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**Prakash Panneer****Member**- Registered: 2006-06-01
- Posts: 110

If R = {(1, 1), (2, 2), (1, 2), (2, 1), (3, 3)} and S = {(1, 1), (2, 2), (2, 3), (3, 2),

(3, 3)} are two relations in the set X = {1, 2, 3}, the incorrect statement is:

(A) R and S are both equivalence relations

(B) R∩S is an equivalence relations

(C)R^(-1)∩S^(-1) is an equivalence relations

(D) R∪ S is an equivalence relations

With Regarks,

Prakash Panneer

*Last edited by Prakash Panneer (2006-06-06 02:30:21)*

Letter, number, arts and science

of living kinds, both are the eyes.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I take it these are permutations and not ordered pairs. For example, R goes from 1 to 1, and 1 to 2, and so on.

To check if they are equivalence relations, you need to check:

a. Every number is related to itself

b. If a is related to b, then b is related to a.

c. If a is related to b, and b is related to c, then a is related to c.

For R and S, these are fairly easy to check.

R∩S = {(1, 1), (2, 2), (3, 3)} which is an equivalence relation

R^(-1)∩S^(-1)

R^(-1) = {(1, 1), (2, 2), (2, 1), (1, 2), (3, 3)}

S^(-1) = {(1, 1), (2, 2), (3, 2), (2, 3), (3, 3)}

Which should look familar.

So that means we are left with D. Determine why D can't be an equivalence relation.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Prakash Panneer****Member**- Registered: 2006-06-01
- Posts: 110

:-) Many thanks for your help Ricky :-)

Letter, number, arts and science

of living kinds, both are the eyes.

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