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February 4th, 2019, 04:59 AM   #1
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Smile 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}"
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February 4th, 2019, 07:33 AM   #2
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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
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Last edited by zylo; February 4th, 2019 at 07:46 AM.
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February 4th, 2019, 01:04 PM   #3
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It's not transitive.
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February 4th, 2019, 01:25 PM   #4
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Quote:
Originally Posted by zylo View Post
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 View Post
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.
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February 4th, 2019, 01:34 PM   #5
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You need two relations to imply transitivity. You only gave 1.
(8,4) implies (4,2) implies (2,1). (8,4) implies (2,1).
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February 4th, 2019, 01:46 PM   #6
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What you've given isn't what's meant by transitivity of R.
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February 4th, 2019, 01:59 PM   #7
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Quote:
Originally Posted by zylo View Post
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.
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February 4th, 2019, 02:05 PM   #8
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Quote:
Originally Posted by skipjack View Post
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.
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February 4th, 2019, 02:16 PM   #9
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Transitivity of implication isn't the same as transitivity of R.
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February 4th, 2019, 02:29 PM   #10
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"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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