RELATIONS AND FUNCTIONS

Exercise 1.1

Q1: 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}

Answers:

(i)
R = {(x, y) : 3x – y = 0} ⟹ y = 3x
Reflexive: No, because (x, x) ∈ R ⇒ 3x – x = 0 ⇒ 2x = 0 ⇒ x = 0, but 0 ∉ A.
Symmetric: No, since if (x, y) ∈ R ⇒ y = 3x, then (y, x) ∈ R ⇒ x = 3y ⇒ not true in general.
Transitive: Yes, because if y = 3x and z = 3y ⇒ z = 3(3x) = 9x ⇒ (x, z) ∉ R unless x = 0. Hence not transitive.
∴ Relation is neither reflexive, nor symmetric, nor transitive.

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

Possible pairs: (1,6), (2,7), (3,8).
Reflexive: No, because (x, x) ∉ R.
Symmetric: No, because if (1,6) ∈ R, then (6,1) ∉ R.
Transitive: No, because no chain like (x, y) and (y, z) exists.
∴ Relation is neither reflexive, nor symmetric, nor transitive.

(iii)
R = {(x, y) : y is divisible by x}

Reflexive: Yes, since every number divides itself.
Symmetric: No, because if y is divisible by x, x need not be divisible by y.
Transitive: Yes, since if y is divisible by x and z is divisible by y, then z is divisible by x.
∴ Relation is reflexive and transitive, but not symmetric.

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

Since x and y are integers, x – y is always an integer.
∴ (x, y) ∈ R for all x, y ∈ Z.
Reflexive: Yes, because x – x = 0 (integer).
Symmetric: Yes, because if x – y is integer, then y – x = –(x – y) is also integer.
Transitive: Yes, because if x – y and y – z are integers, then (x – z) = (x – y) + (y – z) is also integer.
∴ Relation is reflexive, symmetric and transitive.

(v)

(a) R = {(x, y) : x and y work at the same place}
Reflexive – Yes
Symmetric – Yes
Transitive – Yes
∴ Equivalence relation

(b) R = {(x, y) : x and y live in the same locality}
Reflexive – Yes
Symmetric – Yes
Transitive – Yes
∴ Equivalence relation

(c) R = {(x, y) : x is exactly 7 cm taller than y}
Reflexive – No
Symmetric – No
Transitive – No
∴ Neither reflexive, symmetric, nor transitive

(d) R = {(x, y) : x is wife of y}
Reflexive – No
Symmetric – No (if x is wife of y, y is husband of x)
Transitive – No
∴ Neither reflexive, symmetric, nor transitive

(e) R = {(x, y) : x is father of y}
Reflexive – No
Symmetric – No
Transitive – No (if x is father of y and y is father of z, x is grandfather of z)
∴ Neither reflexive, symmetric, nor transitive

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Q2: Show that the relation R in the set R of real numbers, defined as
R = {(a, b) : a ≤ b²} is neither reflexive nor symmetric nor transitive.

Answer:

Reflexive:
For reflexivity, (a, a) must belong to R for all real numbers a.
That means a ≤ a² for all a.
Take a = ½. Then ½ ≤ (½)² = ¼, which is false.
∴ R is not reflexive.

Symmetric:
For symmetry, if (a, b) ∈ R, then (b, a) must also belong to R.
That means if a ≤ b², then b ≤ a² should also hold.
Take a = 0 and b = 1. Then 0 ≤ 1² is true, but 1 ≤ 0² = 0 is false.
∴ R is not symmetric.

Transitive:
For transitivity, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) must also belong to R.
That means if a ≤ b² and b ≤ c², then a ≤ c² should hold.
Take c = 0, b = –1, and a = 0.9.
Then a ≤ b² ⇒ 0.9 ≤ (–1)² = 1 (true)
and b ≤ c² ⇒ –1 ≤ 0² = 0 (true),
but a ≤ c² ⇒ 0.9 ≤ 0 (false).
∴ R is not transitive.

Conclusion:
The relation R is neither reflexive, nor symmetric, nor transitive.

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Q3: 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.

Answer:

Reflexive:
For reflexivity, (a, a) must belong to R for all a.
That means a = a + 1, which is never true for any a.
∴ R is not reflexive.

Symmetric:
For symmetry, if (a, b) ∈ R, then (b, a) must also belong to R.
Here, (a, b) ∈ R means b = a + 1.
For (b, a) to be in R, we need a = b + 1, which is not true.
Example: (1, 2) ∈ R, but (2, 1) ∉ R.
∴ R is not symmetric.

