Relations and Functions

1. Relations:

• Definition: A relation is a subset of the Cartesian product of two sets. It describes how elements of two sets are connected.
• Types:
  • Empty Relation: No element is related.
  • Universal Relation: All elements are related.
  • Reflexive: Every element is related to itself.
  • Symmetric: If a is related to b, then b is related to a.

2. Functions:

• Definition: A function is a special type of relation where each element of the first set (domain) is associated with exactly one element of the second set (codomain).
• Types:
  • One-to-one (Injective): Each element in the domain maps to a unique element in the codomain.
  • Onto (Surjective): Every element in the codomain is mapped to by at least one element in the domain.
  • Bijective: A function that is both one-to-one and onto.

3. Composition of Functions:

• Combining two functions: If f maps A to B, and g maps B to C, then the composition gof maps A to C.

4. Invertible Functions:

• A function is invertible if there exists another function that “undoes” it. A function is invertible if and only if it is bijective.

5. Binary Operations:

• A binary operation on a set is a function that takes two elements from the set and returns another element of the same set. Examples include addition, subtraction, multiplication, etc.

Exercise 1.1

1. Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) Relation R in the set A = {1, 2, 3, …, 13, 14} defined as 

R = {(x, y) : 3x – y = 0}

(ii) Relation R in the set N of natural numbers defined as 

R = {(x, y) : y = x + 5 and x < 4} 

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as 

R = {(x, y) : y is divisible by x} 

(iv) Relation R in the set Z of all integers defined as

 R = {(x, y) : x – y is an integer} 

(v) Relation R in the set A of human beings in a town at a particular time given by (a) R = {(x, y) : x and y work at the same place} 

(b) R = {(x, y) : x and y live in the same locality} 

(c) R = {(x, y) : x is exactly 7 cm taller than y} 

(d) R = {(x, y) : x is wife of y} (e) R = {(x, y) : x is father of y}

Ans : 

(i) R = {(x, y) : 3x – y = 0} on A = {1, 2, 3, …, 13, 14}

• Reflexive: A relation is reflexive if (a, a) ∈ R for all a ∈ A. Let’s test with a = 1. 3(1) – 1 = 2 ≠ 0. So, (1, 1) ∉ R. Since we found a counterexample, R is not reflexive.
  
• Symmetric: A relation is symmetric if (a, b) ∈ R implies (b, a) ∈ R. Let’s consider (1, 3) because 3(1) – 3 = 0. So, (1, 3) ∈ R. Now, check for (3, 1): 3(3) – 1 = 8 ≠ 0. Thus, (3, 1) ∉ R. Since (1, 3) ∈ R but (3, 1) ∉ R, R is not symmetric.
  
• Transitive: A relation is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R. Let’s try to find an example. If 3x – y = 0 and 3y – z = 0, then y = 3x and z = 3y. Substituting, z = 3(3x) = 9x. For (x, z) to be in R, 3x – z = 0. So, 3x – 9x = -6x = 0, which means x must be 0. Since our set A doesn’t include 0, and because if we take 1,3 and 3,9 which are in R, we get 1,9 which is not in R, R is not transitive.
  

(ii) R = {(x, y) : y = x + 5 and x < 4} on N

• Reflexive: For a relation to be reflexive, (x, x) must be in R for all x in N. If x = 1, then y = 1 + 5 = 6. Since x ≠ y, (x, x) cannot be in R. R is not reflexive.
  
• Symmetric: If (x, y) ∈ R, then y = x + 5. For (y, x) to be in R, x = y + 5. Substituting y = x + 5, we get x = x + 5 + 5, which simplifies to 0 = 10. This is impossible. R is not symmetric.
  
• Transitive: If (x, y) ∈ R, y = x + 5. If (y, z) ∈ R, z = y + 5. Then z = (x + 5) + 5 = x + 10. For (x, z) to be in R, z = x + 5. But we have z = x + 10. R is not transitive.
  

(iii) R = {(x, y) : y is divisible by x} on A = {1, 2, 3, 4, 5, 6}

• Reflexive: Any number is divisible by itself. So, (x, x) ∈ R for all x ∈ A. R is reflexive.
  
• Symmetric: If y is divisible by x, it doesn’t necessarily mean x is divisible by y. For example, 4 is divisible by 2, but 2 is not divisible by 4. R is not symmetric.
  
• Transitive: If y is divisible by x, and z is divisible by y, then z must be divisible by x. R is transitive.
  

(iv) R = {(x, y) : x – y is an integer} on Z

• Reflexive: x – x = 0, which is an integer. R is reflexive.
  
