Test by Bikram | Mathematics | Test 2 | Question: 19

Question

Read the statements below and pick the correct ones:

1. $R$ is symmetric if $R == R-1$
2. $R$ on a set $A$ is anti symmetric if and only if $R \cap R-1$ is the subset of diagonal relation.
3. If $R$ is reflexive and transitive then $R_n = R$ for all positive $n$.
4. $\{ xRy : \mid x-y \mid < a \}$ is not a transitive relation where $x$ and $y$ are real numbers and a is a positive constant.

• A.   I,II are correct.
• B.   I,II,III are correct.
• C.   I,III,IV are correct.
• D.   All are correct.

Comments

@hemanth

For, option 3 to be true : 

Read below links

For Reflexive case : 

https://math.stackexchange.com/questions/1537569/let-a-be-a-set-and-let-r-be-a-reflexive-relation-on-a-prove-that-for-every-posi

For Transitive case :

 https://www.quora.com/How-do-I-prove-“If-R-is-transitive-then-R-n-is-transitive-”

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For 3rd

 If R is reflexive and transitive then Rⁿ = R for all positive n. 

R is reflexive means it has all the transitive pairs.If you look into diagraph then we can say it has all length path if it is transitive.Now if we do any power of R,then we will try to get new pairs but as my relation is transitive so it will have the result into same relation.

 transitive closure of the relation is R itself,And transitive closure is R^0 union R^1 ....Now ,my R= R^0 union R^1 ...as relation is transitive and reflexive.

So if we do any power of R,we will get R only.

@@hemanth

Does this answer your query?

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sir i have a doubt on first option

s={1,2,3}

and a binary relation R={(1,1)}

it is not reflexive but R==R^(-1)

than how first stmt correct.