Relations and Functions

The chapter climbs a short, tight ladder of ideas. It opens with the ordered pair and the rule for when two ordered pairs are equal, then builds the Cartesian product $A \times B$ of two sets, counts its elements, and records how it interacts with union, intersection and the empty set. On top of that foundation sits the relation: any subset of $A \times B$, described either by roster or set-builder form, together with the vocabulary of domain, codomain and range and the count of all possible relations between two finite sets. The chapter then isolates the function as the relation obeying the strict "one image per input" rule, and works through a gallery of standard real functions — identity, constant, polynomial, rational, modulus, signum and greatest integer — each defined algebraically and sketched graphically. It closes with the algebra of real functions, the rules for adding, subtracting, scaling, multiplying and dividing two functions defined on a common domain.

Ordered pairs and the Cartesian product

An ordered pair is a pair of elements written in a fixed order and grouped in brackets, $(p,q)$, with $p$ coming from one set and $q$ from another. Order matters absolutely: the licence code $(\text{DL}, 01)$ is not the same object as $(01, \text{DL})$. This gives the defining rule for equality of ordered pairs, and from it the formal definition of the Cartesian product.

where: in equation $(1)$, $a,c$ are the first entries and $b,d$ the second entries of two ordered pairs, and the double arrow means the two conditions on the right are both necessary and sufficient for the pair equality on the left. In equation $(2)$, $A$ and $B$ are any two non-empty sets, and $A \times B$ (read "$A$ cross $B$") is the set of every ordered pair with first coordinate in $A$ and second coordinate in $B$. For $A = \{\text{red}, \text{blue}\}$ and $B = \{b, c, s\}$, listing all such pairs gives $A \times B = \{(\text{red},b), (\text{red},c), (\text{red},s), (\text{blue},b), (\text{blue},c), (\text{blue},s)\}$ — six pairs in all, one for every way of picking a colour and then an object. If either $A$ or $B$ is empty, there is nothing to pair, so $A \times \varphi = \varphi$ for every set $A$. The product does not commute: swapping the roles of $A$ and $B$ generally gives a genuinely different set of pairs, since $(a,b)$ and $(b,a)$ are equal only in the special case $a = b$. The idea extends naturally to three sets: $A \times A \times A = \{(a,b,c) : a,b,c \in A\}$, whose members are called ordered triplets, and $\mathbb{R} \times \mathbb{R} = \{(x,y): x,y \in \mathbb{R}\}$ is precisely the set of coordinates of every point in the Cartesian plane — the reason this construction is named after Descartes.

Relations: domain, codomain and range

A relation turns a raw Cartesian product into something meaningful by selecting only the pairs that satisfy some rule.

Definition. A relation $R$ from a non-empty set $A$ to a non-empty set $B$ is a subset of the Cartesian product $A \times B$. The subset is picked out by describing a relationship between the first element and the second element of each ordered pair. If $(x,y) \in R$, we say $y$ is the image of $x$ under $R$, and $x$ is a preimage of $y$.

where: the domain collects every first element actually used by some pair in $R$, the range collects every second element actually used, and the codomain is the entire target set $B$, whether or not every element of $B$ gets used. Range is always a subset of codomain, and it can be a proper subset — an element of $B$ that no pair points to simply sits in the codomain unused. For $A = \{1,2,3,4,5,6\}$ and $R = \{(x,y): y = x+1\}$, the roster form is $R = \{(1,2),(2,3),(3,4),(4,5),(5,6)\}$, so the domain is $\{1,2,3,4,5\}$ (element $6$ has no partner and drops out), the range is $\{2,3,4,5,6\}$, and the codomain remains the full set $\{1,2,3,4,5,6\}$ by definition, regardless of what is actually hit. A relation can be written in roster form (listing pairs), in set-builder form (stating the rule), or drawn as an arrow diagram, with arrows running from each element of $A$ to its image or images in $B$. Since a relation is any subset of $A \times B$, and a set with $n$ elements has $2^n$ subsets, the total number of relations from a set of size $p$ to a set of size $q$ is $2^{pq}$ — proved in full in the third derivation below.

Functions: the relations that behave

Among all $2^{pq}$ possible relations from $A$ to $B$, functions are the ones that behave like reliable machines: put in an input, always get exactly one predictable output.

