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Worked Examples · Example 4

Q.If P={1,2}P = \{1, 2\}, form the set P×P×PP \times P \times P.

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The Cartesian product P×P×PP \times P \times P is the set of all ordered triples where each coordinate is from P={1,2}P = \{1, 2\}. The result is {(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)}\{(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)\}.

The Cartesian product is a way to combine sets by forming ordered tuples. When you see P×P×PP \times P \times P, think: "Take one element from the first PP, one from the second PP, and one from the third PP, and put them in order." Each coordinate is independent, so the total number of triples is ∣P∣3=23=8|P|^3 = 2^3 = 8.

Why does this work? Because the product builds every possible combination without repetition or omission. The order matters — (1,2,1)(1,2,1) is different from (2,1,1)(2,1,1) — so we list them systematically.

  1. Identify the base set. P={1,2}P = \{1, 2\} has two elements. Every triple will have each entry either 1 or 2.

  2. Think of the structure. P×P×PP \times P \times P means all ordered triples (a,b,c)(a, b, c) where a∈Pa \in P, b∈Pb \in P, c∈Pc \in P. There's no restriction — all combinations are allowed.

  3. Generate systematically. Start with the first coordinate fixed as 1, then vary the second and third:

    • a=1a = 1: then (b,c)(b,c) can be (1,1),(1,2),(2,1),(2,2)(1,1), (1,2), (2,1), (2,2) — that's 4 triples.
    • a=2a = 2: similarly, (b,c)(b,c) gives (1,1),(1,2),(2,1),(2,2)(1,1), (1,2), (2,1), (2,2) — another 4 triples.
  4. List them in order. A clean way is to count in base-2, treating 1 as 0 and 2 as 1, but here's the explicit set:

    • (1,1,1)(1,1,1)
    • (1,1,2)(1,1,2) …

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