Show that rational cosets are either identical or disjoint

Question

Let $\mathbb{Q}$ denote the set of rational numbers.

Let $x,y \in \mathbb{R}$. Let $A_x = x+ \mathbb{Q} , A_y = y+ \mathbb{Q} $

Can someone help me in simple arguments prove that cosets $A_x, A_y$ are either identical or disjoint?

To prove the disjoint case, observe the only possibility for these sets to be disjoint is that $x \in \mathbb{Q}$, and $y \in \mathbb{R}\backslash\mathbb{Q}$.

To prove the identical case, we need to show that if $A_x, A_y$ overlaps, then $x, y \in \mathbb{Q}$. I doesn't know how to do this! Help!

Answer 1

This is a property not just of cosets of $\mathbb Q$ in $\mathbb R$, but of the cosets of any subgroup in any group whatsoever.

Let $(G,\cdot)$ be a group, and let $H \subseteq G$ be a subgroup. Let $a,b \in G$. We have for the (left) cosets that either $a \cdot H = b \cdot H$ or $a \cdot H \cap b \cdot H = \emptyset$. Suppose that $a \cdot H \cap b \cdot H \neq \emptyset$. So we have $g \in a \cdot H \cap b \cdot H$, and so $g = a \cdot h_1 = b \cdot h_2$ for $h_1, h_2 \in H$, and thus $a = b \cdot h_2 \cdot h_1^{-1}$, and similarly $b = a \cdot h_1 \cdot h_2^{-1}$. Now let $x \in a \cdot H$. This means that $x = a \cdot h$ for some $h \in H$. But since $a = b \cdot h_2 \cdot h_1^{-1}$, we have that $x = b \cdot h_2 \cdot h_1^{-1} \cdot h \in b \cdot H$, and so $a \cdot H \subseteq b \cdot H$. Similarly, if $x \in b \cdot H$, we have $x = b \cdot h$ for some $h \in H$, and so $x = a \cdot h_1 \cdot h_2^{-1} \cdot h \in a \cdot H$, and thus $b \cdot H \in a \cdot H$. Therefore if the cosets are not disjoint, they are equal.

Now, if you're not actually familiar with groups or abstract algebra, you can simply replace $H$ with $\mathbb Q$, $G$ with $\mathbb R$, and $\cdot$ with $+$, (and also replace things like "$h^{-1}$" with "$-h$"), and the proof will work out just fine.

Answer 2

If $r \in A_x \cap A_y$ then $r = x+m_1$ and $r = y+m_2$. And so $x+m_1 = y+m_2$. Therefore we have, $x-y = m_2 - m_1 \in \mathbb{Q}$. Suppose without loss $x$ is rational and $y$ is not. Then it follows that $y = x-(m_2 - m_1) \in \mathbb{Q}$ which is a contradiction. We know need to show that $x,y$ can't both be irrational. Suppose $x,y \in \mathbb{R} \setminus \mathbb{Q}$ and $x-y = r \in \mathbb{Q}$. Then $x = r+y$ but then $r = (r+y)+m_1 \Rightarrow y+m_1 = 0 \Rightarrow y = -m_1 \in \mathbb{Q}$; hence a contradiction. Therefore both $x,y \in \mathbb{Q}$.