Question

The relation R defined in N as a R b ⇔ b is divisible by a is
(a) reflexive but not symmetric
(b) symmetric but not transitive
(c) symmetric and transitive
(d) none of these

Answer 1

Answer:

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

Step-by-step explanation:

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Answer 2

Answer:

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