Is R reflexive? symmetric? transtitive? Define a relation on $\mathbb{Z}$ as $xRy$ if $|x-y| < 1$
Is $R$ reflexive? symmetric? transitive?
Okay so my reasoning is as such:
$|x-y| < 1$ only when $x=y$
$R$ exists if $x=y$
If $x=y$ the relationship is reflexive.
If $x=y$ the relationship is symmetric.
If $x=y$ the relationship is not transitive.
Counterexample of non-transitivity:
$(a,a) \wedge (b,b) \Rightarrow (a,b)$ is false.
Is my reasoning correct?
 A: Not quite. It is true that
$$
|x-y| < 1 
$$
if and only if
$$
x = y
$$
for integers.
So $R$ is actually equivalent to $=$, and hence is all three of reflexive, symmetric, and transitive.
For $xRx$, we note that $|x-x| = 0 < 1$. So $R$ is reflexive.
Also if $xRy$ then $x = y$, so $yRx$. So $R$ is symmetric.
Finally if $xRy$ and $yRz$ then $x = y$ and $y = z$, so $x = y = z$ and $xRz$. So this is transitive too.
A: As you note if $x\neq y$ then $|x-y|$ is an integer greater than zero and hence $(x,y)\not\in R$
If $x=y$ then $|x-y|=0\in R$. So $R$ is the equality relation, which is know to be reflexive symmetric and transitive.
A: You have the right idea about the relation in general - it is reflexive and symmetric given that equality is. However, it is also transitive - you seem to be taking transitivity as $(x,y)$ and $(z,w)$ being in $R$ implies that $(x,w)$ is as well - as in your last question, one must remove the middle element for transitivity, but it is necessary that the middle element be the same. So, this fails in the case of trying to use transitivity on $(a,a)$ and $(b,b)$ since we might see the middle elements as being $a$ and $b$ - and removing them gives $(a,b)$, but transitivity only would let us do that if $a$ and $b$ were the same (which they, in general, are not).
To be sure, we'd need something of the form $(a,b)$ and $(b,c)$ being in $R$, but not $(a,c)$ to break transitivity - and since, in this case, $a=b$ and $b=c$, as you show, it does follow that $a=c$ and hence that $(a,c)$ is in $R$. 
