Determine which of the following binary operations on the set N are associative and which are commutative.`(a) a*b = 1AA a , b in N ` `(b) a*b = ((a+b)/2 )AA a , b in N`

Explanation

To solve the question, we will analyze the two binary operations defined on the set of natural numbers \( N \) and determine their properties of associativity and commutativity.

### Part (a): Operation \( a * b = 1 \)

1. **Check for Commutativity**:
- A binary operation is commutative if \( a * b = b * a \) for all \( a, b \in N \).
- Here, \( a * b = 1 \) and \( b * a = 1 \).
- Since both expressions equal 1, we conclude that \( a * b = b * a \).
- **Conclusion**: The operation is **commutative**.

2. **Check for Associativity**:
- A binary operation is associative if \( (a * b) * c = a * (b * c) \) for all \( a, b, c \in N \).
- Calculate \( (a * b) * c \):
- \( a * b = 1 \) (from the operation definition).
- Thus, \( (a * b) * c = 1 * c = 1 \).
- Now calculate \( a * (b * c) \):
- \( b * c = 1 \) (from the operation definition).
- Thus, \( a * (b * c) = a * 1 = 1 \).
- Since both sides equal 1, we have \( (a * b) * c = a * (b * c) \).
- **Conclusion**: The operation is **associative**.

### Part (b): Operation \( a * b = \frac{a + b}{2} \)

1. **Check for Commutativity**:
- A binary operation is commutative if \( a * b = b * a \) for all \( a, b \in N \).
- Calculate \( a * b \):
- \( a * b = \frac{a + b}{2} \).
- Calculate \( b * a \):
- \( b * a = \frac{b + a}{2} = \frac{a + b}{2} \).
- Since both expressions are equal, we conclude that \( a * b = b * a \).
- **Conclusion**: The operation is **commutative**.

2. **Check for Associativity**:
- A binary operation is associative if \( (a * b) * c = a * (b * c) \) for all \( a, b, c \in N \).
- Calculate \( (a * b) * c \):
- First, find \( a * b = \frac{a + b}{2} \).
- Now, calculate \( (a * b) * c = \frac{\frac{a + b}{2} + c}{2} = \frac{a + b + 2c}{4} \).
- Now calculate \( a * (b * c) \):
- First, find \( b * c = \frac{b + c}{2} \).
- Now, calculate \( a * (b * c) = \frac{a + \frac{b + c}{2}}{2} = \frac{2a + b + c}{4} \).
- Now we compare \( (a * b) * c \) and \( a * (b * c) \):
- \( (a * b) * c = \frac{a + b + 2c}{4} \) and \( a * (b * c) = \frac{2a + b + c}{4} \).
- Since \( \frac{a + b + 2c}{4} \neq \frac{2a + b + c}{4} \) in general, the operation is not associative.
- **Conclusion**: The operation is **not associative**.

### Final Summary:
- For operation (a):
- **Commutative**: Yes
- **Associative**: Yes
- For operation (b):
- **Commutative**: Yes
- **Associative**: No