Samacheer Kalvi Class 10 Maths Solutions Chapter 1 Relations and Functions Exercise 1.1

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question 1. Find A × B, A × A and B × A
(i) A = {2, -2, 3} and B = {1, -4}
(ii) A = B = {p, q}
(iii) A = {m, n} ; B = Φ
Answer:
(i) First, we have the sets A = {2, -2, 3} and B = {1, -4}.
To find A × B, we combine each element of A with each element of B as ordered pairs:
\( A \times B = \{ (2, 1), (2, -4), (-2, 1), (-2, -4), (3, 1), (3, -4) \} \)
Next, to find A × A, we combine each element of A with every other element of A:
\( A \times A = \{ (2, 2), (2, -2), (2, 3), (-2, 2), (-2, -2), (-2, 3), (3, 2), (3, -2), (3, 3) \} \)
Finally, for B × A, we combine each element of B with each element of A:
\( B \times A = \{ (1, 2), (1, -2), (1, 3), (-4, 2), (-4, -2), (-4, 3) \} \)
(ii) For this part, set A and set B are the same: A = B = {p, q}.
To find A × B, we combine elements from A with elements from B:
\( A \times B = \{ (p, p), (p, q), (q, p), (q, q) \} \)
For A × A, we combine elements within set A itself:
\( A \times A = \{ (p, p), (p, q), (q, p), (q, q) \} \)
For B × A, we combine elements from B with elements from A:
\( B \times A = \{ (p, p), (p, q), (q, p), (q, q) \} \)
(iii) In this case, A = {m, n} and B = \( \Phi \) (which means B is an empty set).
When one of the sets is empty, the Cartesian product will also be an empty set. Thus:
\( A \times B = \{ m, n \} \times \{ \} = \{ \} \)
For A × A, we combine elements within set A:
\( A \times A = \{ (m, m), (m, n), (n, m), (n, n) \} \)
For B × A, since B is an empty set, the product will be empty:
\( B \times A = \{ \} \times \{ m, n \} = \{ \} \)
In simple words: The Cartesian product means making all possible ordered pairs by taking one item from the first set and one from the second. If a set is empty, then any product with it will also be empty.

🎯 Exam Tip: Remember that a Cartesian product creates ordered pairs. The order of elements in the pair matters, meaning \( (a, b) \) is generally different from \( (b, a) \).

Question 2. Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A × B and B × A.
Answer:
First, we need to list the elements of set B. B contains prime numbers less than 10. Prime numbers are numbers greater than 1 that can only be divided by 1 and themselves. These are 2, 3, 5, and 7.
So, A = {1, 2, 3} and B = {2, 3, 5, 7}.
Now, we find A × B by pairing each element from A with each element from B:
\( A \times B = \{ (1, 2), (1, 3), (1, 5), (1, 7), (2, 2), (2, 3), (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7) \} \)
Next, we find B × A by pairing each element from B with each element from A:
\( B \times A = \{ (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (5, 1), (5, 2), (5, 3), (7, 1), (7, 2), (7, 3) \} \)
In simple words: We first listed all the prime numbers smaller than 10 to get set B. Then, for A × B, we made pairs where the first number came from A and the second from B. For B × A, we did the opposite. Notice that \( A \times B \) and \( B \times A \) are not the same because the order of elements in the pairs is important.

🎯 Exam Tip: Always make sure to list all elements correctly in your sets before performing operations like Cartesian products. Be careful to correctly identify prime numbers.

Question 3. If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4), (3,3),(3, 4)} find A and B.
Answer:
Given the Cartesian product B × A = {(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)}.
In a Cartesian product \( B \times A \), the first element of each ordered pair comes from set B, and the second element comes from set A.
So, to find set B, we collect all the unique first elements from the ordered pairs: -2, 0, 3.
Therefore, \( B = \{ -2, 0, 3 \} \).
To find set A, we collect all the unique second elements from the ordered pairs: 3, 4.
Therefore, \( A = \{ 3, 4 \} \). A set only includes unique elements, so repeating numbers are listed only once.
In simple words: To find set B, look at all the first numbers in the pairs given. To find set A, look at all the second numbers in the pairs. Remember to list each unique number only once.

🎯 Exam Tip: When extracting sets from a Cartesian product, always remember that the first elements belong to the first set mentioned in the product (e.g., B in B × A), and the second elements belong to the second set (A in B × A). Also, sets do not contain duplicate elements.

Question 4. If A ={5, 6}, B = {4, 5, 6} , C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C x C).
Answer:
We are given the sets: A = {5, 6}, B = {4, 5, 6}, and C = {5, 6, 7}.
We need to show that \( A \times A = (B \times B) \cap (C \times C) \).

