Relations and Functions Ex 2.1: Ordered Pairs & Cartesian Product

Core Concepts for Ex 2.1: Ordered Pairs and Cartesian Product

Ordered Pair

An ordered pair consists of two objects or elements in a specific order. Represented as (a, b), the element a is called the first element and b is the second. The order is crucial, so in general, (a, b) ≠ (b, a).

Equality of Ordered Pairs

Two ordered pairs, (a, b) and (c, d), are considered equal if and only if their corresponding elements are equal. That is, (a, b) = (c, d) implies a = c and b = d.

Cartesian Product of Sets

For any two non-empty sets A and B, the Cartesian product, denoted by A × B, is the set of all possible ordered pairs (a, b) where the first element a is from set A and the second element b is from set B. Symbolically: A × B = {(a, b) : a ∈ A, b ∈ B}.

Deep Dive into the Cartesian Product (A × B)

The Cartesian Product is more than just a mechanical process; it's a way of creating a new 'universe' of pairs from existing sets. Let's imagine you have a set of T-shirts, A = {Red, Blue}, and a set of Jeans, B = {Black, Grey}. The Cartesian Product A × B represents all possible outfits you can form. You pair each T-shirt with each pair of jeans: A × B = {(Red, Black), (Red, Grey), (Blue, Black), (Blue, Grey)}.

Notice a few key properties here:

1. Order Matters: (Red, Black) is not the same as (Black, Red). The latter would be an element of B × A, which represents pairing jeans with T-shirts. So, A × B ≠ B × A unless A = B.
2. Number of Elements: If set A has p elements (written as n(A) = p) and set B has q elements (n(B) = q), then the Cartesian product A × B will have p × q elements. In our example, n(A) = 2, n(B) = 2, so n(A × B) = 2 × 2 = 4 elements (pairs).
3. Empty Set: If either A or B is an empty set (∅), then A × B will also be an empty set. You can't form any pairs if one of the sets has no elements! A × ∅ = ∅.
4. Ordered Triplets: The concept can be extended to three sets. A × B × C is the set of all ordered triplets (a, b, c) where a ∈ A, b ∈ B, and c ∈ C. Similarly, A × A × A would be triplets (x, y, z) where all three elements come from set A. This is important for representing points in 3D space.

Solved Examples: Mastering Ex 2.1 Problems

• Problem 1: Equality of Ordered Pairs If (x/3 + 1, y - 2/3) = (5/3, 1/3), find the values of x and y. Step 1: Equate the first elements. According to the rule of equality of ordered pairs, the first elements must be equal. x/3 + 1 = 5/3 Step 2: Solve for x. Subtract 1 from both sides: x/3 = 5/3 - 1 x/3 = 5/3 - 3/3 x/3 = 2/3 Multiply both sides by 3: x = 2. Step 3: Equate the second elements. Similarly, the second elements must be equal. y - 2/3 = 1/3 Step 4: Solve for y. Add 2/3 to both sides: y = 1/3 + 2/3 y = 3/3 y = 1 Final Answer: x = 2 and y = 1.
• Problem 2: Finding the Cartesian Product If Set P = {a, b} and Set Q = {x, y, z}, find P × Q and determine n(P × Q). Step 1: Understand the task. We need to form a set of all ordered pairs where the first element comes from P and the second element comes from Q. Step 2: Form the pairs systematically. Take the first element of P, which is 'a', and pair it with every element of Q. (a, x), (a, y), (a, z) Now, take the second element of P, which is 'b', and pair it with every element of Q. (b, x), (b, y), (b, z) Step 3: Write the final set P × Q. Combine all the pairs into a single set. P × Q = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)}. Step 4: Calculate n(P × Q). We know that n(P) = 2 and n(Q) = 3. Therefore, n(P × Q) = n(P) × n(Q) = 2 × 3 = 6. We can verify this by counting the 6 ordered pairs in our resulting set. Final Answer: P × Q = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)} and n(P × Q) = 6.
• Problem 3: Cartesian Product with Intersections Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify if A × C is a subset of B × D. Step 1: Calculate A × C. A = {1, 2}, C = {5, 6} A × C = {(1, 5), (1, 6), (2, 5), (2, 6)} Step 2: Calculate B × D. B = {1, 2, 3, 4}, D = {5, 6, 7, 8} This will be a large set with 4 × 4 = 16 elements. We don't need to write them all out. We just need to check if the elements of A × C are present in B × D. Step 3: Check if A × C is a subset of B × D. For A × C to be a subset of B × D, every element of A × C must also be an element of B × D. Let's check each element of A × C: - (1, 5): Is 1 ∈ B and 5 ∈ D? Yes, 1 ∈ {1, 2, 3, 4} and 5 ∈ {5, 6, 7, 8}. So (1, 5) ∈ B × D. - (1, 6): Is 1 ∈ B and 6 ∈ D? Yes. So (1, 6) ∈ B × D. - (2, 5): Is 2 ∈ B and 5 ∈ D? Yes. So (2, 5) ∈ B × D. - (2, 6): Is 2 ∈ B and 6 ∈ D? Yes. So (2, 6) ∈ B × D. Step 4: Conclude. Since all elements of A × C are also elements of B × D, we can conclude that A × C is a subset of B × D. Final Answer: Yes, A × C is a subset of B × D.

