Thread: Abstract Math Help if possible

1. Abstract Math Help if possible

Man I've become desperate, 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 (one-to-one) and C subset D are any subsets of A, then f(D-C) = f(D) - f(C).

2. heh, guess I posted in the wrong grade level. Sorry, I'm new

3. Originally Posted by HeirToPendragon
Let A be the set {1,2,3,4}. Prove that a relation R on A with 15 ordered pairs is not transitive.
Just to get you started: How many ordered pairs are there altogether on a set with four elements? Could you have a transitive relation that contains all but one of them?

4. There are 16 ordered pairs on A.

And see that's the problem, I really don't know if you can have 15. The book we're using is absolutely terrible and I assume it's not telling me something important.

5. Originally Posted by HeirToPendragon
There are 16 ordered pairs on A. And see that's the problem, I really don't know if you can have 15. The book we're using is absolutely terrible and I assume it's not telling me something important.

Do you know the definition of transive?
Say that $\displaystyle R$ is a relation on the set with 15 pairs and say $\displaystyle \left( {1,2} \right) \notin R$.
Now every other pair is in $\displaystyle R$. So $\displaystyle \left( {1,3} \right) \in R\,\& \,\left( {3,2} \right) \in R$.
If $\displaystyle R$ were transitive that gives you a contradiction.
Now what other case(s) is(are) to be considered?

6. Yeah that's similar to what I figured out. Here is my answer:

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.

I just wasn't sure if that was enough.

7. may be the next proof appear to be silly but it is a try.

C is subset of D then D=(D-C)union C
f(D)=f(D-C)union f(C) (since the two sets are not intersecting) and given the function is one-to-one then we can say that f(D-C) and f(C) are also non intersecting which means that they are complimenting eachother with respect to f(D)
then f(D)-f(C)=f(D-C)