# Thread: examples of a binary relations

1. ## examples of a binary relations

Is there any certain ways of thinking of these examples? Or just guess and try?

a) Reflexive and symmetric, but not transitive
= abs(a-b) < 1... right? any other example?

b) Reflexive but neither symmetric nor transitive
= a-b < 1, any other type?

c) Symmetric, but neither reflexive nor transitive
= abs(a-b) = 1, any other?

d) Transitive, but neiter reflexive nor symmetric
= a < b

any other examples will be appreciated, thanks

2. Originally Posted by ninano1205
Is there any certain ways of thinking of these examples? Or just guess and try?

a) Reflexive and symmetric, but not transitive
= abs(a-b) < 1... right? any other example?

b) Reflexive but neither symmetric nor transitive
= a-b < 1, any other type?

c) Symmetric, but neither reflexive nor transitive
= abs(a-b) = 1, any other?

d) Transitive, but neiter reflexive nor symmetric
= a < b

any other examples will be appreciated, thanks
a) Let R={(a,b): a,b are real Nos and |a-b|<1}

For R to be reflexive we must have :

For all a belonging to reals,then (a,a)εR ..or |a-a|<1,which is correct since 0<1

For R to be symmetric we must have:

For all a,b reals and (a,b)εR ,then (b,a)εR.But (a,b)εR===> |a,b|<1 which implies that |b,a|<1 ,hence (b,a)εR,SINCE |a-b|=|b-a|

And finally for R TO BE not transitive we must prove that:

There exist x,y,z reals such that (x,y)εR and (y,z)εR BUT ~(x,z)εR ( (x,z) does not belong to R).

This is the negation of the definition for R being transitive in the real Nos:

......for all x,y,z reals (x,y)εR and (y,z)εR ,then (x,z)εR.........................

So let x=8,....y=8,5..........z=9........................ .and:

|x-y|=|8-8,5|=0,5<1 .........hence (x,y)εR........................................... ...................................

|y-z|=|8,5-9|= 0,5<1...........hence (y,z)εR........................................... ...........................................

But.....$\displaystyle |x-z|=|8-9| = 1\geq 1$.................................................. ..................................1

Now for (x,z) not to belong to R we need to have ~|x-z|<1 ,i.e |x-z| must not be less to....1

From the inequalities of real Nos ,if ~|x-z|<1 ,then $\displaystyle |x-z|\geq 1$

Which is exactly what we have proved in (1)............................................... ..................................................

Hence R is not transitive

YOU CAN do the rest of problems more or less in the same way.

To understand what you really doing i suggest you write first the definitions when a relation R in real Nos is reflexive,symmetric, transitive,and then take their negations