Prove that this set (involving fractional part of any rational number) is a partition of the set of rationals.

Question

For any rational number $x$, we can writte $x=q+\,n/m$ where $q$ is an integer and $0\le n/m<1$. Call $n/m$ the fractional part of $x$. For each rational $r\in \{x : 0\le x<1\}$ ,let $A_r = \{ x\in \Bbb Q: \text{the fractional part of $x$ is equal to}\; r\}$.

Prove :$\{A_r : 0\le r<1\}$ is partition of $\Bbb Q$

Answer 1

HINT: To prove that $\{A_r:r\in\Bbb Q\cap[0,1)\}$ is a partition of $\Bbb Q$, you must show that each $A_r$ is a non-empty set of rational numbers, and that every rational number belongs to exactly one $A_r$. It’s clear from the definition that each $A_r\subseteq\Bbb Q$, and you should have no trouble exhibiting a particular rational number that belongs to $A_r$. That leaves you only two things to show:

1. each rational number belongs to some $A_r$ with $r\in\Bbb Q\cap[0,1)$, and
2. if $r,s\in\Bbb Q\cap[0,1)$, and $r\ne s$, then $A_r\cap A_s=\varnothing$.

The first sentence of your question gives the justification for (1), so really all that’s left is to prove (2). I suggest proving the contrapositive: if $x\in A_r\cap A_s$, then $r=s$.