1.2: Exponents and Cancellation

Definition: Inverse

Let $S$ be a set with a binary operation $\star$, and with identity $e$. Let $a \in S$, then $b\in S$ is called an inverse of $a$ if $a \star b= b \star a=e.$

Example $\PageIndex{1}$:

1. For every $a \in \mathbb{Z}$, $-a$ is the inverse of $a$ with the operation $+$.
2. For every $a \in \mathbb{ R} \setminus \{0\}$, $a^{-1}=\frac{1}{a}$ is the inverse of $a$ with the multiplication.

Cancellation law

Let $S$ be a set with a binary operation $\star$. If for any $a, b, c \in S$, $a \star b= a \star c$ then $b=c$

Example $\PageIndex{2}$:

$(1)(0)=(3)(0)=0$, but $1 \ne 3$.

Example $\PageIndex{3}$:

1. For any $a, b, c \in \mathbb{Z}$, $a + b= a + c$ then $b=c$.
2. For any $a, b,c \in \mathbb{Z}$ and $a\ne 0$, $a b= a c$ then $b=c$.

Example $\PageIndex{4}$:

If $ab=0$ then $a=0$ or $b=0$.

Theorem $\PageIndex{1}$

For any integers $a$, and $b$, the following are true.

1. $-(-a)=a.$

2. $0(a)=0.$

3. $(-a)b=-ab.$

4. $(-a)(-b)=ab.$

Proof

1. Let $a \in \mathbb{Z}$. Since $-a$ is the inverse of $a$, $a+(-a)=(-a)+a=0$. Therefore the additive inverse of $-a$ is $a$.

Thus $-(-a)=a.$

2. Let $a \in \mathbb{Z}$. Then by distributive law, $0a+0a=(0+0)a=0a=0a+0.$ Now by cancelations law, $0a=0$.

3. Let $a, b \in \mathbb{Z}$. By distributive law, $((-a)+a)b=(-a)b+ab.$ Since $-a$ is the additive inverse of $a$, $(-a)+a=0$. By (2), $0=(-a)b+ab.$ Thus $(-a)b$ is the additive inverse of $ab$. Hence $-ab= (-a)b$.

4. Let $a, b \in \mathbb{Z}$. Since $(-a)(-b)+(-a)b=(-a)(-b+b)=(-a)(0)=0.$ Hence $(-a)(-b)$ is the additive inverse of $(-a)b$. But $ab$ is the additive inverse of $-ab$. Thus by (3), we have $(-a)(-b)=ab.$.

Definition: Exponentiation

For every $a, n \in \mathbb{Z_+}$, the binary operation exponentiation is denoted as $a^n$, defined as $n$ copies of $a$.

Example $\PageIndex{5}$:

$2^3=8$

Example $\PageIndex{6}$:

1. Determine whether the exponentiation is associative?
2. Determine whether the exponentiation is commutative?

Solution

1. Since $(3^2)^3= 9^3$ is not the same as $3^{2^3}= 3^8$, the exponentiation is not associative.
2. Since $3^2= 9$ is not the same as $2^3= 8$, the exponentiation is not commutative.

Theorem $\PageIndex{2}$

The exponentiation is distributive over multiplication. That is $(ab)^n=a^nb^n, \forall a, b, n \in \mathbb{Z}$.

Proof

Since multiplication is associative, the result follows.

Example $\PageIndex{7}$:

Prove that $a^m a^n= a^{m+n}, \forall a, m, n \in \mathbb{Z}$.

Example $\PageIndex{8}$:

Prove that $(a^m)^n = a^{mn}, \forall a, m, n \in \mathbb{Z}$.