(a) A relation R is defined on Z, the set of integers, as R = {(x, y) : |x - y| is divisible by a prime number 'p', x, y in Z}. Check whether R is an equivalence relation or not. OR (b) A function $f : \mathbb{R} - \{3/5\} \to \mathbb{R} - \{3/5\}$ is defined as $f(x) = \frac{3x + 2}{5x - 3}$ . Show that f is one-one and onto.

Q: (a) A relation R is defined on Z, the set of integers, as R = {(x, y) : |x - y| is divisible by a prime number 'p', x, y in Z}. Check whether R is an equivalence relation or not. OR (b) A function $f : \mathbb{R} - \{3/5\} \to \mathbb{R} - \{3/5\}$ is defined as $f(x) = \frac{3x + 2}{5x - 3}$. Show that f is one-one and onto.

**Option (a):** R is an **equivalence relation** on Z. - **Reflexive:** $|x-x| = 0$, and $0$ is divisible by every prime p ($0 = p\times 0$), so $(x,x) \in R$ for all $x \in Z$. - **Symmetric:** If p divides $|x-y|$, then since $|y-x| = |x-y|$, p also divides $|y-x|$, so $(x,y) \in R \implies (y,x) \in R$. - **Transitive:** If $p | (x-y)$ and $p | (y-z)$, then $p | [(x-y)+(y-z)] = (x-z)$, so $p | |x-z|$, giving $(x,z) \in R$. Since R is reflexive, symmetric, and transitive, **R is an equivalence relation**. **Option (b):** f is **both one-one and onto**. - **One-one:** Let $f(x_1) = f(x_2)$. Cross-multiplying $(3x_1+2)(5x_2-3) = (3x_2+2)(5x_1-3)$ and simplifying gives $19x_2 = 19x_1$, so $x_1 = x_2$. Hence f is one-one. - **Onto:** For any y in the codomain, solving $y = \frac{3x+2}{5x-3}$ for x gives $x = \frac{3y+2}{5y-3}$, which is defined (since $y\ne 3/5$) and lies in the domain $\mathbb{R}-\{3/5\}$ (substituting $x=3/5$ leads to a contradiction). Hence every y has a pre-image, so f is onto. Therefore f is a **bijection**.

Q: Why is divisibility of |x−y| by a prime always an equivalence relation?

Because it is equivalent to congruence modulo p ($x\equiv y \pmod p$), and congruence relations are always reflexive, symmetric, and transitive, since divisibility by p is preserved under negation and addition of multiples of p.

Relations and Functions

Equivalence Relations and One-One Onto Functions

Official Answer

Option (a): R is an equivalence relation on Z.

Reflexive: $|x-x| = 0$ , and $0$ is divisible by every prime p ( $0 = p\times 0$ ), so $(x,x) \in R$ for all $x \in Z$ .
Symmetric: If p divides $|x-y|$ , then since $|y-x| = |x-y|$ , p also divides $|y-x|$ , so $(x,y) \in R \implies (y,x) \in R$ .
Transitive: If $p | (x-y)$ and $p | (y-z)$ , then $p | [(x-y)+(y-z)] = (x-z)$ , so $p | |x-z|$ , giving $(x,z) \in R$ .

Since R is reflexive, symmetric, and transitive, R is an equivalence relation.

Option (b): f is both one-one and onto.

One-one: Let $f(x_1) = f(x_2)$ . Cross-multiplying $(3x_1+2)(5x_2-3) = (3x_2+2)(5x_1-3)$ and simplifying gives $19x_2 = 19x_1$ , so $x_1 = x_2$ . Hence f is one-one.
Onto: For any y in the codomain, solving $y = \frac{3x+2}{5x-3}$ for x gives $x = \frac{3y+2}{5y-3}$ , which is defined (since $y\ne 3/5$ ) and lies in the domain $\mathbb{R}-\{3/5\}$ (substituting $x=3/5$ leads to a contradiction). Hence every y has a pre-image, so f is onto.

