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 February 4th, 2019, 04:59 AM #1 Newbie   Joined: Feb 2019 From: norway Posts: 1 Thanks: 0 Symmetry, transitivity and reflexivity 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}"
 February 4th, 2019, 07:33 AM #2 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 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 Thanks from topsquark Last edited by zylo; February 4th, 2019 at 07:46 AM.
 February 4th, 2019, 01:04 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,298 Thanks: 1971 It's not transitive. Thanks from topsquark
February 4th, 2019, 01:25 PM   #4
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Quote:
 Originally Posted by zylo It is not reflexive: (a,a) is not in set. It is not symmetric: (2a,a) is in set. (a,2a) isn't.
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.

Quote:
 Originally Posted by zylo It is transitive: (2a,a) $\displaystyle \rightarrow$ (4a,2a), (4a,2a) $\displaystyle \rightarrow$ (8a,4a), (2a,a) $\displaystyle \rightarrow$ (8a,4a)
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.

 February 4th, 2019, 01:34 PM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 You need two relations to imply transitivity. You only gave 1. (8,4) implies (4,2) implies (2,1). (8,4) implies (2,1).
 February 4th, 2019, 01:46 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,298 Thanks: 1971 What you've given isn't what's meant by transitivity of R.
February 4th, 2019, 01:59 PM   #7
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Quote:
 Originally Posted by 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).
??? 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.

February 4th, 2019, 02:05 PM   #8
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Quote:
 Originally Posted by skipjack What you've given isn't what's meant by transitivity of R.
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.

 February 4th, 2019, 02:16 PM #9 Global Moderator   Joined: Dec 2006 Posts: 20,298 Thanks: 1971 Transitivity of implication isn't the same as transitivity of R. Thanks from cjem
 February 4th, 2019, 02:29 PM #10 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 "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. Last edited by zylo; February 4th, 2019 at 02:48 PM.

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