Binary Operations.
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Lesson Objectives
- Define a binary operation on a set and determine whether a given operation is binary.
- Test a binary operation for closure, commutativity, and associativity on a given set.
- Find the identity element and inverse of an element under a binary operation.
- Apply the distributive law in the context of binary operations.
- Construct and interpret a Cayley table (operation table) for a binary operation.
Introduction to Binary Operations
A binary operation is a rule that combines any two elements of a set to produce a third element, which may or may not be in the same set. These operations are the building blocks of abstract algebra and appear in everyday arithmetic (addition, subtraction, multiplication) as well as in more advanced areas like matrix multiplication and composition of functions. In this lesson, we will explore the essential properties of binary operations—closure, commutativity, associativity, identity, and inverses—and learn how to use operation tables.
Key Formulas & Definitions for This Lesson
• A binary operation \( * \) on a set \( S \) maps every ordered pair \((a,b) \in S \times S\) to an element \( a * b \).
• Closure: \( a * b \in S \) for all \( a,b \in S \).
• Commutativity: \( a * b = b * a \) for all \( a,b \in S \).
• Associativity: \( (a * b) * c = a * (b * c) \) for all \( a,b,c \in S \).
• Identity: \( e \in S \) such that \( a * e = e * a = a \) for all \( a \in S \).
• Inverse: for \( a \in S \), \( a^{-1} \in S \) such that \( a * a^{-1} = a^{-1} * a = e \).
• A binary operation \( * \) on a set \( S \) maps every ordered pair \((a,b) \in S \times S\) to an element \( a * b \).
• Closure: \( a * b \in S \) for all \( a,b \in S \).
• Commutativity: \( a * b = b * a \) for all \( a,b \in S \).
• Associativity: \( (a * b) * c = a * (b * c) \) for all \( a,b,c \in S \).
• Identity: \( e \in S \) such that \( a * e = e * a = a \) for all \( a \in S \).
• Inverse: for \( a \in S \), \( a^{-1} \in S \) such that \( a * a^{-1} = a^{-1} * a = e \).
Key Definitions:
• Binary Operation: A rule that combines two elements (operands) to produce a result.
• Closure: The property that the result of the operation stays within the set.
• Identity Element: An element that leaves any other element unchanged when combined with it.
• Inverse Element: An element that combines with a given element to produce the identity.
• Binary Operation: A rule that combines two elements (operands) to produce a result.
• Closure: The property that the result of the operation stays within the set.
• Identity Element: An element that leaves any other element unchanged when combined with it.
• Inverse Element: An element that combines with a given element to produce the identity.
Quick Reference: Binary Operation Properties
| Property | Statement | Example |
|---|---|---|
| Closure | \( a * b \in S \) for all \( a,b \in S \) | \( + \) on \( \mathbb{Z} \) is closed. |
| Commutativity | \( a * b = b * a \) | \( + \) and \( \times \) are commutative; \( - \) and \( \div \) are not. |
| Associativity | \( (a * b) * c = a * (b * c) \) | \( + \) and \( \times \) are associative; \( - \) is not. |
| Identity | \( a * e = e * a = a \) | \( 0 \) for addition, \( 1 \) for multiplication. |
| Inverse | \( a * a^{-1} = a^{-1} * a = e \) | \( -a \) for addition, \( 1/a \) for multiplication. |
Scroll sideways on smaller screens to view the full table.
Closure & Commutativity
A binary operation \( * \) on a set \( S \) is closed if for all \( a,b \in S \), the result \( a * b \) is also in \( S \). The operation is commutative if \( a * b = b * a \) for every pair \( a,b \). $$ \text{Closure: } \forall a,b \in S,\; a * b \in S \qquad \text{Commutativity: } \forall a,b \in S,\; a * b = b * a. $$
How to Test Closure and Commutativity:
1. Closure: Take two arbitrary elements from the set and compute the operation. Check that the result is in the set. Repeat for a variety of elements.
2. Commutativity: For a pair \( a,b \), compute \( a*b \) and \( b*a \). If they are equal for all pairs, the operation is commutative.
3. If a single counterexample is found, the property fails.
1. Closure: Take two arbitrary elements from the set and compute the operation. Check that the result is in the set. Repeat for a variety of elements.
2. Commutativity: For a pair \( a,b \), compute \( a*b \) and \( b*a \). If they are equal for all pairs, the operation is commutative.
3. If a single counterexample is found, the property fails.
Example 1: Testing Closure and Commutativity
Problem: Let \( * \) be defined on the set of integers \( \mathbb{Z} \) by \( a * b = a + b + ab \). Is the operation closed? Is it commutative?
