Higher Mathematics · Class 9-10 · Chapter 1 · Sets and functions
Ordered pairs, and the Cartesian product.
This section is short, and not one to skip, because the whole of the next one stands on it. Something arrives that is not a set, and the first rule you learned about sets is turned on its head here: the order matters.
1. An ordered pair: position is identity
Put two elements in a definite order inside round brackets and you have an ordered pair: (a, b). Here a is the first element and b the second. Write down the difference from a set straight away, because it is the thing most often muddled.
(a, b) ≠ (b, a) but {a, b} = {b, a}
There is one exception: if a and b are equal, turning the pair round changes nothing. And equality too goes position by position: (a, b) = (c, d) exactly when a = c and b = d. A first element can never be matched against a second.
The shape of the brackets tells you what you are holding, and that habit is needed on every line from here to the end of the chapter. Tap the six below and see which is which.
bracketstap one
Ordered pair, or set?
A set. Curly braces, so the order means nothing: {3, 5} = {5, 3}. Two members.
An ordered pair. Round brackets, so the order matters: (3, 5) ≠ (5, 3). The same notation can also mean the interval between 3 and 5; which it is, the question settles.
A set whose one member is an ordered pair. n = 1. The two kinds of bracket are doing two jobs: the outer makes a set, the inner makes a pair.
A set of ordered pairs. Two members, and both are ordered pairs. In the next section this very thing gets the name relation.
An ordered triple. Three elements in a fixed order. The idea is the same with one more slot: position is identity.
A set, and it equals {3, 5}. Repeats are not counted in a set, so there are two members. In an ordered pair, though, (3, 3) is perfectly legal, and both its slots are filled.
All six seen. One question settled every one of them: is the outer bracket round, or curly?
It is easy to nod at “the order matters” and another thing to use it. Tap any cell in the grid below to put a point down, then turn it round, and see where on the grid the two points end up.
you place ittap the grid, then turn it round
Turn the order round
check yourself
One question
If (x + 1, 5) = (3, y - 2), what are x and y?
Those are the numbers from the right hand side, with neither equation solved. x + 1 = 3 gives x = 2.
Right. First against first: x + 1 = 3, so x = 2. Second against second: 5 = y - 2, so y = 7. Matching goes position by position, which is the rule for ordered pairs.
A first element has been matched against a second. Ordered pairs do not allow that.
The x is right, the y is not. 5 = y - 2 gives y = 7, adding 2 to both sides.
2. The Cartesian product: every pair at once
Pair every element of A with every element of B and what you get is the Cartesian product: A × B = {(x, y) : x ∈ A and y ∈ B}. In the book's example, with A = {1, 2} and B = {a, b, c}, A × B holds six ordered pairs.
Writing them out as a list makes it very easy to drop one or two. So the picture for this is not a Venn diagram but a grid: A down the side, B across the top, and one ordered pair in every cell. A grid has no empty cells, so nothing gets missed.
gridtap a cell
A = {1, 2}, B = {a, b, c}
A \ B
a
b
c
1
(1, a)
(1, b)
(1, c)
2
(2, a)
(2, b)
(2, c)
Row 1, column a, so the pair is (1, a). The first element always comes from the row, which is to say from A.
(1, b). Staying in a row keeps the first element and changes only the second.
(1, c). Notice that (c, 1) is not in this grid at all; it belongs to the grid for B × A.
(2, a). The second row begins, back at a again.
(2, b). Three cells in each row and two rows, so 2 × 3 = 6 ordered pairs in all.
(2, c), the last cell. Six cells, six ordered pairs, and none missing.
The whole first row: (1, a), (1, b), (1, c). The 1 has been paired with every element of B. Each element of A gets a row like that.
The second row, the same job done with 2. The two rows together are all of A × B.
Every cell seen. Counting the cells gives you the formula: n(A × B) is the number of rows times the number of columns.
So the formula needs no separate remembering; the grid is enough. Any element of A can go in the first slot, and for each of those any element of B can go in the second. Put some numbers in below.
calculatorchange the numbers
How many ordered pairs?
check yourself
Another question
If A = {x, y} and B = {1, 2, 3}, what is n(B × A)?
That is 3 + 2, an addition. Counting cells in a grid means multiplying, because each row holds a full row of cells.
Right. n(B × A) = 3 × 2 = 6. Swapping the order leaves the count alone and changes the members themselves: in B × A every pair has the number first and the letter second.
That is 2 to the 3, the power set count. Here it is grid cells that are wanted, not subsets.
That is 3 × 3. The second set has two elements, not three.
3. A × B and B × A are not the same
The two have the same number of members, because the grid has only been turned on its side. But an equal count does not make equal sets, a point that came up earlier under equal against equivalent. A × B and B × A are equivalent, and not equal.
When can one ordered pair be in both? If (p, q) is in A × B then p ∈ A and q ∈ B; to be in B × A as well needs p ∈ B and q ∈ A. So both p and q have to lie in A ∩ B. Drop the seven pairs below into three bins and see.
sort themchip, then bin
A = {1, 2}, B = {2, 3}: which grid is it in?
Pick up a pair, then drop it in a bin. Which set can the first element come from, and which the second?
Right. Next one.
Not there. In A × B the first must come from A and the second from B.
All in place. Four members each, so the two are equivalent; only one is shared, (2, 2), so they are not equal. And notice that (1, 1) is in neither: 1 is not in B, so it cannot sit in the second slot at all. A × B does not mean any two numbers paired up.
check yourself
Last question
With A = {1, 2} and B = {2, 3}, how many members do A × B and B × A share?
There is one. The 2 is in both sets, so (2, 2) sits in both grids.
Right. For a pair (p, q) to be in both, both p and q must lie in A ∩ B. Here A ∩ B = {2}, so there is only one possibility: (2, 2).
Perhaps (2, 3) and (3, 2) have been taken as one. As ordered pairs they are different, and one is in A × B while the other is in B × A.
Each set has four members, but they do not share four. An equal count does not make equal sets.
Let us leave the door to the next section open. A × B is the set of every possible ordered pair, which is to say every cell of the grid. Now choose any of those cells, in any arrangement you like. What you are holding is called a relation, and that is where functions begin.
cut from the notetap to turn over
Four cards
frontWhat is the one difference between a set and an ordered pair?
backOrder. (a, b) ≠ (b, a), while {a, b} = {b, a}.
frontWhen is (a, b) = (c, d)?
backWhen a = c and b = d. Matching goes position by position.
frontWhat is n(A × B), and why?
backn(A) × n(B). Rows times columns in the grid, with one pair per cell.
frontAre A × B and B × A equal?
backUsually not. The counts match, so they are equivalent; the members differ, so they are not equal.