# Proof question

• Feb 21st 2006, 05:45 AM
DenMac21
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.
• Feb 21st 2006, 02:32 PM
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.
• Feb 22nd 2006, 02:29 AM
DenMac21
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.
• Feb 22nd 2006, 02:37 PM
ThePerfectHacker
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.
• Feb 22nd 2006, 02:50 PM
DenMac21
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).