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Prove that for all real numbers x there is a y n which case x<y

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data
    prove that
    ([itex]\forall[/itex]x[itex]\in[/itex]R) ([itex]\exists[/itex]y[itex]\in[/itex]R) : x < y


    2. Relevant equations



    3. The attempt at a solution
    x < y
    y= x+1
    then x<x+1
    which is correct but i'm kinda not sure of this answer...
     
  2. jcsd
  3. Nov 5, 2012 #2
    y=x+1 satisfies x<y so what wrong with it?
     
  4. Nov 5, 2012 #3
    just wanted to make sure someone actually confused me saying it dosen't
     
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