Question

let R be a relation defined by xRy if 3x+4y divisible by 7 show that R is an equivalent relation on Z​

Answer 1

Answer:

Correct option is

C

R is an equivalence relation

aRa is positive as a+2a=3a is divisible by 3

Hence reflexive

If aRb is positive a+2b is divisible by 3

Hence 3a+3b−(2a+b) is divisible by 3

⇒2a+b is divisible by 3

ie aRb is positive

⇒bRa is positive hence symmetric

aRb,bRc is a+2b,b+2c divisible by 3

⇒a+2b+b+2c divisible by 3

⇒a+3b+2c divisible by 3

So a+2c divisible by 3 hence transitive

So it is an equivalence relation.

Answer 2

Answer:

Answer: To show R is an equivalence relation we will show R is reflexive,symmetric and transitive.

Step-by-step explanation:

1.R is reflexive

For any integer n,

3n+4n=7n which is divisible by 7 as 7n/7=n (an integer).Thus nRn.

Implies R is reflexive.

2.R is symmetric

Let m,n be integers such that 3m+4n is divisible by 7.

=>3m+4n=7p for some integer p (by definition of divisibility)

Then 7m+7n-(3m+4n) is also divisible by 7 ( because 7m+7n-(3m+4n) =7x for some integer x)

=>7n-4n+7m-3n is divisible by 7.

=>3n+4m is divisible by 7

=>nRm

Hence R is symmetric.

3.R is transitive

let l,m and n are integers such that 3l+4m and 3m+4n are divisible by 7.

=>3l+4m=7a and

3m+4n=7b for some integers a and b (by divisibility)

=>by adding

3l+4m+3m+4n=7(a+b)

=>3l+7n= -7m+7(a+b)

=>3l+7n= 7(a+b-m) where a+b-m is an integer

Hence 3l+7n is divisible by 7

Thus R is transitive.

Hence R is an equivalence relation.