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Worked Examples · Example 2

Q.If P={a,b,c}P = \{a, b, c\} and Q={r}Q = \{r\}, form the sets P×QP \times Q and Q×PQ \times P. Are these two products equal?

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✓ Free question

The Cartesian product is ordered: P×QP \times Q pairs each element of PP first with each element of QQ, while Q×PQ \times P reverses the order. Since (a,r)≠(r,a)(a, r) \neq (r, a), the two products are not equal.


The Cartesian product A×BA \times B is the set of all ordered pairs (x,y)(x, y) where x∈Ax \in A and y∈By \in B. The word "ordered" is crucial: the pair (x,y)(x, y) is fundamentally different from (y,x)(y, x) unless x=yx = y. Think of coordinates on a plane — the point (3,5)(3, 5) is not the same as (5,3)(5, 3).

When we form P×QP \times Q, we're asking: "What are all the ways to pick a first component from PP and a second component from QQ?" The reverse product Q×PQ \times P asks the same question with roles swapped. Let's see what happens.


Forming P×QP \times Q

  1. List all ordered pairs (p,q)(p, q) where p∈Pp \in P and q∈Qq \in Q. Since P={a,b,c}P = \{a, b, c\} and Q={r}Q = \{r\}, we pair each element of PP with the single element rr:

P×Q={(a,r),(b,r),(c,r)}.P \times Q = \{(a, r), (b, r), (c, r)\}.

  1. Count the elements. We have ∣P∣=3|P| = 3 and ∣Q∣=1|Q| = 1, so ∣P×Q∣=3×1=3|P \times Q| = 3 \times 1 = 3.

Forming Q×PQ \times P

  1. List all ordered pairs (q,p)(q, p) where q∈Qq \in Q and p∈Pp \in P. Now the first component comes from Q={r}Q = \{r\} and the second from P={a,b,c}P = \{a, b, c\}:

Q×P={(r,a),(r,b),(r,c)}.Q \times P = \{(r, a), (r, b), (r, c)\}.

  1. Count the elements. Again, ∣Q×P∣=1×3=3|Q \times P| = 1 \times 3 = 3.

Comparing the two products

  1. Check element-by-element equality.

    For two sets to be equal, they must contain exactly the same elements. Compare:

    • P×Q={(a,r),(b,r),(c,r)}P \times Q = \{(a, r), (b, r), (c, r)\}
    • Q×P={(r,a),(r,b),(r,c)}Q \times P = \{(r, a), (r, b), (r, c)\}

    The pair (a,r)(a, r) has aa in the first position and rr in the second. The pair (r,a)(r, a) has rr first and aa second. These are different ordered pairs because the order matters in the definition of an ordered pair.

  2. Conclude.

    Since no element of P×QP \times Q appears in Q×PQ \times P (and vice versa), the two sets are disjoint. They have the same cardinality but completely different elements.

Watch out

A common mistake is to think that because both products involve the same "ingredients" {a,b,c,r}\{a, b, c, r\}, they must be equal. But Cartesian products are sets of ordered pairs, and (x,y)≠(y,x)(x, y) \neq (y, x) in general.

Tip

In general, A×B=B×AA \times B = B \times A only when A=BA = B or one of them is empty. The Cartesian product is not commutative.


✓Final answer

The two products are not equal: P×Q={(a,r),(b,r),(c,r)}P \times Q = \{(a, r), (b, r), (c, r)\} and Q×P={(r,a),(r,b),(r,c)}Q \times P = \{(r, a), (r, b), (r, c)\} are disjoint sets.

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