Exercise 4.7

Proof. We show this for $n\in {\mathbb{Z}}_{\text{+}}$ by simple induction on $n$. First, clearly $1^1=1$ by the inductive definition of exponentiation. Next, if $1^n=1$, then we have $1^{n+1}=1^n\cdot 1=1^n=1$ by the inductive definition of exponentiation and the inductive hypothesis. This completes the induction so that the result holds for all $n\in {\mathbb{Z}}_{\text{+}}$.

Clearly if $n=0$ then, by the definition of 0 as an exponent, $1^n=1^0=1$.

Lastly, if $n\in {\mathbb{Z}}_{\text{−}}$ then there is a $k\in {\mathbb{Z}}_{\text{+}}$ where $n=-k$. Then we have

Thus the result has been shown for all the resulting cases when $n\in \mathbb{Z}$. □

Proof. We have

Thus ${(1∕a)}^n$ must be the unique reciprocal of $a^n$, that is ${(1∕a)}^n=1∕a^n$ as desired. □

Proof. We have

as desired. □

Proof. Consider any real $a\ne 0$ and $n,m\in \mathbb{Z}$. We number the following cases for reference:

1.

Case: $n\in {\mathbb{Z}}_{\text{+}}$.

(a)

Case: $m\in {\mathbb{Z}}_{\text{+}}$. Then the result immediately applies by Exercise 4.6.

(b)

Case: $m=0$. Then we have $a^n{a}^m=a^n{a}^0=a^n\cdot 1=a^n=a^{n+0}=a^{n+m}$.

(c)

Case: $m\in {\mathbb{Z}}_{\text{−}}$. Then $m=-k$ for some $k\in {\mathbb{Z}}_{\text{+}}$.

i.

Case: $n>k$. Then $n-k>0$ so that $n-k\in {\mathbb{Z}}_{\text{+}}$ since $n-k\in \mathbb{Z}$ by Exercise 4.5 part (d). We then have $$\begin{array}{llllll}\hfill a^n{a}^m& =a^n{a}^{-k}& \hfill & & \hfill & \\ \hfill & =a^{n-k+k}{a}^{-k}& \hfill \text{(since }n=n+0=n-k+k\text{)}& & \hfill & & \hfill \\ \hfill & =(a^{n-k}{a}^k)a^{-k}& \hfill \text{(by Exercise 4.6 since }k,n-k\in {\mathbb{Z}}_{+}\text{)}& & \hfill & & \hfill \\ \hfill & =a^{n-k}(a^k{a}^{-k})& \hfill \text{(since multiplication is associative)}& & \hfill & & \hfill \\ \hfill & =a^{n-k}\cdot 1& \hfill \text{(by Lemma }\text{3<!-- tex4ht:ref: lem:intreal:negexp -->}\text{ since }k\in {\mathbb{Z}}_{+}\text{)}& & \hfill & & \hfill \\ \hfill & =a^{n-k}& \hfill & & \hfill & \\ \hfill & =a^{n+m}.& \hfill & & \hfill & \end{array}$$

ii.

Case: $n=k$. Then clearly $n+m=n-k=k-k=0$, so that we have $a^n{a}^m=a^k{a}^{-k}=1=a^0=a^{n+m}$ by Lemma 3 and the definition of 0 as an exponent.

iii.

Case: $n<k$. Then $n-k<0$ so that $n-k\in {\mathbb{Z}}_{\text{−}}$ since $n-k\in \mathbb{Z}$ by Exercise 4.5 part (d). Also, clearly $-n\in {\mathbb{Z}}_{\text{−}}$ since $n\in {\mathbb{Z}}_{\text{+}}$. Then we have $$\begin{array}{llllll}\hfill a^n{a}^m& =a^n{a}^{-k}& \hfill & & \hfill & \\ \hfill & =a^n{a}^{-k+n-n}& \hfill \text{(since }-k=-k+0=-k+n-n\text{)}& & \hfill & & \hfill \\ \hfill & =a^n{a}^{n-k-n}& \hfill \text{(since addition is commutative)}& & \hfill & & \hfill \\ \hfill & =a^n(a^{n-k}{a}^{-n})& \hfill \text{(by case 3c below since }n-k,-n\in {\mathbb{Z}}_{-}\text{)}& & \hfill & & \hfill \\ \hfill & =a^n(a^{-n}{a}^{n-k})& \hfill \text{(since multiplication is commutative)}& & \hfill & & \hfill \\ \hfill & =(a^n{a}^{-n})a^{n-k}& \hfill \text{(since multiplication is associative)}& & \hfill & & \hfill \\ \hfill & =1\cdot a^{n-k}& \hfill \text{(by Lemma }\text{3<!-- tex4ht:ref: lem:intreal:negexp -->}\text{ )}& & \hfill & & \hfill \\ \hfill & =a^{n-k}& \hfill & & \hfill & \\ \hfill & =a^{n+m}.& \hfill & & \hfill & \end{array}$$

2.

