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