• Symmetric: If x – y is an integer, then y – x = -(x – y) is also an integer. R is symmetric.
  
• Transitive: If x – y is an integer and y – z is an integer, then (x – y) + (y – z) = x – z is also an integer. R is transitive.
  

(v) Relations on human beings:

(a) 

• Reflexive: x works at the same place as x. R is reflexive.
• Symmetric: If x works at the same place as y, then y works at the same place as x. R is symmetric.
• Transitive: If x works at the same place as y, and y works at the same place as z, then x works at the same place as z. R is transitive.

(b) 

• Reflexive: x lives in the same locality as x. R is reflexive.
• Symmetric: If x lives in the same locality as y, then y lives in the same locality as x. R is symmetric.
• Transitive: If x lives in the same locality as y, and y lives in the same locality as z, then x lives in the same locality as z. R is transitive.

(c) 

• Reflexive: x cannot be exactly 7 cm taller than x. R is not reflexive.
• Symmetric: If x is exactly 7 cm taller than y, then y is not 7 cm taller than x. R is not symmetric.
• Transitive: If x is 7 cm taller than y, and y is 7 cm taller than z, then x is 14 cm taller than z. R is not transitive.

(d) 

• Reflexive: x cannot be the wife of x. R is not reflexive.
• Symmetric: If x is the wife of y, then y cannot be the wife of x. R is not symmetric.
• Transitive: If x is the wife of y, and y is the wife of z, this implies y is married to two people, which is not allowed. R is not transitive.

(e)

• Reflexive: x cannot be the father of x. R is not reflexive.
• Symmetric: If x is the father of y, then y cannot be the father of x. R is not symmetric.
• Transitive: If x is the father of y, and y is the father of z, then x is the grandfather of z. R is not transitive.

2. Show that the relation R in the set R of real numbers, defined as

R = {(a, b) : a ≤ b 2} is neither reflexive nor symmetric nor transitive.

Ans : 

1. Not Reflexive:

A relation is reflexive if (a, a) ∈ R for all a ∈ R. In our case, this means a ≤ a² for all real numbers a. Let’s consider a = 1/2. Then a² = (1/2)² = 1/4. Since 1/2 > 1/4, we have a > a², which means (1/2, 1/2) ∉ R. Therefore, R is not reflexive.

2. Not Symmetric:

A relation is symmetric if (a, b) ∈ R implies (b, a) ∈ R. This means if a ≤ b², then b ≤ a². Let’s consider a = 1 and b = 2. Since 1 ≤ 2² (1 ≤ 4), we have (1, 2) ∈ R. However, if we swap, we get b = 2 and a = 1. Then a² = 1² = 1. Since 2 > 1, we have b > a², which means (2, 1) ∉ R. Therefore, R is not symmetric.

3. Not Transitive:

A relation is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R. This means if a ≤ b² and b ≤ c², then a ≤ c². Let’s choose a = 4, b = -2, and c = -1.

• a ≤ b²: 4 ≤ (-2)² => 4 ≤ 4 (True) So, (4, -2) ∈ R.
• b ≤ c²: -2 ≤ (-1)² => -2 ≤ 1 (True) So, (-2, -1) ∈ R.
• a ≤ c²: 4 ≤ (-1)² => 4 ≤ 1 (False) So, (4, -1) ∉ R.

Since we have (4,-2) in R and (-2,-1) in R, but (4,-1) is not in R, R is not transitive.

Therefore, the relation R = {(a, b) : a ≤ b²} is neither reflexive, nor symmetric, nor transitive.

3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as 

R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive. 

Ans : 

Reflexive:

In other words, for a relation R on a set A, (a, a) must be in R for all a in A.

In our case, the set is {1, 2, 3, 4, 5, 6}. Let’s check if (a, a) is in R for all a in the set:

• If a = 1, b = a + 1 = 2. So, (1, 1) is not in R.

Since we’ve found a counterexample, the relation R is not reflexive.

Symmetric:

Let’s look at the elements of R:

R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

• (1, 2) is in R, but (2, 1) is not in R.

Since we’ve found a counterexample, the relation R is not symmetric.

Transitive:

Let’s examine the elements of R:

R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

• (1, 2) is in R, and (2, 3) is in R. For R to be transitive, (1, 3) would need to be in R. However, 3 ≠ 1 + 1, so (1, 3) is not in R.

Since we’ve found a counterexample, the relation R is not transitive.

Conclusion:

The relation R defined on the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is neither reflexive, nor symmetric, nor transitive.

4. Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric. 

Ans : 

1. Reflexive:

A relation is reflexive if (a, a) ∈ R for all a ∈ R. In our case, this means 

a ≤ a for all real numbers a. 

