# Prove that ¬ is an equivalence relation

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:

1. $\underline{\textbf{Reflexivity}}$

Let $y\in\mathbb{R}$. Then $y-y=0$ and $0 \in\mathbb{Z}$, so $y¬y$.

1. $\underline{\textbf{Symmetry}}$

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, $b¬a$.

1. $\underline{\textbf{Transitivity}}$

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!

• Well done. The idea of problems like this is to verify all the pieces of the definition – Ross Millikan May 9 '18 at 19:30
• Note that if you define f(x) = x-n, where n is the largest integer less than or equal to x, then x¬y is equivalent to f(x) = f(y). Any relation that can be put in the form "x and y get sent to the same output" for some function is an equivalence relation. – Acccumulation May 9 '18 at 19:35
• Looks perfect to me; except for transitivity (and reflexivity), you may first suppose that $a¬b$ and $b¬c$ ($y¬y$), followed by the definition. – poyea May 9 '18 at 19:35
• @poyea I'll change it to reflect that. Thanks! – AustereTiger May 9 '18 at 19:39
• You proof is just fine. It might be interesting to realize that this realation is "have the same fractional part" or "$x - \lfloor x\rfloor = y - \lfloor y\rfloor$" or: if $n \le x < n+1$ and $m \le y < m+1$ where $m$ and $n$ are integers then $x¬y$ means $x -n = y -m$ – fleablood May 9 '18 at 21:35