`a* b =ab+1` `AA a,b in R` then:

Options

A. * is commutative but not associative
B. associative but not commutative
C. both commutative and associative
D. Neither commutative nor associative

Answer

To solve the problem, we need to analyze the binary operation defined as \( a * b = ab + 1 \) for all \( a, b \in \mathbb{R} \). We will check if this operation is commutative and associative.

### Step 1: Check for Commutativity

A binary operation \( * \) is commutative if:
\[ a * b = b * a \]

Let's calculate \( a * b \):
\[ a * b = ab + 1 \]

Now, calculate \( b * a \):
\[ b * a = ba + 1 \]

Since multiplication is commutative (\( ab = ba \)), we have:
\[ b * a = ab + 1 \]

Thus, we can conclude:
\[ a * b = b * a \]

**Conclusion**: The operation is commutative.

### Step 2: Check for Associativity

A binary operation \( * \) is associative if:
\[ (a * b) * c = a * (b * c) \]

First, calculate \( (a * b) * c \):
1. Calculate \( a * b \):
\[ a * b = ab + 1 \]
2. Now calculate \( (a * b) * c \):
\[ (a * b) * c = (ab + 1) * c = (ab + 1)c + 1 = abc + c + 1 \]

Next, calculate \( a * (b * c) \):
1. Calculate \( b * c \):
\[ b * c = bc + 1 \]
2. Now calculate \( a * (b * c) \):
\[ a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a + 1 \]

Now, we compare the two results:
- From \( (a * b) * c \), we have \( abc + c + 1 \).
- From \( a * (b * c) \), we have \( abc + a + 1 \).

Since \( abc + c + 1 \neq abc + a + 1 \) (unless \( a = c \)), we conclude:
\[ (a * b) * c \neq a * (b * c) \]

**Conclusion**: The operation is not associative.

### Final Conclusion

The operation \( * \) is commutative but not associative.

### Summary of Results
- The operation is **commutative**.
- The operation is **not associative**.