Solution:
Step 1: For closure, take any \( a,b \in \mathbb{Z} \). Then \( a + b + ab \) is an integer because sums and products of integers are integers. Hence \( a * b \in \mathbb{Z} \). Closed.
Step 2: For commutativity, compute \( a * b = a + b + ab \) and \( b * a = b + a + ba \). Since addition and multiplication are commutative in \( \mathbb{Z} \), these are equal. Commutative.
Answer: The operation is closed and commutative on \( \mathbb{Z} \).
Problem: Let \( * \) be defined on the set of integers \( \mathbb{Z} \) by \( a * b = a + b + ab \). Is the operation closed? Is it commutative?
Solution:
Step 1: For closure, take any \( a,b \in \mathbb{Z} \). Then \( a + b + ab \) is an integer because sums and products of integers are integers. Hence \( a * b \in \mathbb{Z} \). Closed.
Step 2: For commutativity, compute \( a * b = a + b + ab \) and \( b * a = b + a + ba \). Since addition and multiplication are commutative in \( \mathbb{Z} \), these are equal. Commutative.
Answer: The operation is closed and commutative on \( \mathbb{Z} \).
Example 2: A Non-Commutative Operation
Problem: On the set of real numbers \( \mathbb{R} \), define \( a \circ b = a - b \). Is it closed? Is it commutative?
Solution:
Step 1: For \( a,b \in \mathbb{R} \), \( a - b \in \mathbb{R} \), so it is closed.
Step 2: Check commutativity: \( a \circ b = a - b \), but \( b \circ a = b - a \). These are equal only if \( a = b \). For example, \( 2 \circ 3 = -1 \) and \( 3 \circ 2 = 1 \), so they are not equal. Not commutative.
Answer: The operation is closed but not commutative on \( \mathbb{R} \).
Problem: On the set of real numbers \( \mathbb{R} \), define \( a \circ b = a - b \). Is it closed? Is it commutative?
Solution:
Step 1: For \( a,b \in \mathbb{R} \), \( a - b \in \mathbb{R} \), so it is closed.
Step 2: Check commutativity: \( a \circ b = a - b \), but \( b \circ a = b - a \). These are equal only if \( a = b \). For example, \( 2 \circ 3 = -1 \) and \( 3 \circ 2 = 1 \), so they are not equal. Not commutative.
Answer: The operation is closed but not commutative on \( \mathbb{R} \).
Watch Out!
Do not assume that an operation is commutative just because it is closed. Always test with specific values if the algebraic form isn't obviously symmetric.
Do not assume that an operation is commutative just because it is closed. Always test with specific values if the algebraic form isn't obviously symmetric.
Practice for Closure & Commutativity
- On \( \mathbb{Z} \), define \( a * b = a^2 + b^2 \). Is it closed? Is it commutative?
- On \( \mathbb{N} \), define \( a \star b = a^b \). Is it closed? Is it commutative?
- On \( \mathbb{Q} \), define \( a \oplus b = \frac{a+b}{2} \). Is it closed? Is it commutative?
- On \( \mathbb{R} \), define \( a \otimes b = a^2 - b^2 \). Is it closed? Is it commutative?
- On \( \mathbb{Z} \), define \( a \odot b = a \cdot b - a - b \). Is it closed? Is it commutative?
Associativity
A binary operation \( * \) on a set \( S \) is associative if for all \( a,b,c \in S \), \( (a * b) * c = a * (b * c) \). This property allows us to group operations in any order without changing the result. $$ (a * b) * c = a * (b * c) \quad \forall a,b,c \in S. $$
Testing for Associativity:
1. Pick three elements \( a,b,c \) from the set.
2. Compute \( (a * b) * c \) and \( a * (b * c) \).
3. If they are equal for all choices of \( a,b,c \), the operation is associative.
4. If you find one triple where they differ, the operation is not associative.
1. Pick three elements \( a,b,c \) from the set.
2. Compute \( (a * b) * c \) and \( a * (b * c) \).
3. If they are equal for all choices of \( a,b,c \), the operation is associative.
4. If you find one triple where they differ, the operation is not associative.
Example 3: Associative Operation
Problem: On \( \mathbb{Z} \), define \( a * b = a + b + ab \). Is it associative?
Solution:
Step 1: Compute \( (a * b) * c = (a+b+ab) * c = (a+b+ab) + c + (a+b+ab)c = a+b+c+ab+ac+bc+abc \).
