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Proof Trouble.

  1. Nov 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose that , for any $$ \epsilon > 0, a < b + \epsilon $$ . Then $$ a\le b $$


    3. The attempt at a solution
    I have the proof, its not a question that was assigned to me, it was an example used.
    According to the proof i can choose ANY epsilon greater than 0, so lets choose 10. then 2 < 1 + 10 = 11. Right?
    Then $$ 2 \le 1 $$ . NOT right.

    it's late here in the UK, is it me?
     
  2. jcsd
  3. Nov 3, 2014 #2

    RUber

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    It isn't saying for any one epsilon, but for any epsilon you could possibly choose. Therefore, you can take epsilon as small (positive) as you like and the relation will still be true.
     
  4. Nov 3, 2014 #3

    RUber

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    ##2 < 1 + \epsilon ## for ##1 < \epsilon ## but, you can choose any epsilon, so choose ##\epsilon = 1/2##, then ##2 < 1+ \epsilon ## is false.
     
  5. Nov 3, 2014 #4
    Thanks for clearing that one up! The penny/cent/nickel/kuna has dropped!
     
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