Transitive:
For transitivity, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) must belong to R.
(a, b) ∈ R ⇒ b = a + 1
(b, c) ∈ R ⇒ c = b + 1 = (a + 1) + 1 = a + 2
So (a, c) would mean c = a + 1, which is false because c = a + 2.
Example: (1, 2) and (2, 3) ∈ R, but (1, 3) ∉ R.
∴ R is not transitive.

Conclusion:
The relation R is neither reflexive, nor symmetric, nor transitive.

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Q4: Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.

Answer:

Reflexive
For every real number a, a ≤ a holds.
Thus (a, a) ∈ R for all a.
So R is reflexive.

Symmetric
If (a, b) ∈ R then a ≤ b. Symmetry would require b ≤ a.
Counterexample: a = 0, b = 1. 0 ≤ 1 is true, but 1 ≤ 0 is false.
So R is not symmetric.

Transitive
If (a, b) ∈ R and (b, c) ∈ R then a ≤ b and b ≤ c. By transitivity of ≤ we get a ≤ c.
Hence (a, c) ∈ R for all such a, b, c.
So R is transitive.

Conclusion: R is reflexive and transitive but not symmetric.

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Q5: Check whether the relation R in R defined by
R = {(a, b) : a ≤ b³} is reflexive, symmetric or transitive.

Answer:

Reflexive:
For reflexivity, (a, a) must belong to R for all real numbers a.
That means a ≤ a³ for all a.
Check some values:

• For a = 1 → 1 ≤ 1³ = 1 (true)

• For a = 0 → 0 ≤ 0³ = 0 (true)

• For a = ½ → ½ ≤ (½)³ = ⅛ (false)
  Since it is not true for all a, R is not reflexive.

Symmetric:
For symmetry, if (a, b) ∈ R, then (b, a) must also belong to R.
That means if a ≤ b³, then b ≤ a³ should also hold.
Example: a = 1, b = 2 → 1 ≤ 8 (true), but 2 ≤ 1³ = 1 (false).
∴ R is not symmetric.

Transitive:
For transitivity, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) must belong to R.
That means if a ≤ b³ and b ≤ c³, then a ≤ c³ should hold.
Example: let a = 2, b = 1, c = 2.
a ≤ b³ → 2 ≤ 1³ = 1 (false), so choose valid ones:
Try a = 0.5, b = 1, c = 2.
a ≤ b³ → 0.5 ≤ 1 (true)
b ≤ c³ → 1 ≤ 8 (true)
a ≤ c³ → 0.5 ≤ 8 (true) (so works here)
But check a = 1, b = –1, c = 0.
a ≤ b³ → 1 ≤ (–1)³ = –1 (false).
Try other valid pair: a = –1, b = 0, c = 1.
a ≤ b³ → –1 ≤ 0 (true)
b ≤ c³ → 0 ≤ 1 (true)
a ≤ c³ → –1 ≤ 1 (true).
Though it works in many cases, not always true for all real numbers.
Example: a = 2, b = –1, c = 1
a ≤ b³ → 2 ≤ (–1)³ = –1 (false) ⇒ shows failure cases exist.
Hence relation is not transitive in general.

Conclusion:
The relation R is neither reflexive, nor symmetric, nor transitive.

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Q6: 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.

Answer:

Reflexive:
For reflexivity, (a, a) must belong to R for all a in {1, 2, 3}.
But (1,1), (2,2), (3,3) are not in R.
∴ R is not reflexive.

Symmetric:
For symmetry, if (a, b) ∈ R, then (b, a) must also belong to R.
Here, (1, 2) ∈ R and (2, 1) ∈ R.
∴ Every pair has its reverse pair in R.
So R is symmetric.

Transitive:
For transitivity, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) must belong to R.
Here, (1, 2) ∈ R and (2, 1) ∈ R, so (1, 1) should be in R for transitivity,
but (1, 1) ∉ R.
∴ R is not transitive.

Conclusion:
The relation R is symmetric, but neither reflexive nor transitive.

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Q7: 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 the same number of pages} is an equivalence relation.

Answer:

Reflexive:
Every book has the same number of pages as itself.
So, for every book x, (x, x) ∈ R.
∴ R is reflexive.

Symmetric:
If (x, y) ∈ R, then book x and book y have the same number of pages.
This means book y and book x also have the same number of pages.
∴ (y, x) ∈ R.
So, R is symmetric.

