Examine whether the operation $*$ defined on $\mathbb{R}$, the set of all real numbers, by $a * b = \sqrt{a^2 + b^2}$ is a binary operation or not, and if it is a binary operation, find whether it is associative or not.

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> [!TLDR]
> The operation $*$ defined by $a * b = \sqrt{a^2 + b^2}$ is a binary operation on $\mathbb{R}$ because the result of operating on any two real numbers is always a real number. It is also associative, as $(a * b) * c = a * (b * c) = \sqrt{a^2 + b^2 + c^2}$ for all $a, b, c \in \mathbb{R}$.

When we examine an operation, we first need to understand what it means for it to be a "binary operation" on a given set. Essentially, a binary operation is a rule that combines two elements from a set to produce a third element that *must also be in the same set*. This property is called closure. If the operation is indeed binary, we then investigate further properties, such as associativity. Associativity tells us whether the order in which we group elements for multiple operations affects the final result.

Let's break down the problem into these two parts.

### Part 1: Is $*$ a binary operation on $\mathbb{R}$?

1.  **Understanding Binary Operations:**
    A function $f: S 	imes S 	o S$ is called a binary operation on the set $S$. In simpler terms, for an operation $*$ to be binary on a set $S$, it must satisfy the closure property: for any two elements $a, b \in S$, the result $a * b$ must also be an element of $S$. If the result falls outside the set $S$, then it is not a binary operation on $S$.

2.  **Applying to the given operation:**
    The given operation is $a * b = \sqrt{a^2 + b^2}$ on the set $\mathbb{R}$ (all real numbers). We need to check if for any $a, b \in \mathbb{R}$, the value of $\sqrt{a^2 + b^2}$ is also a real number.
    *   If $a \in \mathbb{R}$, then $a^2$ is a non-negative real number ($a^2 \ge 0$).
    *   If $b \in \mathbb{R}$, then $b^2$ is a non-negative real number ($b^2 \ge 0$).
    *   Therefore, their sum $a^2 + b^2$ is also a non-negative real number ($a^2 + b^2 \ge 0$).
    *   The square root of any non-negative real number is always a real number. For example, $\sqrt{4} = 2 \in \mathbb{R}$, $\sqrt{0} = 0 \in \mathbb{R}$, $\sqr

Operation	Associative?	Quick check
Addition $+$ on $\mathbb{R}$	Yes	$(2+3)+4 = 2+(3+4)$
Multiplication $\times$ on $\mathbb{R}$	Yes	$(2 \times 3) \times 4 = 2 \times (3 \times 4)$
Subtraction $-$ on $\mathbb{R}$	No	$(5-3)-2 \neq 5-(3-2)$
Division $\div$ on $\mathbb{R}^*$	No	$(8 \div 4) \div 2 \neq 8 \div (4 \div 2)$
Matrix multiplication	Yes	$(AB)C = A(BC)$ (if sizes match)
Cross product in $\mathbb{R}^3$	No	$(\hat{i} \times \hat{i}) \times \hat{j} \neq \hat{i} \times (\hat{i} \times \hat{j})$

Q.Examine whether the operation $$ defined on $\mathbb{R}$ , the set of all real numbers, by $a b = \sqrt{a^2 + b^2}$ is a binary operation or not, and if it is a binary operation, find whether it is associative or not.

Concept understanding — Binary Operation Associativity

The idea in plain words

The precise definition

What this really means

Common examples at a glance

Step by step: testing associativity …

Q.Examine whether the operation ∗*∗ defined on R\mathbb{R}R, the set of all real numbers, by a∗b=a2+b2a * b = \sqrt{a^2 + b^2}a∗b=a2+b2​ is a binary operation or not, and if it is a binary operation, find whether it is associative or not.