1.1: Binary operations
- Page ID
- 7419
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Binary operation
Definition: Binary operation
Let \(S\) be a non-empty set, and \( \star \) said to be a binary operation on \(S\), if \(a \star b \) is defined for all \(a,b \in S\). In other words, \( \star\) is a rule for any two elements in the set \(S\).
Example \(\PageIndex{1}\):
The following are binary operations on \(\mathbb{Z}\):
- The arithmetic operations, addition \(+\), subtraction \(-\), multiplication \( \times \), and division \(\div \).
- Define an operation oplus on \(\mathbb{Z}\) by \(a \oplus b =ab+a+b, \forall a,b \in\mathbb{Z}\).
- Define an operation ominus on \(\mathbb{Z}\) by \(a \ominus b =ab+a-b, \forall a,b \in\mathbb{Z}\).
- Define an operation otimes on \(\mathbb{Z}\) by \(a \otimes b =(a+b)(a+b), \forall a,b \in\mathbb{Z}\).
- Define an operation oslash on \(\mathbb{Z}\) by \(a \oslash b =(a+b)(a-b), \forall a,b \in\mathbb{Z} \).
- Define an operation min on \(\mathbb{Z}\) by \(a \vee b =\min \{a,b\}, \forall a,b \in\mathbb{Z}\).
- Define an operation max on \(\mathbb{Z}\) by \(a \wedge b =\max \{a,b\}, \forall a,b \in\mathbb{Z}\).
- Define an operation defect on \(\mathbb{Z}\) by \(a \ast_3 b = a+b-3, \forall a,b \in\mathbb{Z}\).
Lets explore the binary operations, before we proceed:
Example \(\PageIndex{2}\):
- \(2 \oplus 3=(2)(3)+2+3=11\).
- \(2 \otimes 3=(2+3)(2+3)=25\).
- \(2 \oslash 3=(2+3)(2-3)=-5\).
- \(2 \ominus 3=(2)(3)+2-3=5\).
- \(2 \vee 3= 2\).
- \(2 \wedge 3 =3\).
Exercise \(\PageIndex{2}\)
- \(-2 \oplus 3\).
- \(-2 \otimes 3\).
- \(-2 \oslash 3\).
- \(-2 \ominus 3\).
- \(-2 \vee 3\).
- \(-2 \wedge 3 \).
- Answer
-
\(-5, 1,-5,-2,3\)
Properties:
Closure property
Definition: Closure
Let \(S\) be a non-empty set. A binary operation \( \star \) on \(S\) is said to be a closed binary operation on \(S\), if \(a \star b \in S, \forall a, b \in S\).
Below we shall give some examples of closed binary operations, that will be further explored in class.
Example \(\PageIndex{3}\): Closed binary operations
The following are closed binary operations on \(\mathbb{Z}\).
- The addition \(+\), subtraction \(-\), and multiplication \( \times \).
- Define an operation oplus on \(\mathbb{Z}\) by \(a \oplus b =ab+a+b, \forall a,b \in\mathbb{Z}\).
- Define an operation ominus on \(\mathbb{Z}\) by \(a \ominus b =ab+a-b, \forall a,b \in\mathbb{Z}\).
- Define an operation otimes on \(\mathbb{Z}\) by \(a \otimes b =(a+b)(a+b), \forall a,b \in\mathbb{Z}\).
- Define an operation oslash on \(\mathbb{Z}\) by \(a \oslash b =(a+b)(a-b), \forall a,b \in\mathbb{Z} \).
- Define an operation min on \(\mathbb{Z}\) by \(a \vee b =\min \{a,b\}, \forall a,b \in\mathbb{Z}\).
- Define an operation max on \(\mathbb{Z}\) by \(a \wedge b =\max \{a,b\}, \forall a,b \in\mathbb{Z}\).
- Define an operation defect on \(\mathbb{Z}\) by \(a \ast_3 b = a+b-3, \forall a,b \in\mathbb{Z}\).
Exercise \(\PageIndex{1}\)
Determine whether the operation ominus on \(\mathbb{Z_+}\) is closed?
- Answer
-
The operation ominus on \(\mathbb{Z_+}\) is closed.
Example \(\PageIndex{4}\): Counter Example
Division (\( \div \) ) is not a closed binary operations on \(\mathbb{Z}\).
\( 2, 3 \in \mathbb{Z} \) but \( \frac{2}{3} \notin \mathbb{Z} \).
