Relations and Functions

An ordered pair is a pair of elements written in a fixed order, denoted (a, b), where a is called the first component and b the second component. Unlike a set, order matters here: (2, 3) and (3, 2) are different ordered pairs, whereas {2, 3} and {3, 2} are the same set.

Equality of ordered pairs: Two ordered pairs (a, b) and (c, d) are equal if and only if a = c and b = d.

Given two non-empty sets A and B, the Cartesian product A × B is the set of all ordered pairs (a, b) such that a ∈ A and b ∈ B. In set-builder form, A × B = {(a, b) : a ∈ A, b ∈ B}. If A = ∅ or B = ∅, then A × B = ∅ by definition.

If A and B are finite sets with n(A) = p and n(B) = q, then A × B has exactly pq ordered pairs, so n(A × B) = n(A) × n(B). In general A × B ≠ B × A unless A = B or one of the sets is empty, because the order of components is different. More generally, for three sets A, B, C, the Cartesian product A × B × C consists of ordered triplets (a, b, c).

Worked example. Let A = {1, 2} and B = {3, 4}. Then:

• A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
• B × A = {(3, 1), (3, 2), (4, 1), (4, 2)}

Clearly A × B ≠ B × A, and n(A × B) = n(A) × n(B) = 2 × 2 = 4, which matches the 4 pairs listed.

A relation R from a non-empty set A to a non-empty set B is defined as any subset of the Cartesian product A × B. If (a, b) ∈ R, we say "a is related to b" and write a R b, where a ∈ A and b ∈ B.

Three sets are associated with a relation R from A to B:

• Domain of R is the set of all first components of the ordered pairs in R.
• Co-domain of R is the whole set B, fixed in advance whether or not every element of B is used.
• Range of R is the set of all second components actually occurring in the ordered pairs of R; the range is always a subset of the co-domain.

If n(A) = p and n(B) = q, the total number of possible relations from A to B equals the number of subsets of A × B, which is 2^pq, since A × B itself has pq elements.

Worked example. Let A = {1, 2, 3, 4} and define R = {(x, y) : y = x + 1, x ∈ A, y ∈ A}. Testing each x ∈ A: x = 1 gives y = 2 (valid); x = 2 gives y = 3 (valid); x = 3 gives y = 4 (valid); x = 4 gives y = 5, which is not in A, so it is rejected. Hence R = {(1, 2), (2, 3), (3, 4)}.

Domain of R = {1, 2, 3}, Co-domain of R = A = {1, 2, 3, 4}, and Range of R = {2, 3, 4}. Notice Range ⊆ Co-domain, and here Range ≠ Co-domain since 1 is never a second component.

A function f from a non-empty set A to a non-empty set B is a special relation in which every element of A is associated with exactly one element of B. Formally, f : A → B is a function if:

1. every a ∈ A has an image in B (nothing in A is left out), and
2. each a ∈ A has only one image in B (no element of A is related to two or more elements of B).

Since a function is a relation, it also has a domain (= A, and here every element of A must be used), co-domain (= B), and range (the set of actual images, a subset of B). If f(a) = b, then b is called the image of a and a is called a pre-image of b.

Every function is a relation, but a relation is a function only if it satisfies both conditions above; in particular, no element of A may repeat as a first component with two different second components.

Worked example. Let A = {1, 2, 3} and B = {4, 5, 6, 7}.

• R₁ = {(1, 4), (2, 5), (3, 6)} is a function, since every element of A appears exactly once as a first component.
• R₂ = {(1, 4), (1, 5), (2, 6), (3, 7)} is not a function, because 1 is related to both 4 and 5, violating uniqueness of image.
• R₃ = {(1, 4), (2, 5)} is not a function from A to B, because 3 ∈ A has no image at all.

For real functions given by a formula, such as f(x) = x², the domain is taken to be the largest subset of R for which the formula gives a well-defined real value, unless a smaller domain is explicitly specified.

A real-valued function of a real variable has both its domain and co-domain as subsets of R. The following standard functions form the basic vocabulary used throughout the chapter.

