Prove that $ exp_{a}(\frac{p}{q}) = \sqrt[q]{a^{p}} \space \forall \space p,q \in \mathbb{Z} $ 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 ?
 A: 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 $\mathbb{C}$?)
