# Thread: Let R1 and R2 be relations from {1, 2, 3} to {1, 2, 3, 4}. Find

1. ## Let R1 and R2 be relations from {1, 2, 3} to {1, 2, 3, 4}. Find

Let R1 = {(1, 2), (2, 3), (3, 4)} and R2 = {(1, 1), (1, 2),(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)} be relations from {1, 2, 3} to {1, 2, 3, 4}. Find

b) R1 ∩ R2.
c) R1 − R2.

So with b). Does this mean the ordered pairs in R1 and R2?
And with c) does this mean the sets in R1 that are not in R2?

What does the part of the problem, " be relations
from {1, 2, 3} to {1, 2, 3, 4}" mean?

2. ## Re: Let R1 and R2 be relations from {1, 2, 3} to {1, 2, 3, 4}. Find

Originally Posted by lamentofking
What does the part of the problem, " be relations
from {1, 2, 3} to {1, 2, 3, 4}" mean?
Any subset of $A\times B$ is a relation $A\to B$. (some authors do not allow the empty relation)

3. ## Re: Let R1 and R2 be relations from {1, 2, 3} to {1, 2, 3, 4}. Find

So for b) R1 ∩ R2. The answer is all the ordered pairs in R1 (Since they are in R2)?

4. ## Re: Let R1 and R2 be relations from {1, 2, 3} to {1, 2, 3, 4}. Find

Originally Posted by lamentofking
Let R1 = {(1, 2), (2, 3), (3, 4)} and R2 = {(1, 1), (1, 2),(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)} be relations from {1, 2, 3} to {1, 2, 3, 4}. Find

b) R1 ∩ R2.
c) R1 − R2.

So with b). Does this mean the ordered pairs in R1 and R2?
Yes.

And with c) does this mean the sets in R1 that are not in R2?
Well, "ordered pairs", not "sets" but yes.

What does the part of the problem, " be relations
from {1, 2, 3} to {1, 2, 3, 4}" mean?
A "relation from set A to set B" is a set of ordered pairs in which the first member of each pair is in A and the second member is in B.
So, here, "relations from {1, 2, 3} to {1, 2, 3, 4}" are sets of ordered pair where the first member of each pair one of 1, 2, or 3 and the second member is one of 1, 2, 3, or 4. You should see that this is true for the given relations.

5. ## Re: Let R1 and R2 be relations from {1, 2, 3} to {1, 2, 3, 4}. Find

Originally Posted by lamentofking
So for b) R1 ∩ R2. The answer is all the ordered pairs in R1 (Since they are in R2)?
Yes, $R1\cap R2= R1$. And because, as you say, $R1\subset R2$, $R1- R2$ is just as easy. (I think Plato's parenthetical statement must not apply here.)

6. ## Re: Let R1 and R2 be relations from {1, 2, 3} to {1, 2, 3, 4}. Find

So then because R1 and R2 share the same ordered pairs, R1 - R2 is equal to the empty set correct?

7. ## Re: Let R1 and R2 be relations from {1, 2, 3} to {1, 2, 3, 4}. Find

Yes.

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### be relation from {1,2,3} to {1,2,3,4} find

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