Ordered Pairs
Just as we saw how sets can be built via propositional logic, we’ll see how relations and functions are built up using sets, a clear example of how multiple seemingly disparate concepts can be accumulated into a cohesive whole. But before we delve into what relations and functions are exactly, there is one more set operation we should get acquainted with: the set product. Ironically, even though order doesn’t matter when considering elements in a set, constructing a set product does require that we know in what order we are doing things. In this section, we discuss a new type of mathematical object that takes order into account.
Ordered Pairs
When we were discussing sets, we noted that order and repetition did not matter. All that mattered was whether or not an element was included in the set or not. Sometimes, we do care about the order in which elements are listed. In such a case, we list the elements between parentheses ( ) instead of curly braces { }.
An ordered pair is an ordered collection of two elements a and b, and is normally written using parentheses as in
$$\left(a, b\right)$$where element a is the first item in the pair, and element b is the second item.
Notice that the definition of ordered pair does not require that elements a and b have to come from the same universe (set). We can insert anything we want in an ordered pair.
Contrast ordered pairs with sets. Because order is irrelevant in sets, we have that
$$ \{a, b\} = \{b, a\} $$whereas for ordered pairs, we have that (a, b) and (b, a) are different. In order for two different ordered pairs to be the same, we would need the corresponding parts to be the same.
Two ordered pairs (a, b) and (c, d) are called equal and we write
$$\left(a, b\right) = \left(c, d\right)$$if (and only if) a = c and b = d.
The following are all examples of ordered pairs:
\[ \begin{array}{ l l l } (1, -1) & (0, 12) & (12, 2.0013) \\ (5, \text{tokyo}) & (3, 4) & (\text{red}, -13) \\ (-\pi^e, \sqrt{2}) & (0, 12) & (0, -9) \end{array} \]Let’s let (a, b) refer to the upper-middle ordered pair above (0, 12), and let’s refer to the lower-middle ordered pair above (0, 12) as (c, d). Doing this tells us that
\[ \begin{array}{ l l } a = 0 & b = 12 \\ c = 0 & d = 12 \end{array} \]Of course, we can easily see that since
$$ a = c = 0 $$$$ b = d = 12 $$we must then have that (a, b) = (c, d), meaning that we really have that
$$ (0, 12) = (0, 12) $$as expected. This is true by definition, but this is also easily seen because the corresponding elements of both ordered pairs are equal.
Now compare (0, 12) to (0, -9). Let’s refer to these ordered pairs as
$$ (0, 12) = (a, b) $$$$ (0, -9) = (c, d) $$Here, we see that while
$$ a = c = 0 $$we also have that
$$ b \neq d $$because b = 12 and d = -9. Hence, we have that
$$ (0, 12) \neq (0, -9) $$as expected.
Ordering Multiple Elements
Naturally, we aren’t limited to two elements. We can collect as many elements as we want in a list. Equality of two n-tuples is also straightforward.
An n-tuple is an ordered list of n elements and is normally written using parentheses as
$$ (a_1, a_2,\ldots, a_n) $$where $a_1$ is the first element in the n-tuple, $a_2$ is the second element, and so on as expected.
Two n-tuples $(a_1, a_2, \ldots, a_n)$ and $(b_1, b_2, \ldots, b_m)$ are called equal when they have the same number of elements (n = m), and their corresponding elements are equal to each other; that is, we have that
$$ (a_1, a_2, \ldots, a_n) = (b_1, b_2, \ldots, b_m) $$if (and only if)
$$ n = m $$and
\[ \begin{array}{ c } a_1 = b_1 \\ a_2 = b_2 \\ \vdots \\ a_n = b_m \end{array} \]It is common to refer to n-tuples with only two elements simply as tuples (in addition to ordered pairs), and n-tuples with three elements as triples (or ordered triples).
None of the following n-tuples are equal to each other because the order in which the elements appear is different between each ordered pair:
\[ \begin{array}{ l l l} (1, 2, 3) & (2, 1, 3) & (3, 1, 2) \\ (1, 3, 2) & (2, 3, 1) & (3, 2, 1) \end{array} \]The following n-tuples are not equal to each other because they have a different number of elements:
\[ \begin{array}{ l l } (1, 2, 3) & (1, 1, 2, 3) \end{array} \]