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Worked Examples · Example 3

Q.Let A={a,b,c}A = \{a, b, c\}, B={1,2}B = \{1, 2\}. Find A×BA \times B and B×AB \times A. Is A×B=B×AA \times B = B \times A?

Chandigarh CbseNCERTSubjective· 2mImportance★★★★★est
38% · 8/21 Questions
✓ Free question

Listing all ordered pairs shows A×B≠B×AA\times B\neq B\times A because ordered pairs are not symmetric.

The Cartesian product of sets AA and BB is

A×B={(a,b):a∈A, b∈B}A\times B=\{(a,b): a\in A,\ b\in B\}

and in general A×B≠B×AA\times B\neq B\times A unless A=BA=B, since (a,b)=(b,a)(a,b)=(b,a) only when a=ba=b.

  1. Given sets. A={a,b,c}A=\{a,b,c\} (3 elements), B={1,2}B=\{1,2\} (2 elements).

  2. List A×BA\times B by pairing every element of AA with every element of BB:

A×B={(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)}A\times B=\{(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)\}

This has n(A)×n(B)=3×2=6n(A)\times n(B)=3\times2=6 elements.

  1. List B×AB\times A by pairing every element of BB with every element of AA:

B×A={(1,a),(1,b),(1,c),(2,a),(2,b),(2,c)}B\times A=\{(1,a),(1,b),(1,c),(2,a),(2,b),(2,c)\}

This also has n(B)×n(A)=2×3=6n(B)\times n(A)=2\times3=6 elements.

  1. Compare. An ordered pair (a,1)∈A×B(a,1)\in A\times B but (a,1)∉B×A(a,1)\notin B\times A (since B×AB\times A's pairs have their first component in B={1,2}B=\{1,2\}, not aa). So no element of A×BA\times B matches any element of B×AB\times A (except in the trivial case A=BA=B).

∴ A×B≠B×A\therefore\ A\times B\neq B\times A

Self-check: Both sets have the same cardinality (6=66=6), confirming n(A×B)=n(B×A)=n(A)n(B)n(A\times B)=n(B\times A)=n(A)n(B) always holds even though the sets themselves differ.

✓Final answer

A×B={(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)}A\times B=\{(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)\}; B×A={(1,a),(1,b),(1,c),(2,a),(2,b),(2,c)}B\times A=\{(1,a),(1,b),(1,c),(2,a),(2,b),(2,c)\}. Since ordered pairs differ, A×B≠B×AA\times B\neq B\times A.

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