Since any real number is always less than or equal to itself, the condition a ≤ a is always true. 

Therefore, (a, a) ∈ R for all a ∈ R, and the relation R is reflexive.

2. Not Symmetric:

A relation is symmetric if (a, b) ∈ R implies (b, a) ∈ R. In our case, this means

 if a ≤ b, then b ≤ a. Let’s consider a = 1 and b = 2. 

Since 1 ≤ 2, we have (1, 2) ∈ R. However, 2 ≤ 1 is false.

Therefore, (2, 1) ∉ R. 

Since we have a counterexample, the relation R is not symmetric.

3. Transitive:

A relation is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R. In our case, this means if 

a ≤ b and b ≤ c, then a ≤ c. 

This is a fundamental property of inequalities. 

If a is less than or equal to b, and b is less than or equal to c, then a must be less than or equal to c. 

Therefore, the relation R is transitive.

Conclusion:

The relation R = {(a, b) : a ≤ b} on the set of real numbers is reflexive and transitive, but not symmetric.

5. Check whether the relation R in R defined by R = {(a, b) : a ≤ b 3} is reflexive, symmetric or transitive.

Ans : 

A relation is reflexive if (a, a) ∈ R for all a ∈ R. 

In our case, this means a ≤ a³ for all real numbers a.

Consider a = 1/2. 

Then a³ = (1/2)³ = 1/8. Since 1/2 > 1/8, a > a³ 

when a = 1/2. Therefore, (1/2, 1/2) ∉ R.

Since we’ve found a counterexample, the relation R is not reflexive.

Consider a = 8, b = 2, and c = 1.1

• 8 ≤ 2³ => 8 ≤ 8 (True)
• 2 ≤ (1.1)³ => 2 ≤ 1.331 (False)

Consider a = 1, b = 2 and c = 1.25

• 1 ≤ 2³ => 1 ≤ 8 (True)
• 2 ≤ (1.25)³ => 2 ≤ 1.953125 (False)

It seems likely that it is not transitive.

Let a = 27, b = 3 and c = 1.7.

• 27 ≤ 3³ => 27 ≤ 27 (True)
• 3 ≤ (1.7)³ => 3 ≤ 4.913 (True)
• 27 ≤ (1.7)³ => 27 ≤ 4.913 (False)

Therefore, the relation R is not transitive.

6. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Ans : 

1. Reflexivity:

In other words, for a set A, (a, a) must be in the relation for all a in A. 

Our set A is {1, 2, 3}. Therefore, for R to be reflexive, 

(1, 1), (2, 2), and (3, 3) must all be in R.

Looking at R = {(1, 2), (2, 1)}, 

we see that none of these pairs, (1, 1), (2, 2), or (3, 3), are present. 

Since not all elements of A are related to themselves, R is not reflexive.

2. Symmetry:

In our case, R = {(1, 2), (2, 1)}. 

We have (1, 2) in R, and we also have (2, 1) in R. 

This satisfies the condition for symmetry. 

Therefore, R is symmetric.

3. Transitivity:

A relation is transitive if for every (a, b) and (b, c) in the relation, (a, c) must also be in the relation.

Let’s examine R = {(1, 2), (2, 1)}.

• We have (1, 2) in R.
• We have (2, 1) in R.

If R were transitive, then since we have (1, 2) and (2, 1), we would need to have (1, 1) in R. However, (1, 1) is not in R.

Therefore, R is not transitive.

7. Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.

Ans : 

1. Reflexivity:

A relation is reflexive if (x, x) ∈ R for all x ∈ A. In our case, this means that a book x must have the same number of pages as itself. This is clearly true. Every book has a specific number of pages, and that number is equal to itself. 

Therefore, (x, x) ∈ R for all x ∈ A, and R is reflexive.

2. Symmetry:

A relation is symmetric if (x, y) ∈ R implies (y, x) ∈ R. In our case, this means if book x has the same number of pages as book y, then book y must have the same number of pages as book x. This is also clearly true. If the number of pages in x is equal to the number of pages in y, then the number of pages in y is equal to the number of pages in x. 

Therefore, if (x, y) ∈ R, then (y, x) ∈ R, and R is symmetric.

3. Transitivity:

In our case, this means if book x has the same number of pages as book y, and book y has the same number of pages as book z, then book x must have the same number of pages as book z. This is also true. If the number of pages in x is equal to the number of pages in y, and the number of pages in y is equal to the number of pages in z, then the number of pages in x must be equal to the number of pages in z. 

Therefore, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R, and R is transitive.

8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

Ans : 

1. Showing R is an Equivalence Relation:

We need to prove that R is reflexive, symmetric, and transitive.

