logic - Transitivity on relations -

i have question concerning proving properties of relations. question this: how go proving that, if r , s (r , s both being different relations) transitive, r union s transitive?

the answer false, , counter example given solution in book.

i understand how counterexample works explained in book, don't understand is, how arrive conclusion statement false.

basically can see myself giving proof if values (x,y,z) in r , s, if (x,y) in r , (y,z) in r, (x, z) in r since r transitive. , if (x,y) in s , (y,z) in s, (x,z) in s since s transitive. since (x,z) in both r , s, intersection true. why wouldn't union of r , s true well?

is because proof cannot ended "since (x,z) in both r , s, (x,z) can in r or s"? basically, proof can't ended or statement @ end?

i understand how counterexample works explained in book, don't understand is, how arrive conclusion statement false.

given there's (presumably valid) counterexample, statement has false. trying apply proof counterexample can reveal error.

that's not it's never case union of 2 transitive relations transitive. indeed, there obvious examples such union of transitive relation or union of less-than , less-than-or-equal-to (which equal less-than-or-equal-to reasonable definition). original statement asserts case any 2 transitive relations. single counterexample disproves it. if provide (valid) proof of statement, you'd have discovered paradox. causes mathematicians reevaluate system's axioms remove paradox. in case there no paradox.

let t union of r , s (for sake of simplicity, let's assume domain equals range , it's same set both). trying prove if xty , ytz, must case xtz. part of proof outline, state following:

if (x,y) in r , (y,z) in r, (x, z) in r since r transitive

this true it's definition of transitivity. point out, can used prove transitivity of intersection of 2 transitive relations. however, since t union, there's no reason assume xry; might xsy only. since can't prove antecedent (that xry , yrz), consequent (that xrz) irrelevant. similarly, can't show xsz. if can't show xrz or xsz, there's no reason believe xtz.

this implies sort of situation gives counter example statement: when 1 half of transitive pair comes r , other comes s. simple, contrived, example, define relations on set {1,2,3}:

r={(1,2)}
s={(2,3)}

clearly, both r , s transitive (as there no x, y, z such xry , yrz or xsy , ysz). on other hand,

t={(1,2),(2,3)}

is not transitive. while 1t2 (since 1r2) , 2t3 (since 2s3), not case 1t3. textbook gives more natural counterexample, feel gives understanding of can cause assertion fail.