Determine whether this relation is reflexive, symmetric... Determine whether this relation $R$ on the set of all integers is reflexive, symmetric, anti-symmetric and/or transitive where $x\,R\,y$ iff 
$x = y + 1$  or $x = y-1$


*

*It is not reflexive: 
Let $x = 2$: $2\neq 2 + 1$  and $2 \neq 2 - 1$.

*It is symmetric:
If $x = y + 1$ then $y = x - 1$
and if $x = y - 1$ then $y = x + 1$.

*It is not anti-symmetric:
Let $x = 3$ and $y = 2$;
then $3 = 2 + 1$ ($x\,R\,y$) and $2 = 3 - 1$ ($y\,R\,x$) 
And let $x = 2$ and $y = 3$;
then $2 = 3 - 1$ ($x\,R\,y$) and $3 = 2 + 1$ ($y\,R\,x$)
but $3\neq 2$.
Can anyone prove whether this relation is transitive or not?
thanks.
 A: You did fine with reflexivity, and with symmetry and antisymmetry.
Now, let's look at transitivity:
We can summarize the relation as follows: $xRy$ if and only if $x$ and $y$ differ by $1$. 
So, suppose $xRy$ ($x$ and $y$ differ by one) and $yRz$ ($y$ and $z$ differ by one), 
What may be the case about the difference between $x$ and $z$? 


*

*(Suspect a counterexample exists: just find $x, y, z$ such that $x = y - 1, y = z - 1 \implies x = z - 2$.
Or, vice versa, $x = y+1, y = z+1 \implies x = z+2)$


Let $x = 0$, $y = 1$, and $z = 2$, so we certainly have $x, y, z \in \mathbb Z$


*

*Clearly, $x = y - 1$ since $0 = 1-1$, so $x R y$, 

*And $y = z - 1$, since $1 = 2 - 1$, so $y R z$. 

*But it is not true that $x = z + 1 $, since $0 \neq  2+1 = 3$  nor
does $x = z - 1$, since $0 \neq 2 - 1 = 1$.


Hence, $x$ is not related to $z$, and transitivity fails.
All we need is one counterexample to prove that a relation is non-transitive, and we've just found one such couterexample
A: To disprove, you only need to find a counter example.
Clearly $R 0 1$ and $R 1 0$ hold, but then it would also have to be the case that $R 0 0$ holds, which is not true.
In general, transitive and symmetric would imply reflexive (on its domain) [which would be a contradiction to $R$ not reflexive for 2, which is in the domain]: $Rxy$ implies $Rxy$ by symmetry, from which $Rxy \wedge Ryx \implies Rxx$ by transitivity.