Transitive:
If (x, y) ∈ R and (y, z) ∈ R,
then x and y have the same number of pages, and y and z have the same number of pages.
Therefore, x and z also have the same number of pages.
So, (x, z) ∈ R.
∴ R is transitive.

Conclusion:
Since the relation R is reflexive, symmetric, and transitive,
R is an equivalence relation.

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Q8: 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}.

Answer:

Reflexive
For every a in A, |a − a| = 0, and 0 is even.
So (a, a) ∈ R for all a.
Therefore R is reflexive.

Symmetric
If (a, b) ∈ R then |a − b| is even. But |b − a| = |a − b|, so |b − a| is even.
Hence (b, a) ∈ R.
Therefore R is symmetric.

Transitive
Assume (a, b) ∈ R and (b, c) ∈ R. Then |a − b| and |b − c| are even.
Even numbers have the form 2k, 2m, so |a − b| = 2k and |b − c| = 2m.
Then |a − c| = |(a − b) + (b − c)| = |2k + 2m| = 2(k + m), which is even.
Thus (a, c) ∈ R.
Therefore R is transitive.

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

Relation among elements by parity
Numbers 1, 3, 5 are odd; 2, 4 are even.

• For any x, y ∈ {1, 3, 5}, both x and y are odd so |x − y| is even. Thus every pair in {1, 3, 5} is related.

• For any x, y ∈ {2, 4}, both x and y are even so |x − y| is even. Thus every pair in {2, 4} is related.

• If x is odd (from {1, 3, 5}) and y is even (from {2, 4}), then |x − y| is odd, so x is not related to y.

Hence the equivalence classes are {1, 3, 5} and {2, 4}, with all elements within each class related and no cross-relations between the classes.

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Q9: Show that each of the relations 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.

Answer:

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

• Reflexive: Yes, because |a – a| = 0, which is a multiple of 4.

• Symmetric: Yes, because |a – b| = |b – a|.

• Transitive: Yes, because if |a – b| and |b – c| are multiples of 4, then |a – c| is also a multiple of 4.

Equivalence relation: Yes.

Elements related to 1:
Numbers x such that |x – 1| is a multiple of 4 in A = {0, 1, …, 12}
1 + 0 = 1, 1 + 4 = 5, 1 + 8 = 9 → {1, 5, 9}

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

• Reflexive: Yes, a = a

• Symmetric: Yes, if a = b then b = a

• Transitive: Yes, if a = b and b = c then a = c

Equivalence relation: Yes.

Elements related to 1:
Only 1 itself → {1}

Final Answer:
(i) {1, 5, 9}
(ii) {1}

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Q10: 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.

Answer:

(i) Symmetric but neither reflexive nor transitive
Set: A = {1, 2, 3}
R = {(1, 2), (2, 1)}

• Symmetric: Yes, (1,2) ↔ (2,1)

• Reflexive: No, (1,1), (2,2), (3,3) ∉ R

• Transitive: No, (1,2) and (2,1) ∈ R, but (1,1) ∉ R

(ii) Transitive but neither reflexive nor symmetric
Set: A = {1, 2, 3}
R = {(1, 2), (2, 3), (1, 3)}

• Transitive: Yes, (1,2) & (2,3) ⇒ (1,3) ∈ R

• Reflexive: No, (1,1), (2,2), (3,3) ∉ R

• Symmetric: No, e.g., (1,2) ∈ R but (2,1) ∉ R

(iii) Reflexive and symmetric but not transitive
Set: A = {1, 2, 3}
R = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}

• Reflexive: Yes, all (a,a) ∈ R

• Symmetric: Yes, all pairs have their reverse

• Transitive: No, (1,2) & (2,3) ∈ R, but (1,3) ∉ R

(iv) Reflexive and transitive but not symmetric
Set: A = {1, 2, 3}
R = {(1,1), (2,2), (3,3), (1,2)}

• Reflexive: Yes, all (a,a) ∈ R

• Transitive: Yes, no chain violates transitivity

• Symmetric: No, (1,2) ∈ R but (2,1) ∉ R

(v) Symmetric and transitive but not reflexive
Set: A = {1, 2, 3}
R = {(1,1), (2,2)}

• Symmetric: Yes, all pairs have reverse

• Transitive: Yes, no chain violates transitivity

• Reflexive: No, (3,3) ∉ R

Final Answer (examples only):
(i) {(1,2), (2,1)}
(ii) {(1,2), (2,3), (1,3)}
(iii) {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}
(iv) {(1,1),(2,2),(3,3),(1,2)}
(v) {(1,1),(2,2)}

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Q11: 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.

