For the relation R = {(x,y) : x + 2y <= 3}, defined by A = {0,1,2,3}, determine if it is reflexive, symmetric, antisymmetric and transitive.

Aleah Booth 2022-07-16 Answered
For the relation R = ( x , y ) : x + 2 y 3 , defined by A = { 0 , 1 , 2 , 3 }, determine if it is reflexive, symmetric, antisymmetric and transitive.
The 2y is throwing me a bit here. To determine if it is reflexive, I have done the following: x + 2 x 3, which is a no with counter example (1,2), as that would give me (1,4) and be greater than 3. Is every second number doubled? Ie. (x,2x) or (x,2y)?
I have concluded this relation is not symmetric, as it does not imply y + 2 x 3, on the basis that is x = 2   a n d   y = 1, this would result in 2 + 2 ( 1 )   f o r   x + 2 y   b u t   1 + 2 ( 2 ) for y + 2 x, which is greater than 3. I have no confidence in this answer however.
I'm floundering on this one. Every time it was touched on during the lecture, it was simple examples, like x + y is even, or if there was an equality, there wasn't a defined set for it.
For determining transitive, what would I use as z?
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Answers (2)

Reese King
Answered 2022-07-17 Author has 13 answers
Step 1
It might be more helpful to think of the relation like this. Don't forget that, here, R is a subset of A × A . Then
R = { ( x , y ) A × A | x + 2 y 3 }
With that established, the (perhaps) cleaner way of thinking of this:
x and y are related by R (i.e. xRy) ( x , y ) R A × A x + 2 y 3
In testing reflexivity, we seek ( x , x ) R for all x A. By the above, that would require x + 2 x = 3 x 3. As you have seen, there is a counterexample.
Step 2
For symmetry, whenever ( x , y ) R, we want ( y , x ) R. That would mean that, whenever x + 2 y 3, that y + 2 x 3. Again, you have a counterexample to this.
For transitivity, you want whenever ( x , y ) , ( y , z ) R that ( x , z ) R. This would mean that
x + 2 y 3 y + 2 z 3 } x + 2 z 3
I'll tell you right away that there is a counterexample to this as well.
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Tamara Bryan
Answered 2022-07-18 Author has 2 answers
Step 1
I think you got the first two.
Let's try transitive: if ( x , y ) R and ( y , z ) R , then x + 2 y 3 and y + 2 z 3. We need to check if x + 2 z 3. Well, after trying a few things, I think I have a counter example. Namely ( 3 , 0 ) R and ( 0 , 1 ) R . Clearly though, ( 3 , 1 ) R .
Step 2
Thus R isn't transitive.
Finally for antisymmetry, note that ( 0 , 1 ) R . And ( 1 , 0 ) R . But 0 1.
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