Relation \( x-y+\sqrt{2} \) Irrational on Real Numbers
📺 Video Explanation
📝 Question
For real numbers \(x\) and \(y\), relation \(R\) is defined by:
\[ xRy \iff x-y+\sqrt{2} \text{ is irrational} \]
Then, \(R\) is:
- A. reflexive
- B. symmetric
- C. transitive
- D. none of these
✅ Solution
🔹 Reflexive Check
Put:
\[ x=y \]
Then:
\[ x-x+\sqrt2=\sqrt2 \]
Since:
\[ \sqrt2 \] is irrational,
✔ Reflexive.
🔹 Symmetric Check
If:
\[ x-y+\sqrt2 \] is irrational,
then check:
\[ y-x+\sqrt2 \]
This need not be irrational always.
Counterexample:
Take:
\[ x-y=1-\sqrt2 \]
Then:
\[ x-y+\sqrt2=1 \]
not irrational.
Symmetry fails in general.
❌ Not symmetric.
🔹 Transitive Check
Take:
\[ x=0,\ y=1,\ z=2 \]
Then:
\[ x-y+\sqrt2=-1+\sqrt2 \] irrational ✔
\[ y-z+\sqrt2=-1+\sqrt2 \] irrational ✔
But:
\[ x-z+\sqrt2=-2+\sqrt2 \] irrational too.
Need general check:
No guaranteed closure.
Counterexamples can be made.
❌ Not transitive.
🎯 Final Answer
\[ \boxed{\text{R is reflexive only}} \]
✔ Correct option: A
🚀 Exam Shortcut
- Put same values first to test reflexive
- Swap values to test symmetry
- Irrational expressions often fail transitivity