Step 2: Compute \( a * (b * c) = a * (b+c+bc) = a + (b+c+bc) + a(b+c+bc) = a+b+c+bc+ab+ac+abc \).
Step 3: Both expressions are equal. Associative.
Answer: The operation is associative on \( \mathbb{Z} \).
Problem: On \( \mathbb{Z} \), define \( a * b = a + b + ab \). Is it associative?
Solution:
Step 1: Compute \( (a * b) * c = (a+b+ab) * c = (a+b+ab) + c + (a+b+ab)c = a+b+c+ab+ac+bc+abc \).
Step 2: Compute \( a * (b * c) = a * (b+c+bc) = a + (b+c+bc) + a(b+c+bc) = a+b+c+bc+ab+ac+abc \).
Step 3: Both expressions are equal. Associative.
Answer: The operation is associative on \( \mathbb{Z} \).
Example 4: Non-Associative Operation
Problem: On \( \mathbb{Z} \), define \( a \star b = a - b \). Is it associative?
Solution:
Step 1: Compute \( (a \star b) \star c = (a - b) - c = a - b - c \).
Step 2: Compute \( a \star (b \star c) = a - (b - c) = a - b + c \).
Step 3: These are equal only if \( c = 0 \) for all \( a,b \), which is not true. For example, \( (2-3)-4 = -5 \) and \( 2-(3-4) = 3 \). Not associative.
Answer: The operation is not associative on \( \mathbb{Z} \).
Problem: On \( \mathbb{Z} \), define \( a \star b = a - b \). Is it associative?
Solution:
Step 1: Compute \( (a \star b) \star c = (a - b) - c = a - b - c \).
Step 2: Compute \( a \star (b \star c) = a - (b - c) = a - b + c \).
Step 3: These are equal only if \( c = 0 \) for all \( a,b \), which is not true. For example, \( (2-3)-4 = -5 \) and \( 2-(3-4) = 3 \). Not associative.
Answer: The operation is not associative on \( \mathbb{Z} \).
Watch Out!
Many students assume subtraction and division are associative because addition and multiplication are. Always test with three distinct numbers to confirm.
Many students assume subtraction and division are associative because addition and multiplication are. Always test with three distinct numbers to confirm.
Practice for Associativity
- On \( \mathbb{R} \), define \( a * b = \frac{a+b}{2} \). Is it associative?
- On \( \mathbb{Z} \), define \( a \star b = a \cdot b + 1 \). Is it associative?
- On \( \mathbb{Q} \), define \( a \oplus b = a - b + 1 \). Is it associative?
- On \( \mathbb{R} \), define \( a \odot b = a^2 + b \). Is it associative?
- On \( \mathbb{Z} \), define \( a \diamond b = a + 2b \). Is it associative?
Identity & Inverse Elements
An identity element \( e \) for a binary operation \( * \) on a set \( S \) satisfies \( a * e = e * a = a \) for all \( a \in S \). If such an element exists, it is unique. For an element \( a \in S \), its inverse \( a^{-1} \) satisfies \( a * a^{-1} = a^{-1} * a = e \).
Finding Identity and Inverse:
1. To find the identity, solve \( a * e = a \) and \( e * a = a \) for \( e \).
2. The identity must be the same for both equations and belong to \( S \).
3. For a given \( a \), solve \( a * x = e \) for \( x \). This \( x \) is the inverse of \( a \).
4. Verify that \( x * a = e \) as well.
1. To find the identity, solve \( a * e = a \) and \( e * a = a \) for \( e \).
2. The identity must be the same for both equations and belong to \( S \).
3. For a given \( a \), solve \( a * x = e \) for \( x \). This \( x \) is the inverse of \( a \).
4. Verify that \( x * a = e \) as well.
Example 5: Finding Identity and Inverse
Problem: On \( \mathbb{Z} \), define \( a * b = a + b + ab \). Find the identity element and the inverse of 2, if they exist.
Solution:
Step 1: Find \( e \) such that \( a * e = a \). So \( a + e + ae = a \). Simplify: \( e + ae = 0 \Rightarrow e(1+a) = 0 \). For this to hold for all \( a \), we need \( e = 0 \). Check \( e * a = 0 + a + 0 = a \). So \( e = 0 \).
Step 2: Find inverse of 2. Solve \( 2 * x = 0 \): \( 2 + x + 2x = 0 \Rightarrow 2 + 3x = 0 \Rightarrow x = -\frac{2}{3} \). But \( -\frac{2}{3} \notin \mathbb{Z} \). So 2 has no inverse in \( \mathbb{Z} \).
