confused:Given the simple LTE (less then equal) relation on S= {1,2,3,4} defined by [less and equal ], we define a complex NTG (not grater then) relation on S x S by (w,x) NTG (y,z) if w[less and equal) y(adsbygoogle = window.adsbygoogle || []).push({}); orx [less and equal z. (thisorconfusing me )

Show that NTG is (R) reflexive, but not (T) transitive and not (AS) antisymmetric.

After I list the pairs: (1,1) (1,2) (1,3) (1,4) (2,1) (2,2), (2,3) (2,4) (3,1) (3,2) (3,3) (3,4)

(4,1) (4,2) (4,3) (4,4)

Now I don't know how to start at all. Or may be...

Reflexiveproperty means that (x,x) is in the realation for any x in S.

Antisymmetric (AS)(x,y) and (y,x) both in the relation implies that x=y, but I need to shownotAS. Does that mean I have to show non-symetric?

Transitiveproperty means that if (x,y) and (y,z) are in the realation, then (x,z) is also.

How about if I start with (1,1) NGT (2,1) b/c {1 is < and = to 2}

I would appreciate any suggestions.

Thank you again

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# Homework Help: More Discrete Math's

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