# Ordered Pair of Real Numbers

### Sets of Ordered Pairs and Their Graphs

By an **ordered pair** of real numbers we shall mean a pair of numbers with one of them designated as the “first.” Then (4, 5) represents an ordered pair in which the first number is 4. We shall often call the numbers in an ordered pair the **components** of that pair. In (-5, 3) the first component is -5 and the second component is 3. The ordered pair (3, -5) on the other hand is a *different* pair, for in it the first component is 3 and the second component is -5.

Let A = {1, 2, 3} and B = {4, 5}. Let us form the set of all possible ordered pairs whose first component is a member of A and whose second component is a member of B: {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}. We shall call this set the **Cartesian product** of sets A and B and designate it by A×B. Cartesian products can be found for other sets in exactly the same way.

**Example 1** Find the Cartesian products.

(a) {1,2}×{3} (b) {1,2}×{0, 1, 2} (c) {0,1}×{0, 1}

**Solution**

(a) {(1, 3), (2, 3)}

(b) {(1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}

(c) {(0, 0), (0, 1), (1, 0), (1, 1)}

We say that the set. {(1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)} has six members since there are exactly six ordered *pairs* in this set. Notice that this set is the Cartesian product of C = {1, 2} and D = {0, 1, 2} which we found in Example 1(b). It is interesting to observe that sets C and D have two and three members, respectively, and the *product* of two and three is six, the number of members in C×D. As Example 1 illustrates, it is generally true that for finite sets the number of elements in their Cartesian product can be found by multiplying the number of elements in one set by the number of elements in the other. Note, too, that in Example 1(c) we have found the Cartesian product of a set with itself.

Often a Cartesian product will contain many members. If A = {-5, -4, -3, ..., 3, 4, 5}, then A×A will contain exactly 121 ordered pairs (since there are 11 members in A and 11∙11 = 121)! The set, B = {(1, 4), (-2, 3), (3, -2), (0, -1)} which contains four of these ordered pairs is a subset of A×A. We shall often be interested in subsets of a Cartesian product rather than the Cartesian product itself. We note, of course, that each such subset is itself a set whose members are ordered pairs.

Let us next find the **graph** of the set B = {(1, 4), (-2, 3), (3, -2), (0, -1)}. We begin by considering two real number lines, one horizontal, the other vertical, which intersect at right angles at their origins. See Figure 1.1. Note that the positive direction on the horizontal line (called the horizontal axis) is to the right and the positive direction on the vertical line (vertical axis) is upward. We shall call the plane which these two axes determine, the coordinate plane; and we shall call their point of intersection, the origin of this plane.

Each element (ordered pair) of set B will correspond to a single point in the coordinate plane. The element (1, 4) will correspond to the point determined by two intersecting lines. On the horizontal axis, at the point whose coordinate there is 1, we draw a perpendicular to this axis; on the vertical axis, at the point whose coordinate there is 4, we draw another perpendicular line. The required point is the intersection of these two lines as shown in Figure 1.2. We shall say that this *point corresponds* to the ordered pair (1, 4) and that the components of this ordered pair, 1 and 4, are the coordinates of the point. And in Figure 1.2 we see that the point which corresponds to (-2, 3) is not the same as the point which corresponds to (3, -2). These are different ordered pairs and they correspond to different points. Notice that it is always the *first* component of the ordered pair which determines the location of the perpendicular line to the *horizontal* axis ; the *second* component determines the position of the perpendicular to the *vertical* axis. To find the point corresponding to the pair (0, -1) we find that point on the horizontal axis whose coordinate is 0 and observe that there is no need to draw a perpendicular there—the vertical axis already has been drawn. The point which corresponds to (0, -1) is the intersection of the vertical axis and the perpendicular to the vertical axis at -1, and it is shown in Figure 1.2. Finally, the graph of the set B = {(1, 4), (-2, 3), (3, -2), (0, -1)} is the set of the four points which we have found, and it is shown in Figure 1.3.

The same technique can be used to find the graph of any set of ordered pairs of numbers, although this may be a tedious task if such a set contains many elements.

**Example 2 **Find the graph of each of these sets:

(a) {(-1,-1), (0,1)} (b) {(0,1), (-1,-1)}

(c) {(3,3), (-3,3), (-3,-3), (3, -3), (0, 0)}

(d) {(1,0), (1,1), (1,2), (2,1), (2,2), (3,2)}

**Solution**

** **

Since {(-1, -1), (0, 1)} = {(0, 1), (-1, -1)} (note that they have the *same* ordered pairs as members), it is not surprising that their graphs are identical.

As Example 2 illustrates, to every set of ordered pairs of numbers corresponds a set of points in the coordinate plane. And this correspondence is reversible. To every set of points in the coordinate plane there also corresponds a set of ordered pairs.

