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..... .................................................. ..................................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

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