [(0,a)] is a straight line with y-intercept a and slope 1. So your function f takes a in R and maps it to that straight line. Now can you see why it's a bijection?
     
  8. Dec 13, 2012 #7
    Yes. Thank you so, so much.
     
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