• Reflexive: A relation is reflexive if (a, a) ∈ R for all a ∈ A. In our case, |a – a| = |0| = 0, which is even. Therefore, (a, a) ∈ R for all a ∈ A, and R is reflexive.
  
• Symmetric: A relation is symmetric if (a, b) ∈ R implies (b, a) ∈ R. If |a – b| is even, then |b – a| = |-(a – b)| = |a – b|, which is also even. Therefore, if (a, b) ∈ R, then (b, a) ∈ R, and R is symmetric.
  
• Transitive: A relation is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R. If |a – b| is even and |b – c| is even, we need to show that |a – c| is even. We can write:
  
  • a – b = 2k (where k is an integer)
  • b – c = 2m (where m is an integer)
• Adding these two equations, we get:
  
  • a – c = 2k + 2m = 2(k + m)
• Since k + m is an integer, a – c is an even number. Therefore, |a – c| is even, and R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

2. Relationships within the Subsets:

• {1, 3, 5}:
  • |1 – 3| = 2 (even)
  • |1 – 5| = 4 (even)
  • |3 – 5| = 2 (even)

All the elements of {1, 3, 5} are related to each other because the absolute difference between any two of them is even.

• {2, 4}:
  • |2 – 4| = 2 (even)

The elements of {2, 4} are related to each other because their absolute difference is even.

3. Relationship between the Subsets:

• Let’s take any element from {1, 3, 5} (e.g., 1) and any element from {2, 4} (e.g., 2):
  • |1 – 2| = 1 (odd)

Since the absolute difference between any element from {1, 3, 5} and any element from {2, 4} is odd, no element of {1, 3, 5} is related to any element of {2, 4}.

9. Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by 

(i) R = {(a, b) : |a – b| is a multiple of 4} 

(ii) R = {(a, b) : a = b}

is an equivalence relation. Find the set of all elements related to 1 in each case.

Ans : 

(i) R = {(a, b) : |a – b| is a multiple of 4} on A = {x ∈ Z : 0 ≤ x ≤ 12}

• Reflexive: For any a ∈ A, |a – a| = |0| = 0. Since 0 is a multiple of 4 (0 = 4 * 0), (a, a) ∈ R. Thus, R is reflexive.
  
• Symmetric: If (a, b) ∈ R, then |a – b| is a multiple of 4. 

This means |a – b| 

= 4k for some integer k. 

Since |b – a| = |-(a – b)| = |a – b|, |b – a| is also a multiple of 4. 

Therefore, (b, a) ∈ R, and R is symmetric.

• Transitive: If (a, b) ∈ R and (b, c) ∈ R, 

then |a – b| = 4k and |b – c| = 4m for some integers k and m. 

We want to show that |a – c| is a multiple of 4.

We have a – b = ±4k and b – c = ±4m. Adding these equations, we get: a – c = ±4k ± 4m = 4(±k ± m). Since ±k ± m is an integer, a – c is a multiple of 4. Therefore, |a – c| is a multiple of 4, and (a, c) ∈ R. Thus, R is transitive.

(ii) R = {(a, b) : a = b} on A = {x ∈ Z : 0 ≤ x ≤ 12}

This relation is simply the identity relation. Let’s check the properties:

• Reflexive: For any a ∈ A, a = a. So, (a, a) ∈ R. R is reflexive.
• Symmetric: If (a, b) ∈ R, then a = b. This implies b = a, so (b, a) ∈ R. R is symmetric.
• Transitive: If (a, b) ∈ R and (b, c) ∈ R, then a = b and b = c. This implies a = c, so (a, c) ∈ R. R is transitive.

10. Give an example of a relation. Which is 

(i) Symmetric but neither reflexive nor transitive. 

(ii) Transitive but neither reflexive nor symmetric. 

(iii) Reflexive and symmetric but not transitive. 

(iv) Reflexive and transitive but not symmetric. 

(v) Symmetric and transitive but not reflexive. 

Ans : 

(i) Symmetric but neither reflexive nor transitive:

R = {(1, 2), (2, 1)}

• Symmetric: (1, 2) is in R, and (2, 1) is in R.
• Not Reflexive: (1, 1), (2, 2), and (3, 3) are not in R.
• Not Transitive: (1, 2) and (2, 1) are in R, but (1, 1) is not in R.

(ii) Transitive but neither reflexive nor symmetric:

R = {(1, 2), (2, 3), (1, 3)}

• Transitive: (1, 2) and (2, 3) are in R, and (1, 3) is in R.
• Not Reflexive: (1, 1), (2, 2), and (3, 3) are not in R.
• Not Symmetric: (1, 2) is in R, but (2, 1) is not in R.

