Assignment 1: Problem 3a

[Anne S]

This is just a starting place...

The set (x,y) belongs to R if x+2y is divisible by 3 [that's 3|(x+2y)
in this week's notation]
is composed of 3 subsets (or cases).
-- x=y belongs because x+2x = 3x and is divisible by 3.
-- 3|x && 3|y belongs because:
3|x --> x=3x' and 3|y --> y=3y'
(3x') + 2(3y') = 3(x'+2y') and is divisible by 3.
-- Let z be a member of the set of integers.
for any x, if y = 3z+x, then (x,y) is a member of the set
because x+2(3z+x) = x+2(3z)+2x = 3(x + 2z) and is divisible by
3.

[Aaron Albers]

Set A is reflexive if (a, a)  R for all a  A

A = { (x, y)  R | 3 divides x + 2y }    ->  A =  { (x, x)  R | 3 divides x + 2x } = { (x, x)  R | 3 divides x(1 + 2) } = { (x, x)  R | 3 divides 3x }  for all x  A

[Anne S]

Set A is symmetric because x=y is symmetric; (x=3x', y=3y') is
symmetric; and (x, y=3z+x) is symmetric.

On Sep 15, 8:43 pm, Aaron Albers <aaroncalb...@gmail.com> wrote:
> Set A is reflexive if (a, a) [image: \in \!\,] R for all a [image: \in \!\,]
>  A
>
> A = { (x, y) [image: \in \!\,] R | 3 divides x + 2y }    ->  A =  {
> (x, x) [image:
> \in \!\,] R | 3 divides x + 2x } = { (x, x) [image: \in \!\,] R | 3 divides
> x(1 + 2) } = { (x, x) [image: \in \!\,] R | 3 divides 3x }  for all x [image:
> \in \!\,] A
>

[Aaron Albers]

So it is both reflexive and symmetric. If it is also transitive then it is equivalent.

[Diane]

3|x+2y is not anti-symmetric.  By example:

x = 5 and y = 2;

then 3|(5+2(2)) and 3|(2+2(5)) but a not equal to b.