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Epsilon argument

  1. Sep 24, 2009 #1
    Hi,

    Want to show: a<=r

    However I ended up with such an argument: For each eps>0, a<r+eps. Does this statement imply a<=r?
     
  2. jcsd
  3. Sep 24, 2009 #2
    No, this implies that you can find any epsilon that when added to r makes it larger than a. This would work even when a is not equal to r, and is lesser. for eg a = 1, r = 2.
     
  4. Sep 24, 2009 #3

    Office_Shredder

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    This does show a <= r. If a>r, then a-r is a positive number. In particular, you then showed that

    a<r+(a-r) (picking eps=a-r)

    Hence a<a. That doesn't make any sense, so it must have been a<=r the whole time
     
  5. Sep 24, 2009 #4
    hey, it helped a lot. thank you all
     
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