# 2nd PUC Mathematics Notes

# Chapter 1

__Relation & Functions__

**1. What is a relation? **

A relation can be defined as a collection of ordered pairs where the order pairs are formed based on certain condition.

**Eg.** R = {(a,b) [latex]\in[/latex] A X B: a = b}, here a [latex]\in[/latex] A and b [latex]\in[/latex] B

**2. What are the different types of relations?**

The different types of relations are:

- Reflexive relation
- Symmetric relation
- Transitive relation
- Equivalence relation
- Empty relation
- Universal relation

**3. Explain reflexive relation**.

A relation is said to be reflexive if for every a [latex]\in[/latex] A there exists (a, a) in relation R.

**Eg**. Let R be a relation in set {1,2,3} given by {(1,1), (2,2), (3,3)}. Thus R is Reflexive

**4. Explain symmetric relation.**

If ( [latex] a_1 , a_2 [/latex]) [latex]\in[/latex] R implies [latex]\in[/latex] R for All [latex]\in[/latex] A , then such a relation is said to be symmetric relation.

**Eg.** Let R be a relation in set A={1,2,3} given by {(1,2),(2,1),(1,3),(3,1) }. Thus, R is Symmetric.

**5. Explain transitive relation**.

Ans. If ([latex] a_1 , a_2 [/latex])[latex]\in[/latex]R and ([latex] a_2 , a_3 [/latex])[latex]\in[/latex]R implies ([latex] a_1 , a_3 [/latex])[latex]\in[/latex]R for all [latex] a_1 , a_2 , a_3[/latex][latex]\in[/latex]A

then such a relation is said to be transitive relation.

**Eg**. Let R be a relation in set A= {1,2,3} given by {(1,2), (2,1), (1,1)}. Thus, R is Transitive

**6. What is an equivalence relation?**

A relation which is reflexive, symmetric and transitive is called an Equivalence Relation.

**7. What is an empty relation?**

It is a relation where each element of set is not related to any element of the same set.

**Eg.** R = {(a, b) A x A: a = -b}. A is set of positive real numbers.

**8.What is universal relation?**

It is a relation where each element of set is related to every element of same set.

**Eg. **R = {(a, b) A x A: a – b is a number}. A is set of positive real numbers

**9. Determine whether the following function is reflexive, symmetric and transitive. Relation R in the set N of natural numbers defined as R = {(x,y): y=x+7 and x<3}.**

Here set A = Set of Natural Numbers = N = {1,2,3, 4….}

R = {(x, y):y = x+7 and x<3} = {(x,x+7): x = 1,2,3}

R = {(1,8),(2,9),(3,10)}

**Reflexive:** For 1 in set A, (1,1) does not belong to relation R

Similarly for 2,3 in set A, (2,2),(3,3) does not belong to relation R. As (x,x) [latex]\notin[/latex] R [latex]\forall[/latex] x [latex]\in[/latex] A, R is not reflexive.

**Symmetric:** (x,x+7)[latex]\in[/latex] R implies that (x+7,x) [latex]\notin[/latex] R, like (2,9) [latex]\in[/latex] R but (9,2) [latex]\notin[/latex] R, R is not symmetric

**Transitive:** (x,x+7) [latex]\in[/latex] R, (x+7,x+14) [latex]\in[/latex] R but (x,x+14) [latex]\notin[/latex] R, R is not transitive.

Hence R is not reflexive, not symmetric and not transitive

**10. Determine whether the following function is reflexive, symmetric and transitive. Relation R in the set A = {1,2,3,4,5,6} defined as R = {(p,q):q is divisible by p}.**

Here set A = {1,2,3,4,5,6} R = {(p, q): q is divisible by p}

**Reflexive: ** For p in set A, (p,p) belongs to relation R as p is divisible by p itself. As (p,p) [latex]\in[/latex] R [latex]\forall[/latex] x [latex]\in[/latex] A, R is reflexive.

**Symmetric:** (p,q)[latex]\in[/latex] R implies that (q,p) [latex]\notin[/latex]R like q may be divisible by p but p may not be divisible by q.

**Example:**10 may be divisible by 2 but 2 is not divisible by 10. R is not symmetric

**Transitive:** (p,q)[latex]\in[/latex] R, (q,r) [latex]\in[/latex]R

Thus q is divisible by p and r is divisible by q. This implies that r is divisible by p

(p,r) [latex]\in[/latex] R, R is transitive.