Summary of arithmetic operations and corresponding sets:
\(+\) | \(\times\) | \(-\) | \(\div\) | |
\(\mathbb{Z_+}\) | closed | closed | not closed | not closed |
\(\mathbb{Z}\) | closed | closed | closed | not closed |
\(\mathbb{Q}\) | closed | closed | closed | closed (only when \(0\) is not included) |
\(\mathbb{R}\) | closed | closed | closed | closed (only when \(0\) is not included) |
Associative property
Definition: Associative
Let \(S\) be a subset of \(\mathbb{Z}\). A binary operation \( \star \) on \(S\) is said to be associative , if \( (a \star b) \star c = a \star (b \star c) , \forall a, b,c \in S\).
We shall assume the fact that the addition (\(+\)) and the multiplication (\( \times \)) are associative on \(\mathbb{Z_+}\). (You don't need to prove them!).
Below is an example of proof when the statement is True.
Example \(\PageIndex{5}\): Associative
Determine whether the binary operation oplus is associative on \(\mathbb{Z}\).
We shall show that the binary operation oplus is associative on \(\mathbb{Z}\).
- Proof:
-
Let \(a,b,c \in \mathbb{Z}\). Then consider, \((a \oplus b) \oplus c = (ab+a+b) \oplus c = (ab+a+b)c+(ab+a+b)+c= (ab)c+ac+bc+ab+a+b+c\).
On the other hand, \(a \oplus (b \oplus c)=a \oplus (bc+b+c)= a(bc+b+c)+a+(bc+b+c)=a(bc)+ab+ac+a+bc+b+c. \)
Since multiplication is associative on \(\mathbb{Z}\), \((a \oplus b) \oplus c =a \oplus (b \oplus c). \)
Thus, the binary operation oplus is associative on \(\mathbb{Z}\). \( \Box\)
Below is an example of how to disprove when a statement is False.
Example \(\PageIndex{6}\): Not Associative
Determine whether the binary operation subtraction (\( -\)) is associative on \(\mathbb{Z}\).
Answer: The binary operation subtraction (\( -\)) is not associative on \(\mathbb{Z}\).
- Counterexample:
-
Choose \( a=2,b=3, c=4,\) then \((2-3)-4=-1-4=-5 \), but \(2-(3-4)=2-(-1)=2+1=3\).
Hence the binary operation subtraction (\( -\)) is not associative on \(\mathbb{Z}\).
Commutative property
Definition: Commutative property
Let \(S\) be a non-empty set. A binary operation \( \star \) on \(S\) is said to be commutative, if \( a \star b = b \star a,\forall a, b \in S\).
We shall assume the fact that the addition (\(+\)) and the multiplication( \( \times \)) are commutative on \(\mathbb{Z_+}\). (You don't need to prove them!).
Below is the proof of subtraction (\( -\)) NOT being commutative.
Example \(\PageIndex{7}\): NOT Commutative
Determine whether the binary operation subtraction \( -\) is commutative on \(\mathbb{Z}\).
- Counterexample:
-
Choose \(a=3\) and \(b=4\).
Then \(a-b=3-4=-1\), and \(b-a= 4-3=1\).
Hence the binary operation subtraction \( -\) is not commutative on \(\mathbb{Z}\).
Example \(\PageIndex{8}\): Commutative
Determine whether the binary operation oplus is commutative on \(\mathbb{Z}\).
We shall show that the binary operation oplus is commutative on \(\mathbb{Z}\).
- Proof:
-
Let \(a,b \in \mathbb{Z}\).
Then consider, \((a \oplus b) = (ab+a+b).\)
On the other hand, \( (b \oplus a) = ba+b+a. \)
Since multiplication is associative on \(\mathbb{Z}\), \((a \oplus b) = (b \oplus a). \)
Thus, the binary operation oplus is commutative on \(\mathbb{Z}\). \( \Box\)
Identity
Definition: Identity
A non-empty set \(S\) with binary operation \( \star \), is said to have an identity \(e \in S\), if \( e \star a=a\star e=a, \forall a \in S.\)
Note that \(0\) is called additive identity on \(( \mathbb{Z}, +)\), and \(1\) is called multiplicative identity on \(( \mathbb{Z}, \times )\).
Example \(\PageIndex{9}\): Is identity unique?
Let \(S\) be a non-empty set and let \(\star\) be a binary operation on \(S\). If \(e_1\) and \(e_2\) are two identities in \((S,\star) \), then \(e_1=e_2\).
Proof:
Suppose that \(e_1\) and \(e_2\) are two identities in \((S,\star) \).
Then \(e_1=e_1 \star e_2=e_2.\)
Hence identity is unique. \( \Box\)
Example \(\PageIndex{10}\): Identity
Does \(( \mathbb{Z}, \oplus )\) have an identity?