Solution:

Let A be the set of all points in the plane.
Let P(x₁, y₁) and Q(x₂, y₂) be any two points in A, and let O(0, 0) be the origin.

For any point R(x, y), its distance from the origin O is
d(R, O) = √(x² + y²).

Define the relation R on A as follows:
(P, Q) ∈ R if and only if d(P, O) = d(Q, O).

1. Reflexive:
For every point P in A, we have d(P, O) = d(P, O).
Therefore, (P, P) ∈ R for all P.
Hence, R is reflexive.

2. Symmetric:
If (P, Q) ∈ R, then d(P, O) = d(Q, O).
Since equality is symmetric, d(Q, O) = d(P, O).
Therefore, (Q, P) ∈ R.
Hence, R is symmetric.

3. Transitive:
If (P, Q) ∈ R and (Q, R) ∈ R, then
d(P, O) = d(Q, O) and d(Q, O) = d(R, O).
By the transitive property of equality, d(P, O) = d(R, O).
Therefore, (P, R) ∈ R.
Hence, R is transitive.

Since the relation R is reflexive, symmetric, and transitive,
therefore R is an equivalence relation on A.

4. Description of the equivalence class:

Let P₀(a, b) be a fixed point in A.
The equivalence class [P₀] = { Q ∈ A : (Q, P₀) ∈ R }
is the set of all points Q whose distance from O equals the distance of P₀ from O.

Let the distance of P₀ from O be
r = d(P₀, O) = √(a² + b²).

If P₀ ≠ (0, 0), then r > 0.
The condition d(Q, O) = r means
√(x² + y²) = r
⇒ x² + y² = r².

This is the equation of a circle with centre at the origin and radius r.

Hence, for any P₀ ≠ (0, 0), the set of all points related to P₀ is the circle passing through P₀ with the origin as centre.

Remark:
If P₀ = (0, 0), then r = 0, and the equivalence class of the origin is the single point {(0, 0)}.

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Q12: Show that the relation R defined in the set A of all triangles as R = {(T₁, T₂) : T₁ is similar to T₂}, is an equivalence relation. Consider three right-angled 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?

Solution:

Let A be the set of all triangles.
The relation R is defined on A as follows:
(T₁, T₂) ∈ R if and only if T₁ is similar to T₂.

1. Reflexive:
Every triangle is similar to itself.
Therefore, (T, T) ∈ R for every triangle T in A.
Hence, R is reflexive.

2. Symmetric:
If (T₁, T₂) ∈ R, then T₁ is similar to T₂.
Since similarity of triangles is a symmetric property, T₂ is also similar to T₁.
Therefore, (T₂, T₁) ∈ R.
Hence, R is symmetric.

3. Transitive:
If (T₁, T₂) ∈ R and (T₂, T₃) ∈ R, then T₁ is similar to T₂ and T₂ is similar to T₃.
By the transitive property of similarity, T₁ is similar to T₃.
Therefore, (T₁, T₃) ∈ R.
Hence, R is transitive.

Since R is reflexive, symmetric, and transitive,
therefore R is an equivalence relation on the set A of all triangles.

Now, consider the given triangles:

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

All are right-angled triangles.

To check which triangles are similar, compare the ratios of their corresponding sides.

For T₁ and T₂:
Ratios of corresponding sides are
3/5, 4/12, 5/13
= 3/5, 1/3, and 5/13

These ratios are not equal.
Therefore, T₁ and T₂ are not similar.

For T₁ and T₃:
Ratios of corresponding sides are
3/6, 4/8, 5/10
= 1/2, 1/2, and 1/2

All ratios are equal.
Therefore, T₁ and T₃ are similar.

For T₂ and T₃:
Ratios of corresponding sides are
5/6, 12/8, 13/10
= 0.83, 1.5, and 1.3 (approximately)

These are not equal.
Therefore, T₂ and T₃ are not similar.

Conclusion:
The relation R is an equivalence relation.
Among the given triangles, T₁ and T₃ are related because they are similar triangles.

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Q13: Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂) : P₁ and P₂ have the same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right-angled triangle T with sides 3, 4 and 5?

Solution:

Let A be the set of all polygons.

Define the relation R on A by:
(P₁, P₂) ∈ R if and only if the number of sides of P₁ equals the number of sides of P₂.

1. Reflexive:
   Every polygon has the same number of sides as itself.
   Therefore (P, P) ∈ R for every polygon P in A.
   Hence R is reflexive.

