Reflexive = transistive relation?

[jwxie]

Given A = {1,2,3}
R1 = {1,1 2,2 3,3}

I know it is reflexive, and I know it is symmetric. But what about its transitivity?

Def of transitive: a,b in R, b,c in R, then a,c is also in R

let a = 1
let b = 1
let c = 1

(1,1) and (1,1)
So yes, the book says it is an equivalence relation, so its transitivity is also valid.

therefore, are all reflexive relations transitive?

but what if the questions asks: constructs a reflexive and sysmmetric but not transitive?

I can do 11 22 33 and add 12 21 to make them both reflexive and symmetric. since i added 12 and 21, thus these 2 ordered pairs destroyed the transitivity?

 

[CRGreathouse]

therefore, are all reflexive relations transitive?

No. You need to construct a relation where (a, b) and (b, c) are in the relation, but (a, c) is not.

 

[jwxie]

No. You need to construct a relation where (a, b) and (b, c) are in the relation, but (a, c) is not.

Well can't I make a = 1 b = 1 and c =1?

The book said 11 22 33 is an equivalence relations on A, so I am guessing that's how he did it.

 

[Landau]

Def of transitive: a,b in R, b,c in R, then a,c is also in R

let a = 1
let b = 1
let c = 1

(1,1) and (1,1)
So yes

This is wrong. You can't just "take" a,b,c=1. You need to check that FOR ALL a,b,c in R the implication "if (a,b) and (b,c) are in R, then (a,c) is in R" is valid.

Btw, you need to be careful with dropping brackets.
R1 = {(1,1), (2,2), (3,3)} is correct,
R1 = {1,1 2,2 3,3} is nonsensical.

 

[Redbelly98]

... are all reflexive relations transitive?

No. See the second example here:
http://en.wikipedia.org/wiki/Equivalence_relation#Relations_that_are_not_equivalences

The relation "has a common factor greater than 1 with" between natural numbers greater than 1, is reflexive and symmetric, but not transitive. (Example: The natural numbers 2 and 6 have a common factor greater than 1, and 6 and 3 have a common factor greater than 1, but 2 and 3 do not have a common factor greater than 1).

(Bold emphasis added by me.)

 

[HallsofIvy]

Given A = {1,2,3}
R1 = {1,1 2,2 3,3}

This is not even a relation! A relation is a collection of order pairs from A, a subset of A\times A. Did you mean {(1,1), (2,2), (3,3)}? If that is the case, then, yes it is transitive. Transitive means "if (a, b) and (b, c) are in the relation, then (a, c) is also in the relation. Here, in order that (a, b) and (b, c) be in the relation, we must have a= b= c. so that (a, b)= (a, c) is in the set.

I know it is reflexive, and I know it is symmetric. But what about its transitivity?

Def of transitive: a,b in R, b,c in R, then a,c is also in R

let a = 1
let b = 1
let c = 1

(1,1) and (1,1)
So yes, the book says it is an equivalence relation, so its transitivity is also valid.

therefore, are all reflexive relations transitive?

but what if the questions asks: constructs a reflexive and sysmmetric but not transitive?

I can do 11 22 33 and add 12 21 to make them both reflexive and symmetric. since i added 12 and 21, thus these 2 ordered pairs destroyed the transitivity?