Hence R is reflexive and transitive but not symmetric.

**11. Show that relation Q in set of real numbers R defined as Q = {(x,y): x ****y**^{2}) is neither reflexive nor symmetric nor transitive.

^{2}) is neither reflexive nor symmetric nor transitive.

Here set A = set of real numbers R

Q = {(x,y): x y^{2}}

**Reflexive:** For x in set A, (x,x) does not belongs to relation Q as x x^{2}

As (x,x) Q x A, Q is not reflexive.

**Symmetric**: (x,y) Q implies that (y,x) Q, like x y^{2} but y x^{2} thus (y,x

Q is not symmetric

**Transitive:** (x,y) Q, (y,z) Q

Thus x [latex]\leq[/latex] y^{2} and y [latex]\leq[/latex] z^{2}. This implies that x [latex]\not\leq[/latex] z^{2}, (x,z) [latex]\notin[/latex] Q,

**Example:** Consider (3,-2) [latex]\in[/latex] Q and (-2,1)[latex]\in[/latex] Q, ie 3 [latex]\leq[/latex] 4 and -2 [latex]\leq[/latex] 1 but (3,1) [latex]\notin[/latex] R since 3 [latex]\not\leq[/latex] Q is not transitive.

Hence Q is not reflexive, not symmetric and not transitive.

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

Here set A = Set of all books in library

R = {(x,y): x and y have same number of chapters}

**Reflexive:** For x in set A, (x,x) belongs to relation R

As x and x will have same number of chapters since both the books are the same. As (x,x) [latex]\in[/latex] R [latex]\forall[/latex] x [latex]\in[/latex] A, R is reflexive.

**Symmetric:** (x,y) [latex]\in[/latex] R implies that (y,x) [latex]\in[/latex] R, like if x and y have same number of chapters then y and x should also have same number of chapters R is symmetric.

**Transitive:** (x,y) [latex]\in[/latex] R, (y,z) [latex]\in[/latex] R

Thus x and y have same number of chapters and y and z have the same number

of chapters. This implies that x and z have same number of chapters.

(x,z) [latex]\in[/latex] R, R is transitive.

Hence R is reflexive, symmetric and transitive. Hence R is an Equivalence Relation.

**13. Show that the relation R in set A = {x[latex]\in[/latex]****Z:0[latex]\leq[/latex]****x****[latex]\leq[/latex]****12}, given by R = {(x,y) : |x-y| is multiple of 4} is an equivalence relation. **

Here set A = {0,1,2,3,4,5,6,7,8,9,10,11,12}

R = {(x,y): |x-y| is multiple of 4}

**Reflexive:** |x-x| = 0. Zero is multiple of 4.

[latex]\forall[/latex] x [latex]\in[/latex] A, (x,x) [latex]\in[/latex] R. Hence R is Reflexive

**Symmetric:** If |x-y| is multiple of 4, then |y-x| = |-(x-y)| = |x-y| is again multiple of 4.

(x,y) [latex]\in[/latex] R implies (y,x) [latex]\in[/latex] R. Hence R is Symmetric.

**Transitive:** If |x-y| is multiple of 4, and |y-z| is multiple of 4, then

|x-y| + |y-z| is also multiple of 4,

|x-y+y-z| = |x-z| is also multiple of 4.

(x,y) [latex]\in[/latex]R, (y,z) [latex]\in[/latex]Rthus (x,z) [latex]\in[/latex]R, R is Transitive.

Hence R is Symmetric, Symmetric and Transitive. Hence R is an Equivalence Relation.

**14. Give an example of a relation which is symmetric and transitive but not reflexive.**

Consider relation R = {{1,1),(1,2),(2,1)} defined in set {1,2,3}.

**Reflexive:** (1,1) [latex]\in[/latex]R but (2,2) and (3,3) [latex]\notin[/latex] R.

Thus R is not Reflexive.

**Symmetric****:** (1,2) [latex]\in[/latex] R and (2,1) [latex]\in[/latex] R.

Thus R is Symmetric.

**Transitive:** (1,2) [latex]\in[/latex] R, (2,1) [latex]\in[/latex] R and (1,1) [latex]\in[/latex] R

Thus R is Transitive.