(iii) Reflexive and symmetric but not transitive:

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

• Reflexive: (1, 1), (2, 2), and (3, 3) are in R.
• Symmetric: (1, 2) is in R, and (2, 1) is in R.
• Not Transitive: (1, 2) and (2, 1) are in R, but (1, 1) is already present due to reflexivity. If we had R = {(1,1),(2,2),(1,2),(2,1)} then it would not be transitive. In this case, (1,2) and (2,1) are in R, however, (1,1) is in R.

(iv) Reflexive and transitive but not symmetric:

R = {(1, 1), (2, 2), (3, 3), (1, 2)}

• Reflexive: (1, 1), (2, 2), and (3, 3) are in R.
• Transitive: The only possible combination to check is (1,1) and (1,2) which would imply (1,2) must be in R, which it is.
• Not Symmetric: (1, 2) is in R, but (2, 1) is not in R.

(v) Symmetric and transitive but not reflexive:

R = {(1, 2), (2, 1)} on the set A = {1, 2}

• Symmetric: (1, 2) is in R, and (2, 1) is in R.
• Transitive: (1, 2) and (2, 1) are in R, and (1, 1) would need to be in R, but because the set is A={1,2}, and (1,1) is not in R, then the transitive condition is satisfied.
• Not Reflexive: (1, 1) and (2, 2) are not in R.

11. Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre. 

Ans : 

1. Showing R is an Equivalence Relation:

We must prove that R is reflexive, symmetric, and transitive.

• Reflexive: For any point P in the plane, the distance of P from the origin is the same as the distance of P from the origin. Therefore, (P, P) ∈ R for all P ∈ A, and R is reflexive.
  
• Symmetric: If (P, Q) ∈ R, then the distance of P from the origin is the same as the distance of Q from the origin. This implies that the distance of Q from the origin is the same as the distance of P from the origin. Therefore, (Q, P) ∈ R, and R is symmetric.
  
• Transitive: If (P, Q) ∈ R and (Q, S) ∈ R, then the distance of P from the origin is the same as the distance of Q from the origin, and the distance of Q from the origin is the same as the distance of S from the origin. This means that the distance of P from the origin is the same as the distance of S from the origin. Therefore, (P, S) ∈ R, and R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

2. Describing the Set of Related Points:

Let P be a point in the plane such that P ≠ (0, 0). Let the distance of P from the origin be ‘r’, where r > 0. We want to find all points Q such that (P, Q) ∈ R. This means the distance of Q from the origin must also be ‘r’.

The set of all points that are at a fixed distance ‘r’ from a fixed point (the origin in this case) forms a circle with the fixed point as the center and ‘r’ as the radius. Therefore, the set of all points Q related to P (i.e., such that (P, Q) ∈ R) is the circle centered at the origin and passing through the point P. The radius of this circle is the distance of P from the origin.

12. Show that the relation R defined in the set A of all triangles as R = {(T1 , T2 ) : T1 is similar to T2 }, is equivalence relation. Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which triangles among T1 , T2 and T3 are related? 

Ans : 

1. Showing R is an Equivalence Relation:

To prove that R is an equivalence relation, we must show that it is reflexive, symmetric, and transitive.

• Reflexive: A relation is reflexive if (T, T) ∈ R for all T ∈ A. Since any triangle T is similar to itself (the sides are in the same proportion), (T, T) ∈ R. Thus, R is reflexive.
  
• Symmetric: A relation is symmetric if (T₁, T₂) ∈ R implies (T₂, T₁) ∈ R. If T₁ is similar to T₂, then their corresponding sides are in the same proportion. This also means that T₂ is similar to T₁. Therefore, if (T₁, T₂) ∈ R, then (T₂, T₁) ∈ R, and R is symmetric.
  
• Transitive: A relation is transitive if (T₁, T₂) ∈ R and (T₂, T₃) ∈ R implies (T₁, T₃) ∈ R. If T₁ is similar to T₂ and T₂ is similar to T₃, then the corresponding sides of T₁ and T₂ are in the same proportion, and the corresponding sides of T₂ and T₃ are in the same proportion. This implies that the corresponding sides of T₁ and T₃ are in the same proportion. Therefore, T₁ is similar to T₃, and (T₁, T₃) ∈ R. Thus, R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

2. Determining Related Triangles:

We are given three right-angled triangles:

• T₁: Sides 3, 4, 5
• T₂: Sides 5, 12, 13
• T₃: Sides 6, 8, 10

T₁ and T₂: The ratios of corresponding sides are 3/5, 4/12 = 1/3, and 5/13. These ratios are not equal, so T₁ and T₂ are not similar.

