Question

If \[ R=\{(x,y)\mid x,y\in \mathbb{R},\ x^2+y^2=1\} \] is a relation in \(\mathbb{R}\), then \(R\) is: 

• A. Reflexive
• B. Symmetric
• C. Transitive
• D. Equivalence Relation

Hint:

For relation-property questions:
• Reflexive \(\Rightarrow\) check whether \((x,x)\) belongs to the relation for all elements.
• Symmetric \(\Rightarrow\) interchange \(x\) and \(y\) and verify whether the condition remains unchanged.
• Transitive \(\Rightarrow\) usually test with a counterexample.
• Equivalence relation requires all three properties simultaneously. Relations involving expressions like \(x^2+y^2\) are often symmetric because swapping variables does not change the equation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To classify a relation, we must test it against the three fundamental properties:
- Reflexivity: A relation R on set A is reflexive if \((x, x) \in R\) for every \(x \in A\).
- Symmetry: A relation R is symmetric if \((x, y) \in R \implies (y, x) \in R\).
- Transitivity: A relation R is transitive if \((x, y) \in R\) and \((y, z) \in R \implies (x, z) \in R\).
The relation given here is the unit circle equation \(x^2 + y^2 = 1\) on the set of real numbers \(\mathbb{R}\).
Step 2: Key Formula or Approach:
We evaluate the properties using logic and counterexamples:
1. For reflexivity, check if \(x^2 + x^2 = 1\) for all \(x \in \mathbb{R}\).
2. For symmetry, check if swapping \(x\) and \(y\) in \(x^2 + y^2 = 1\) changes the validity of the equation.
3. For transitivity, assume \(x^2 + y^2 = 1\) and \(y^2 + z^2 = 1\), then check if \(x^2 + z^2 = 1\) is always true.
Step 3: Detailed Explanation:
Let's analyze each property in depth:
Reflexivity:
For R to be reflexive, \((x, x)\) must satisfy the condition for all \(x \in \mathbb{R}\).
Substitute \(y = x\): \(x^2 + x^2 = 1 \implies 2x^2 = 1 \implies x = \pm 1/\sqrt{2}\).
This condition is only met for two specific values of \(x\).
For other values, such as \(x = 0\), \(0^2 + 0^2 = 0 \neq 1\).
Thus, \((0, 0) \notin R\).
Since it does not hold for all \(x \in \mathbb{R}\), R is not reflexive.
Symmetry:
Suppose \((x, y) \in R\). This means \(x^2 + y^2 = 1\).
We know that addition is commutative in real numbers, so \(x^2 + y^2 = y^2 + x^2\).
Therefore, if \(x^2 + y^2 = 1\), it is necessarily true that \(y^2 + x^2 = 1\).
This implies \((y, x) \in R\).
Since the implication holds for all pairs, the relation is symmetric.
Transitivity:
Let \((x, y) \in R\) and \((y, z) \in R\).
This means \(x^2 + y^2 = 1\) and \(y^2 + z^2 = 1\).
Let's use specific values as a counterexample:
Take \(x = 1, y = 0\). Then \(1^2 + 0^2 = 1\), so \((1, 0) \in R\).
Take \(y = 0, z = 1\). Then \(0^2 + 1^2 = 1\), so \((0, 1) \in R\).
Now check for \((x, z)\), which is \((1, 1)\):
\(1^2 + 1^2 = 1 + 1 = 2 \neq 1\).
So, \((1, 1) \notin R\).
The transitive property fails. Thus, the relation is not transitive.
Step 4: Final Answer:
The relation is only symmetric. It is neither reflexive nor transitive. Therefore, it is also not an equivalence relation.
The correct answer is (B).