2. Symmetric:
   If (P₁, P₂) ∈ R, then P₁ and P₂ have the same number of sides.
   Equality of the number of sides is symmetric, so P₂ and P₁ have the same number of sides.
   Therefore (P₂, P₁) ∈ R.
   Hence R is symmetric.

3. Transitive:
   If (P₁, P₂) ∈ R and (P₂, P₃) ∈ R, then P₁ and P₂ have the same number of sides, and P₂ and P₃ have the same number of sides.
   Thus P₁ and P₃ have the same number of sides.
   Therefore (P₁, P₃) ∈ R.
   Hence R is transitive.

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

Equivalence class of the given triangle T (sides 3, 4, 5):

The triangle T is a polygon with 3 sides. The set of all polygons related to T is the set of all polygons that also have 3 sides. In other words, the equivalence class of T is the set of all triangles (this includes all types of triangles: equilateral, isosceles, scalene, acute, obtuse, right-angled, etc.).

Therefore, the set of all elements in A related to T is the set of all triangles.

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Q14: Let L be the set of all lines in the 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.

Solution:

Let L be the set of all lines in the XY-plane.
Define a relation R on L as follows:
(L₁, L₂) ∈ R if and only if L₁ is parallel to L₂.

1. Reflexive:
Every line is parallel to itself.
Therefore, (L, L) ∈ R for every line L in L.
Hence, R is reflexive.

2. Symmetric:
If (L₁, L₂) ∈ R, then L₁ is parallel to L₂.
The relation “is parallel to” is symmetric, so L₂ is also parallel to L₁.
Therefore, (L₂, L₁) ∈ R.
Hence, R is symmetric.

3. Transitive:
If (L₁, L₂) ∈ R and (L₂, L₃) ∈ R, then L₁ is parallel to L₂ and L₂ is parallel to L₃.
Therefore, L₁ is parallel to L₃.
Hence, (L₁, L₃) ∈ R.
Thus, R is transitive.

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

Finding the set of all lines related to y = 2x + 4:

The given line is
y = 2x + 4.

This line has slope m = 2.

Two lines are parallel if and only if they have the same slope.
Therefore, all lines parallel to y = 2x + 4 will have slope 2.

The general equation of a line with slope 2 is
y = 2x + c, where c is any real number.

Thus, the set of all lines related to y = 2x + 4 is
{ y = 2x + c : c ∈ ℝ }.

Hence, the set of all lines related to y = 2x + 4 consists of all lines having slope 2, i.e., all lines parallel to it.

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Q15: 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

Let A = {1, 2, 3, 4} and
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}.

1. Reflexive:
To be reflexive we need (1,1), (2,2), (3,3), (4,4) in R.
All four appear in R.
Therefore R is reflexive.

2. Symmetric:
R would be symmetric if whenever (a,b) ∈ R then (b,a) ∈ R.
But (1,2) ∈ R while (2,1) ∉ R.
Therefore R is not symmetric.

3. Transitive:
We must check: whenever (a,b) ∈ R and (b,c) ∈ R, then (a,c) ∈ R.

Check relevant compositions:

· (1,2) and (2,2) ⇒ (1,2) which is in R.

· (1,3) and (3,2) ⇒ (1,2) which is in R.

· (3,2) and (2,2) ⇒ (3,2) which is in R.

· (1,1) with (1,2) ⇒ (1,2) (in R); (1,1) with (1,3) ⇒ (1,3) (in R).

· (2,2) with (2,2) ⇒ (2,2) (in R).

· (3,3) with (3,2) ⇒ (3,2) (in R).

· (4,4) only composes to (4,4) (in R).

No required transitive consequence is missing. Hence R is transitive.

Conclusion: R is reflexive and transitive but not symmetric. The correct option is (B).

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Q16: 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

Solution:

The relation R is defined as:
(a, b) ∈ R if and only if

1. a = b – 2, and

2. b > 6.

Now, check each option:

(A) (2, 4):
Here a = 2, b = 4.
Check: a = b – 2 → 2 = 4 – 2 → 2 = 2 (True)
But b > 6 → 4 > 6 (False).
So (2, 4) ∉ R.

(B) (3, 8):
a = 3, b = 8.
Check: a = b – 2 → 3 = 8 – 2 → 3 = 6 (False).
So (3, 8) ∉ R.

(C) (6, 8):
a = 6, b = 8.
Check: a = b – 2 → 6 = 8 – 2 → 6 = 6 (True)
and b > 6 → 8 > 6 (True).
Both conditions are true.
So (6, 8) ∈ R.