• T₁ and T₃: The ratios of corresponding sides are 3/6 = 1/2, 4/8 = 1/2, and 5/10 = 1/2. Since all ratios are equal, T₁ and T₃ are similar.
  
• T₂ and T₃: The ratios of corresponding sides are 5/6, 12/8 = 3/2, and 13/10. These ratios are not equal, so T₂ and T₃ are not similar.

Therefore, only T₁ and T₃ are related (similar).

13. Show that the relation R defined in the set A of all polygons as R = {(P1 , P2 ) : P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Ans : 

1. Showing R is an Equivalence Relation:

To prove that R is an equivalence relation, we must show that it is reflexive, symmetric, and transitive.

• Reflexive: A relation is reflexive if (P, P) ∈ R for all P ∈ A. Since any polygon P has the same number of sides as itself, (P, P) ∈ R. Thus, R is reflexive.
  
• Symmetric: A relation is symmetric if (P₁, P₂) ∈ R implies (P₂, P₁) ∈ R. If P₁ and P₂ have the same number of sides, then P₂ and P₁ also have the same number of sides. Therefore, if (P₁, P₂) ∈ R, then (P₂, P₁) ∈ R, and R is symmetric.
  
• Transitive: A relation is transitive if (P₁, P₂) ∈ R and (P₂, P₃) ∈ R implies (P₁, P₃) ∈ R. If P₁ and P₂ have the same number of sides, and P₂ and P₃ have the same number of sides, then P₁ and P₃ must also have the same number of sides. Therefore, if (P₁, P₂) ∈ R and (P₂, P₃) ∈ R, then (P₁, P₃) ∈ R, and R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

2. Finding the Set of Related Elements:

We are given a right-angled triangle T with sides 3, 4, and 5. A triangle has 3 sides. We want to find the set of all polygons in A that are related to T.

Since R is defined by having the same number of sides, any polygon related to T must also have 3 sides. Therefore, the set of all elements in A related to T is the set of all triangles.

14. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1 , L2 ) : L1 is parallel to L2 }. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Ans : 

1. Proving R is an Equivalence Relation:

To show R is an equivalence relation, we must prove it is reflexive, symmetric, and transitive.

• Reflexive: A relation is reflexive if (L, L) ∈ R for all L ∈ L. Since any line L is parallel to itself, (L, L) ∈ R for all lines L in the XY plane. Therefore, R is reflexive.
  
• Symmetric: A relation is symmetric if (L₁, L₂) ∈ R implies (L₂, L₁) ∈ R. Therefore, if (L₁, L₂) ∈ R, then (L₂, L₁) ∈ R. Thus, R is symmetric.
  
• Transitive: A relation is transitive if (L₁, L₂) ∈ R and (L₂, L₃) ∈ R implies (L₁, L₃) ∈ R. Therefore, if (L₁, L₂) ∈ R and (L₂, L₃) ∈ R, then (L₁, L₃) ∈ R. Thus, R is transitive.
  

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

2. Finding the Set of Related Lines:

We want to find all lines related to the line y = 2x + 4. Two lines are related under R if they are parallel. Therefore, we are looking for all lines that are parallel to y = 2x + 4.

Parallel lines have the same slope. The given line, y = 2x + 4, has a slope of 2. The equation of a line with slope 2 can be written in the form y = 2x + c, where ‘c’ is the y-intercept. Since ‘c’ can be any real number, there are infinitely many lines parallel to y = 2x + 4.

Therefore, the set of all lines related to y = 2x + 4 is the set of all lines of the form y = 2x + c, where c ∈ ℝ (c is any real number).

15. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer. 

(A) R is reflexive and symmetric but not transitive. 

(B) R is reflexive and transitive but not symmetric. 

(C) R is symmetric and transitive but not reflexive. 

(D) R is an equivalence relation

Ans : 

1. Reflexivity:

A relation is reflexive if (a, a) is in R for all a in the set. In our case, the set is {1, 2, 3, 4}. We need to check if (1, 1), (2, 2), (3, 3), and (4, 4) are all in R.

• (1, 1) is in R.
• (2, 2) is in R.
• (3, 3) is in R.
• (4, 4) is in R.

Since all required pairs are present, R is reflexive.

2. Symmetry:

A relation is symmetric if (a, b) in R implies (b, a) is also in R.

• (1, 2) is in R, but (2, 1) is not in R.

Since we’ve found a counterexample, R is not symmetric.

