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How to prove x^a >= 1?

  1. Feb 11, 2012 #1
    (Sorry for the wrong title!)

    Let a>1 (a in R). Then how do you prove for all x in R, a^x > 0 ?

    And also that a^y>1 for all y>0, how do you prove?

    And also how do you prove that 0^b = 0 when b is a real number?

    (For me this is so difficult. So please just help me. Enough hints will suffice, no full solution.)
     
    Last edited by a moderator: Feb 22, 2012
  2. jcsd
  3. Feb 11, 2012 #2

    micromass

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    This depends on what you can use. I would prove it by the following scheme:

    1) First prove it for all natural numbers x (by induction).

    2) Use [itex]a^{-x}=(a^x)^{-1}[/itex] to prove it for all integers.

    3) Use [itex]a^{m/n}=\sqrt[n]{a^m}[/itex] to prove it for all rational numbers.

    4) Use continuity of [itex]a^x[/itex] to prove it for all real numbers.
     
  4. Feb 11, 2012 #3
    Could you help me bit more on proving the part 3)?

    So by the theorem in Rudin's Mathematical Analysis in page 10 thm 1.21, I can prove this right?
     
    Last edited by a moderator: Feb 22, 2012
  5. Feb 12, 2012 #4
    b must be restricted, as in b > 0, so that 0^b is not

    undefined and/or is not indeterminate.
     
    Last edited by a moderator: Feb 22, 2012
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