Thread: Introduction to Discrete Mathematics

Hey I'm in my first term for Discrete Mathematics and still fairly confused with most of it. Anyway I came into some trouble on my homework, and I don't necessarily need anyone to solve it for me but kind of give me a push in right direction with some examples or explanations, because I really need to get it done and can't get hold of my instructor that quickly. I have an idea as to what I need to do, but can't really bring it all together right now so I'm really hoping someone can help me a little more than my book.

Let A = {1, 2, 3, 4} and B = {x ∈ ℤ | x mod 2 = 0}. Let R: A → B be a relation such that (x, y) ∈ R → x/y ∈ ℤ.

a) Write R in set roster notation.
b) Draw an arrow diagram for R.
c) Show why R is or is not a function.

Are you saying that or .
If it is the second then the question makes no sense.

They're arrows. If A, then B.

Originally Posted by Varine

They're arrows. If A, then B.

Well then there are only three pairs . WHY?
So answer the rest.

Hi Varine,

Just a few points:

'R: A -> B' is usually used to denote a function from A to B. As question (c) is whether R is functional or not, this is not a very good notation. In any case, the arrow used here is much different from the 'if, then' arrow. That arrow connects sentences to yield other sentences.

Usually, 'R: A -> B' indicates not only that R is functional but that it is defined for all A (i.e. it is a function /from/ A). That is to say, for each member x of A, there should be a y in B such that (x, y) is in R.

But notice that there is no even number (i.e. member of B) which divides 3 by an integer. So although some pair of the form (3, y) must be in R, none of that form will satisfy the condition (x, y) ∈ R → x/y ∈ ℤ.

On the other hand, suppose that 'R: A -> B' does not require R to be defined for all x in A. Then let R = {} (i.e. the empty set). As R has no members, every member of it satisfies (x, y) ∈ R → x/y ∈ ℤ (trivially!).

If in the condition (x, y) ∈ R → x/y ∈ ℤ we have a double arrow (if and only if) instead, then R = {(2, 2), (4, 2), (4, 4)} as the reply suggested.

For (c), remember that a function is a relation R (i.e. set of pairs (x, y)) such that no thing x is related to two things y and z (i.e. there are no pairs (x, y) and (x, z) such that z is not equal to y and (x, y) and (x, z) are both in R).

Hope this helps.

Best,

Sam