# I dont understand how R1 = { (a,a),(a,c) } is transitive

• Mar 5th 2013, 01:54 PM
erockdontstop
I dont understand how R1 = { (a,a),(a,c) } is transitive
I am learning about equivalence relations, and in my text book it says

R1 = { (a,a), (a,c) }, How is this set transitive?

Also a totally separate question
how is R = { (a,b), (a,c) } transitive?

• Mar 5th 2013, 02:04 PM
Plato
Re: I dont understand how R1 = { (a,a),(a,c) } is transitive
Quote:

Originally Posted by erockdontstop
I am learning about equivalence relations, and in my text book it says
R1 = { (a,a), (a,c) }, How is this set transitive?
Also a totally separate question
how is R = { (a,b), (a,c) } transitive?

Look at the negation: $\mathcal{R}$ is not transitive if and only if
$(\exists (a,b)\in\mathcal{R}~\&~(b,c)\in\mathcal{R})$ but $(a,c)\notin\mathcal{R}$.

Are the two you posted not transitive? If not, then they are transitive.
• Mar 7th 2013, 05:47 AM
HallsofIvy
Re: I dont understand how R1 = { (a,a),(a,c) } is transitive
A general logic rule: If, in the statement "x implies y", x is false then then entire statement is true whether y is true or false.

A set of pairs (a relation) is transitive if and only if the statement "if (a, b) is in the set and (b, c) is in the set then (a, c) is in the set". In both examples you give the hypothesis, "(a, b) is in the set and (b, c) is in the set" is NEVER true so the entire statement is true.