**Example 3** Find the set of ordered pairs corresponding to each of the following sets of points.

**Solution**

(a) {(3, 2)} (b) {(2,0)}

(c) {(0, 1), (-1, 1), (-1, 0), (-2, 1), (-2, 0), (-2, -1)}

Algebraic expressions such as 2x+3y-1, x^2-3xy+2y^2 and (x-3y)/xy contain two variables. As was true for expressions in one variable, expressions with two variables represent numbers, but unless we know the replacement for *each variable* the value of such expressions cannot be determined. Of course, if the replacements for x and y are known, the value of such expressions can be found by making the indicated replacements and performing the arithmetic operations in the resulting numerical expressions.

**Example 4** Find the value of each of the following expressions in two variables for the replacements indicated.

(a) 2x + 3y -1 when x = 1 and y = -2

(b) x^2-3xy+2y^2 when x = 2 and y= 2

(c) (x-3y)/xy when x= -4 and y = -1**Solution**

(a) 2∙1+3(-2)-1=2-6-1=-5

(b) 2^2-3(2)(2)+2∙2^2=4-12+8=0

(c) (-4-3(-1))/((-4)(-1))=(-4+3)/4=-1/4

An algebraic expression in one variable has associated with it a set of numbers called its domain. The members of the domain are those numbers which may be replacements for the variable in the expression. For example, 3x^2-2x+1, {0, 1} describes the expression 3x^2-2x+1 whose domain is {0, 1}. Associated with such an expression is another set called its range. The range is the set of values of the expression when all replacements of the variable by members of the domain are made. The range of 3x^2-2x+1, {0, 1} is the set {1, 2}. When an expression in *two* variables is considered, its domain is *not* a set of numbers. As Example 4 shows, to find the value of an expression in two variables requires that we have two numbers—a replacement for x and a replacement for y. Let us use an ordered *pair* of numbers to indicate *one* member of the domain of an expression in two variables, with the understanding that the first component of the pair is the replacement for x and the second component, the replacement for y. Then by x-y + 2, {(1, 2), (2, 4), (3, 6)} we shall mean the expression x-y+2 with domain {(1, 2), (2, 4), (3, 6)}. And the range of this expression shall (again) be the set of all values it has over the given domain. In this case when x=1 and y=2, x-y+2=1-2+2= 1; when x = 2 and y = 4, x-y+2=2-4+2=0; when x = 3 and y=6, x-y+2= 3 -6+ 2 = -1; and so the range is {-1, 0, 1}.

**Example 5** For each of the following expressions in two variables find the range for the given domain.

(a) 3x + 4y-2, {(1, 1), (2, 2), (-1, 3)}

(b) x^2+4xy-5y^2, {(-1, -1), (3, 1)}

**Solution**

(a) 3x + 4y-2

(1,1) : 3∙1 + 4∙1-2 = 5

(2,2) : 3∙2 + 4∙2-2 = 12

(-1,3) : 3(-1) + 4∙3-2 = 7

The range is {5, 7, 12}.

(b) x^2+4xy-5y^2

(-1, -1) : (-1)^2+4(-1)(-1)-5(-1)^2=0

(3, 1) : 3^2+4∙3∙1-5∙1^2=16

The range is {0, 16}.

An open sentence in one variable is neither true nor untrue. After the variable in such a sentence is replaced by some number in the domain ; a numerical statement results which is either true or false. That subset of the domain which results in true statements we have called the solution set of the sentence. We shall treat open sentences in two variables in exactly the same way. The domain of a sentence in two variables is a set of ordered pairs of numbers, and that subset of the domain which yields a true statement will again be the solution set.

**Example 6** For each open sentence find the solution set for the given domain.

(a) x-2y = 5, {(1, -1), (2, -2), (5, 0), (3, -1)}

(b) y= x^2- 4, {(1, 3), (-2, 0), (0, 4)}

(c) y <= 2x + 1, {(1, 4), (2, 5), (1, -1), (0, 0)}

**Solution**

(a) x-2y = 5

(1, -1) : 1-2(-1) =5

3=5 False

(2, -2) : 2-2(-2) =5

6=5 False

(5, 0) : 5 -2(0) = 5

5=5 True

(3, -1) : 3 -2(-1) =5

5=5 True

The solution set is {(5, 0), (3, -1)}.

(b) y= x^2- 4

(1,3) : 3=1^2-4

3 =-3 False

(-2,0) : 0 = (-2)^2-4

0=0 True

(0,4) : 4=0^2-4

4= -4 False.

The solution set is {(-2, 0)}.

(c) y <= 2x + 1

(1, 4) : 4<=2∙1 + 1

4<=3 False

(2, 5) : 5<=2∙2 + 1

5<=5 True

(1,-1) : -1<=2∙1 + 1

-1<=3 True

(0, 0) : 0<=2∙0 + 1

0<=1 True

The solution set is {(2, 5), (1, -1), (0, 0)}.

We can find the graph of the solution set of an open sentence in two variables as we have found the graph of solution sets of sentences in one variable.

**Example 7** Find the graph of the solution set for each of the open sentences in Example 6.

**Solution**

(a) {(5, 0), (3, -1)} (b) {(-2,0)} (c) {(2, 5), (1, -1), (0, 0)}

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