mrsfox

I need some help on how to approach this problem. I can't seem to find any examples that help me understand this, so if anyone have an approach example to post I would be very grateful

"Consider a relation R defined on the set of integers. Determine for the following if the relations are reflexive, symmetric, transitive.

R = {(a, b)|a = 2b}"

zylo

You have posted a relation (a,b).

It is not reflexive: (a,a) is not in set.

It is not symmetric: (2a,a) is in set. (a,2a) isn't.

It is transitive: (2a,a) $\displaystyle \rightarrow$ (4a,2a), (4a,2a) $\displaystyle \rightarrow$ (8a,4a), (2a,a) $\displaystyle \rightarrow$ (8a,4a)



https://www.csee.umbc.edu/~stephens/203/PDF/10-2.pdf

skipjack

It's not transitive.

cjem

You've got the right idea here, but you need to be careful. The statement "(a,a) is not in set" isn't true in general. Indeed, (0,0) is in R. Fortunately, you don't need it to be true in general. To show the relation is not reflexive, you just need one example of a pair of integers of the form (a,a) that isn't in R. (1,1) works.

Similarly for symmetry: if a = 0, then (2a,a) and (a,2a) are both in R. Again, we just need a single example of symmetry failing. e.g. (2,1) is in R but (1,2) isn't.

I'm not sure what you're trying to say here, but this relation is obviously not transitive. Transitivity says: if $(a,b)$ and $(b,c)$ are in $R$, then so is $(a,c)$. But, for example, we see that $(4,2)$ and $(2,1)$ in $R$ while $(4,1)$ isn't.

zylo

You need two relations to imply transitivity. You only gave 1.
(8,4) implies (4,2) implies (2,1). (8,4) implies (2,1).

skipjack

What you've given isn't what's meant by transitivity of R.

cjem

??? What on earth does "(8,4) implies (4,2)" even mean??? Refer to the link you posted to see a definition of transitivity:

A binary relation R on a set A is transitive if "for all x,y,z in A, (x,y) in R and (y,z) in R implies (x,z) in R."

In our case, A is the set of integers and R is as given in the OP. Take x = 4, y = 2, z = 1. Then we have (x,y) in R and (y,z) in R, but (x,z) is not in R. This shows that the relation is not transitive.

zylo

See link definition in my first post.

If I can multiply by 2 twice, I can multiply by 4, and if I can divide by 2 twice, I can divide by 4.

I can see where that would be a problem for someone who doesn't think zero is excluded is obvious.

skipjack

Transitivity of implication isn't the same as transitivity of R.

zylo

"In mathematics, a binary relation R over a set X is transitive if whenever an element a is related to an element b and b is related to an element c then a is also related to c"

Quoted from wiki, "transitive relation."

But I see the problem now:

If aRb and bRc then aRc.

But if b appears on right side it can't appear on left side unless (b,b) is in set, which it isn't. Interesting.