Answer: Identity is \( 0 \); 2 has no inverse in \( \mathbb{Z} \).
Problem: On \( \mathbb{Z} \), define \( a * b = a + b + ab \). Find the identity element and the inverse of 2, if they exist.
Solution:
Step 1: Find \( e \) such that \( a * e = a \). So \( a + e + ae = a \). Simplify: \( e + ae = 0 \Rightarrow e(1+a) = 0 \). For this to hold for all \( a \), we need \( e = 0 \). Check \( e * a = 0 + a + 0 = a \). So \( e = 0 \).
Step 2: Find inverse of 2. Solve \( 2 * x = 0 \): \( 2 + x + 2x = 0 \Rightarrow 2 + 3x = 0 \Rightarrow x = -\frac{2}{3} \). But \( -\frac{2}{3} \notin \mathbb{Z} \). So 2 has no inverse in \( \mathbb{Z} \).
Answer: Identity is \( 0 \); 2 has no inverse in \( \mathbb{Z} \).
Example 6: Identity and Inverse in a Different Set
Problem: On \( \mathbb{Q} \), define \( a \circ b = a + b - ab \). Find the identity and the inverse of \( \frac{1}{2} \).
Solution:
Step 1: Find \( e \) such that \( a \circ e = a \): \( a + e - ae = a \Rightarrow e - ae = 0 \Rightarrow e(1-a) = 0 \). For this to hold for all \( a \), we need \( e = 0 \). Check \( e \circ a = 0 + a - 0 = a \). So \( e = 0 \).
Step 2: Find inverse of \( \frac{1}{2} \). Solve \( \frac{1}{2} \circ x = 0 \): \( \frac{1}{2} + x - \frac{1}{2}x = 0 \Rightarrow \frac{1}{2} + \frac{1}{2}x = 0 \Rightarrow \frac{1}{2}x = -\frac{1}{2} \Rightarrow x = -1 \).
Step 3: Verify \( x \circ \frac{1}{2} = -1 + \frac{1}{2} - (-1)\frac{1}{2} = -\frac{1}{2} + \frac{1}{2} = 0 \). So inverse is \(-1\).
Answer: Identity is \( 0 \); inverse of \( \frac{1}{2} \) is \( -1 \).
Problem: On \( \mathbb{Q} \), define \( a \circ b = a + b - ab \). Find the identity and the inverse of \( \frac{1}{2} \).
Solution:
Step 1: Find \( e \) such that \( a \circ e = a \): \( a + e - ae = a \Rightarrow e - ae = 0 \Rightarrow e(1-a) = 0 \). For this to hold for all \( a \), we need \( e = 0 \). Check \( e \circ a = 0 + a - 0 = a \). So \( e = 0 \).
Step 2: Find inverse of \( \frac{1}{2} \). Solve \( \frac{1}{2} \circ x = 0 \): \( \frac{1}{2} + x - \frac{1}{2}x = 0 \Rightarrow \frac{1}{2} + \frac{1}{2}x = 0 \Rightarrow \frac{1}{2}x = -\frac{1}{2} \Rightarrow x = -1 \).
Step 3: Verify \( x \circ \frac{1}{2} = -1 + \frac{1}{2} - (-1)\frac{1}{2} = -\frac{1}{2} + \frac{1}{2} = 0 \). So inverse is \(-1\).
Answer: Identity is \( 0 \); inverse of \( \frac{1}{2} \) is \( -1 \).
Watch Out!
An identity element may not exist for every operation. Even if it does, some elements may not have an inverse within the set. Always check whether the inverse belongs to the set.
An identity element may not exist for every operation. Even if it does, some elements may not have an inverse within the set. Always check whether the inverse belongs to the set.
Practice for Identity & Inverse
- On \( \mathbb{R} \), define \( a * b = a + b - 3 \). Find the identity.
- On \( \mathbb{Q} \), define \( a \star b = a + b + 2ab \). Find the identity and the inverse of 1.
- On \( \mathbb{Z} \), define \( a \oplus b = a + b - 2 \). Find the identity and the inverse of 5.
- On \( \mathbb{R} \setminus \{1\} \), define \( a \circ b = a + b - ab \). Find the identity and the inverse of 3.
- On \( \mathbb{Q} \setminus \{0\} \), define \( a \odot b = \frac{a}{b} \). Find the identity and the inverse of 2.