Exam Traps and Key Points for Ex 2.1

1. Order Matters! The most common error is treating an ordered pair (a, b) the same as (b, a) or a set {a, b}. Always remember A × B ≠ B × A unless A = B. The position of each element is fixed.

2. Use Correct Notation. The Cartesian product A × B is a set of ordered pairs. Your final answer must be written correctly: A × B = { (pair1), (pair2), ... }. Using () for the set or {} for the pairs will lose you marks.

3. The Empty Set Property. A frequently asked MCQ question involves the empty set. Remember: if either set A or set B is empty (∅), their Cartesian product A × B is also the empty set. Don't get confused and write {∅}.

4. Reconstructing a Set. If you are given a few elements of A × A and asked to find set A, remember that set A consists of all the unique first and second elements from the given pairs. For example, if (1, 2) and (3, 1) are in A × A, then A must contain {1, 2, 3}.

Practice Questions with Solutions

• Q: If (3x - 2, y + 3) = (7, 5), find the values of x and y. A: Step 1: By the definition of equality of ordered pairs, we equate the corresponding components. 3x - 2 = 7 y + 3 = 5 Step 2: Solve the first equation for x. 3x = 7 + 2 3x = 9 x = 3 Step 3: Solve the second equation for y. y = 5 - 3 y = 2 Final answer: x = 3 and y = 2.
• Q: If A = {x, y} and B = {1, 2, 3}, find A × B and B × A. Are they equal? A: Step 1: Find A × B by pairing each element of A with each element of B. A × B = {(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)} Step 2: Find B × A by pairing each element of B with each element of A. B × A = {(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)} Step 3: Compare the two sets. For example, (x, 1) is in A × B, but it is not in B × A (unless x=1). Since the ordered pairs are not identical, the sets are not equal. Final answer: A × B and B × A are not equal.
• Q: Let A = {1, 2}. Find A × A × A. A: Step 1: First, find A × A. A × A = {1, 2} × {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)} Step 2: Now, find (A × A) × A. We will pair each ordered pair from A × A with each element of A to form an ordered triplet. Pairing (1, 1) with elements of A gives: (1, 1, 1), (1, 1, 2) Pairing (1, 2) with elements of A gives: (1, 2, 1), (1, 2, 2) Pairing (2, 1) with elements of A gives: (2, 1, 1), (2, 1, 2) Pairing (2, 2) with elements of A gives: (2, 2, 1), (2, 2, 2) Step 3: Combine all the triplets into a single set. Final answer: A × A × A = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}.
• Q: The Cartesian product P × P has 9 elements, among which are found (a, b) and (c, a). Find the set P and the remaining elements of P × P. A: Step 1: Determine the number of elements in P. Since n(P × P) = 9, we have n(P) × n(P) = 9, which means n(P) = 3. Step 2: Find the elements of set P. The elements of P are the set of all unique first and second components of the given ordered pairs. The given pairs are (a, b) and (c, a). The components are a, b, and c. Since n(P) = 3, these must be all the elements. So, P = {a, b, c}. Step 3: Find all the elements of P × P. P × P = {a, b, c} × {a, b, c} P × P = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}. Step 4: Identify the remaining elements. The given elements are (a, b) and (c, a). The remaining elements are the other seven pairs. Final answer: Set P = {a, b, c}. The remaining elements of P × P are {(a, a), (a, c), (b, a), (b, b), (b, c), (c, b), (c, c)}.

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