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Mathematics LibreTexts

1.2 Exponents and Cancellation

  • Page ID
    14707
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

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