Using the theorem stated in the solution to Exercise 5, we can again, similar to Exercise 6, define $A$ , $B$ and $C$ to be the sets of all integers $m$ such that the first, second and third law of exponents, respectively, holds for all $n\in\mathbb{Z}$ and $a,b\in\mathbb{R}-\{0\}$ . Then, $0\in A\cap B\cap C$ , as, by definition, for every $n\in\mathbb{Z}$ and $a,b\in\mathbb{R}-\{0\}$ , $a^{n}a^{0}=a^{n}\cdot1=a^{n+0}$ , $(a^{n})^{0}=1=a^{n\cdot0}$ , and $a^{0}b^{0}=1\cdot1=1=(ab)^{0}$ .

To prove the “induction” step stated in the theorem of Exercise 5, we may find it useful to note that $a^{n}a=a^{n+1}$ for all $n\in\mathbb{Z}$ , regardless of whether $n$ is positive, zero or negative. Indeed, if $n$ is positive, then this is just by definition, if $n=0$ then we have $a=a$ , if $n=-1$ then we have $(1/a)\cdot a=1=a^{0}$ , and if $n<-1<0$ then $n+1<0$ and, by denoting $m=-n$ , $m>1>0$ , we have $(1/a^{m})\cdot a=1/a^{m-1}$ , which follows from $(1/a^{m})\cdot a\cdot a^{m-1}=1$ . Similarly, for every $n\in\mathbb{Z}$ , $a^{n}a^{-1}=a^{n}(1/a)=a^{n-1}$ . This follows from the fact that $n-1\in\mathbb{Z}$ and $a^{n-1}a=a^{n}$ .

Now, if $m\in A$ , then for every $n\in\mathbb{Z}$ and $a\in\mathbb{R}-\{0\}$ , $a^{n}a^{m+1}=a^{n}a^{m}a$ $=a^{n+m}a=a^{n+m+1}$ , and $a^{n}a^{m-1}=a^{n}a^{m-1}aa^{-1}$ $=a^{n}a^{m}a^{-1}$ $=a^{n+m}a^{-1}$ $=a^{n+m-1}$ , hence, $m-1,m+1\in A$ , $A$ satisfies the conditions of the theorem, and $A=\mathbb{Z}$ . If $m\in B$ , then for every $n\in\mathbb{Z}$ and $a\in\mathbb{R}-\{0\}$ , $(a^{n})^{m+1}=(a^{n})^{m}a^{n}$ $=a^{nm}a^{n}$ [using the fact that $A=\mathbb{Z}$ ] $=a^{nm+n}$ $=a^{n(m+1)}$ , and $(a^{n})^{m-1}=(a^{n})^{m}(a^{n})^{-1}$ $=a^{nm}(a^{n})^{-1}$ $=a^{n(m-1)+n}(a^{n})^{-1}$ [using $A=\mathbb{Z}$ again] $=a^{n(m-1)}a^{n}(a^{n})^{-1}$ $=a^{n(m-1)}$ . Hence, $B=\mathbb{Z}$ . If $m\in C$ , then for every $n\in\mathbb{Z}$ and $a,b\in\mathbb{R}-\{0\}$ , $a^{m+1}b^{m+1}=a^{m}ab^{m}b$ $=a^{m}b^{m}ab$ $=(ab)^{m}(ab)$ $(ab)^{m+1}$ , and $a^{m-1}b^{m-1}=a^{m}a^{-1}b^{m}b^{-1}$ $=a^{m}b^{m}a^{-1}b^{-1}$ $=(ab)^{m}a^{-1}b^{-1}$ $=(ab)^{m-1}(ab)a^{-1}b^{-1}$ $=(ab)^{m-1}$ . Therefore, $C=\mathbb{Z}$ .