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Exercise 2.2 · Q8

Q.Let A={x,y,z}A = \{x, y, z\} and B={1,2}B = \{1, 2\}. Find the number of relations from AA to BB.

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A relation from AA to BB is any subset of the Cartesian product A×BA \times B. Since AA has 3 elements and BB has 2, ∣A×B∣=6|A \times B| = 6, and the number of subsets of a 6-element set is 26=642^6 = 64. So there are 64 relations.

The key idea is simple: a relation is just a set of ordered pairs where the first element comes from AA and the second from BB. That means every relation is a subset of A×BA \times B. So the problem reduces to: how many subsets does A×BA \times B have?

Let’s unpack why that’s the right way to think.

What is a relation?

In mathematics, a relation from set AA to set BB is any collection of ordered pairs (a,b)(a, b) with a∈Aa \in A and b∈Bb \in B. There’s no extra condition — you can include any pair you like, and leave out any you don’t. That’s exactly the definition of a subset of A×BA \times B.

So the number of relations equals the number of subsets of A×BA \times B.

Now we just need two things: the size of A×BA \times B, and how many subsets a set of that size has.

  1. Find ∣A×B∣|A \times B|. AA has 3 elements: x,y,zx, y, z. BB has 2 elements: 1,21, 2. The Cartesian product A×BA \times B is the set of all ordered pairs:

A×B={(x,1),(x,2),(y,1),(y,2),(z,1),(z,2)}A \times B = \{(x,1), (x,2), (y,1), (y,2), (z,1), (z,2)\}

That’s 3×2=63 \times 2 = 6 pairs. So ∣A×B∣=6|A \times B| = 6.

  1. Number of subsets of a set with nn elements. For any set with nn elements, the number of subsets is 2n2^n. Why? Because each element can either be in or out of a given subset — that’s 2 choices per element, and choices are independent. So 2×2×⋯×22 \times 2 \times \dots \times 2 (nn times) gives 2n2^n. …

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