Proving $x^{r} \cdot x^{s} = x^{r + s}$ provided two facts

Question

I have the following two facts:

• For positive numbers $a$ and $b$ and natural numbers $n$ and $m$, we have $a = b$ if and only if $a^{n} = b^{n}$ if and only if $a^{1/m} = b^{1/m}$.

• For a positive number $x$ and integers $m$ and $n$, with $n$ positive, $(x^{1/n})^{m} = (x^{m})^{1/n}$.

Using these statements, I would like to prove for rationals $r$ and $s$, we have $x^{r} \cdot x^{s} = x^{r + s}$. I would also like to show $(x^{r})^{s} = (x^{s})^{r}$.

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I really cannot make much progress on either of these. I tried the typical approach that one takes when they see the word "rational," and I expressed $r = a/b$ and $s = c/d$ for integers $a, b, c, d$, but I couldn't get anything from there on. Any help at all would be much appreciated.

EDIT: I misinterpreted the question. I also know that $x^{m} * x^{n} = x^{m+n}$ holds for integers (note that this is different because I want to prove the fact for rationale). I also know $(x^{m})^{n} = x^{m*n}$ holds foe integers $m$ and $n$.

Finally, my book has the following definition: for a rational $r = m/n$, we define $x^{r}$ by $(x^{m})^{1/n}$

Answer 1

You can't with the given facts, as they tell nothing about rational exponents. (Only about naturals or inverses of naturals.)

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With the extended rules:

$$(x^{p/q}x^{m/n})^{qn}=(x^{p/q})^{qn}(x^{m/n})^{qn}=x^{p/q\,qn}x^{mn/n\,qn}=x^{pn}x^{mq}=x^{pn+mq}$$

which implies

$$x^{p/q}x^{m/n}=x^{(pn+mq)/(qn)}.$$

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We used $$(x^{1/n})^m=(x^m)^{1/n}=x^{m/n}.$$

Answer 2

You cannot show the first equality using these facts alone:

Define $\widetilde{a^b}:=a^{b^2}$. Then we have

For positive numbers $a$ and $b$ and natural numbers $n$ and $m$, we have $a=b$ if and only if $\widetilde{a^n}=\widetilde{b^n}$ if and only if $\widetilde{a^{1/m}}=\widetilde{b^{1/m}}$.

and

For a positive number $x$ and integers $m$ and $n$, with $n$ positive, $\widetilde{\left(\widetilde{x^{1/n}}\right)^m}=\widetilde{\left(\widetilde{x^{m}}\right)^{1/n}}$.

Hence any proof for the first equality from the yellow properties alone would also show $$x^{r^2+s^2}=\widetilde{x^r}\cdot\widetilde{x^r}=\widetilde{x^{r+s}}=x^{(r+s)^2}=x^{r^2+s^2}\cdot x^{2rs} $$