1.2: Exponents and Cancellation
- Page ID
- 14707
This page is a draft and is under active development.
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 \((a^m)^n = a^{mn}, \forall a, m, n \in \mathbb{Z} \).