We are doing some more with relations and this time we are given a relation and told that it is not equivalent. We need to find out which property it does not fulfill. So we at least know there is going to be one, maybe more.
Here is the relation: $aRb$ iff either $a$ mod 4 = $b$ mod 4 or $a$ mod 6 = $b$ mod 6, over $\mathbb{N}$.
So I assume we are supposed to show whether or not its reflexive, symmetric, and transitive?
Reflexive: Suppose $x \in \mathbb{N}$. Then, either $x$ mod 4 = $x$ mod 4 or $x$ mod 6 = $x$ mod 6. Either way, this shows that $xRx$, therefore the relation is reflexive.
Symmetric: Here is where I think I am getting confused. Suppose $x, y \in \mathbb{N}$. Then $x$ mod 4 = $y$ mod 4 and $y$ mod 4 = $x$ mod 4. The same thing applies for $x$ mod 6 = $y$ mod 6. Therefore, the relation is symmetric? (Is that right?)
Transitive: If $xRy$ and $yRz$ then, $xRz$. So, if $x$ mod 4 = $y$ mod 4 and $y$ mod 4 = $z$ mod 4, then would $x$ mod 4 = $z$ mod 4? If so, then this property is transitive as well.
So to me, I am getting that is an equivalence relation since I got it to be reflexive, symmetric, and transitive. So I know I am doing something wrong since the directions clearly state that it is not one. I am not sure on the symmetric portion. I think that's where I am messing up. Any help?
But for the transitive part, remember what your relation says. It says that $a\pmod 4 = b\pmod 4$ OR $a\pmod 6 = b\pmod 6$. So if $xRy$ and $yRz$, you can have that $x\pmod 4 = y\pmod 4$ and that $y\pmod 6 = z\pmod 6$. Do you then have that $xRz$?
I hope this clears it up, otherwise let me know.