Distributive Law
When a set has two binary operations, say \( * \) and \( \circ \), we say that \( * \) is distributive over \( \circ \) if for all \( a,b,c \): $$ a * (b \circ c) = (a * b) \circ (a * c) \quad \text{and} \quad (b \circ c) * a = (b * a) \circ (c * a). $$ This is analogous to the distributive law of multiplication over addition.
Testing the Distributive Law:
1. Choose arbitrary elements \( a,b,c \) from the set.
2. Compute \( a * (b \circ c) \) and \( (a * b) \circ (a * c) \).
3. If they are equal for all choices, then \( * \) distributes over \( \circ \).
4. If a counterexample exists, the law does not hold.
1. Choose arbitrary elements \( a,b,c \) from the set.
2. Compute \( a * (b \circ c) \) and \( (a * b) \circ (a * c) \).
3. If they are equal for all choices, then \( * \) distributes over \( \circ \).
4. If a counterexample exists, the law does not hold.
Example 7: Distributive Law (Success)
Problem: On \( \mathbb{Z} \), let \( * \) be addition and \( \circ \) be multiplication. Does addition distribute over multiplication?
Solution:
Step 1: We need to check \( a + (b \times c) = (a + b) \times (a + c) \).
Step 2: Try \( a=2, b=3, c=4 \). LHS = \( 2 + 12 = 14 \). RHS = \( 5 \times 6 = 30 \). Not equal.
Step 3: So addition does NOT distribute over multiplication. (Multiplication distributes over addition, but the reverse is not true.)
Answer: Addition does not distribute over multiplication on \( \mathbb{Z} \).
Problem: On \( \mathbb{Z} \), let \( * \) be addition and \( \circ \) be multiplication. Does addition distribute over multiplication?
Solution:
Step 1: We need to check \( a + (b \times c) = (a + b) \times (a + c) \).
Step 2: Try \( a=2, b=3, c=4 \). LHS = \( 2 + 12 = 14 \). RHS = \( 5 \times 6 = 30 \). Not equal.
Step 3: So addition does NOT distribute over multiplication. (Multiplication distributes over addition, but the reverse is not true.)
Answer: Addition does not distribute over multiplication on \( \mathbb{Z} \).
Example 8: Distributive Law (Failure)
Problem: On \( \mathbb{R} \), define \( a * b = a + b \) and \( a \circ b = a b \). Does \( * \) distribute over \( \circ \)?
Solution:
Step 1: We need \( a * (b \circ c) = (a * b) \circ (a * c) \). That is \( a + (bc) = (a+b)(a+c) \).
Step 2: Expand RHS: \( a^2 + a c + a b + bc \). For this to equal \( a + bc \) for all \( a,b,c \), we would need \( a^2 + ac + ab = a \) for all, which is false. For example, \( a=1,b=2,c=3 \): LHS = \( 1+6=7 \), RHS = \( 3 \times 4 = 12 \).
Answer: \( * \) does not distribute over \( \circ \).
Problem: On \( \mathbb{R} \), define \( a * b = a + b \) and \( a \circ b = a b \). Does \( * \) distribute over \( \circ \)?
Solution:
Step 1: We need \( a * (b \circ c) = (a * b) \circ (a * c) \). That is \( a + (bc) = (a+b)(a+c) \).
Step 2: Expand RHS: \( a^2 + a c + a b + bc \). For this to equal \( a + bc \) for all \( a,b,c \), we would need \( a^2 + ac + ab = a \) for all, which is false. For example, \( a=1,b=2,c=3 \): LHS = \( 1+6=7 \), RHS = \( 3 \times 4 = 12 \).
Answer: \( * \) does not distribute over \( \circ \).
Watch Out!
The distributive law is not symmetric; an operation may distribute over another but not vice versa (e.g., multiplication distributes over addition, but addition does not distribute over multiplication).
The distributive law is not symmetric; an operation may distribute over another but not vice versa (e.g., multiplication distributes over addition, but addition does not distribute over multiplication).
Practice for Distributive Law
- On \( \mathbb{Z} \), let \( * \) be multiplication and \( \circ \) be addition. Does \( * \) distribute over \( \circ \)?
- On \( \mathbb{R} \), define \( a * b = a + b - 1 \) and \( a \circ b = a + b \). Does \( * \) distribute over \( \circ \)?
- On \( \mathbb{Z} \), define \( a * b = a \) and \( a \circ b = b \). Does \( * \) distribute over \( \circ \)?
- On \( \mathbb{R} \), define \( a * b = a^2 + b \) and \( a \circ b = a - b \). Does \( * \) distribute over \( \circ \)?