(D) (8, 7):
a = 8, b = 7.
Check: a = b – 2 → 8 = 7 – 2 → 8 = 5 (False).
So (8, 7) ∉ R.

Correct answer: (C) (6, 8) ∈ R

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Exercise 1.2

Q1: 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*?

Solution:

Let R* be the set of all non-zero real numbers. Define f : R* → R* by f(x) = 1/x.

1. One-one (injective):
Assume f(x1) = f(x2). Then 1/x1 = 1/x2. Multiplying both sides by x1 x2 (non-zero) gives x2 = x1. Hence x1 = x2. Therefore f is one-one.

2. Onto (surjective):
Let y be any element of R*. We must find x in R* such that f(x) = y. Solve 1/x = y. Then x = 1/y, which is a non-zero real number since y ≠ 0. Thus for every y ∈ R* there exists x = 1/y ∈ R* with f(x) = y. Therefore f is onto.

Thus f is both one-one and onto (a bijection).

3. If domain is replaced by N (natural numbers) with co-domain R*:
Define g : N → R* by g(n) = 1/n. g is one-one, since 1/n1 = 1/n2 implies n1 = n2. However g is not onto R* because many non-zero real numbers (for example 1/π, √2, 0.3, etc.) are not of the form 1/n for any natural number n. Hence g is not surjective.

Final answer: No. The result is not true if the domain is replaced by N (the function is not onto R*).

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Q2: Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x²
(ii) f : Z → Z given by f(x) = x²
(iii) f : R → R given by f(x) = x²
(iv) f : N → N given by f(x) = x³
(v) f : Z → Z given by f(x) = x³

Solution:

(i) f : N → N given by f(x) = x²
For any x₁, x₂ ∈ N, if f(x₁) = f(x₂), then x₁² = x₂² ⇒ x₁ = x₂ (since both are positive).
Hence, f is injective.
However, not every natural number is a perfect square (for example, 2, 3, 5 are not squares of any natural number).
Therefore, f is not surjective.
→ f is injective but not surjective.

(ii) f : Z → Z given by f(x) = x²
For x₁ = 2 and x₂ = –2, f(2) = 4 and f(–2) = 4, so f(x₁) = f(x₂) but x₁ ≠ x₂.
Thus, f is not injective.
Also, negative integers such as –1, –2, etc., are not squares of any integer.
Hence, f is not surjective.
→ f is neither injective nor surjective.

(iii) f : R → R given by f(x) = x²
For x₁ = 2 and x₂ = –2, f(2) = 4 and f(–2) = 4, so f is not injective.
Also, no real number x gives f(x) = –1 (since the square of any real number is non-negative).
Hence, f is not surjective.
→ f is neither injective nor surjective.

(iv) f : N → N given by f(x) = x³
If f(x₁) = f(x₂), then x₁³ = x₂³ ⇒ x₁ = x₂.
So f is injective.
But not every natural number is a perfect cube (for example, 2, 3, 4, 5, 6, etc., are not cubes).
Hence, f is not surjective.
→ f is injective but not surjective.

(v) f : Z → Z given by f(x) = x³
If f(x₁) = f(x₂), then x₁³ = x₂³ ⇒ x₁ = x₂.
Thus, f is injective.
However, not every integer is a perfect cube (for example, 2 or 4 are not cubes of any integer).
Hence, f is not surjective.
→ f is injective but not surjective.

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Q3: 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.

Solution:

Definition reminder: For any real x, [x] (the greatest integer function or floor function) is the largest integer less than or equal to x. For example, [2.3] = 2, [−1.4] = −2, [3] = 3.

1. Not one-one (not injective).
To show f is not one-one we need two distinct real numbers x1 ≠ x2 with f(x1) = f(x2).

Example: take x1 = 1.2 and x2 = 1.7. Both are real and x1 ≠ x2.
Compute f(x1) = [1.2] = 1 and f(x2) = [1.7] = 1.
Hence f(x1) = f(x2) although x1 ≠ x2. Therefore f is not one-one.

(More generally, for any non-integer t, all x in the interval [t, t+1) have the same floor. So many different inputs give the same output.)

2. Not onto (not surjective).
The codomain is R (all real numbers). To show f is not onto we must find some real number y such that there is no x with f(x) = y.

