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Exercise 2.1 · Q5

Q.If A={−1,1}A = \{-1, 1\}, find A×A×AA \times A \times A.

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The Cartesian product A×A×AA \times A \times A is the set of all ordered triples where each coordinate is either −1-1 or 11. There are 23=82^3 = 8 such triples, and the answer is {(−1,−1,−1),(−1,−1,1),(−1,1,−1),(−1,1,1),(1,−1,−1),(1,−1,1),(1,1,−1),(1,1,1)}\{(-1,-1,-1), (-1,-1,1), (-1,1,-1), (-1,1,1), (1,-1,-1), (1,-1,1), (1,1,-1), (1,1,1)\}.

The Cartesian product is a way to build new sets from existing ones by forming ordered tuples. When you see A×A×AA \times A \times A, think: "Take one element from the first AA, one from the second AA, and one from the third AA, and put them together in an ordered triple." The order matters — (−1,1,−1)( -1, 1, -1) is different from (1,−1,−1)(1, -1, -1).

Here AA has just two elements: −1-1 and 11. So for each of the three positions in the triple, you have exactly 2 choices. That gives 2×2×2=82 \times 2 \times 2 = 8 possible triples. The job is simply to list them all systematically.

  1. Fix the first coordinate. Start with −1-1 in the first slot. Then the second and third coordinates each can be −1-1 or 11, independently. That gives four triples:

    (−1,−1,−1)(-1, -1, -1), (−1,−1,1)(-1, -1, 1), (−1,1,−1)(-1, 1, -1), (−1,1,1)(-1, 1, 1).

  2. Now take 11 as the first coordinate. Again, the second and third coordinates each have two possibilities, giving another four triples:

    (1,−1,−1)(1, -1, -1), (1,−1,1)(1, -1, 1), (1,1,−1)(1, 1, -1), (1,1,1)(1, 1, 1).

  3. Combine both groups. The full set is the union of these eight ordered triples. No duplicates appear because each triple is uniquely determined by its three coordinates. …

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