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Analysis - Powers of real numbers

  1. Apr 18, 2006 #1
    Hey everyone!

    This question came up in my analysis assignment, we're studying continuity and differentiability at the moment so I'm unsure of my answer! It seems too short :(

    For a), it seems to follow immediately,

    If [itex]mq = np[/itex] then [itex]m = \frac{np}{q}[/itex] and so we have
    [itex](x^{\frac{1}{n}})^m = (x^{\frac{1}{n}})^{\frac{np}{q}} = ((x^{\frac{1}{n}})^n)^{\frac{p}{q}} = (x^{\frac{p}{q}}) = x^{\frac{1}{n}} = (x^{\frac{1}{q}})^p[/itex] as required.

    And b) is similar (there are too many exponents to latexify before my stuff gets stolen upstairs in the study floors).
    I went about it proving if [itex]a \in \mathbb{R}[/itex] and [itex]m,n \in \mathbb{N}[/itex] then [itex]a^{m+n} = {a^m}{a^n}[/itex] and [itex](a^m)^n = a^{mn}[/itex] by induction. (Can you use induction over m and n by doing it twice fixing m the first time and inducting over n and the second time fixing n?).

    Or it's probably better to first prove, if [itex]m\in \mathbb{Z}, n \in \mathbb{N}, x \in (0,\infty)[/itex] then [itex]x^{\frac{m}{n}} = (x^m)^{\frac{1}{n}}[/itex].
    If [itex]x \in (0,\infty)[/itex] and [itex]m,n \in \mathbb{Z}[/itex] then [itex](x^m)^n = x^{mn} = (x^n)^m[/itex]. Now let [itex] y:=x^{\frac{m}{n}} = (x^{\frac{1}{n}})^m > 0[/itex]. So [itex]y^n = ((x^{\frac{1}{n}})^m)^n = ((x^{\frac{1}{n}})^n)^m = x^m[/itex]. Therefore it follows that [itex]y = (x^m)^{\frac{1}{n}}[/itex].

    If we proved it for the naturals, how can we prove it for the rationals? It seems to follow naturally.

    These analysis proofs seem so empty, although subtle. I have huge trouble sponging it all together into a valid proof.

    Thanks a lot!
    -Gareth
     
    Last edited: Apr 18, 2006
  2. jcsd
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