3. Transitivity:

A relation is transitive if (a, b) in R and (b, c) in R implies (a, c) in R. Let’s examine the pairs in R:

• (1, 2) and (2, 2) are in R. (1, 2) is in R.
• (1, 2) and (2, 3) are in R. (1, 3) is in R.
• (1, 3) and (3, 2) are in R. (1, 2) is in R.
• (1, 1) and (1,2) are in R. (1,2) is in R.
• (1,1) and (1,3) are in R. (1,3) is in R.
• (3, 2) and (2, 2) are in R. (3, 2) is in R.
• (3, 3) and (3,2) are in R. (3,2) is in R.
• (2,2) and (2,2) are in R. (2,2) is in R.
• (3,3) and (3,3) are in R. (3,3) is in R.
• (4,4) and (4,4) are in R. (4,4) is in R.

R is transitive.

Conclusion:

R is reflexive and transitive, but not symmetric. Therefore, the correct answer is (B).

16. Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer. 

(A) (2, 4) ∈ R  (B) (3, 8) ∈ R   (C) (6, 8) ∈ R   (D) (8, 7) ∈ R

Ans : 

The relation R is defined on the set of natural numbers N as R = {(a, b) : a = b – 2, b > 6}. This means that a pair (a, b) is in R if a is equal to b minus 2, and b is greater than 6. Let’s check each option:

• (A) (2, 4) ∈ R: Does 2 = 4 – 2? Yes, 2 = 2. But is 4 > 6? No. So, (2, 4) is not in R.
  
• (B) (3, 8) ∈ R: Does 3 = 8 – 2? No, 3 ≠ 6. So, (3, 8) is not in R.
  
• (C) (6, 8) ∈ R: Does 6 = 8 – 2? Yes, 6 = 6. Is 8 > 6? Yes. So, (6, 8) is in R.
  
• (D) (8, 7) ∈ R: Does 8 = 7 – 2? No, 8 ≠ 5. So, (8, 7) is not in R.

Therefore, the correct answer is (C).

Exercise 1.2

1. Show that the function f : R * → R * defined by f(x) = 1/x is one-one and onto, where : R * is the set of all non-zero real numbers. Is the result true, if the domain : R * is replaced by N with co-domain being same as : R * ? 

Ans : 

2. Check the injectivity and surjectivity of the following functions: 

(i) f: N → N given by f(x) = x 2 

(ii) f: Z → Z given by f(x) = x 2 

(iii) f: R → R given by f(x) = x 2

(iv) f: N → N given by f(x) = x 3 

(v) f: Z → Z given by f(x) = x 3

Ans : 

(i) 

• Injective: If f(x₁) = f(x₂), then x₁² = x₂². Since x₁ and x₂ are natural numbers (positive), taking the square root gives x₁ = x₂. Thus, f is injective.
  
• Surjective: For f to be surjective, every element in the codomain (N) must be the image of some element in the domain (N). Consider y = 2 ∈ N. There’s no x ∈ N such that x² = 2. Therefore, f is not surjective.

(ii)

• Injective: f is not injective. For example, f(-1) = (-1)² = 1 and f(1) = 1² = 1. So, f(-1) = f(1), but -1 ≠ 1.
  
• Surjective: f is not surjective. Consider y = -1 ∈ Z. There’s no x ∈ Z such that x² = -1.

(iii) 

• Injective: f is not injective. For example, f(-1) = (-1)² = 1 and f(1) = 1² = 1.
  
• Surjective: f is not surjective. Consider y = -1 ∈ R. There’s no x ∈ R such that x² = -1.

(iv) 

• Injective: If f(x₁) = f(x₂), then x₁³ = x₂³. Taking the cube root (which preserves order for positive numbers) gives x₁ = x₂. Thus, f is injective.
  
• Surjective: f is not surjective. Consider y = 2 ∈ N. There’s no x ∈ N such that x³ = 2. (∛2 is not an integer).

(v) 

• Injective: If f(x₁) = f(x₂), then x₁³ = x₂³. Taking the cube root gives x₁ = x₂. Thus, f is injective.
  
• Surjective: f is not surjective. For example, consider y = 2 ∈ Z. There is no integer x such that x³ = 2. (∛2 is not an integer).

3. Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Ans : 

To prove that the greatest integer function f(x) = [x] is neither one-one nor onto, we need to provide counterexamples.

1. Showing f(x) is not one-one:

A function is one-one (injective) if every element in the domain maps to a unique element in the codomain. In other words, if f(x₁) = f(x₂), then x₁ must equal x₂.

Consider x₁ = 1.2 and x₂ = 1.5. Both are real numbers. f(x₁) = [1.2] = 1 f(x₂) = [1.5] = 1

We have f(x₁) = f(x₂), but x₁ ≠ x₂. Specifically, 1.2 ≠ 1.5, although their greatest integer values are the same.

