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:

3. Reflexive relation
4. Symmetric relation
5. Transitive relation
6. Equivalence relation
7. Empty relation
8. 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 y²) is neither reflexive nor symmetric nor transitive.

Here set A = set of real numbers R

Q = {(x,y): x  y²}

Reflexive:  For x in set A, (x,x) does not belongs to relation Q as x  x²

As (x,x)  Q  x A, Q is not reflexive.

Symmetric: (x,y) Q implies that (y,x)  Q, like x  y² but y  x² thus (y,x

Q is not symmetric

Transitive: (x,y) Q, (y,z)  Q

Thus x [latex]\leq[/latex] y² and y [latex]\leq[/latex] z². This implies that x [latex]\not\leq[/latex] z², (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 = {(T₁,T₂) : T₁ is congruent to T₂} is an equivalence relation.

Here set T = Set of all triangles in a plane

P = {(T₁,T₂): T₁ is congruent to T₂}

Reflexive:  For T₁ in set T, (T₁,T₁) belongs to relation P . 

T₁ and T₁ will be congruent, since both the triangles are the same. As (T₁,T₁) [latex]\in[/latex] R [latex]\forall[/latex] T₁ [latex]\in[/latex] A, P is reflexive.

Symmetric: (T₁,T₂) [latex]\in[/latex] P implies that (T₂,T₁) [latex]\in[/latex] P, like if T₁ and T₂ are congruent then T₂ and T₁ should also be congruent P is symmetric.

Transitive: (T₁,T₂) [latex]\in[/latex] P, (T₂,T₃) [latex]\in[/latex] P 

Thus T₁ and T₂ are congruent and T₂ and T₃ are congruent.

This implies that T₁ and T₃ are also congruent.

(T₁,T₃) [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 = {(P₁,P₂) : P₁ and P₂ have the same number of sides} is an equivalence relation.

Here set A = Set of all polygons

R = {(P₁,P₂): P₁ have same number of sides P₂}

Reflexive: For P₁ in set A, (P₁,P₁) belongs to relation R

P₁ and P₁ will have same number of sides, since both the polygons are same.

As (P₁,P₁) [latex]\in[/latex] R [latex]\forall[/latex] P₁ [latex]\in[/latex] A, R is reflexive.

Symmetric: (P₁,P₂) [latex]\in[/latex] R implies that (P₂,P₁) [latex]\in[/latex] R, like if P₁ and P₂ have same number of sides, then P₂ and P₁ will be having same number of sides.

R is symmetric.

Transitive: (P₁,P₂) [latex]\in[/latex] R, (P₂,P₃) [latex]\in[/latex] R

If P₁ and P₂ are polygons having same number if sides and P₂ and P₃ are

Polygons having same number of sides.

This implies that P₁ and P₃ are also have same number of sides.

(P₁,P₃) [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:-

1. One-One or Injective Function
2. Many-One Function
3. Onto or Surjective Function
4. 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).