Symmetric relation and definition of even numbers Let a relation $\rho$ be defined on $\mathcal{Z}$(set of integers) by '$a \rho b$ if and only if $a-b$ is even' for $ a,b \in \mathcal{Z}$.
Then, is the above relation symmetric or anti-symmetric?
Can the negative integers be treated as even or odd numbers? Or, is the definition of even and odd numbers only for the positive integers? What actually is the definition of even numbers? Because if I use the definition 'divisible by $2$' for even numbers, then the negative integers cannot be neglected.
Here is where I am getting stuck. Because, based on this basic definition, the above relation will be symmetric or anti-symmetric.
Thanks in advance for anyone who will make this topic clear.
 A: The basis of your question seems to be what the definition of even/odd is.
Given an integer $a$ (regardless of if it is positive, negative, or zero), the following are equivalent staements:


*

*$a$ is even

*$a$ is divisible by $2$

*$a$ is a multiple of $2$

*The remainder of $a$ when divided by $2$ is $0$

*There exists some integer $k$ (possibly positive, negative, or zero) such that $a = 2\times k$

*$a\equiv 0\pmod{2}$

*$\vdots$
Meanwhile, given an integer $a$ (regardless of if it is positive, negative, or zero), the following are equivalent statements:


*

*$a$ is odd

*$a$ is not divisible by $2$

*$a$ is not a multiple of $2$

*The remainder of $a$ when divided by $2$ is $1$

*There does not exist any integer $k$ such that $a=2\times k$

*There does exist some integer $k$ such that $a=2\times k + 1$

*$a\equiv 1\pmod{2}$

*$\vdots$
You have that $\{\dots,-6,-4,-2,0,2,4,6,\dots\}$ are all even numbers (yes, zero too) and you have that $\{\dots,-5,-3,-1,1,3,5,7,\dots\}$ are all odd numbers.
Your relation will indeed be an equivalence relation, and the equivalence classes will be the set of even numbers and the set of odd numbers respectively.  (Note: even minus even is always even, while odd minus odd is always even).
To see the symmetry property, suppose that $a-b$ is even.  Then that means that there is some $k$ such that $(a-b) = 2\times k$.  Then noting that $(b-a)=-(a-b)=-(2\times k) = 2\times (-k)$ you have that $b-a$ is also two times an integer (this time $-k$ instead of $k$, but that is fine).  So, if $a-b$ is even, it follows that $b-a$ must also be even and so the relation is symmetric.
The relation will be transitive as well.  I encourage you to check this yourself, but the proof for it begins: "Suppose that $a-b$ is even as well as $b-c$ is even."  The proof will end "And so we learn that $a-c$ must also be even."
A: If an integer $n$ is even then $-n$ is also even. So if $a-b$ is even then $b-a$ is also even because $b-a=-(a-b)$. So the relation $\rho$ is symmetric.
