1.2: Exponents and Cancellation

Last updated
Save as PDF

Page ID: 14707

This page is a draft and is under active development.

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

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 \(\PageIndex{1}\):

For every \(a \in \mathbb{Z}\), \(-a\) is the inverse of \(a\) with the operation \(+\).
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 \(\PageIndex{3}\):

For any \(a, b, c \in \mathbb{Z}\), \(a + b= a + c\) then \(b=c\).
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 \(\PageIndex{6}\):

Determine whether the exponentiation is associative?
Determine whether the exponentiation is commutative?

Solution

Since \((3^2)^3= 9^3\) is not the same as \(3^{2^3}= 3^8\), the exponentiation is not associative.
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 \((a^m)^n = a^{mn}, \forall a, m, n \in \mathbb{Z} \).