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I have managed to prove the usual exponent laws for integer exponents, using induction on the natural numbers. I am now trying to prove that the laws also hold for rational exponents. I'm starting by trying to prove the law $x^nx^m = x^{n+m}$, where $n = \frac a b$ and $m = \frac c d$.

Using the following definition:

$x = y^{\frac a b} \iff x^b = y^a$

I can prove it as follows:

$x^{\frac a b}x^{\frac c d} = x^{\frac {ad} {bd}}x^{\frac {bc} {bd}} = (x^{ad})^{\frac 1 {bd}}(x^{bc})^{\frac 1 {bd}} = (x^{ad}x^{bc})^{\frac 1 {bd}} = (x^{ad+bc})^{\frac 1 {bd}} = x^{\frac {ad + bc} {bd}} = x^{{\frac a b}+{\frac c d}}$

Aside from the definition, I have proved no other things about rational exponents. So, the problem I have is that I use the following assumptions in this proof:

  1. $x^{\frac a b} = (x^a)^{\frac 1 b}$
  2. $(xy)^{\frac a b} = x^{\frac a b}y^{\frac a b}$

Both assumptions I need to prove but I cannot figure out how. Can anyone help?

Edit: I now realize the proof of (1.) is trivial: $y = (x^a)^{\frac 1 b} \iff y^b = (x^a)^1 = x^a \iff y = x^{\frac a b}$. For (2.) as far as I get is proving that $(xy)^{\frac a b} = (x^ay^a)^{\frac 1 b}$

esotechnica
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  • Remember that you also need to prove that rational exponentiation is well-defined. That is, if $x^{\frac a b}=(\sqrt[b] x)^a$, then why, for instance, does $(\sqrt[12] x)^{18}=(\sqrt[2] x)^3$? – Jack M Mar 28 '17 at 05:33
  • Wouldn't this be provable from the fact that any two rational numbers ${\frac a b}$ and ${\frac c d}$ are equal iff $ad = bc$? – esotechnica Mar 28 '17 at 05:45
  • WLOG $bd>0.$ For $x\geq 0$ we have $0\leq (x^nx^m)^{bd}=x^{ad+bc}=(x^{n+m})^{bd}.$... If $y_1,y_2\geq 0$ and $e\in \mathbb N$ and $y_1^e=y_2^e$ then $y_1=y_2.$ – DanielWainfleet Mar 28 '17 at 10:25
  • In my previous comment I meant to say for positive $a,b,c,d$ and $x\geq 0$, – DanielWainfleet Mar 28 '17 at 10:32
  • @esotechnica Sure, something like that. Raise both sides of the equation to the $12\cdot 2$-th power, then regroup things in the right way to get both sides equal. – Jack M Mar 28 '17 at 11:02

1 Answers1

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Let $x, y$ be not negative, $a=1$ and $b$ odd. $$y=x^{\frac{1}{b}} \overset{def.}\Leftrightarrow y^b=x \Leftrightarrow y= \sqrt[b]{x}.$$

Now let $x, y$ not negative, $a$ be arbitrary and $b$ odd. $$y=x^{\frac{a}{b}} \overset{def.}\Leftrightarrow y^b = x^a \Leftrightarrow \sqrt[b]{y^b} = \sqrt[b]{x^a} \Leftrightarrow y = \sqrt[b]{x^a} $$

Since $(\sqrt[b]{x^a})^b = x^a$ and by your proof for integer exponents $((\sqrt[b]{x})^a)^b = (\sqrt[b]{x})^{ab} = x^a$, we have $(\sqrt[b]{x^a})=(\sqrt[b]{x})^a$.

Now let $x, y$ be not negative and $b, d$ odd. $$x^{\frac{a}{b}}x^{\frac{c}{d}}=x^{\frac{ad}{bd}}x^{\frac{bc}{bd}} = (\sqrt[bd]{x})^{ad} (\sqrt[bd]{x})^{bc} \overset{\text{form the integer result}}=(\sqrt[bd]{x})^{ad+bc}= x^{\frac{ad+bc}{bd}}.$$ For even denominators you basically do the same, but pay attention because even roots are not unique ($x^2 = 2 \Rightarrow x= \sqrt{2}$ or $- \sqrt{2}$).