- On \( \mathbb{Q} \), define \( a * b = a + b \) and \( a \circ b = a - b \). Does \( * \) distribute over \( \circ \)?
Binary Operation Tables
A binary operation on a finite set can be represented by an operation table (Cayley table). The rows and columns are labelled by the elements of the set, and the entry in row \( a \) and column \( b \) is \( a * b \). These tables make it easy to test closure, commutativity, and to find identity and inverses.
Using an Operation Table:
1. Closure: Check that every entry in the table belongs to the set.
2. Commutativity: The table is symmetric about the main diagonal.
3. Identity: Look for a row that is identical to the heading row and a column that is identical to the heading column.
4. Inverse: For each element, find the column where the product is the identity.
1. Closure: Check that every entry in the table belongs to the set.
2. Commutativity: The table is symmetric about the main diagonal.
3. Identity: Look for a row that is identical to the heading row and a column that is identical to the heading column.
4. Inverse: For each element, find the column where the product is the identity.
Example 9: Constructing and Interpreting an Operation Table
Problem: Let \( S = \{1, 2, 3, 4\} \) and define \( * \) by \( a * b = \) the remainder when \( a \cdot b \) is divided by 5. Construct the operation table and determine if the operation is closed, commutative, and find the identity and inverses.
Solution:
Step 1: Compute each product modulo 5:
\( 1*1=1, 1*2=2, 1*3=3, 1*4=4 \);
\( 2*1=2, 2*2=4, 2*3=1, 2*4=3 \);
\( 3*1=3, 3*2=1, 3*3=4, 3*4=2 \);
\( 4*1=4, 4*2=3, 4*3=2, 4*4=1 \).
Step 2: Table:
\[ \begin{array}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{array} \]
Step 3: All entries are in \( S \), so closed. The table is symmetric, so commutative. The row and column for 1 match the headings, so identity is 1. Inverses: \( 1^{-1}=1, 2^{-1}=3, 3^{-1}=2, 4^{-1}=4 \).
Answer: Closed, commutative, identity 1, inverses as listed.
Problem: Let \( S = \{1, 2, 3, 4\} \) and define \( * \) by \( a * b = \) the remainder when \( a \cdot b \) is divided by 5. Construct the operation table and determine if the operation is closed, commutative, and find the identity and inverses.
Solution:
Step 1: Compute each product modulo 5:
\( 1*1=1, 1*2=2, 1*3=3, 1*4=4 \);
\( 2*1=2, 2*2=4, 2*3=1, 2*4=3 \);
\( 3*1=3, 3*2=1, 3*3=4, 3*4=2 \);
\( 4*1=4, 4*2=3, 4*3=2, 4*4=1 \).
Step 2: Table:
\[ \begin{array}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{array} \]
Step 3: All entries are in \( S \), so closed. The table is symmetric, so commutative. The row and column for 1 match the headings, so identity is 1. Inverses: \( 1^{-1}=1, 2^{-1}=3, 3^{-1}=2, 4^{-1}=4 \).
Answer: Closed, commutative, identity 1, inverses as listed.
Example 10: Non-Commutative Table
Problem: Let \( S = \{a, b\} \) with operation \( * \) defined by the table. Determine if it is commutative and find the identity.
\[ \begin{array}{c|cc} * & a & b \\ \hline a & a & b \\ b & a & b \end{array} \]
Solution:
Step 1: Check symmetry: \( a*b = b \) but \( b*a = a \). So not commutative.
Step 2: Check identity: row a matches headings, column a does not (a*a = a, but b*a = a). So no identity element.
Answer: Not commutative, no identity.
Problem: Let \( S = \{a, b\} \) with operation \( * \) defined by the table. Determine if it is commutative and find the identity.
\[ \begin{array}{c|cc} * & a & b \\ \hline a & a & b \\ b & a & b \end{array} \]
Solution:
Step 1: Check symmetry: \( a*b = b \) but \( b*a = a \). So not commutative.
Step 2: Check identity: row a matches headings, column a does not (a*a = a, but b*a = a). So no identity element.
Answer: Not commutative, no identity.
Watch Out!
An operation table can be used to test properties, but it is easy to misread the row/column labels. Always double-check that the row and column headings are correct.
An operation table can be used to test properties, but it is easy to misread the row/column labels. Always double-check that the row and column headings are correct.
Practice for Binary Operation Tables
- Construct the table for \( * \) on \( \{0,1,2\} \) defined by \( a * b = a + b \pmod{3} \). Is it commutative? Find the identity.