First, let's calculate \( A \times A \):
\( A \times A = \{ (5, 5), (5, 6), (6, 5), (6, 6) \} \) ..... (1)

Next, calculate \( B \times B \):
\( B \times B = \{ (4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6) \} \) ..... (2)

Then, calculate \( C \times C \):
\( C \times C = \{ (5, 5), (5, 6), (5, 7), (6, 5), (6, 6), (6, 7), (7, 5), (7, 6), (7, 7) \} \) ..... (3)

Now, find the intersection of \( (B \times B) \) and \( (C \times C) \). This means finding the ordered pairs that are present in both sets:
\( (B \times B) \cap (C \times C) = \{ (5, 5), (5, 6), (6, 5), (6, 6) \} \) ..... (4)

Comparing the result from (1) and (4), we can see that both are identical:
\( A \times A = (B \times B) \cap (C \times C) \).
This proves the given statement.
In simple words: We first found all possible pairs for A with itself. Then we found all pairs for B with itself and C with itself. After that, we looked for pairs that were common to both \( (B \times B) \) and \( (C \times C) \). Both sets of pairs matched, so the statement is proven true.

🎯 Exam Tip: To prove set equalities, always calculate each side of the equation separately, simplifying them until you can clearly show they are the same. Show all intermediate steps for clarity.

Question 5. Given A = {1,2,3}, B = {2,3,5}, C = {3,4} and D = {1,3,5}, check if (A ∩C) × (B ∩ D) = (A × B) ∩ (C × D) is true?
Answer:
We are given the sets: A = {1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}.
We need to check if \( (A \cap C) \times (B \cap D) = (A \times B) \cap (C \times D) \) is true.

First, let's calculate the Left Hand Side (LHS): \( (A \cap C) \times (B \cap D) \).
Find the intersection of A and C:
\( A \cap C = \{ 1, 2, 3 \} \cap \{ 3, 4 \} = \{ 3 \} \)
Find the intersection of B and D:
\( B \cap D = \{ 2, 3, 5 \} \cap \{ 1, 3, 5 \} = \{ 3, 5 \} \)
Now, calculate the Cartesian product of these intersections:
\( (A \cap C) \times (B \cap D) = \{ 3 \} \times \{ 3, 5 \} = \{ (3, 3), (3, 5) \} \) ..... (1)

Next, let's calculate the Right Hand Side (RHS): \( (A \times B) \cap (C \times D) \).
Calculate \( A \times B \):
\( A \times B = \{ 1, 2, 3 \} \times \{ 2, 3, 5 \} = \{ (1, 2), (1, 3), (1, 5), (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5) \} \)
Calculate \( C \times D \):
\( C \times D = \{ 3, 4 \} \times \{ 1, 3, 5 \} = \{ (3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5) \} \)
Now, find the intersection of \( (A \times B) \) and \( (C \times D) \). This means finding the common ordered pairs:
\( (A \times B) \cap (C \times D) = \{ (3, 3), (3, 5) \} \) ..... (2)

Comparing the results from (1) and (2), we see that both sides are equal.
Therefore, \( (A \cap C) \times (B \cap D) = (A \times B) \cap (C \times D) \) is true.
In simple words: We checked both sides of the equation. On the left side, we first found the common elements in sets A and C, and in sets B and D. Then we multiplied these results. On the right side, we first multiplied A with B, and C with D. Then we found the common pairs from these two multiplications. Since both sides ended up with the same pairs, the statement is true.

🎯 Exam Tip: This problem demonstrates a distributive property of Cartesian products over set intersection. Always calculate intersections and unions first before performing Cartesian products, or vice-versa, depending on the structure of the equation.

Question 6. Let A = {x ∈ W | x < 2}, B = {x ∈ N |1 < x < 4} and C = {3, 5}. Verify that
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A x C)
(iii) (A ∪ B) × C = (A × C) ∪ (B x C)
Answer:
First, let's list the elements for each set:
A = {x ∈ W | x < 2} means A contains whole numbers less than 2. So, \( A = \{ 0, 1 \} \).
B = {x ∈ N | 1 < x < 4} means B contains natural numbers greater than 1 and less than 4. So, \( B = \{ 2, 3, 4 \} \).
C = {3, 5}.

