1. ## relations

Hello,
1. How do I prove this eaulity? or I do I refute this claim?
R∘(S∪T) = (R∘S)∪(R∘T)

2.
xRy ⇔ 8|(x+7y) , ∀x,y∈ℤ
what are the equivaliance classes and what is the divison of
ℤ according to this relation?

Thank you!

2. ## Re: relations

What is the definition of "R∘S" and what kinds of things are "R", "S", and "T"?

(2) is saying that x and y are equivalent if and only if x+ 7y is divisible by 8. That is equivalent to saying that x+ 7y is a multiple of 8. So lets start with y= 0. x+ 7y is just x. That's a multiple of y if and only if x is a multiple of 8: The equivalence class containing 0 is just the set of all multiples of 8. What about y= 1? You want x+ 7 to be a multiple of 8. What numbers are 7 less than a multiple or 8? 1 itself, of course, 9, 17, 24, … And don't forget the negative numbers. 0 is a multiple of 8 and -7 is 7 less than that. -8- 7= -15, etc. What about y= 2? If y= 2, x+ 7y= x+ 14 must be a multiple or 8. What numbers are 14 less than a multiple of 8? Do you see that, since 14= 8+ 6, any number is 14 less than a multiple of 8 if and only if it is 6 less than a multiple of 8? Such numbers are 2 itself, 14- 6= 8, 24- 6= 18, etc. as well as 0- 6= -6, -8- 6 -14, etc. What about x= 3? Continue that reasoning until you find yourself getting the same numbers again.

3. ## Re: relations

Hi,
thank you so much for your respond. It helped me alot.

4. ## Re: relations

Originally Posted by ronrong7
Hi,
thank you so much for your respond. It helped me alot.
Please answer the question about what R, S & T are in #1. BE COMPLETE!

5. ## Re: relations

Hi,
sorry I didn't answer but if those signs are not universal I have no Idea on how to translate them into english.
sorry

6. ## Re: relations

Originally Posted by ronrong7
Hello,
O.K. Lets say that $R,~S,~\&~T$ are relations on the same set $U$.
$\displaystyle \begin{array}{l} (x,y) \in R \circ (S \cup T)\\ (\exists z \in U)\left[ {(x,z) \in R \wedge (z,y) \in (S \cup T)} \right]\\ (\exists z \in U)\left[ {(x,z) \in R \wedge \left( {(z,y) \in (S) \vee (z,y) \in T)} \right)} \right]\\ \Rightarrow (x,y) \in R \circ S \vee (x,y) \in R \circ T \end{array}$