Question

For the following relation on the set {x:x∈Ζ and 1 ≤ x ≤ 12}. 

List:

(a) the ordered pairs belonging to the relation:
(b) the transitive closure

S = {(x, y): 2x = 3y}

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Answer 1

We know that x can be any number from 1 to 12. If 2x must be equal to 3y, then 2x must be a multiple of 3. Here's the list of all possible values:

`\begin{array}cx&2x\\1&2\\2&4\\3&6\\4&8\\5&10\\6&12\\7&14\\8&16\\9&18\\10&20\\11&22\\12&24\\\end{array}`

So, the only values for x when 2x is a multiple of 3 are

`x=3 \implies 2x=6=3\cdot 2\\x=6\implies 2x=12=3\cdot 4\\x=9\implies 2x=18=3\cdot 6\\x=12\implies 2x=24=3\cdot 8`

So, the ordered pairs are

`(3,2),\ (6,4),\ (9,6),\ (12,8)`

For the transitive closure, you have to look for pairs like (a,b), (b,c), and add the pair (a,c), so that the relation becomes transitive. In our example, we only have the pairs (9,6) and (6,4), so we must add the pair (9,4) in order to make the relation transitive. The transitive closure is thus

`(3,2),\ (6,4),\ (9,6),\ (12,8),\ (9,4)`