(i) Verify: \( A \times (B \cup C) = (A \times B) \cup (A \times C) \)
Left Hand Side (LHS): \( A \times (B \cup C) \)
First, find \( B \cup C \):
\( B \cup C = \{ 2, 3, 4 \} \cup \{ 3, 5 \} = \{ 2, 3, 4, 5 \} \)
Now, calculate \( A \times (B \cup C) \):
\( A \times (B \cup C) = \{ 0, 1 \} \times \{ 2, 3, 4, 5 \} = \{ (0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5) \} \) ..... (1)

Right Hand Side (RHS): \( (A \times B) \cup (A \times C) \)
First, calculate \( A \times B \):
\( A \times B = \{ 0, 1 \} \times \{ 2, 3, 4 \} = \{ (0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4) \} \)
Next, calculate \( A \times C \):
\( A \times C = \{ 0, 1 \} \times \{ 3, 5 \} = \{ (0, 3), (0, 5), (1, 3), (1, 5) \} \)
Now, find the union of \( (A \times B) \) and \( (A \times C) \):
\( (A \times B) \cup (A \times C) = \{ (0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4), (0, 5), (1, 5) \} \) ..... (2)
Comparing (1) and (2), we see that LHS = RHS. Hence, it is proved.

(ii) Verify: \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)
Left Hand Side (LHS): \( A \times (B \cap C) \)
First, find \( B \cap C \):
\( B \cap C = \{ 2, 3, 4 \} \cap \{ 3, 5 \} = \{ 3 \} \)
Now, calculate \( A \times (B \cap C) \):
\( A \times (B \cap C) = \{ 0, 1 \} \times \{ 3 \} = \{ (0, 3), (1, 3) \} \) ..... (1)

Right Hand Side (RHS): \( (A \times B) \cap (A \times C) \)
From part (i), we already have:
\( A \times B = \{ (0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4) \} \)
\( A \times C = \{ (0, 3), (0, 5), (1, 3), (1, 5) \} \)
Now, find the intersection of \( (A \times B) \) and \( (A \times C) \):
\( (A \times B) \cap (A \times C) = \{ (0, 3), (1, 3) \} \) ..... (2)
Comparing (1) and (2), we see that LHS = RHS. Hence, it is verified.

(iii) Verify: \( (A \cup B) \times C = (A \times C) \cup (B \times C) \)
Left Hand Side (LHS): \( (A \cup B) \times C \)
First, find \( A \cup B \):
\( A \cup B = \{ 0, 1 \} \cup \{ 2, 3, 4 \} = \{ 0, 1, 2, 3, 4 \} \)
Now, calculate \( (A \cup B) \times C \):
\( (A \cup B) \times C = \{ 0, 1, 2, 3, 4 \} \times \{ 3, 5 \} = \{ (0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5) \} \) ..... (1)

Right Hand Side (RHS): \( (A \times C) \cup (B \times C) \)
From part (i), we have \( A \times C = \{ (0, 3), (0, 5), (1, 3), (1, 5) \} \).
Now, calculate \( B \times C \):
\( B \times C = \{ 2, 3, 4 \} \times \{ 3, 5 \} = \{ (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5) \} \)
Now, find the union of \( (A \times C) \) and \( (B \times C) \):
\( (A \times C) \cup (B \times C) = \{ (0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5) \} \) ..... (2)
Comparing (1) and (2), we see that LHS = RHS. Hence, it is verified.
In simple words: We first wrote out the elements of sets A and B based on their definitions. Then, for each of the three parts, we calculated the left side and the right side of the equation separately. We found that the results for both sides were identical in all three cases, which means the given statements are true. These are examples of how Cartesian products interact with union and intersection operations.

🎯 Exam Tip: Always define your sets clearly by listing their elements first. When verifying set equalities, it's best to compute each side (LHS and RHS) independently and then compare the final results. This helps in avoiding errors and keeping the solution organized.

Question 7. Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that
(i) (A ∩ B) x C = (A × C) ∩ (B x C)
(ii) A × (B – C) = (A × B) – (A × C)
Answer:
First, let's list the elements for each set:
A = The set of all natural numbers less than 8: \( A = \{ 1, 2, 3, 4, 5, 6, 7 \} \).
B = The set of all prime numbers less than 8: \( B = \{ 2, 3, 5, 7 \} \).
C = The set of even prime numbers. The only even prime number is 2. So, \( C = \{ 2 \} \).