Case: $n=0$.

(a)

Case: $m\in {\mathbb{Z}}_{\text{+}}$. Since $a^n{a}^m=a^m{a}^n$ and $a^{n+m}=a^{m+n}$, this the same as case 1b above.

(b)

Case: $m=0$. Then we have $a^n{a}^m=a^0{a}^0=1\cdot 1=1=a^0=a^{0+0}=a^{n+m}$.

(c)

Case: $m\in {\mathbb{Z}}_{\text{−}}$. Then there is a $k\in {\mathbb{Z}}_{\text{+}}$ such that $m=-k$, and $a^n{a}^m=a^0{a}^{-k}=1\cdot (1∕a^k)=1∕a^k=a^{-k}=a^m=a^{0+m}=a^{n+m}$.

3.

Case: $n\in {\mathbb{Z}}_{\text{−}}$.

(a)

Case: $m\in {\mathbb{Z}}_{\text{+}}$. This is the same as case 1c above.

(b)

Case: $m=0$. This is the same as case 2c above.

(c)

Case: $m\in {\mathbb{Z}}_{\text{−}}$. Here we have that $n=-k$ and $m=-l$ for some $k,l\in {\mathbb{Z}}_{\text{+}}$. Hence we have $$\begin{array}{llllll}\hfill a^n{a}^m& =a^{-k}{a}^{-l}& \hfill & & \hfill & \\ \hfill & =\left(\frac{1}{a^k}\right)\left(\frac{1}{a^l}\right)& \hfill \text{(by the definition of negative exponents)}& & \hfill & & \hfill \\ \hfill & ={\left(\frac{1}{a}\right)}^k{\left(\frac{1}{a}\right)}^l& \hfill \text{(by Lemma }\text{2<!-- tex4ht:ref: lem:intreal:recpow -->}\text{ )}& & \hfill & & \hfill \\ \hfill & ={\left(\frac{1}{a}\right)}^{k+l}& \hfill \text{(by Exercise 4.6 since }k,l\in {\mathbb{Z}}_{+}\text{)}& & \hfill & & \hfill \\ \hfill & =\frac{1}{a^{k+l}}& \hfill \text{(by Lemma }\text{2<!-- tex4ht:ref: lem:intreal:recpow -->}\text{ )}& & \hfill & & \hfill \\ \hfill & =a^{-(k+l)}& \hfill \text{(by the definition of negative exponents)}& & \hfill & & \hfill \\ \hfill & =a^{-k-l}& \hfill & & \hfill & \\ \hfill & =a^{n+m}.& \hfill & & \hfill & \end{array}$$

Thus in all cases we have shown the result. □

Proof. Consider any real $a\ne 0$ and $n,m\in \mathbb{Z}$. We again number the cases for reference:

1.

Case: $n\in {\mathbb{Z}}_{\text{+}}$.

(a)

Case: $m\in {\mathbb{Z}}_{\text{+}}$. Then the result immediately applies by Exercise 4.6.

(b)

Case: $m=0$. Then we have ${(a^n)}^m={(a^n)}^0=1=a^0=a^{n\cdot 0}=a^{\mathit{nm}}$ by the definition of a 0 exponent.

(c)

Case: $m\in {\mathbb{Z}}_{\text{−}}$. Then there is a $k\in {\mathbb{Z}}_{\text{+}}$ such that $m=-k$. Then we have $$\begin{array}{llllll}\hfill {(a^n)}^m& ={(a^n)}^{-k}& \hfill & & \hfill & \\ \hfill & =\frac{1}{{(a^n)}^k}& \hfill \text{(by the definition of negative exponents)}& & \hfill & & \hfill \\ \hfill & =\frac{1}{a^{\mathit{nk}}}& \hfill \text{(by Exercise 4.6 since }n,k\in {\mathbb{Z}}_{\text{+}}\text{)}& & \hfill & & \hfill \\ \hfill & =a^{-(\mathit{nk})}& \hfill \text{(by the definition of negative exponents)}& & \hfill & & \hfill \\ \hfill & =a^{n(-k)}& \hfill & & \hfill & \\ \hfill & =a^{\mathit{nm}}.& \hfill & & \hfill & \end{array}$$

2.

