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The basis of the Real numbers over the Irrationals

  1. Sep 24, 2012 #1
    1. What can be said of the dimension of the basis of the Reals over the Irrationals



    2. Relevant equations



    3. I believe the basis is infinite because any real number can be made out of the combination of irrational vectors multiplied by the same irrational coefficient to make any real number. ie. sqrt(2)*sqrt(2)=2 + sqrt(51)*sqrt(51)=53, etc. Could you please help me figure out if this is the correct solution or what a better way to phrase the proof would be? Thanks!
     
    Last edited by a moderator: Sep 25, 2012
  2. jcsd
  3. Sep 24, 2012 #2

    Dick

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    The irrationals aren't a vector subspace of R, so the question is pretty meaningless. Did you maybe mean the rationals?
     
  4. Sep 24, 2012 #3
    Actually, I think it's a trick question because I thought that at first but I wasn't sure. How would I prove that? Does it break the multiply by zero axium since zero is not an irrational? Thanks for your help!
     
  5. Sep 24, 2012 #4

    Dick

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    It sure does. 0 is not an irrational. Lot's of other things break. You can also add two irrationals and get a rational etc etc.
     
  6. Sep 25, 2012 #5
    Hey thanks, how would adding two irrationals to get a rational break the vector space rule?
     
  7. Sep 25, 2012 #6

    Dick

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    To be a vector space you first have to be an additive group. The addition operation has to be closed. I.e. if x and y are irrational then x+y has to be irrational. Give a counterexample.
     
  8. Sep 25, 2012 #7

    Mark44

    Staff: Mentor

    What are you doing here? √2 + √2 = 2, and 2 + √51 + √51 = 53, but in connecting expressions as you did above, you are saying that 2 = 53, which is clearly not true.
     
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