
# 1.2 Exponents and Cancellation

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Definition

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

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