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.

The key idea here involves understanding the definition of a binary operation and its property of associativity.

For an operation $*$ to be a binary operation on $\mathbb{R}$, for any $a, b \in \mathbb{R}$, the result $a * b$ must also be in $\mathbb{R}$. Given $a * b = \sqrt{a^2 + b^2}$:

1. For any real numbers $a$ and $b$, $a^2 \ge 0$ and $b^2 \ge 0$.
2. Therefore, $a^2 + b^2 \ge 0$.
3. The square root of any non-negative real number is a unique non-negative real number, which is always an element of $\mathbb{R}$.
4. Thus, $*$ is a binary operation on $\mathbb{R}$.

To check for associativity, we must verify if $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{R}$.

1. Calculate $(a * b) * c$: $(a * b) * c = (\sqrt{a^2 + b^2}) * c = \sqrt{(\sqrt{a^2 + b^2})^2 + c^2} = \sqrt{a^2 + b^2 + c^2}$. …

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 $f: S \times S \to S$ 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, $\sqrt{4} = 2 \in \mathbb{R}$, $\sqrt{0} = 0 \in \mathbb{R}$, $\sqrt{2} \in \mathbb{R}$. It will never be an imaginary number like $\sqrt{-1}$.

3. Conclusion for Binary Operation:

   Since for any $a, b \in \mathbb{R}$, the result $a * b = \sqrt{a^2 + b^2}$ is always a real number, the operation $*$ is indeed a binary operation on $\mathbb{R}$.

Part 2: Is $*$ associative?

1. Understanding Associativity:

   An operation $*$ on a set $S$ is associative if for all $a, b, c \in S$, the following condition holds:

   $(a * b) * c = a * (b * c)$

   This means that when performing the operation on three elements, the grouping of the elements does not change the final result.

2. Evaluating the Left Hand Side (LHS): $(a * b) * c$

   First, we calculate $a * b$:

   $a * b = \sqrt{a^2 + b^2}$

   Now, we apply the operation again, using this result with $c$:

   $(a * b) * c = (\sqrt{a^2 + b^2}) * c$ …

Method: Checking closure (binary operation) then associativity

For "is $*$ a binary operation, and is it associative?" work in two clean stages.

Steps

Step 1: Closure test — does the output stay in the set?

$*$ is a binary operation on a set $S$ if $a*b\in S$ for every $a,b\in S$. For $a*b=\sqrt{a^2+b^2}$: since $a^2+b^2\ge 0$, its square root is a real number, so $*$ maps $\mathbb{R}\times\mathbb{R}$ into $\mathbb{R}$.

Step 2: Compute the left grouping $(a*b)*c$

Substitute step by step, using $(\sqrt{a^2+b^2})^2=a^2+b^2$:

Step 3: Compute the right grouping $a*(b*c)$

$a*(b*c)=\sqrt{a^2+(b^2+c^2)}=\sqrt{a^2+b^2+c^2}.$ …

Common Mistakes

Mistake 1: Mishandling the nested square root

Why it's wrong: students often forget that $(\sqrt{a^2+b^2})^2=a^2+b^2$, leaving a messy un-simplified expression. Correct approach: square the inner root cleanly before adding $c^2$.

Mistake 2: "Proving" associativity with a single numeric triple

Why it's wrong: associativity must hold for all $a,b,c$; one example that works is not a proof. Correct approach: derive $(a*b)*c$ and $a*(b*c)$ symbolically and compare. …