- Construct the table for \( \star \) on \( \{1,2,3\} \) defined by \( a \star b = \) the maximum of \( a \) and \( b \). Is it commutative? Find the identity.
- Construct the table for \( \circ \) on \( \{0,1\} \) defined by \( a \circ b = a \cdot b \). Is it closed? Is it commutative?
- Given the table below, find the identity and the inverse of each element. \[ \begin{array}{c|cccc} \oplus & 1 & 2 & 3 & 4 \\ \hline 1 & 4 & 3 & 2 & 1 \\ 2 & 3 & 2 & 1 & 4 \\ 3 & 2 & 1 & 4 & 3 \\ 4 & 1 & 4 & 3 & 2 \end{array} \]
- Construct a table for \( * \) on \( \{p,q,r\} \) that is commutative but has no identity.
Cumulative Practice Exercises
- On \( \mathbb{Z} \), define \( a * b = a^2 + b \). Is it closed? Is it commutative? Is it associative?
- On \( \mathbb{Q} \), define \( a \star b = a + b - 2 \). Find the identity and the inverse of 3.
- On \( \mathbb{R} \), define \( a \circ b = a + b + 1 \). Is it closed? Is it commutative? Is it associative? Find the identity and inverse of 2.
- On \( \mathbb{Z} \), define \( a \odot b = a + b - ab \). Is it closed? Is it commutative? Is it associative? Find the identity and inverse of 1.
- On \( \mathbb{R} \setminus \{0\} \), define \( a \otimes b = \frac{a}{b} \). Is it closed? Is it commutative? Is it associative? Find the identity.
- On \( \mathbb{Z} \), define \( a * b = a + 2b \). Is it commutative? Is it associative? Find the identity, if it exists.
- On \( \mathbb{Z} \), define \( a \star b = a + b + 3 \). Find the identity and the inverse of 5.
- Given \( S = \{1,2,3,4\} \) with \( * \) defined by \( a * b = \text{LCM}(a,b) \). Is it closed? Is it commutative? Find the identity, if it exists.
- On \( \mathbb{Q} \), define \( a \oplus b = a + b - ab \). Does \( \oplus \) distribute over addition? Does addition distribute over \( \oplus \)?
- Construct the operation table for \( * \) on \( \{a,b,c\} \) defined by \( a * b = c \), \( b * c = a \), \( c * a = b \), and \( x * x = x \) for all \( x \). Is it commutative? Is it associative? Find the identity.
Solutions to Cumulative Exercises
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Step 1: Closure: \( a^2 + b \in \mathbb{Z} \), so closed.
Step 2: Commutativity: \( a*b = a^2+b \), \( b*a = b^2+a \). Not equal generally (e.g., 1*2=3, 2*1=5). Not commutative.
Step 3: Associativity: \( (a*b)*c = (a^2+b)^2+c \), \( a*(b*c) = a^2 + (b^2+c) \). Not equal generally (e.g., 1,2,3: LHS= (1+2)^2+3=12, RHS=1+(4+3)=8). Not associative.
Answer: Closed; not commutative; not associative. -
Step 1: Find \( e \): \( a+e-2 = a \Rightarrow e=2 \). Check \( e*a = 2+a-2 = a \). So \( e=2 \).
Step 2: Inverse of 3: \( 3 \star x = 2 \Rightarrow 3+x-2=2 \Rightarrow x=1 \). Check \( 1 \star 3 = 1+3-2=2 \).
Answer: Identity is 2; inverse of 3 is 1. -
Step 1: Closure: \( a+b+1 \in \mathbb{R} \), closed.
Step 2: Commutativity: \( a+b+1 = b+a+1 \), so commutative.
Step 3: Associativity: \( (a*b)*c = (a+b+1)+c+1 = a+b+c+2 \), \( a*(b*c) = a+(b+c+1)+1 = a+b+c+2 \). Associative.
Step 4: Identity: \( a+e+1 = a \Rightarrow e = -1 \). Check \( e*a = -1+a+1 = a \). So \( e=-1 \).
Step 5: Inverse of 2: \( 2*x = -1 \Rightarrow 2+x+1 = -1 \Rightarrow x = -4 \).
Answer: Closed, commutative, associative; identity -1; inverse of 2 is -4. -
Step 1: Closure: \( a+b-ab \in \mathbb{Z} \), closed.
Step 2: Commutativity: \( a+b-ab = b+a-ba \), commutative.