Definition. A relation $f$ from a non-empty set $A$ to a non-empty set $B$ is a function if every element of $A$ has one and only one image in $B$. Equivalently, $f$ is a function if its domain is all of $A$, and no two distinct ordered pairs in $f$ share the same first element. If $(a,b) \in f$, we write $f(a) = b$, calling $b$ the image of $a$ and $a$ the preimage of $b$, and denote the function $f : A \to B$.

where: the symbol $\forall$ reads "for every," $\exists!$ reads "there exists exactly one," so equation $(4)$ is the precise translation of the definition: every single element of $A$, with no exception, must be matched to exactly one element of $B$ — not zero, not two or more. This single sentence is the sharpest possible test of "is this relation a function?" Two failure modes cover every non-function you will ever meet: some element of $A$ has no image at all (domain is not all of $A$), or some element of $A$ has two or more images (the "one and only one" clause is violated). A real valued function is one whose range is $\mathbb{R}$ or a subset of $\mathbb{R}$; if its domain is also $\mathbb{R}$ or a subset of $\mathbb{R}$, it is called a real function — the kind this chapter graphs.

Standard real functions and their graphs

Seven functions recur so often that you must recognise their formulas and their graphs on sight.

where: in the polynomial function $n$ is a non-negative integer and $a_0, a_1, \ldots, a_n \in \mathbb{R}$ are constants; in the rational function $p(x)$ and $q(x)$ are polynomial functions with $q(x)$ never allowed to vanish. The identity function graphs as the straight line $y=x$ through the origin at $45°$; the constant function $f(x)=c$ graphs as a horizontal line at height $c$, with domain $\mathbb{R}$ but range the single value $\{c\}$. The modulus function folds the negative $x$-axis up onto the positive $y$-values, producing a symmetric "V" with vertex at the origin; the signum function collapses every positive input to $1$ and every negative input to $-1$, with a single defined value $0$ exactly at the origin, so its range is the three-point set $\{-1,0,1\}$. The greatest integer function $[x]$ (also called the floor function) stays constant at each integer value $n$ across the half-open interval $n \le x < n+1$, then jumps up by exactly $1$ at the next integer — so $[2.3] = 2$, $[-0.5] = -1$ and $[4] = 4$; its graph is a staircase, rising in unit steps and constant between consecutive integers, with each step including its left endpoint but not its right.

Algebra of real functions

Two real functions defined on the same domain $X \subseteq \mathbb{R}$ can be combined pointwise — evaluate each function at $x$ separately, then combine the two numbers with ordinary arithmetic.

where: $f, g : X \to \mathbb{R}$ are real functions sharing the common domain $X$, $\alpha$ is any real scalar, and every combined function is again defined for $x \in X$, except the quotient, which additionally needs $g(x) \neq 0$ — so the domain of $f/g$ is $X$ with the zeros of $g$ removed. Each rule is simply "do the arithmetic operation to the two output numbers, at every input separately" — nothing here is new arithmetic, only new bookkeeping about where the resulting function is defined. For $f(x) = x^2$ and $g(x) = 2x+1$: $(f+g)(x) = x^2+2x+1$, $(f-g)(x) = x^2-2x-1$, $(fg)(x) = x(2x+1) = 2x^2+x$, and $(f/g)(x) = \dfrac{x^2}{2x+1}$ for $x \neq -\tfrac12$ — the one point where the quotient's domain shrinks relative to $f$ and $g$ individually.

Board level - equal ordered pairs and the size of a Cartesian product (Example 1 pattern)

If $(x+1, y-2) = (3,1)$, find the values of $x$ and $y$. Also, if a set $A$ has $3$ elements and $B = \{3,4,5\}$, find the number of elements in $A \times B$.

Step 1. Since the two ordered pairs are equal, their corresponding entries must be equal individually, by equation $(1)$: $x+1 = 3$ and $y-2 = 1$.

Step 2. Solve each small equation on its own: from $x+1=3$, $x = 2$; from $y-2=1$, $y=3$.

Step 3. For the second part, $n(A) = 3$ and $n(B) = 3$ (since $B=\{3,4,5\}$ has three listed elements). By equation $(2)$ and the counting derivation above, $n(A \times B) = n(A) \times n(B) = 3 \times 3 = 9$.

Step 4. No further simplification is needed; both parts are independent applications of the two basic rules of this section. Answer: x = 2, y = 3, and n(A × B) = 9.

Board level - testing relations for the function property (Example 11 pattern)

Examine each relation below and state, with reasons, whether it is a function. For each one that is, give its domain and range. (i) $R_1 = \{(2,1),(3,1),(4,2)\}$ (ii) $R_2 = \{(2,2),(2,4),(3,3),(4,4)\}$ (iii) $R_3 = \{(1,2),(2,3),(3,4),(4,5),(5,6),(6,7)\}$.

Step 1. Apply the function test from equation $(4)$ to $R_1$: the first elements used are $2, 3, 4$, all distinct, and each is paired with exactly one image ($1, 1, 2$ respectively — note that two different inputs, $2$ and $3$, are allowed to share the same output $1$; the rule only forbids one input having two outputs). So $R_1$ is a function, with domain $\{2,3,4\}$ and range $\{1,2\}$.