Also note that you have to place proper restrictions on the domain of $x$ and $y$, as for example you might not want to have $\sqrt{-1}$.

cdwe
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  • I'm unsure how you justified the equivalence of $y^b = x^a \equiv (y^b)^{\frac 1 b} = (x^a)^{\frac 1 b}$? (going from second to third equivalence in your second line. I've eliminated the radicals as I find it confusing). That is, how do you prove that $x = y \equiv a^n = b^n$ in general for rational exponents? – esotechnica Mar 28 '17 at 01:09
  • If you know $x^a = y^b$, you apply $()^{\frac{1}{b}}$ on both sides. I'm not sure if I understand the second part of your comment, if it's still unclear could you explain that again? – cdwe Mar 28 '17 at 01:20
  • I guess I'm asking, what rule or principle allows you to take any equation, and apply an exponent to both sides. That is, if $x^a = y^b$, how do you conclude that $(x^a)^{\frac 1 b} = (y^b)^{\frac 1 b}$? I know that you can do this, but since I'm trying to derive the exponential laws with rational exponents from first principles, I'm not sure how to prove that such a statement holds. – esotechnica Mar 28 '17 at 01:42
  • $x^a =y^b \Rightarrow (x^a)^{\frac{1}{b}} = (y^b)^{\frac{1}{b}}$ holds because exponentiation is a well defined function. $x^a =y^b \Leftarrow (x^a)^{\frac{1}{b}} = (y^b)^{\frac{1}{b}}$ holds because in this case the exponentiation is injectice (because i choose b odd). To see this, you can prove that $()^{b}$ is injective for odd b. $\sqrt[b]{}$ defined as the inverse function to $()^b$ then automatically becomes injective. See Henry Swansons answer here: http://math.stackexchange.com/a/569120/347301 – cdwe Mar 28 '17 at 07:07
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    I'm not really sure what you mean by well-defined function. How do you prove this? – esotechnica Mar 28 '17 at 08:51
  • Are you aware of any good books that describe how to construct these proofs from definitions rigorously? I feel like I'm missing some understanding. – esotechnica Mar 28 '17 at 08:55
  • Also, I have edited my original post above, and think I have found a proof, but I need to be able to prove that $(xy)^{\frac a b} = x^{\frac a b}y^{\frac a b}$. Can you help? I only get as far as proving $(xy)^{\frac a b} = (x^ay^a)^{\frac 1 b}$... – esotechnica Mar 28 '17 at 08:57
  • @last com: we have shown that $(xy)^{\frac{a}{b}} = \sqrt[b]{(xy)^a}$. Now $(\sqrt[b]{(xy)^a})^b = (xy)^a$. On the other hand, look at $(x^{\frac{a}{b}}y^{\frac{a}{b}})^b =(x^{\frac{a}{b}})^b (y^{\frac{a}{b}})^b $ because of the law exponent law for integers. $(x^{\frac{a}{b}})^b (y^{\frac{a}{b}})^b = (xy)^a$, so putting things together we get $((xy)^{\frac{a}{b}})^b = (x^{\frac{a}{b}}y^{\frac{a}{b}})^b$. Now again, if $()^b$ is injective, we can conclude $(x^{\frac{a}{b}}) (y^{\frac{a}{b}}) = (xy)^{\frac{a}{b}}$. – cdwe Mar 28 '17 at 09:38
  • @comments before that: Unfortuneately I don't own any literature in english, so I don't think I can provide any helpful tips there. My "well defined function", I mean the necessary property of a function, that "the output is unique for the input values provided", as you put it in your question here: http://math.stackexchange.com/questions/1917581/how-do-i-define-exactly-what-a-function-is If we know $x^a = y^b$, then $x^a$ and $y^b$ are the same number, i.e. the same "input value", so if we put in in the function $()^{\frac{1}{b}}$ we get one unique "output value". – cdwe Mar 28 '17 at 09:55
  • I'm so thoroughly confused... Don't you need to prove $x = y \to x^{\frac a b} = y^{\frac a b}$ in order to show it is a function? I don't know how to do this?? – esotechnica Mar 28 '17 at 11:25
  • I'm going to try to explain it in chat later today! – cdwe Mar 28 '17 at 12:24
  • Let us continue this discussion in chat. – cdwe Mar 28 '17 at 16:19