Question

From class: 11

Chapter: 1 (Sets, Functions and Relations)

Answer any 2 questions

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Answer 1

QUESTION :-

Let assume that F is a relation on the set R real numbers defined by xFy if and only if x-y is an integer. Prove that F is an equivalence relation on R.

SOLUTION :-

Reflexive: Consider x belongs to R,then x – x = 0 which is an integer. Therefore xFx.

Symmetric: Consider x and y belongs to R and xFy. Then x – y is an integer. Thus, y – x = – ( x – y), y – x is also an integer. Therefore yFx.

Transitive: Consider x and y belongs to R, xFy and yFz. Therefore x-y and y-z are integers. According to the transitive property, ( x – y ) + ( y – z ) = x – z is also an integer. So that xFz.

Thus, R is an equivalence relation on R.

QUESTION :-

Show that the relation R is an equivalence relation in the set A = { 1, 2, 3, 4, 5 } given by the relation R = { (a, b):|a-b| is even }.

SOLUTION :-

R = { (a, b):|a-b| is even }. Where a, b belongs to A

Reflexive Property :

From the given relation,

|a – a| = | 0 |=0

And 0 is always even.

Thus, |a-a| is even

Therefore, (a, a) belongs to R

Hence R is Reflexive

Symmetric Property :

From the given relation,

|a – b| = |b – a|

We know that |a – b| = |-(b – a)|= |b – a|

Hence |a – b| is even,

Then |b – a| is also even.

Therefore, if (a, b) ∈ R, then (b, a) belongs to R

Hence R is symmetric.

Transitive Property :

If |a-b| is even, then (a-b) is even.

Similarly, if |b-c| is even, then (b-c) is also even.

Sum of even number is also even

So, we can write it as a-b+ b-c is even

Then, a – c is also even

So,

|a – b| and |b – c| is even , then |a-c| is even.

Therefore, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) also belongs to R

Hence R is transitive.

QUESTION :-

Let m be a positive integer A relation R is defined on the set Z by 'aRb'. a-b is divisible by m a, b∈Z, show that R is an. Equivalence relation on set R

SOLUTION :-

Reflexive: For any a∈N, we have a−a=0=0×n

⇒−a is divisible by n⇒(a,a)∈R

Thus (a,a)∈R for all a∈Z. So, R is reflexive.

Symmetry: Let (a,b)∈R. Then,

⇒(a,b)∈R⇒(a−b) is divisible by n.

⇒(a−b)=np for some p∈Z

⇒b−a=n(−p)

⇒b−a is divisible by n

Thus, (a,b)∈R⇒(b,a)∈R for all a,b∈Z

So, R is symmetric on Z.

Transitive: Let a,b,c∈Z such that (a,b)∈R and (b,c)∈R. Then (a,b)∈R⇒(a−b) is divisible by n.

⇒a−b=np for some p∈Z

(b,c)∈R⇒(b−c) is divisible by n.

⇒b−c=nq forsome q∈Z

therefore,(a,b)∈R and b−c∈R

⇒a−b=npb−c=nq

⇒(a−b)+(b−c)=np+nq

a−c=n(p+q)

⇒(a−c)∈R

Thus, (a,b)∈R and (b,c)∈R

⇒(a,c)∈R for all a,b,c∈Z

So, R is transitive relation on Z.

Thus, R being reflexive, symmetric and transitive, is an equivalence relation

TO MORE INFORMATION :-

1. Reflexive: A relation is said to be reflexive, if (a, a) ∈ R, for every a ∈ A.
2. Symmetric: A relation is said to be symmetric, if (a, b) ∈ R, then (b, a) ∈ R.
3. Transitive: A relation is said to be transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.

Answer 2

Refer the attachment...