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Exercise 2.1 · Q9

Q.Let AA and BB be two sets such that n(A)=3n(A) = 3 and n(B)=2n(B) = 2. If (x,1)(x, 1), (y,2)(y, 2), (z,1)(z, 1) are in A×BA \times B, find AA and BB, where xx, yy and zz are distinct elements.

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The Cartesian product A×BA \times B consists of all ordered pairs (a,b)(a, b) where a∈Aa \in A and b∈Bb \in B. Since the second coordinates of the given pairs tell us what's in BB, and the first coordinates tell us what's in AA, we find A={x,y,z}A = \{x, y, z\} and B={1,2}B = \{1, 2\}.

The Cartesian product A×BA \times B is the set of all ordered pairs where the first element comes from AA and the second from BB. This means every ordered pair (a,b)∈A×B(a, b) \in A \times B satisfies a∈Aa \in A and b∈Bb \in B.

When we're told that certain pairs belong to A×BA \times B, we can work backwards: the first coordinates must be elements of AA, and the second coordinates must be elements of BB. This reverse-engineering is the key to finding the sets.

Finding set BB:

  1. Look at the second coordinates of all given pairs: (x,1)(x, 1), (y,2)(y, 2), (z,1)(z, 1).

    The second coordinates are 11, 22, and 11. Since these pairs are in A×BA \times B, each second coordinate must be an element of BB.

  2. The distinct second coordinates are 11 and 22, so we know {1,2}⊆B\{1, 2\} \subseteq B.

  3. We're told n(B)=2n(B) = 2, meaning BB has exactly two elements. Since we've already identified two elements that must be in BB, we have B={1,2}B = \{1, 2\}.

Finding set AA:

  1. Now examine the first coordinates: xx, yy, and zz. Since each pair is in A×BA \times B, we know x∈Ax \in A, y∈Ay \in A, and z∈Az \in A.

  2. The problem states that xx, yy, and zz are distinct elements, so {x,y,z}⊆A\{x, y, z\} \subseteq A and these are three different elements. …

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