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Thread: 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 $\displaystyle r,s \in Q,a \in R$ is $\displaystyle (a^r )^s = a^{rs}$ then author of book wants to show that $\displaystyle a$ must be positive real number.

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

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

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

    My question is why $\displaystyle a^r$ must be positive number?
    If $\displaystyle r$ is rational number then $\displaystyle \frac{r}{3}$ is also a rational number, so:
    $\displaystyle a^r = a^{\frac{r}{3} \cdot 3} = (a^{\frac{r}{3}} )^3$ which means that $\displaystyle a^r$ can be also negative number.
    Based on that, then $\displaystyle a = a^{r \cdot \frac{1}{r}} = (a^r )^{\frac{1}{r}} $ can mean that $\displaystyle a$ can be also negative number.
    Last edited by DenMac21; Feb 21st 2006 at 06:15 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, $\displaystyle (a^r)^s=a^{rs}$ is only defined if $\displaystyle a^r$ is positive. But $\displaystyle a^r$ can only be defined if $\displaystyle a$ 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, $\displaystyle (a^r)^s=a^{rs}$ is only defined if $\displaystyle a^r$ is positive. But $\displaystyle a^r$ can only be defined if $\displaystyle a$ 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 $\displaystyle a^b$ for all complex numbers (imaginary numbers). For now that is not necessary. Thus, only positive real numbers (non-imaginary) are defined. For example $\displaystyle (-1)^2$ still has meaning but $\displaystyle (-1)^{1/2}$ 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 $\displaystyle a^b$ for all complex numbers (imaginary numbers). For now that is not necessary. Thus, only positive real numbers (non-imaginary) are defined. For example $\displaystyle (-1)^2$ still has meaning but $\displaystyle (-1)^{1/2}$ 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 $\displaystyle (-1)^{1/2}$ but there is rational solution of $\displaystyle (-1)^{1/3}$ which author didn't mention. He only took in consideration $\displaystyle a^r, r=\frac{1}{2}$ but not $\displaystyle a^r, r=\frac{1}{3}$. He said that $\displaystyle a$ must be positive real number, but it can be also negative real number (under right condition of course).
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