Expert: Sherry Wallin - 1/3/2009

Question
Let A={1,2,3}  . let R be a relation on A.
Define R={(x,y)}є AXA:x|y}  find R
Soluction .    in this case ,we observe that  R=Φ,which is a empty relation

please describe it .i can not understand this problem  

Answer
Hi Pratap~
    A is the set containing the natural numbers 1,2, and 3. A relation is just a rule that operates on a set. This particular relation R is the set that takes each of 1,2,and 3 and makes all the ordered pairs (x,y): (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3) and where x divides y. In other words 1 divides 1 is true, 1 divides 2 is true, 1 divides 3 is true, 2 divides 1 is false, 2 divides 2 is true, 2 divides 3 is false, 3 divides 1 is false, 3 divides 2 is false, and 3 divides 3 is true. So the only elements in RxR that "fit" are the ones that are true and they are (1,1),(1,2),(1,3),(2,2),(3,3). The notation x|y says "x divides y" and the way you know  this is true is if you can write
y = nx where n is an integer. So from (1,1) you have 1 = 1(1), and from (1,2) you have 2 = 2(1), and from (1,3) you have 3 = 3(1), in (2,2) you have 2 = 1(2), and finally in (3,3) you have 3 = 1(3). (2,1) is false because 1 = (1/2)2 and (1/2) is not an integer so 2 does not divide 1.
Hopefully this explains the concept of relation and the notation divides.

Math Prof