Let $¬$ be a relation on $\mathbb{R}$ defined by $x¬y$ if $$y-x\in\mathbb{Z}$$
Prove that $¬$ is an equivalence relation on $\mathbb{R}$.
How would one go about doing this? My gut instinct tells me to use the definition of an equivalence relation which in my book is that it's reflexive, symmetric and transitive. Thing is, how would I show any of those to be true?
I have a proposed solution here to this:
- $\underline{\textbf{Reflexivity}}$
Let $y\in\mathbb{R}$. Then $y-y=0$ and $0 \in\mathbb{Z}$, so $y¬y$.
- $\underline{\textbf{Symmetry}}$
Let $a,b \in \mathbb{R}$ and $m\in\mathbb{Z}$. Suppose $a¬b$, then $a-b \in \mathbb{Z}$. Say $a-b=m$ implying $b-a=-(-a-b)=-m$ and $-m\in\mathbb{Z}$. Thus, $b¬a$.
- $\underline{\textbf{Transitivity}}$
Let $a,b,c\in\mathbb{R}$ and $m,n\in\mathbb{Z}$. Suppose $a¬b$ and $b¬c$ meaning $a-b$ and $b-c$ are both in $\mathbb{Z}$. Also suppose that $a-b=m$ and $b-c=n$. Then $a-c=(a-b)+(b-c)=m+n$. Now, $m+n\in\mathbb{Z}$; that is, $a-c\in\mathbb{Z}$. Therefore, $a¬c$.
Any improvements would be welcome. Got an exam tomorrow!
Your proof is correct and well-presented. Good job!
Mostly posting this so this question can be finally considered to have an answer and thus be removed from the unanswered queue. Made it Community Wiki since I have nothing much of substance to add.