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Inequality a^p>b^p

  1. May 17, 2008 #1
    It struck me when reading about the axioms of inequalities, that this inequality wasnt that easy to straight ahead deduce from the axioms.
    the inequality to be proven is given p>0, a>b>0, prove that a^p>b^p.
    the axioms are:
    1. for two real numbers x and y exactly one of the following cases i true:
    a) x=y b) x<y c)x>y
    2. if x<y, then for all z x+z<y+z
    3. x<y, z>0, then xz<yz
    4. if x<y and y<z, then x<z
    lemma: for x,y,z,w>0, if x<y, w<z, then xw<yz.
    By axiom 2 we have xw<yw, and again by axiom 2 wy<yz, and by axiom 4 xw<yz. Now assume p is natural, then for p=1 the inequaliy holds. Assume it holds for p=n, ie a^n>b^n. then by the lemma we can multiply left side with a, and right side with b and establishing a^(n+1)>b^(n+1), thus by the induction principle it holds for all p [tex]\in[/tex]N. It also follow by axiom 1, that if a^p>b^p for p[tex]\in[/tex]N, then a>b. Now write:
    ad by earlier result, we get that a^p>b^p for rational p.
    Now, let the sequence {p_k} of rational numbers be the numbers of the decimal expansion of the real number p, ie [tex]\sum_{k=1}^n p_k = p[/tex] to n decimal places.
    thus we have:
    [tex]a^p=\lim_{n\to \infty}a^{\sum_{k=1}^n p_k}>\lim_{n\to \infty}b^{\sum_{k=1}^n p_k}=b^p[/tex]
    thus a>b implies that a^p>b^p for all and p,a,b>0.

    Now my questions are:
    Is this proof correct? Im mostly concerned about the last step when introducing a limit. Do I need to state anything about continuity of the function a^p? Or is it sufficient to use the proved lemma, thus I can continue to multiply the inequality with a^(pk)>b^(pk) and continue to infinity to establish the inequality.
    Also, thus the introduction of a limit need extra proofs of the limit, if I want to deduce the inequality straight down from the axioms?
    I know that I can just state the the function a^x is strictly increasing for a,x>0 and thus a>b -> a^x>b^x, but that does need extra proofs conecrning derivatives etc, right? (and its not so much fun :smile:)
    Or have I missed a really obvious way to prove this kind of intuitive inequality?:rolleyes:
    and last, a challenge for you, find a simplier proof. :smile:
  2. jcsd
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