Identity function: f : R → R defined by f(x) = x for every x ∈ R. Domain = R, Range = R. Its graph is the straight line through the origin inclined at 45° to the axes.

Constant function: f : R → R defined by f(x) = c for every x ∈ R, where c is a fixed real number. Domain = R, Range = {c}, a single value.

Polynomial function: f(x) = a₀ + a₁x + a₂x² + … + a_nxⁿ, where n is a non-negative integer and a₀, a₁, …, a_n are real constants. Every polynomial expression is defined for all real x, so its domain is always R; the range depends on the degree (for example, the range of f(x) = x² is [0, ∞)).

Rational function: f(x) = g(x)/h(x), where g(x) and h(x) are polynomial functions and h(x) ≠ 0. Its domain is R minus the roots of h(x), i.e. all real x for which the denominator does not vanish.

Worked example. Find the domain of f(x) = (x² + 2x + 1)/(x² − 5x + 6). The denominator factors as (x − 2)(x − 3), which is zero when x = 2 or x = 3. So the domain is R − {2, 3}, that is, all real numbers except 2 and 3.

A closely related task is finding the domain of functions involving square roots, such as f(x) = √(x − 3): here the expression under the root must be non-negative, so x − 3 ≥ 0, giving domain [3, ∞).

Modulus (absolute value) function: f : R → R defined by f(x) = |x|, where |x| = x if x ≥ 0, and |x| = −x if x < 0. It measures the distance of x from 0 on the number line, so it is never negative. Domain = R, Range = [0, ∞).

Signum function: f : R → R defined by f(x) = 1 if x > 0; f(x) = 0 if x = 0; f(x) = −1 if x < 0. It only records the sign of x. Domain = R, Range = {−1, 0, 1}, a set of just three values.

Greatest integer function (floor function): f : R → R defined by f(x) = [x], where [x] denotes the greatest integer less than or equal to x. For example, [4] = 4, [4.7] = 4, [−4] = −4, and [−4.7] = −5, since for negative non-integers the greatest integer function rounds further down, not towards zero. Domain = R, Range = Z, the set of all integers.

Worked example. Evaluate [3.8], [−2.3], sgn(−7) and |−7|.

• [3.8]: the greatest integer ≤ 3.8 is 3.
• [−2.3]: the greatest integer ≤ −2.3 is −3, not −2, since −2 > −2.3.
• sgn(−7) = −1, since −7 < 0.
• |−7| = 7, since the modulus discards the sign.

Let f and g be two real-valued functions with domains Dom(f) and Dom(g) respectively, both subsets of R. New functions can be built from f and g using ordinary arithmetic, applied pointwise, i.e. one value of x at a time:

• Sum: (f + g)(x) = f(x) + g(x)
• Difference: (f − g)(x) = f(x) − g(x)
• Product: (fg)(x) = f(x) · g(x)
• Quotient: (f/g)(x) = f(x)/g(x), defined only where g(x) ≠ 0
• Scalar multiple: (kf)(x) = k · f(x), for a real constant k

In every case, the domain of the new function is Dom(f) ∩ Dom(g); for the quotient, we further remove from this intersection every x where g(x) = 0.

Worked example. Let f(x) = x + 1 and g(x) = x² − 1, both with domain R. Find (f + g)(x), (f − g)(x), (fg)(x) and (f/g)(x), along with their domains.

• (f + g)(x) = (x + 1) + (x² − 1) = x² + x, domain R.
• (f − g)(x) = (x + 1) − (x² − 1) = −x² + x + 2, domain R.
• (fg)(x) = (x + 1)(x² − 1) = (x + 1)²(x − 1), domain R.
• (f/g)(x) = (x + 1)/(x² − 1) = 1/(x − 1) for x ≠ −1, x ≠ 1; domain is R − {−1, 1}. Both values must be excluded: g(1) = 0 keeps x = 1 out because it is a genuine pole of 1/(x − 1), while g(−1) = 0 keeps x = −1 out even though the factor (x + 1) cancels algebraically, since the original quotient (f/g)(x) was never defined there.