Observe that f(x) always produces an integer value (f(x) ∈ Z for every x ∈ R). Therefore any non-integer real number, for example y = 0.5 or y = π/2, cannot be hit by f: there is no x with [x] = 0.5 because [x] is always an integer. Thus such y ∈ R are not in the image of f. Hence f is not onto R.

Remark (optional): The actual range (image) of f is the set of all integers Z. If the codomain were taken to be Z instead of R, then f would be onto Z (surjective onto Z) but would still not be one-one.

Conclusion: The greatest integer function f : R → R, f(x) = [x], is neither one-one nor onto.

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Q4: 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.

Solution:

Let f : R → R be defined by f(x) = |x|.

That is,
f(x) =
• x, if x ≥ 0
• –x, if x < 0

1. f is not one-one (not injective):
To be one-one, different values of x should give different values of f(x).

Take x₁ = 2 and x₂ = –2.
Then, f(2) = |2| = 2 and f(–2) = |–2| = 2.
Thus, f(x₁) = f(x₂) even though x₁ ≠ x₂.

Hence, f is not one-one.

2. f is not onto (not surjective):
The codomain is R (all real numbers).

We know that |x| ≥ 0 for all real x. So f(x) can never be negative.
That means there is no x ∈ R such that f(x) = –1 or any other negative value.

Hence, values like –1, –2, –3, etc., are not obtained by f(x).

Therefore, f is not onto R.
(However, if the codomain were taken as [0, ∞), then f would be onto that set.)

Conclusion:
The modulus function f : R → R defined by f(x) = |x| is neither one-one nor onto.

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Q5: Show that the Signum Function f : R → R, given by

   1, if x > 0
f(x) =  0, if x = 0
   –1, if x < 0

is neither one-one nor onto.

Solution:

Let f : R → R be defined as
 f(x) = 1, if x > 0
 f(x) = 0, if x = 0
 f(x) = –1, if x < 0

1. f is not one-one (not injective):
For f to be one-one, distinct values of x must give distinct values of f(x).

Take x₁ = 2 and x₂ = 3.
Then f(2) = 1 and f(3) = 1.
Although x₁ ≠ x₂, we have f(x₁) = f(x₂).

Therefore, f is not one-one.

2. f is not onto (not surjective):
The codomain is R (all real numbers).
The function f(x) takes only three values: –1, 0, and 1.

Hence, for any y ∈ R such that y ≠ –1, 0, or 1 (for example, y = 2 or y = ½), there is no x ∈ R with f(x) = y.

Therefore, f is not onto R.

Conclusion:
The signum function f : R → R defined by

 f(x) = 1, if x > 0
 f(x) = 0, if x = 0
 f(x) = –1, if x < 0

is neither one-one nor onto.

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Q6: 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.

Solution:
Given, f : A → B is defined as
f = {(1, 4), (2, 5), (3, 6)}

We know that a function is one-one (injective) if different elements of the domain map to different elements of the codomain.
That is, if f(x₁) = f(x₂) ⇒ x₁ = x₂.

Here,
f(1) = 4
f(2) = 5
f(3) = 6

We observe that 4, 5, and 6 are all distinct.
Hence, no two different elements of A have the same image in B.

Therefore, f is one-one (injective).

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Q7: 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 + x²

Solution:

(i) f(x) = 3 – 4x

To check one-one:
Let f(x₁) = f(x₂)
⇒ 3 – 4x₁ = 3 – 4x₂
⇒ –4x₁ = –4x₂
⇒ x₁ = x₂

Hence, f is one-one.

To check onto:
Let y ∈ R.
Then, y = 3 – 4x
⇒ x = (3 – y)/4, which is a real number for every real y.

Thus, for every y ∈ R, there exists x ∈ R such that f(x) = y.
Hence, f is onto.

Therefore, f is bijective (both one-one and onto).

(ii) f(x) = 1 + x²

To check one-one:
Let f(x₁) = f(x₂)
⇒ 1 + x₁² = 1 + x₂²
⇒ x₁² = x₂²
⇒ x₁ = ±x₂

Since f(2) = f(–2) = 5, f is not one-one.

To check onto:
For any y ∈ R,
f(x) = y ⇒ 1 + x² = y ⇒ x² = y – 1

x is real only if y – 1 ≥ 0 ⇒ y ≥ 1.
Hence, there is no x ∈ R such that f(x) = y when y < 1.

So, f is not onto (its range is [1, ∞), not all of R).

Therefore, f is neither one-one nor onto.

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Q8: Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is a bijective function.

Solution:

Given:
f : A × B → B × A is defined by f(a, b) = (b, a).

