Question

On the set \(\mathbb{R}\) of real numbers the relation \(\rho\), defined by \(x\rho y\) \((x,y\in\mathbb{R})\) iff:

• A. $|x-y|<2$ is reflexive but neither symmetric nor transitive
• B. $|x|\ge y$ is reflexive and transitive but not symmetric
• C. $x>|y|$ is transitive but neither reflexive nor symmetric
• D. $x-y<2$ is reflexive and symmetric but not transitive

Hint:

To quickly test relation properties on exams, use simple numbers like $1, 0,$ and $-1$ as test cases. Negative numbers are especially useful for breaking symmetry and reflexivity checks when absolute value brackets are involved!

Answer 1

The Correct Option is C

Concept: A binary relation $\rho$ defined over a set is analyzed by checking three independent properties against counterexamples or proofs:
• Reflexive: $x\rho x$ must be true for all elements $x$.
• Symmetric: If $x\rho y$ is true, then $y\rho x$ must also be true.
• Transitive: If $x\rho y$ and $y\rho z$ are both true, then $x\rho z$ must be true. Step 1: Evaluate choice (C) for Reflexivity.
The relation rule is given as $x\rho y \iff x > |y|$. Let us check reflexivity by replacing $y$ with $x$: $$x\rho x \implies x > |x|$$ By definition, the absolute value of any real number is always greater than or equal to the number itself ($|x| \ge x$). Therefore, the inequality $x > |x|$ is never true for any real number. Since it fails for all inputs, the relation is not reflexive.

Step 2: Evaluate choice (C) for Symmetry.
Let us check if the relation can be flipped symmetrically. Suppose $x\rho y$ is true, which gives the inequality: $$x > |y|$$ This does not mean that $y > |x|$ will be true. Let us verify this with a simple counterexample. If we pick $x = 5$ and $y = -1$:
• $5 > |-1| \implies 5 > 1$ (This is true, so $5\rho(-1)$ holds)
• Now check the flipped pair: $-1 > |5| \implies -1 > 5$ (This is completely false) Since $5\rho(-1)$ is true but $(-1)\rho 5$ is false, the relation is not symmetric.

Step 3: Evaluate choice (C) for Transitivity.
Let us assume that both $x\rho y$ and $y\rho z$ are true. This gives us two simultaneous inequalities: $$x > |y| \quad \text{and} \quad y > |z| \quad \cdots (1)$$ Since absolute values are always non-negative, we know that $|y| \ge y$. Let us chain our inequalities together using equation (1): $$x > |y| \ge y > |z| \quad \Rightarrow \quad x > |z|$$ This satisfies the definition for the relation pair $x\rho z$. Since this chain holds true for all real variables, the relation is definitively transitive. This means statement (C) is completely accurate.