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# Let $\ast$ be the binary operation on $N$ defined by $a \ast b=H.C.F.$ of a and b. Is $\ast$ commutative? Is $\ast$ associative? Does there exist identity for this binary operation on $N$?

$\begin{array}{1 1} \text{Not Commutative, Not Associative, Identity Exists} \\ \text{Not Commutative, Not Associative, Identity Does not Exist} \\ \text{Commutative, Associative, Identity Exists} \\ \text{Commutative, Associative, Identity Does not Exist} \end{array}$

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A)
Toolbox:
• An operation $\ast$ on $A$ is commutative if $a\ast b = b \ast a\; \forall \; a, b \in A$
• An operation $\ast$ on $A$ is associative if $a\ast ( b \ast c) = (a \ast b) \ast c\; \forall \; a, b, c \in A$
• An element $e \in N$ is an identify element for operation * if $a*e=e*a$ for all $a \in N$
• The highest common factor (HCF), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.
Given that the binary operation on N given by $a \ast b =$ HCF $(a,b)$:
$\textbf {Step 1: Checking if the operation is commutative}$:
We know $HCF (a,b)$ = $HCF (b,a)$.
$\Rightarrow a*b=b*a$ for all $a,b \in N \rightarrow \ast$ is commutative.
$\textbf {Step 2: Checking if the operation is associative}$:
For an operation $\ast$ to be associative $a\ast ( b \ast c) = (a \ast b) \ast c\; \forall \; a, b, c \in N$
$(a*b)*c =$ HCF $(a,b) \ast c =$ HCF $(a,b,c)$
$a*(b*c)= a*($ HCF $(b,c)) =$ HCF $(a,b,c)$
$\Rightarrow (a*b)*c=a*(b*c) \forall a,b,c \in N \rightarrow \ast$ is associative.
$\textbf {Step 2: Checking if the operation has an identity}$:
We know that the element $e \in N$ is an identify element for operation * if $a*e=e*a$ for all $a \in N$
We can clearly see that this is NOT true for all $a \in N$ through a small example:
Let $a=5$, $5 \ast 1 = 1$ and so is $6 \ast 1 = 1$.
So, this relation is not true for al $a \in N$. Therefore $\ast$ does not have an identity element in $N$.