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Prove that $ exp_{a}(\frac{p}{q}) = \sqrt[q]{a^{p}} \space \forall \space p,q \in \mathbb{Z} $ with $ q \geq 2 $

I'm not sure how to approach this question. I was thinking through in induction with base case $ q = 2 $ but there is also the $p$ variable.

Any ideas ?

Antonio Vargas
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Gregory Peck
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  • I take it that your notation $exp_a(p/q)$ means $a^{p/q}$. In which case, the question arises, how are you defining $a^{p/q}$? One way to define it is as $\root q\of{a^p}$, in which case your problem disappears. – Gerry Myerson Feb 12 '15 at 11:39

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As noted by @Gerry Myerson the problem disappear if you simply define: $$ a^{\frac{p}{q}}:= \sqrt[q]{a^p} =x \quad such \;that \quad x^q=a^p $$

but this seems to be a trick to avoid the problem by a suitable definition that is justified by the exponential rules for integer exponents. If we are satisfied with this: okay!

But if we want a more deeper answer then ve have to justify that formula starting from a more general definition of an exponential with rational exponent.

For positve integers exponents a good definition can be: $$ \forall a \in \mathbb{R}: \quad a^0=1, \quad a^n =a\cdot a^{n-1} \quad \forall n \in \mathbb {N} $$

From this definition we prove easily that $$ a^{n+m}=a^n\cdot a^m \quad \forall n,m \in \mathbb{N} ; \qquad a=1=a $$ So we have a function $\exp_a:\mathbb{K} \rightarrow \mathbb{K}$ such that: $$ (1)\;\forall a\in \mathbb{K}:\quad \exp_a(0)=1, \quad \exp_a(1)=a, \quad \exp_a(n+m)=\exp_a(n)\exp_a(m) \; \forall n,m \in \mathbb{N} $$ Now we can adopt $(1)$ as a definition of the exponential function and use it to extend such function to other kinds of exponents, chosing $n,m \in \mathbb{K} $.

The definition is:

$$ (2)\;\forall a\in \mathbb{K}:\quad \exp_a(0)=1, \quad \exp_a(1)=a, \quad \exp_a(x+y)=\exp_a(x)\exp_a(y) \; \forall x,y \in \mathbb{K} $$

E.g., for an integer $n<0$ we see that our definition give: $$ \exp_a\left( n+(-n) \right)=\exp_a(0) = \exp_a(n)\exp_a(-n)=1 $$ so $\exp_a(-n)$ must be the inverse of $\exp_a(n)$, and this fact justifies : $$ a^{-n}=\dfrac{1}{a^n} $$ Note that from $(2)$ we have also the rule: $$ (a^n)^m = a^{nm} \; \forall n,m \in \mathbb{Z} $$ Now, if we want an exponent $p \in \mathbb{Q}$, we have, for $p=m/n \Rightarrow np=m $: $$ \exp_a(np)=\exp_a (m) \Rightarrow x^n=\left[exp_a(p)\right]^{\,n}=\exp_a(m) $$ so that $x$ is the number such that $x^n=a^m$ i.e. $x=\sqrt[n]{a^m}$.

At last note that it does matter the choice of the field $\mathbb{K}$ in def. $(2)$, because the exponential function must have value in $\mathbb{K}$ so, for negative integer exponents, $\mathbb{K}$ must be at least $\mathbb{Q}$, and for exponents in $\mathbb{Q}$ the field $\mathbb{K}$ must be a suitable subfield of $\mathbb{R}$ ( see The exponential extension of $\mathbb{Q}$ is a proper subset of $\mathbb{C}$?)

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Emilio Novati
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  • I don't know about the "only one continuous function" bit. Given a function defined only on the integers, there are many ways to extend it to a continuous function on the reals (or on the rationals, or on the complex numbers). – Gerry Myerson Feb 12 '15 at 23:12
  • @Gerry :The claim refers to the fact that there exists only one ( non-trivial) solution of the functional equation $f(x+y)=f(x)f(y)$ with $f(1)=a$, See e.g. :J.Aczél, Lectures on functional equations and their applications. The continuity being necessary when $\mathbb{K}$ is a complete space to avoid ''wild'' solutions. – Emilio Novati Feb 13 '15 at 08:51
  • Sure, but for that you want the functional equation to hold for all $x,y$ in the field. What you have written in your answer is only that it holds for all $x,y$ in the naturals. – Gerry Myerson Feb 13 '15 at 09:42
  • good point! I wrote that we want to ''extend to other kinds of exponents'', but I've not specified what kinds. I edit and change a bit the definition. – Emilio Novati Feb 13 '15 at 10:18
  • But after the edit, it still says there's only one continuous function on the field that satisfies the functional relation on the naturals, and that's plainly false. (Also, one of those $n$s on the right side of the equation wants to be an $m$.) – Gerry Myerson Feb 13 '15 at 11:52
  • @GerryMyerson: Thanks,corrected typo.But I don't understand your first question. The condition $f(1)=a$ fix only one exponential function that, for any integer $n$, is exactly $a^n$. – Emilio Novati Feb 13 '15 at 12:17
  • I'm saying that, for any $a$, there are many functions, continuous on the reals, satisfying $f(n)=a^n$ for all natural numbers $n$. – Gerry Myerson Feb 13 '15 at 21:29
  • but, as far as I know, only one of those functions is such that $f(x+y)=f(x)f(y) \forall x,y \in mathbb{R}$. – Emilio Novati Feb 15 '15 at 09:39
  • No kidding. But what you actually write is that there is only one continuous function satisfying ($f(1)=a$ and) $f(m+n)=f(m)f(n)$ for all natural numbers $m,n$, and that's just nonsense. – Gerry Myerson Feb 15 '15 at 10:52
  • Thank you @Gerry. Now i've understand the point! (maybe I had some trouble because my english is not fluent). I have edit and I hope that now work properly. – Emilio Novati Feb 15 '15 at 11:08