Relation \( ab\geq0 \) on Real Numbers
📺 Video Explanation
📝 Question
Let relation \( S \) on the set of real numbers \( \mathbb{R} \) be defined by:
\[ (a,b)\in S \iff ab\geq0 \]
Then, \( S \) is:
- A. symmetric and transitive only
- B. reflexive and symmetric only
- C. antisymmetric relation
- D. an equivalence relation
✅ Solution
🔹 Reflexive Check
For every real number \(a\):
\[ a\cdot a=a^2\geq0 \]
✔ Reflexive.
🔹 Symmetric Check
If:
\[ ab\geq0 \]
Then:
\[ ba\geq0 \]
✔ Symmetric.
🔹 Transitive Check
Suppose:
\[ ab\geq0 \quad \text{and} \quad bc\geq0 \]
This means:
- \(a\) and \(b\) have same sign or one is zero
- \(b\) and \(c\) have same sign or one is zero
Therefore:
- \(a\) and \(c\) also have same sign or zero
So:
\[ ac\geq0 \]
✔ Transitive.
🔹 Antisymmetric Check
Take:
\[ (2,3)\in S \quad \text{and} \quad (3,2)\in S \]
But:
\[ 2\neq3 \]
❌ Not antisymmetric.
🎯 Final Answer
Since relation is reflexive, symmetric, and transitive:
\[ \boxed{\text{S is an equivalence relation}} \]
✔ Correct option: D
🚀 Exam Shortcut
- Same sign product gives non-negative value
- Square is always non-negative → reflexive
- Sign logic helps test transitivity quickly