Step 2. Apply the same test to $R_2$: the first element $2$ appears twice, once paired with $2$ and once paired with $4$. A single input cannot have two different images and still satisfy "one and only one image," so $R_2$ fails the function test — it is a relation, but not a function.

Step 3. Apply the test to $R_3$: the first elements $1,2,3,4,5,6$ are all distinct, and each is paired with exactly one image. So $R_3$ is a function, with domain $\{1,2,3,4,5,6\}$ and range $\{2,3,4,5,6,7\}$.

Step 4. Collect the verdicts: two of the three relations qualify as functions, and one does not, purely because of a repeated first element with conflicting images. Answer: R1 is a function (domain {2,3,4}, range {1,2}); R2 is not a function (2 has two images); R3 is a function (domain {1,2,3,4,5,6}, range {2,3,4,5,6,7}).

Competitive level (JEE / NEET) - domain of a function combining a square root and a factorable denominator

Find the domain of the real function $f(x) = \dfrac{\sqrt{x-1}}{x^2 - 5x + 6}$.

Step 1. Two separate restrictions apply simultaneously, and the domain is the set of $x$ satisfying both at once. The square root in the numerator forces its argument to be non-negative: $x - 1 \ge 0$, i.e. $x \ge 1$.

Step 2. The denominator must never be zero. Factorise it exactly as in Example 21 of the chapter: $x^2 - 5x + 6 = (x-2)(x-3)$, which vanishes when $x=2$ or $x=3$. So $x \neq 2$ and $x \neq 3$ are required, on top of the condition from Step 1.

Step 3. Combine both restrictions on the real line: start with $x \ge 1$ (a closed ray from $1$ onward), then delete the two isolated points $x=2$ and $x=3$ that would make the denominator vanish. This is the classic JEE domain pattern — an inequality restriction from an even root, intersected with finitely many excluded points from a factorable denominator.

Step 4. Write the resulting domain in interval notation as a union of intervals: $[1,2) \cup (2,3) \cup (3,\infty)$. Answer: domain of f = [1, 2) ∪ (2, 3) ∪ (3, ∞). Timing note: budget about ninety seconds. The efficient order is always "handle the root's inequality first, then factor the denominator for excluded points, then intersect" — doing the factoring before the inequality wastes no time either, but skipping the inequality altogether (a very common slip when a numerator also has algebraic structure) silently admits values like $x=0$ that make the square root undefined. Always write the final domain as an explicit interval, since JEE and NEET options are frequently designed to catch a student who reports "all reals except 2 and 3" and forgets the $x \ge 1$ restriction entirely.

The single most frequent error in this chapter is treating "relation" and "function" as interchangeable — every function is a relation, but most relations are not functions, and examiners exploit this by presenting a relation with a repeated or missing first element and asking whether it qualifies. A related slip is forgetting that codomain and range differ: the codomain is the whole target set $B$ fixed by the problem, while the range is only the part actually hit, and range $\subsetneq$ codomain is normal, not an error. Students also mishandle $A \times B$ versus $B \times A$, assuming commutativity that fails except in trivial cases. In the algebra of real functions, the common oversight is forgetting that $f/g$ needs its own smaller domain — every point where $g(x)=0$ must be excluded even if $f$ is defined there. With the greatest integer function, students round $[x]$ to the nearest integer instead of always rounding down: $[-1.2] = -2$, not $-1$.

• State the precise rule for when two ordered pairs are equal, and use it to solve for unknowns inside ordered pairs.
• Define the Cartesian product $A \times B$, compute $n(A \times B) = pq$, and explain why $A \times B \neq B \times A$ in general.
• Write a relation in roster form, set-builder form, and as an arrow diagram, and correctly identify its domain, codomain and range.
• Count the total number of relations between two finite sets as $2^{pq}$, and explain why this counts subsets of $A \times B$.
• State the precise definition of a function, and correctly classify any given relation as a function or not, with a stated reason.
• Recognise and sketch the graphs of the identity, constant, polynomial, rational, modulus, signum and greatest integer functions.
• Evaluate $[x]$ correctly for negative non-integer inputs, without confusing "greatest integer $\le x$" with simple truncation.
• Combine two real functions using the algebra of functions — sum, difference, scalar multiple, product, quotient — including the extra domain restriction for a quotient.
• Find the domain of an algebraic real function by combining even-root non-negativity conditions with denominator-nonzero conditions.
• Prove simple set-theoretic identities about Cartesian products, such as distributivity over intersection, using the element-chasing double-inclusion method.

Ask the tutor: Want to work through classifying a tricky relation as a function, or finding the domain of a harder algebraic function, step by step together?