Let ‘&’ be a binary operation defined on the set N. Which of the following definitions is commutative but not associative?

The correct option is a & b = ab – 8.

To understand why this is correct, we need to test the binary operation for two properties: commutativity and associativity on the set of natural numbers $N$.

1. Checking for Commutativity:
An operation $\&$ is commutative if for all $a,b\in N$:
$a\&b=b\&a$

Let us compute both sides using the definition $a\&b=ab–8$:
Left-hand side (LHS):
$a\&b=ab–8$
Right-hand side (RHS):
$b\&a=ba–8$

Since multiplication of natural numbers is commutative ($ab=ba$), it follows that:
$ab–8=ba–8$
Therefore, $a\&b=b\&a$. The operation is commutative.

2. Checking for Associativity:
An operation $\&$ is associative if for all $a,b,c\in N$:
$(a\&b)\&c=a\&(b\&c)$

Let us compute the LHS:
$(a\&b)\&c=(ab–8)\&c$
Applying the operation definition again where the first element is now $(ab–8)$:
$(a\&b)\&c=(ab–8)c–8=abc–8c–8$

Now, let us compute the RHS:
$a\&(b\&c)=a\&(bc–8)$
Applying the operation definition where the second element is $(bc–8)$:
$a\&(b\&c)=a(bc–8)–8=abc–8a–8$

Comparing the LHS and the RHS, we see that:
$abc–8c–8\ne abc–8a–8$ (in general, unless $a=c$).
For example, let $a=1$, $b=2$, and $c=3$:
LHS: $(1\&2)\&3=(1·2–8)\&3=(–6)\&3=(–6)·3–8=–18–8=–26$
RHS: $1\&(2\&3)=1\&(2·3–8)=1\&(–2)=1·(–2)–8=–2–8=–10$
Since $–26\ne –10$, the operation is not associative.

Thus, the definition $a\&b=ab–8$ satisfies the conditions of being commutative but not associative.