Math, asked by sachinkumar9327, 6 months ago

Check whether the relation R in the set N of natural numbers given by R = {( a,b) : a is divisor of b} is reflexive symmetric,or transitive. Also,determine whether R is an equivalence relation

Answers

Answered by haroderadhika282
10

Answer:

Let there be a natural number n,

We know that n divides n, which implies nRn.

So, Every natural number is related to itself in relation R.

Thus relation R is reflexive .

Let there be three natural numbers a,b,c and let aRb,bRc

aRb implies a divides b and bRc implies b divides c, which combinedly implies that a divides c i.e. aRc.

So, Relation R is also transitive .

Let there be two natural numbers a,b and let aRb,

aRb implies a divides b but it can't be assured that b necessarily divides a.

For ex, 2R4 as 2 divides 4 but 4 does not divide 2 .

Thus Relation R is not symmetric .

Answered by ishwaryam062001
1

Answer:

It is no longer an equivalence relation due to the fact it is no longer symmetric.

Step-by-step explanation:

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Let's analyze the relation R in the set N of herbal numbers given through R = {(a,b): a is a divisor of b}.

Reflexive: For the relation R to be reflexive, (a, a) must be in R for all a in N. Here, a is a divisor of itself, so (a, a) is in R for all a in N. Hence, R is reflexive.

Symmetric: For the relation R to be symmetric, if (a, b) is in R, then (b, a) ought to additionally be in R for all a, b in N. Here, if a is a divisor of b, it does now not always imply that b is a divisor of a. Hence, R is now not symmetric.

Transitive: For the relation R to be transitive, if (a, b) and (b, c) are in R, then (a, c) have to additionally be in R for all a, b, c in N. Here, if a is a divisor of b and b is a divisor of c, then a is a divisor of c. Hence, R is transitive.

Since R is reflexive and transitive, however now not symmetric, it is a partial order relation. It is no longer an equivalence relation due to the fact it is no longer symmetric.

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