$aRb$ if and only if $a=b$ or $a-b=2^n$ for a natural number $n$ Is $R$ reflexive?
Is $R$ symmetric?
Is $R$ transitive?
I know $a=b$ is an equivalence relation so it is reflexive, symmetric, and transitive. I know $a-b=2^n$ is reflexive but not symmetric or transitive.
Not symmetric because:
$6-2=2^2$ but $2-6$ cannot be expressed as $2^{\text{natural number}}$.
Not transitive because:
$6-4=2^1$ and $4-3=2^0$ but $6-3$ cannot be expressed as $2^{\text{natural number}}$.
I did the truth tables for $P$ iff $Q$ or $R$ and the statement is true if either $Q$ and $R$ are both true, $Q$ is true and $R$ is false, or $R$ is true and $Q$ is false.
Therefore, I thought it would be enough to show that because $a=b$ is an equivalence relation, $aRb$ iff $a=b$ or $a-b=2^n$ is an equivalence relation too.
However, I got this question wrong on an exam. 
Do you have to show that both $a=b$ and $a-b=2^n$ are equivalence relations to show $aRb$ is one?
 A: Hint $\ $ You have $\ a\,R\,b \iff a-b \in  \{0\}\cup \{2^n\} =: S.\,$ Suppose more generally that $\,S\subset \Bbb Z\,$ is arbitrary set of integers containing $\,0.\,$ Then checking the equivalence relation properties


*

*reflexive: $\quad\ \ \ a\, R\, a \iff 0\in S,\ $ which is true by hypothesis

*symmetric: $\,\ (a\, R\, b\,\Rightarrow\, b\, R\, a)\iff (a\!-\!b\in S\,\Rightarrow\, b\!-a\!\in  S)$

*transitive: $\ \ \ (a\, R\, b,\ b\,R\,c\,\Rightarrow\, a\, R\, c)\iff (a\!-\!b,b\!-c\in S\,\Rightarrow\,a\!-\!c\in S)$
If so then $\,b=0\,$ in "symmetric" yields $\,a\in S\,\Rightarrow\, -a\in S,\,$ i.e. $\,S\,$ is closed under the operation of negation. Thus $\,a,c\in S\,\Rightarrow\,-c\in S\,$ so substituting $b,c \to 0,-c\,$ in "transitive" yields $\,a+c\in S,\,$ i.e. $\,S\,$ is closed under addition. Conversely, if $\,S\,$ is closed under addition and negation then  then it satisfies the listed implications, so $\,R\,$ is an equivalence relation. If you know group theory then you will recognize this as the subgroup test, i.e. $\,R\,$ is an equivalence relation iff $\,S\,$ is a subgroup of the additive group of integers.
A: You cannot "split" this into two relations, you have to treat it as a whole.  Put it in brackets if it helps:
$$a\,R\,b\quad\hbox{if and only if}\quad (a=b\ \hbox{or}\ a-b=2^n\ \hbox{for some $n\in\Bbb N$})\ .$$
Now if $a=b$ then the RHS is true (because the first part is true), so $a\,R\,b$.  That is, $a\,R\,a$ for all $a$, and $R$ is reflexive.
For symmetry you might think through the problem like this: suppose $a\,R\,b$.  Does it necessarily follow that $b\,R\,a$?  Well, since $a\,R\,b$, one possibility is that $a=b$: then certainly $b=a$ and so $b\,R\,a$.  But what if $a\ne b$?  Then $a-b$ is a power of $2$: does this mean that either $b=a$ or $b-a$ is a power of $2$?  Obviously not for the first, and not for the second either because $b-a$ will be negative.  This should suggest the example that $3\,R\,1$ but $1\,\not R\,3$.  Then all you need to say for the answer is:

$R$ is not symmetric because, for example, $3\,R\,1$ but $1\,\not R\,3$.

In the same way you can show that $R$ is not transitive.  Try it for yourself.