Therefore f is a bijection.

equivalence relationreflexive symmetric transitivecongruence modulo pone-one functiononto functionbijectionrational function

Marking Scheme

11 mark: correctly proving reflexivity of R (or stating the $f(x_1)=f(x_2)$ set-up for one-one).
21 mark: correctly proving symmetry of R (or correct cross-multiplication and simplification for one-one).
31 mark: correctly proving transitivity of R with valid justification using $p|(x-y)$ and $p|(y-z)$ (or concluding $x_1=x_2$ and stating one-one for b).
41 mark: correct conclusion that R is an equivalence relation (or correctly solving $y=f(x)$ for x in terms of y for onto).
51 mark: (for b) verifying $x\ne 3/5$ and concluding f is onto, hence bijective; (for a) clear final summary statement.

Hint

For (a), check reflexivity, symmetry, and transitivity using $x\equiv y \pmod{p}$ . For (b), prove one-one by cross-multiplication and onto by solving $y=f(x)$ for x in terms of y.

Quick Oral Answer

Part (a): R is congruence modulo the prime p, which is reflexive, symmetric, and transitive, hence an equivalence relation. Part (b): f is one-one by cross-multiplication forcing $x_1=x_2$ , and onto since $x=\frac{3y+2}{5y-3}$ always gives a valid pre-image in the domain.

Analysis & Explanation

This 5-mark question tests two of the most heavily examined proof techniques in the Relations and Functions chapter: verifying the three properties of an equivalence relation, and proving bijectivity of a rational function.

Concept (a): The relation |x−y| divisible by a prime p is exactly congruence modulo p ( $x \equiv y \pmod{p}$ ). Congruence relations are always equivalence relations because divisibility is closed under addition of multiples — a general pattern worth recognising, not just a one-off computation.

Concept (b): For rational functions of the form $\frac{ax+b}{cx+d}$ , one-one is proved by cross-multiplying $f(x_1)=f(x_2)$ and showing it forces $x_1=x_2$ , while onto is proved by algebraically inverting $y=f(x)$ to express x in terms of y and confirming the result stays within the restricted domain.

Exam trap: In part (a), students often try to prove transitivity using only the absolute values $|x-y|$ and $|y-z|$ without justifying that p divides the signed differences $(x-y)$ and $(y-z)$ , which is essential to conclude p divides their sum. In part (b), a common slip is forgetting to verify that the computed pre-image x is not the excluded point $3/5$ , which is required since the domain is R−{3/5}.

Real-world link: Congruence classes modulo a prime (part a) underpin modern cryptographic systems like RSA, while bijective rational functions (part b) are used in Möbius transformations in complex analysis and computer graphics.

Common Mistakes

1In part (a), proving transitivity using only $|x-y|$ and $|y-z|$ without linking to the signed differences $(x-y)$ and $(y-z)$ , leaving the divisibility of $(x-z)$ unjustified.
2In part (b), forgetting to verify that the pre-image $x=\frac{3y+2}{5y-3}$ is not equal to the excluded value $3/5$ .
3Skipping the cross-multiplication expansion step and asserting $x_1=x_2$ without full algebraic justification, losing method marks.

Interesting Facts

Congruence modulo a prime number, the basis of part (a)'s relation, is the mathematical foundation of RSA encryption, which secures much of the internet's data transmission today.

Functions of the form $f(x)=\frac{ax+b}{cx+d}$ , as in part (b), are called Möbius transformations and are fundamental in complex analysis and computer graphics for their conformal (angle-preserving) properties.

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Frequently Asked Questions

Why is divisibility of |x−y| by a prime always an equivalence relation?

Because it is equivalent to congruence modulo p ( $x\equiv y \pmod p$ ), and congruence relations are always reflexive, symmetric, and transitive, since divisibility by p is preserved under negation and addition of multiples of p.

What must be checked in addition to cross-multiplication when proving a rational function is onto?

After solving $y=f(x)$ for x in terms of y, you must confirm the resulting x is not equal to any value excluded from the domain — here, $x\ne 3/5$ — otherwise the function would not truly map onto the stated codomain.