
#1
Dec1208, 12:50 AM

P: 17

Man I've become desperate. I just signed up needing help on this homework. Can anyone help me with these two problems?
Let A be the set {1,2,3,4}. Prove that a relation R on A with 15 ordered pairs is not transitive. I've got no clue on that one. And this second one, which I know the proof, but I need some help wording it correctly: If f is injective (onetoone) and C subset D are any subsets of A, then f(DC) = f(D)  f(C). I know the proof, but every time I try and word it, it just sounds wrong. 



#2
Dec1208, 04:05 AM

HW Helper
P: 2,155

Since there are 16 possible ordered pairs, if 15 are in the relation only 1 is omited. Show that there exist a pair of pairs that implicate the omited pair.
hint can you do Let A be the set {1,2,3}. Prove that a relation R on A with 8 ordered pairs is not transitive. or Let A be the set {1,2}. Prove that a relation R on A with 3 ordered pairs is not transitive. Since the function is injective you can work with f(x) instead of x. if f(x) is in f(DC) what can we say about x. if f(x) is in f(D)f(C) what can we say about x. 



#3
Dec1208, 07:05 AM

P: 13

EDIT: Sorry, I'm just being nitpicky. The general case and idea is sound. It works for 3 elements and above. 



#4
Dec1208, 07:25 AM

HW Helper
P: 2,155

Abstract math help if possible 



#5
Dec1208, 07:41 AM

P: 17

{(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)} 



#6
Dec1208, 08:01 AM

P: 13

If you pick one ordered pair and remove it from that relation you have listed, can you find an example of ordered pairs such that (a,b) and (b,c) is in the relation but (a,c) isn't? I.e. the relation isn't transitive.
At first pick an actual pair. You should then see that the case is the same whichever pair you remove, which answers your question. 



#7
Dec1208, 08:09 AM

P: 17

Ok, I see the problem at least. I couldn't visualize with symmetric originally. But if you have (4,3) and (3,4) but not (4,4) you aren't transitive. Well at least that group isn't, but it's supposed to be for any a,b,c right?
Now the problem is... how do I put that in words? 



#8
Dec1208, 08:24 AM

P: 13

If you remove (a,b) then since you have four elements you can pick an element c which isn't equal to a or b. Try going from there.
This is the condition which doesn't hold with only two elements. 



#9
Dec1208, 08:30 AM

P: 17

So does this seem sufficient?:
A relation R on A has 16 possible ordered pairs. Let R be a relation on A with 15 ordered pairs excluding aRc. Since all the other remaining pairs are in R, then aRb and bRc. However, since a does not relate to c, R is not transitive. 



#10
Dec1208, 11:13 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

Yes, that is perfectly good.
By the way, this should have been posted in the "homework and coursework" section. 


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