- Answer
-
Let \(e\) be the identity on \(( \mathbb{Z}, \oplus )\).
Then \( e \oplus a=a\oplus e=a, \forall a \in \mathbb{Z}.\)
Thus \(ea+e+a=a\), and \(ae+a+e=a\) \(\forall a \in \mathbb{Z}.\)
Since \(ea+e+a=a\) \(\forall a \in \mathbb{Z},\) \(ea+e=0 \implies e(a+1)=0\) \(\forall a \in \mathbb{Z}.\)
Therefore \(e=0\).
Now \( 0 \oplus a=a\oplus 0=a, \forall a \in \mathbb{Z}.\)
Hence \(0\) is the identity on \(( \mathbb{Z}, \oplus )\).
Example \(\PageIndex{11}\):
Does \(( \mathbb{Z}, \otimes )\) have an identity?
- Answer
-
Let \(e\) be the identity on \(( \mathbb{Z}, \otimes )\).
Then \( e \otimes a=a \otimes e=a, \forall a \in \mathbb{Z}.\)
Thus \((e+a)(e+a)=(a+e)(a+e) =a, \forall a \in \mathbb{Z}.\)
Now, \( (a+e)(a+e) =a,\forall a \in \mathbb{Z}.\)
\(\implies a^2+2ea+e^2=a,\forall a \in \mathbb{Z}.\)
Choose \(a=0\) then \(e=0\).
If \(e=0\) then \( a^2=a,\forall a \in \mathbb{Z}.\)
This is a contradiction. Thus \(e=0\) is not an identity. Hence \(e\ne 0.\)
Choose \(a=1\). Then \(2e+e^2=0 \imples e(2+e)=0.\) Since \(e\ne 0\), \(e=-2\) This will not work for \(a=0.\)
For any other values of \(e\) will not work \(a=0\).
Hence, \(( \mathbb{Z}, \otimes )\) has no identity.
Distributive Property
Definition: Distributive property
Let \(S\) be a non-empty set. Let \(\star_1\) and \( \star_2\) be two different binary operations on \(S\).
Then \(\star_1\) is said to be distributive over \( \star_2\) on \(S \) if \( a \star_1 (b \star_2 c)= (a\star_1 b) \star_2 (a \star_1 c), \forall a,b,c,\in S \).
Note that the multiplication distributes over the addition on \(\mathbb{Z}.\) That is, \(4(10+6)=(4)(10)+(4)(6)=40+24=64\).
Further, we extend to \( (a+b)(c+d) =ac+ad+bc+bd\) (FOIL).
F-First
O-Outer
I-Inner
L-Last
This property is very useful to find \((26)(27)\) as shown below:
Example \(\PageIndex{12}\): Find \((26)(27)\)
\(20\) | \(6\) | |
---|---|---|
\(20\) | \(400\) | \(120\) |
\(7\) | \(140\) | \(42\) |
Hence \((26)(27) =400+120+140+42=702\).
Let's play a game!
Example \(\PageIndex{13}\):
Does multiplication distribute over subtraction?
Example \(\PageIndex{14}\):
Does division distribute over addition?
- Answer
-
Counterexample:
Choose \(a = 2, b = 3, \) and \(c = 4.\)
Then \(a \div (b + c) = 2 \div(3+4)\)
\(= 2 \div 7.\)
\(= \frac{2}{7}\).
and \( (a \div b) + (a \div c) = \frac{2}{3} + \frac{2}{4}\).
\(= \frac{7}{6}\).
Since \(\frac{2}{7} \ne \frac{7}{6}\), the binary operation \(\div\) is not distributive over \(+.\)
Example \(\PageIndex{15}\):
Does \( \otimes\) distribute over \(\oplus\) on \(\mathbb{Z}\) ?
- Answer
-
Counterexample:
Choose \(a = 2, b = 3, \) and \(c = 4.\)
Then \(2\otimes (3 \oplus 4) = 2\otimes [(3)(4)+3+4]\)
\(= 2\otimes 19\)
\(= (2+19)(2+19)\)
\(= 441\)
and \( (2\otimes 3)\oplus (2 \otimes 4)=[(2+3)(2+3)] \oplus [(2+4)(2+4)]\)
\(= 25 \oplus 36\)
\(= (25)(36)+25+36\)
\(= 961.\)
Since \(441 \ne 961,\) the binary operation \( \otimes\) is not distributive over \(\oplus \) on \(\mathbb{Z}\).
Summary
In this section, we have learned the following for a non-empty set \(S\):
- Binary operation,
- Closure property,
- Associative property,
- Commutative property,
- Distributive property, and
- Identity.