Case: $n=0$. Then we have ${(a^n)}^m={(a^0)}^m=1^m=1=a^0=a^{0\cdot m}=a^{\mathit{nm}}$ by the definition of 0 as an exponent and Lemma 1 .

3.

Case: $n\in {\mathbb{Z}}_{\text{−}}$. Then $n=-k$ for some $k\in {\mathbb{Z}}_{\text{+}}$.

(a)

Case: $m\in {\mathbb{Z}}_{\text{+}}$. Then we have $$\begin{array}{llllll}\hfill {(a^n)}^m& ={(a^{-k})}^m& \hfill & & \hfill & \\ \hfill & ={\left(\frac{1}{a^k}\right)}^m& \hfill \text{(by the definition of negative exponents)}& & \hfill & & \hfill \\ \hfill & ={\left[{\left(\frac{1}{a}\right)}^k\right]}^m& \hfill \text{(by Lemma }\text{2<!-- tex4ht:ref: lem:intreal:recpow -->}\text{ )}& & \hfill & & \hfill \\ \hfill & ={\left(\frac{1}{a}\right)}^{\mathit{km}}& \hfill \text{(by Exercise 4.6 since }k,m\in {\mathbb{Z}}_{+}\text{)}& & \hfill & & \hfill \\ \hfill & =\frac{1}{a^{\mathit{km}}}& \hfill \text{(by Lemma }\text{2<!-- tex4ht:ref: lem:intreal:recpow -->}\text{ )}& & \hfill & & \hfill \\ \hfill & =a^{-(\mathit{km})}& \hfill \text{(by the definition of negative exponents)}& & \hfill & & \hfill \\ \hfill & =a^{(-k)m}& \hfill & & \hfill & \\ \hfill & =a^{\mathit{nm}}.& \hfill & & \hfill & \end{array}$$

(b)

Case: $m=0$. The same argument as in case 1b above applies here as it does not depend on $n$ being positive.

(c)

Case: $m\in {\mathbb{Z}}_{\text{−}}$. Then $m=-l$ for some $l\in {\mathbb{Z}}_{\text{+}}$, and we have $$\begin{array}{llllll}\hfill {(a^n)}^m& ={(a^{-k})}^{-l}& \hfill & & \hfill & \\ \hfill & ={\left(\frac{1}{a^k}\right)}^{-l}& \hfill \text{(by the definition of negative exponents)}& & \hfill & & \hfill \\ \hfill & =\frac{1}{{(1∕a^k)}^l}& \hfill \text{(by the definition of negative exponents)}& & \hfill & & \hfill \\ \hfill & =\frac{1}{{[{(1∕a)}^k]}^l}& \hfill \text{(by Lemma }\text{2<!-- tex4ht:ref: lem:intreal:recpow -->}\text{ )}& & \hfill & & \hfill \\ \hfill & =\frac{1}{{(1∕a)}^{\mathit{kl}}}& \hfill \text{(by Exercise 4.6 since }k,l\in {\mathbb{Z}}_{\text{+}}\text{)}& & \hfill & & \hfill \\ \hfill & ={\left(\frac{1}{1∕a}\right)}^{\mathit{kl}}& \hfill \text{(by Lemma }\text{2<!-- tex4ht:ref: lem:intreal:recpow -->}\text{ )}& & \hfill & & \hfill \\ \hfill & =a^{\mathit{kl}}& \hfill & & \hfill & \\ \hfill & =a^{(-k)(-l)}& \hfill & & \hfill & \\ \hfill & =a^{\mathit{nm}}.& \hfill & & \hfill & \end{array}$$

Thus in all cases we have shown the result. □

Proof. We have the following cases:

Case: $m\in {\mathbb{Z}}_{\text{+}}$. The result then follows immediately from Exercise 4.6.

Case: $m=0$. Then we have $a^m{b}^m=a^0{b}^0=1\cdot 1=1={(\mathit{ab})}^0={(\mathit{ab})}^m$ by the definition of a 0 exponent.

Case: $m\in {\mathbb{Z}}_{\text{−}}$. Then there is a $k\in {\mathbb{Z}}_{\text{+}}$ such that $m=-k$. Then we have

Therefore in all cases the result has been shown. □