Showing transitivity of a relation for a set

Question

I have to check if the following set is an equivalence relation:

Set: $A=\mathbb{R}$
Relation: $x \sim y$ if $x=ay$ for some $a \in \mathbb{Q}\backslash \{0\}$

I know I have to use the reflexive, symmetric and transitive properties here.

It's reflexive since for $x \sim x$, we have $x=ax$ which is true if $a=1$. It's symmetric because for all $x,y \in A$ and $a,b \in \mathbb{Q}$, we have: $x=ay$, $y=bx$which implies that $y=(ba)y$. Since $0$ is not included in $\mathbb{Q}$, this is always possible since we will never have a division by $0$ error or some kind of statement which does not make sense.

How do I show transitivity though?

I believe this means I have to show that if $x=ay$ and $y=bz$ then $x=cz$ for $a,b,c\in\mathbb{Q}$.

Is my understanding right? How would I proceed from here?

Answer 1

Yes, you are right. You need only to use existence quantifier: if $\exists a, x=ay$ and $\exists b,y=bz$, then $\exists c = a\cdot b$ for which $x=ab \cdot z=c\cdot z$.

For symmetry: if $\exists a, x=ay,a\ne 0$, then $y = \frac{1}{a}x$.

Answer 2

Well, for all $s,t\in\Bbb R$, we define $s\sim t$ as $\exists a\in\Bbb Q{\smallsetminus}\{0\}~(s=at)$.   So obviously we will be looking to affirm the existence of such a non-zero rational factor relating real terms.

Reflexivity: $\checkmark$

• The multiplicative identity is a non-zero rational number. That is $1\in\Bbb Q{\smallsetminus}\{0\}$.
• So for any $x\in\Bbb R$ we have $x=1x$.
• Therefore $\forall x\in \Bbb R~.(x\sim x)$

Symmetry: $\checkmark$

• When $a\in\Bbb Q{\smallsetminus}\{0\}$ then its multiplicative inverse, $a^{-1}$, is also an non-zero-rational number. That is: $a^{-1}\in\Bbb Q{\smallsetminus}\{0\}$.
• Thus for any $x,y\in\Bbb R$ and such $a$, where $x=ay$, then we have $y=a^{-1}x$.
• Therefore: $\forall x\in\Bbb R~\forall y\in\Bbb R~.(x\sim y\to y\sim x).$

Transitivity: $\checkmark$

• For any $a,b\in\Bbb Q{\smallsetminus}\{0\}$ then their product is also a non-zero rational. That is: $ab\in\Bbb Q{\smallsetminus}\{0\}$.
• Thus for any $x,y,z\in\Bbb R$ and such $a,b$, where $x=ay$ and $y=bz$, then we have $x=ab\,z$.
• Therefore $\forall x\in\Bbb R~\forall y\in\Bbb R~\forall z\in\Bbb R~.((x\sim y\land y\sim z)\to x\sim z).$