Question

Assume that R and S are relation on a set A. If R and S are transitive is R U S transitive? why?

Answer 1

what is this, abstract algebra?

Answer 2

umm discrete math/ Relations

Answer 3

again, not something I've studied
you are a CS major, right? you sure are hitting the maths hard

Answer 4

nooo I am a math major hahahah ik i am too stupid to be a math major hahaha

Answer 5

I somehow thought you said you were a CS major
now I feel better that you are surpassing me :)
obviously I can't help here though, sorry :/

Answer 6

Thanks Turing :D

Answer 7

good luck pipp-sam... whatever :D

Answer 8

hahahaha

Answer 9

it is given that R{r1, r2, r3 .. } and S{s1, s2, s3 .. } have relation on A{a1, a2, a3 ...}
RxA = {(r,a)}
SxA = {(s,a)}

Answer 10

sorry i had another window open

Answer 11

ohhh i found an answer

Answer 12

lol ... what was it??

Answer 13

I am betting R U S = {r, s}
which will give the same result as RUS x A = {(r,a),(s,a)}

Answer 14

ummm no u show smth diff. wait let me printscreen

Answer 15

Hmm ...
is R ans S really transitive??

Answer 16

no its not and the guy provided a counter example

Answer 17

s false. For example, let A be the integers and xSy be defined as 'x-y is divisible by 2' and xRy be defined as 'x-y is divisible by 3.' Then let T=R U S. Then xTy if x-y is divisible by 3 or divisible 2. So 1T3 and 3T6, but not 1T6.

Answer 18

This is another answer