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Math Help - Proof question

  1. #1
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    Proof question

    I have questions about one proof.

    It concerns numbers with rational exponents.

    Having in mind that for all r,s \in Q,a \in R is (a^r )^s  = a^{rs} then author of book wants to show that a must be positive real number.

    Proof is:
    If r is rational number then \frac{r}{2} is also a rational number, so:
    a^r  = a^{\frac{r}{2} \cdot 2}  = (a^{\frac{r}{2}} )^2  \ge 0
    Similarly, if r is rational number then -r is rational number, so:
    a^{ - r}  = a^{r \cdot ( - 1)}  = (a^r )^{ - 1} which gives us that a^r  \ne 0. Then is a^r  > 0

    If r  \ne 0 then is \frac{1}{r} \ne 0.

    Based on above, we get that a = a^{r \cdot \frac{1}{r}}  = (a^r )^{\frac{1}{r}} which means that a must be positive number because a^r is positive number so is (a^r )^{\frac{1}{r}} positive number.

    My question is why a^r must be positive number?
    If r is rational number then \frac{r}{3} is also a rational number, so:
    a^r  = a^{\frac{r}{3} \cdot 3}  = (a^{\frac{r}{3}} )^3 which means that a^r can be also negative number.
    Based on that, then a = a^{r \cdot \frac{1}{r}}  = (a^r )^{\frac{1}{r}} can mean that a can be also negative number.
    Last edited by DenMac21; February 21st 2006 at 07:15 AM.
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  2. #2
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    I do not know what your author wants to do; but exponents are definied ONLY for positive real numbers on this level.
    Thus, (a^r)^s=a^{rs} is only defined if a^r is positive. But a^r can only be defined if a is positive.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    I do not know what your author wants to do; but exponents are definied ONLY for positive real numbers on this level.
    Thus, (a^r)^s=a^{rs} is only defined if a^r is positive. But a^r can only be defined if a is positive.
    Can you explain me this?
    I don't understand.
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  4. #4
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    Quote Originally Posted by DenMac21
    Can you explain me this?
    I don't understand.
    In more advanced math we define a^b for all complex numbers (imaginary numbers). For now that is not necessary. Thus, only positive real numbers (non-imaginary) are defined. For example (-1)^2 still has meaning but (-1)^{1/2} does not. Thus, we only consider postive real numbers. Thus, I do not know what your author is trying to do because as I understand it, it has only meaning for positives numbers thus there is nothing to prove.

    Maybe I am wrong and missing something what your author is trying to say.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker
    In more advanced math we define a^b for all complex numbers (imaginary numbers). For now that is not necessary. Thus, only positive real numbers (non-imaginary) are defined. For example (-1)^2 still has meaning but (-1)^{1/2} does not. Thus, we only consider postive real numbers. Thus, I do not know what your author is trying to do because as I understand it, it has only meaning for positives numbers thus there is nothing to prove.

    Maybe I am wrong and missing something what your author is trying to say.
    Neither do I!
    Yes I know that there is no rational solution of (-1)^{1/2} but there is rational solution of (-1)^{1/3} which author didn't mention. He only took in consideration a^r, r=\frac{1}{2} but not a^r, r=\frac{1}{3}. He said that a must be positive real number, but it can be also negative real number (under right condition of course).
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