Higher Mathematics · Classes 9-10 · Chapter 1

Relation.

Greater and lesser, brother and sister, a student and an age: each one is a pairing. Mathematics collects the pairings into one set and calls it a relation. Domain, range and the inverse all come from there.

1. From a relation to ordered pairs

We often need to look at how the elements of one set, or of two sets, relate to each other. Which number of N is greater, who is whose brother or sister in a family, how old each student in a class is: all of these are relations.

Example 18. Take the relation x < y in A = {0, 1, 2, 3}. Only the pairs whose first entry is smaller than the second stay: S = {(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)}. It is a subset of A × A.

Example 19. In a family, a is the father, b the mother, c the elder son, d the younger son, e the daughter and f the elder son's wife. In F = {a, b, c, d, e, f} the relation “x is the brother of y” is B = {(c, d), (c, e), (d, c), (d, e)}. (c, e) is there but (e, c) is not, because e is a girl, not a brother.

2. Relation, domain and range

Definition 4. If X and Y are nonempty sets, any subset of their Cartesian product X × Y is called a relation from X to Y. So R ⊆ X × Y is a relation from X to Y.

Every relation is a set of ordered pairs. The set of first entries is the domain and the set of second entries is the range. S = {(2, 4), (2, 2), (2, 3)} has domain {2} and range {2, 3, 4}.

Example 24. Let A = {2, 3, 5, 7} and B = {1, 2, 4, 7, 10}. The relation “y is divisible by x” is D = {(2, 2), (2, 4), (2, 10), (5, 10), (7, 7)} ⊂ A × B. No member of B is divisible by 3, so no pair starts with 3. Tap each arrow below and lift its pair into the set.

Example 25. L = {(x, y) : x ∈ R, y ∈ R and x < y}. We have a < b if and only if (a, b) ∈ L. So L describes the lesser-greater relation of the real numbers.

3. The inverse relation

Reverse every ordered pair (x, y) of a relation S to (y, x) and you get the inverse relation S⁻¹. So dom S⁻¹ = range S, and range S⁻¹ = dom S. If S = {(1, 4), (2, 1), (3, 0), (4, 1), (5, 4)}, then S⁻¹ = {(4, 1), (1, 2), (0, 3), (1, 4), (4, 5)}.

4. From a rule to a list

To list a relation given by a rule, put each x of A into the rule to find y, and keep the pair only if y ∈ A. In A = {−2, −1, 0, 1, 2} with y = x²: for x = ±2, y = 4 ∉ A, so those drop out. We get S = {(−1, 1), (0, 0), (1, 1)}, dom S = {−1, 0, 1}, range S = {0, 1}.

See A × A as dots on a grid. Pick an x and the point lands at (x, x²). Inside the grid, the pair is in the relation; outside, it is left out.

Now sort eight pairs yourself: which are in y = x²?

In the same A take y² = x. At x = 0, y = 0. At x = 1, y = ±1. At x = 2, y = ±√2 ∉ A, so it drops out. (For negative x there is no real y with y² = x.) We get S = {(0, 0), (1, 1), (1, −1)}, dom S = {0, 1}, range S = {−1, 0, 1}.

In the same A, x + y = 1 gives S = {(−1, 2), (0, 1), (1, 0), (2, −1)}. And x − y = 1 gives S = {(−1, −2), (0, −1), (1, 0), (2, 1)}.

Swap the small set for Z and the list never ends. The relation “x is the square of y” is {(y², y) : y ∈ Z} = {(0, 0), (1, 1), (1, −1), (4, 2), (4, −2), (9, 3), ···}.

5. Relations from Exercise 1.2

In Exercise 1.2 No.4, S = {(1, 5), (2, 10), (3, 15), (4, 20)} has domain {1, 2, 3, 4} and range {5, 10, 15, 20}. Another, S = {(2, 1), (2, 2), (2, 3)}, has domain {2}, range {1, 2, 3} and S⁻¹ = {(1, 2), (2, 2), (3, 2)}. Now reverse three relations yourself: S on the front of each card, S⁻¹ on the back.

6. Check yourself

Three questions. Recall which place in a pair is which, then take the first.

The second.

In the last, check the rule and the set both.

The whole note on seven cards.

Source: Higher Mathematics, Classes 9-10 (NCTB), Chapter 1 “Set and Function”, the “Relation” part.