Hence R is symmetric, transitive but not reflexive.

**15. Show that relation P in set T of all triangles in plane given by P = {(T**_{1},T_{2}) : T_{1} is congruent to T_{2}} is an equivalence relation.

_{1},T

_{2}) : T

_{1}is congruent to T

_{2}} is an equivalence relation.

Here set T = Set of all triangles in a plane

P = {(T_{1},T_{2}): T_{1} is congruent to T_{2}}

**Reflexive:** For T_{1} in set T, (T_{1},T_{1}) belongs to relation P .

T_{1} and T_{1} will be congruent, since both the triangles are the same. As (T_{1},T_{1}) [latex]\in[/latex] R [latex]\forall[/latex] T_{1 }[latex]\in[/latex] A, P is reflexive.

**Symmetric****:** (T_{1},T_{2}) [latex]\in[/latex] P implies that (T_{2},T_{1}) [latex]\in[/latex] P, like if T_{1} and T_{2} are congruent then T_{2 }and T_{1} should also be congruent P is symmetric.

**Transitive:** (T_{1},T_{2}) [latex]\in[/latex] P, (T_{2},T_{3}) [latex]\in[/latex] P

Thus T_{1} and T_{2} are congruent and T_{2} and T_{3} are congruent.

This implies that T_{1} and T_{3} are also congruent.

(T_{1},T_{3}) [latex]\in[/latex] P, P is transitive.

Hence P is reflexive, symmetric and transitive. Hence P is an Equivalence Relation.

**16. Show that relation R defined on set A of all polygons as R = {(P**_{1},P_{2}) : P_{1} and P_{2} have the same number of sides} is an equivalence relation.

_{1},P

_{2}) : P

_{1}and P

_{2}have the same number of sides} is an equivalence relation.

Here set A = Set of all polygons

R = {(P_{1},P_{2}): P_{1} have same number of sides P_{2}}

**Reflexive:** For P_{1} in set A, (P_{1},P_{1}) belongs to relation R

P_{1} and P_{1} will have same number of sides, since both the polygons are same.

As (P_{1},P_{1}) [latex]\in[/latex] R [latex]\forall[/latex] P_{1} [latex]\in[/latex] A, R is reflexive.

**Symmetric:** (P_{1},P_{2}) [latex]\in[/latex] R implies that (P_{2},P_{1}) [latex]\in[/latex] R, like if P_{1} and P_{2} have same number of sides, then P_{2} and P_{1} will be having same number of sides.

R is symmetric.

**Transitive:** (P_{1},P_{2}) [latex]\in[/latex] R, (P_{2},P_{3}) [latex]\in[/latex] R

If P_{1} and P_{2} are polygons having same number if sides and P_{2} and P_{3} are

Polygons having same number of sides.

This implies that P_{1} and P_{3} are also have same number of sides.

(P_{1},P_{3}) [latex]\in[/latex] R, R is transitive.

Hence R is reflexive, symmetric and transitive. Hence R is an Equivalence Relation.

**17. What is a function?**

It is a relation between two sets of numbers. The two sets are called Domain and

Co Domain. Range is the set of all the outputs of the function.

**18. What are the different types of functions?**

The different types of functions are:-

- One-One or Injective Function
- Many-One Function
- Onto or Surjective Function
- Bijective Function

**19. Explain one-one function.**

A function f:X → Y is said to be injective if each element in X has distinct images in Y.

**20. Explain many-one function.**

A function f: A→ B is said to be many-one if each element in A does not have distinct images in B.

**21. Explain onto function.**

A function X→ Y is said to be Surjective if every element in Y is an image of some element in X.

**22. What is a bijective function?**

A function which is both injective (one-one) and Surjective is called Bijective function.

**23. What is the criteria for a function to be one-one?**

The criteria for a function to be one-one is that for every [latex]x_1[/latex] and [latex]x_2[/latex] [latex]\in[/latex] X, f([latex]x_1[/latex]) = f([latex]x_2[/latex]) implies that [latex]x_1[/latex] = [latex]x_2[/latex]

**24. What is the criteria for a function to be onto?**

The criteria for a function to be onto is that for every y∈ Y there exists an element x ∈ X such that

Y = f(x).