(i) Verify: \( (A \cap B) \times C = (A \times C) \cap (B \times C) \)
Left Hand Side (LHS): \( (A \cap B) \times C \)
First, find \( A \cap B \):
\( A \cap B = \{ 1, 2, 3, 4, 5, 6, 7 \} \cap \{ 2, 3, 5, 7 \} = \{ 2, 3, 5, 7 \} \)
Now, calculate \( (A \cap B) \times C \):
\( (A \cap B) \times C = \{ 2, 3, 5, 7 \} \times \{ 2 \} = \{ (2, 2), (3, 2), (5, 2), (7, 2) \} \) ..... (1)

Right Hand Side (RHS): \( (A \times C) \cap (B \times C) \)
First, calculate \( A \times C \):
\( A \times C = \{ 1, 2, 3, 4, 5, 6, 7 \} \times \{ 2 \} = \{ (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2) \} \)
Next, calculate \( B \times C \):
\( B \times C = \{ 2, 3, 5, 7 \} \times \{ 2 \} = \{ (2, 2), (3, 2), (5, 2), (7, 2) \} \)
Now, find the intersection of \( (A \times C) \) and \( (B \times C) \):
\( (A \times C) \cap (B \times C) = \{ (2, 2), (3, 2), (5, 2), (7, 2) \} \) ..... (2)
Comparing (1) and (2), we see that LHS = RHS. Hence, it is verified.

(ii) Verify: \( A \times (B - C) = (A \times B) - (A \times C) \)
Left Hand Side (LHS): \( A \times (B - C) \)
First, find \( B - C \). This means elements in B but not in C:
\( B - C = \{ 2, 3, 5, 7 \} - \{ 2 \} = \{ 3, 5, 7 \} \)
Now, calculate \( A \times (B - C) \):
\( A \times (B - C) = \{ 1, 2, 3, 4, 5, 6, 7 \} \times \{ 3, 5, 7 \} \)
\( = \{ (1, 3), (1, 5), (1, 7), (2, 3), (2, 5), (2, 7), (3, 3), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 3), (5, 5), (5, 7), (6, 3), (6, 5), (6, 7), (7, 3), (7, 5), (7, 7) \} \) ..... (1)

Right Hand Side (RHS): \( (A \times B) - (A \times C) \)
First, calculate \( A \times B \):
\( A \times B = \{ 1, 2, 3, 4, 5, 6, 7 \} \times \{ 2, 3, 5, 7 \} \)
\( = \{ (1, 2), (1, 3), (1, 5), (1, 7), (2, 2), (2, 3), (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7), (4, 2), (4, 3), (4, 5), (4, 7), (5, 2), (5, 3), (5, 5), (5, 7), (6, 2), (6, 3), (6, 5), (6, 7), (7, 2), (7, 3), (7, 5), (7, 7) \} \)
Next, calculate \( A \times C \):
\( A \times C = \{ 1, 2, 3, 4, 5, 6, 7 \} \times \{ 2 \} \)
\( = \{ (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2) \} \)
Now, find the difference \( (A \times B) - (A \times C) \). This means pairs in \( A \times B \) that are not in \( A \times C \):
\( (A \times B) - (A \times C) = \{ (1, 3), (1, 5), (1, 7), (2, 3), (2, 5), (2, 7), (3, 3), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 3), (5, 5), (5, 7), (6, 3), (6, 5), (6, 7), (7, 3), (7, 5), (7, 7) \} \) ..... (2)
Comparing (1) and (2), we see that LHS = RHS. Hence, it is verified.
In simple words: We first wrote out the numbers in each set (A, B, C) based on the rules given. Then, for both parts (i) and (ii), we calculated the left side and the right side of the equations separately. For example, to find \( B - C \), we took numbers that are in B but not in C. In both parts, the final list of pairs for the left side matched the final list of pairs for the right side, which means the statements are true.

🎯 Exam Tip: When defining sets based on conditions (like "natural numbers less than 8" or "prime numbers less than 8"), be very careful to include or exclude boundary values correctly. For verification problems, compute each side of the equality separately to maintain clarity and accuracy.

Relations
Let A and B be any two non-empty sets. A "relation” R from A to B is a subset of A × B satisfying some specified conditions.
In simple words: A relation between two sets A and B is basically a collection of some chosen pairs from all the possible pairs you can make between A and B. It connects certain elements from A to certain elements from B based on a rule.

🎯 Exam Tip: The definition of a relation is fundamental. Remember that a relation is always a subset of the Cartesian product, meaning every pair in the relation must also exist in the Cartesian product of the two sets.

Note:

• The domain of a relation R from A to B, written as \( R = \{x \in A | xRy, \text{ for some } y \in B\} \), is the set of all first elements of the ordered pairs in the relation.
• The co-domain of the relation R is the entire set B.
• The range of the relation is the set of all second elements of the ordered pairs in the relation, written as \( R = \{y \in B | xRy, \text{ for some } x \in A\} \).

In simple words: For any relationship between two sets: the 'domain' is all the starting items that are used, the 'co-domain' is the whole second set, and the 'range' is only the ending items that actually get paired up with something from the first set.

🎯 Exam Tip: Differentiate clearly between co-domain and range. The co-domain is the entire target set, while the range is only the subset of the co-domain that actually gets "mapped to" by elements from the domain.