We have to show that f is one-one and onto, i.e., bijective.

1. To show that f is one-one (injective):

Let
f(a₁, b₁) = f(a₂, b₂).

Then, by definition,
(b₁, a₁) = (b₂, a₂).

This means,
b₁ = b₂ and a₁ = a₂.

Hence,
(a₁, b₁) = (a₂, b₂).

Therefore, f is one-one.

2. To show that f is onto (surjective):

Let (b, a) be any element in B × A.

We must find (a′, b′) ∈ A × B such that f(a′, b′) = (b, a).

Now,
f(a′, b′) = (b′, a′).

For f(a′, b′) = (b, a),
we must have
b′ = b and a′ = a.

Since a ∈ A and b ∈ B,
(a′, b′) = (a, b) ∈ A × B.

Thus, for every (b, a) ∈ B × A, there exists (a, b) ∈ A × B such that f(a, b) = (b, a).

Therefore, f is onto.

3. Since f is both one-one and onto, it is bijective.

Conclusion:
The function f : A × B → B × A defined by f(a, b) = (b, a) is bijective.

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Q9: Let f : N → N be defined by

f(n) = (n + 1) / 2 , if n is odd
   n / 2  , if n is even

for all n ∈ N.

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

Solution:

Define f : N → N by
f(n) = (n + 1)/2 if n is odd, and f(n) = n/2 if n is even.

1. Check injectivity (one-one):
If f were one-one, different inputs must give different outputs.
But f(1) = (1 + 1)/2 = 1 and f(2) = 2/2 = 1. Here 1 ≠ 2 but f(1) = f(2).
Therefore f is not one-one.

2. Check surjectivity (onto):
Let m be any natural number. Choose n = 2m (which is even). Then
f(2m) = (2m)/2 = m.
Thus every m ∈ N has a preimage (take n = 2m).
Therefore f is onto.

Conclusion about bijectivity:
A function is bijective only if it is both one-one and onto. Here f is onto but not one-one.
So f is not bijective.

Final answer: No. The function is not bijective (onto yes, one-one no).

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Q10: 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.

Solution:

We have f(x) = (x – 2) / (x – 3), where x ≠ 3.

1. To check whether f is one-one (injective):

Let f(x₁) = f(x₂).

Then,
(x₁ – 2) / (x₁ – 3) = (x₂ – 2) / (x₂ – 3).

Cross-multiplying,
(x₁ – 2)(x₂ – 3) = (x₂ – 2)(x₁ – 3).

Expanding both sides:
x₁x₂ – 3x₁ – 2x₂ + 6 = x₁x₂ – 3x₂ – 2x₁ + 6.

Simplify:
–3x₁ – 2x₂ = –3x₂ – 2x₁.

Rearranging terms,
x₁ = x₂.

Hence, f is one-one.

2. To check whether f is onto (surjective):

Let y ∈ B = R – {1}.

We must find x ∈ A = R – {3} such that f(x) = y.

That is,
y = (x – 2) / (x – 3).

Multiply both sides by (x – 3):
y(x – 3) = x – 2.

Simplify:
yx – 3y = x – 2.

Rearrange:
yx – x = 3y – 2.

x(y – 1) = 3y – 2.

So,
x = (3y – 2) / (y – 1).

For x to exist, denominator ≠ 0, i.e., y ≠ 1.
But y ≠ 1 (since y ∈ B = R – {1}).

Also, when y = 1, x would be undefined, which is excluded.

Hence, for every y ∈ B, there exists x ∈ A such that f(x) = y.

Therefore, f is onto.

Conclusion:

The function f(x) = (x – 2) / (x – 3) is one-one and onto.

Answer: Yes.

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Q11: Let f : R → R be defined as f(x) = x⁴. 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:
For x = 1 and x = –1,
f(1) = 1⁴ = 1 and f(–1) = (–1)⁴ = 1.
Thus, different inputs give the same output ⇒ not one-one.

Also, f(x) = x⁴ ≥ 0 for all x ∈ R,
so negative real numbers have no preimage ⇒ not onto.

Correct answer: (D) f is neither one-one nor onto.

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Q12: 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:
For x₁, x₂ ∈ R,
if f(x₁) = f(x₂), then 3x₁ = 3x₂ ⇒ x₁ = x₂.
Hence, f is one-one.

Also, for any y ∈ R,
take x = y/3 ⇒ f(x) = 3(y/3) = y.
So, every y has a preimage ⇒ f is onto.

Correct answer: (A) f is one-one onto.

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