Step 3: Associativity: \( (a*b)*c = (a+b-ab)+c - (a+b-ab)c = a+b+c-ab-ac-bc+abc \). \( a*(b*c) = a+(b+c-bc) - a(b+c-bc) = a+b+c-bc-ab-ac+abc \). Equal. Associative.
Step 4: Identity: \( a+e-ae = a \Rightarrow e(1-a)=0 \Rightarrow e=0 \). Check \( e*a = 0+a-0 = a \). So \( e=0 \).
Step 5: Inverse of 1: \( 1*x = 0 \Rightarrow 1+x-x=0 \Rightarrow 1=0 \), impossible. So 1 has no inverse.
Answer: Closed, commutative, associative; identity 0; 1 has no inverse. -
Step 1: Closure: for \( a,b \neq 0 \), \( a/b \neq 0 \), closed.
Step 2: Commutativity: \( a/b \neq b/a \) generally (e.g., 2/1=2, 1/2=0.5). Not commutative.
Step 3: Associativity: \( (a/b)/c = a/(bc) \), \( a/(b/c) = ac/b \). Not equal generally (e.g., 2,4,2: LHS=0.25, RHS=1). Not associative.
Step 4: Identity: \( a/e = a \Rightarrow e=1 \). Check \( e/a = 1/a \), which is not always a. So no identity.
Answer: Closed; not commutative; not associative; no identity. -
Step 1: Commutativity: \( a+2b \) vs \( b+2a \). Not equal generally (e.g., 1,2: 1*2=5, 2*1=4). Not commutative.
Step 2: Associativity: \( (a* b)*c = (a+2b)+2c = a+2b+2c \), \( a*(b*c) = a+2(b+2c) = a+2b+4c \). Not equal generally. Not associative.
Step 3: Identity: \( a+e = a \Rightarrow e=0 \). Check \( e*a = 0+2a \neq a \). No identity.
Answer: Not commutative; not associative; no identity. -
Step 1: Identity: \( a+e+3 = a \Rightarrow e = -3 \). Check \( e*a = -3+a+3 = a \). So \( e=-3 \).
Step 2: Inverse of 5: \( 5 \star x = -3 \Rightarrow 5+x+3 = -3 \Rightarrow x = -11 \).
Answer: Identity is -3; inverse of 5 is -11. -
Step 1: Closure: LCM of two numbers from S may not be in S (e.g., LCM(2,3)=6). Not closed.
Step 2: Commutativity: LCM is commutative.
Step 3: Identity: \( a * e = a \) means LCM(a,e)=a, so e must be a divisor of all a. Only 1 works. So identity is 1.
Answer: Not closed; commutative; identity is 1. -
Step 1: Check \( \oplus \) over addition: \( a \oplus (b+c) = a+(b+c)-a(b+c) = a+b+c-ab-ac \). \( (a \oplus b) + (a \oplus c) = (a+b-ab)+(a+c-ac) = 2a+b+c-ab-ac \). Not equal generally, so \( \oplus \) does not distribute over addition.
Step 2: Check addition over \( \oplus \): \( a + (b \oplus c) = a + b + c - bc \). \( (a+b) \oplus (a+c) = (a+b)+(a+c) - (a+b)(a+c) = 2a+b+c - (a^2+ab+ac+bc) \). Not equal generally.
Answer: Neither distributes over the other. -
Step 1: Table:
\[ \begin{array}{c|ccc} * & a & b & c \\ \hline a & a & c & b \\ b & c & b & a \\ c & b & a & c \end{array} \]
Step 2: Symmetric, so commutative.
Step 3: Check associativity: \( (a*b)*c = c*c = c \), \( a*(b*c) = a*a = a \). Not equal. Not associative.
Step 4: Identity: row a matches headings, column a matches headings, so \( a \) is identity.
Answer: Commutative; not associative; identity is \( a \).
Conclusion & Summary
Binary operations extend the familiar arithmetic operations to abstract sets. By testing for closure, commutativity, associativity, identity, and inverses, we can classify operations and understand their structure. These concepts are foundational for group theory and abstract algebra.
Key Takeaways:
1. Closure: \( a * b \in S \) for all \( a,b \in S \).
2. Commutativity: \( a * b = b * a \).
3. Associativity: \( (a * b) * c = a * (b * c) \).
4. Identity: \( a * e = e * a = a \).
5. Inverse: \( a * a^{-1} = a^{-1} * a = e \).
6. Distributive: \( a * (b \circ c) = (a * b) \circ (a * c) \).
Keep practicing: abstract algebra is like learning a new language—every new operation is a chance to think differently!
Video Resource
Watch this video for more examples of binary operations and their properties.
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