February 4th, 2019, 03:59 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, 06:33 AM  #2 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
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/102.pdf Last edited by zylo; February 4th, 2019 at 06:46 AM. 
February 4th, 2019, 12:04 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,633 Thanks: 2080 
It's not transitive.

February 4th, 2019, 12:25 PM  #4  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  Quote:
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.  
February 4th, 2019, 12:34 PM  #5 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
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, 12:46 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,633 Thanks: 2080 
What you've given isn't what's meant by transitivity of R.

February 4th, 2019, 12:59 PM  #7  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  Quote:
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, 01:05 PM  #8 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  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, 01:16 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,633 Thanks: 2080 
Transitivity of implication isn't the same as transitivity of R.

February 4th, 2019, 01:29 PM  #10 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
"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 01:48 PM. 

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reflexivity, symmetry, transitivity 
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