prove that $R$ is an equivalence relation

Question

$$\forall a,b \in \mathbb{Q} \quad aRb \Leftrightarrow \quad \exists k \in \mathbb{Z}: \quad b=2^ka$$ 1) Reflexivity:

$\forall a \in \mathbb{Q}\quad aRa \Leftrightarrow \quad \exists k \in \mathbb{Z}: \quad a=2^ka $

choosing $k=0 \quad \Rightarrow a=2^0a=a \Rightarrow aRa \Rightarrow R \text{ is reflexive}$

2) Symmetry

3) Transitivity:

$\forall a,b,c \in \mathbb{Q}:$ $$ aRb \Leftrightarrow \quad \exists k \in \mathbb{Z}: \quad b=2^ka $$ $$ bRc \Leftrightarrow \quad \exists h \in \mathbb{Z}: \quad c=2^hb $$ Then $$ aRc \Leftrightarrow \quad \exists p \in \mathbb{Z}: \quad c=2^pa $$

so $aRb,bRc \Rightarrow c=2^hb=2^h2^ka=2^{h+k}a$

choosing $p=k+h \Rightarrow c=2^pa \Rightarrow aRc \Rightarrow \text{ R is transitive}$

Can anyone confirm that 1) and 3) are correct? I tried to prove 2) but is ended like transitive proof and I think that is entirely wrong, I have no idea how to succeed, can anyone provide some hints/proof/solution?. thanks in advance

Answer 1

Your proofs for parts (1) and (3) are correct. For symmetry, suppose $aRb$, so that \begin{equation} b = 2^ka \end{equation} for some $k\in\mathbf{Z}$. Can you think of an integer $l$ so that \begin{equation} a = 2^lb? \end{equation} (Hint: Remember that negative integers are integers too!) Once you have such an integer, you can conclude that $bRa$, meaning that $R$ is symmetric.

Answer 2

Looks correct to me! (Except for the exponents on the right hand side of the proof for transitivity, which are probably just typos.)

For symmetry, note that if $a R b$, then there is some $k$ for which $b = 2^k a$. But then $a = 2^{-k} b$; hence $b R a$. (Indeed, there exists an $\ell \equiv -k$ for which $a = 2^\ell b$.)