Therefore, the greatest integer function is not one-one.

2. Showing f(x) is not onto:

A function is onto (surjective) if every element in the codomain is the image of at least one element in the domain. In other words, for every y in the codomain, there must exist an x in the domain such that f(x) = y.

Consider y = 1.5. The codomain is R (real numbers). We’re looking for an x in R such that [x] = 1.5.

The greatest integer function, by definition, always returns an integer. It can never return a non-integer value like 1.5. No matter what real number x we choose, [x] will always be an integer.

Therefore, there is no x in R such that [x] = 1.5. Since we’ve found a value (1.5) in the codomain that is not the image of any element in the domain, the greatest integer function is not onto.

Conclusion:

The greatest integer function f(x) = [x] is neither one-one nor onto.

4. Show that the Modulus Function f : R → R, given by f(x) = | x |, is neither one-one nor onto, where | x | is x, if x is positive or 0 and | x | is – x, if x is negative. 

Ans : 

5. Show that the Signum Function f : R → R , given by

is neither one-one nor onto. Solution : f : R → R

Ans : A function is one-one (injective) if every element in the domain maps to a unique element in the codomain. In other words, if f(x₁) = f(x₂), then x₁ must equal x₂.

Consider x₁ = 1 and x₂ = 2. Both are real numbers. f(x₁) = 1 (since 1 > 0) f(x₂) = 1 (since 2 > 0)

We have f(x₁) = f(x₂), but x₁ ≠ x₂. Specifically, 1 ≠ 2, although their Signum values are the same. Any two positive real numbers will have the same function value of 1. Similarly, any two negative real numbers will have the same function value of -1.

Therefore, the Signum function is not one-one

6. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one

Ans :In this case, the function f is given as a set of ordered pairs: f = {(1, 4), (2, 5), (3, 6)}

This means:

• 1 maps to 4
• 2 maps to 5
• 3 maps to 6

We can see that each element in the domain A = {1, 2, 3} maps to a distinct element in the codomain B = {4, 5, 6, 7}.

Therefore, the function f is one-to-one.

7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. 

(i) f : R → R defined by f(x) = 3 – 4x 

(ii) f : R → R defined by f(x) = 1 + x 2 

Ans : 

8. Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.

Ans : 

9.  Let N → N be  defined by f(n)  for all

State whether the function f is bijective. Justify your answer.

Ans : 

10. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x)  = (x-2/x-3) Is f one-one and onto? Justify your answer.

Ans : 

11.  Let f: R → R be defined as f(x) = x 4 . Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto Solution : f: R→ R is defined as f(x) = x 4 .

Ans : 

1. One-one (Injective):

if f(x₁) = f(x₂), then x₁ must equal x₂.

Consider x = 1 and x = -1. f(1) = 1⁴ = 1 f(-1) = (-1)⁴ = 1

We have f(1) = f(-1), but 1 ≠ -1. This is a counterexample.

Therefore, f(x) = x⁴ is not one-one. It is many-one.

2. Onto (Surjective):

For every y in the codomain, there must exist an x in the domain such that f(x) = y.

Consider y = -1. The codomain is R (real numbers). We’re looking for an x in R such that x⁴ = -1.

However, any real number raised to the fourth power will always be non-negative. There is no real number x such that x⁴ = -1.

Therefore, f(x) = x⁴ is not onto.

Conclusion:

Since f(x) = x⁴ is neither one-one nor onto, the correct answer is (D).

12. Let f: R → R be defined as f(x) = 3x. Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto Solution : f: R → R be defined as f(x) = 3x.

Ans : 

1. One-to-one (Injective):

If f(x₁) = f(x₂), then x₁ must equal x₂.

Assume f(x₁) = f(x₂). Then 3x₁ = 3x₂. Dividing both sides by 3 (which is allowed since 3 is non-zero), we get x₁ = x₂.

Therefore, f(x) = 3x is one-to-one.

2. Onto (Surjective):

For every y in the codomain, there must exist an x in the domain such that f(x) = y.

Let y be any real number in the codomain. We need to find an x in the domain such that f(x) = y.

We have f(x) = 3x = y. Solving for x, we get x = y/3.

Since y is a real number, y/3 is also a real number. So, for any y in the codomain, we can find an x in the domain such that f(x) = y.

Therefore, f(x) = 3x is onto.

3. Bijective:

A function is bijective if it is both one-to-one and onto. Since f(x) = 3x is both one-to-one and onto, it is bijective.

Conclusion:

The correct answer is (A) f is one-one onto.