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

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  • $\begingroup$ (1) and (3) look correct to me. Also, for the symmetric proof, note that $aRb \to b = 2^k a \to a = 2^{-k}b$, also note that if $k$ is an integer, then $-k$ is also an integer, and therefore $aRb \to bRa$ $\endgroup$ – Rob Bland Sep 2 '16 at 21:30
  • $\begingroup$ In (3) you don't necessarily have the same $\;k\in\Bbb Z\;$ for both cases. Yet afterwards you uszxe $\;h,k\;$ so it is fine. $\endgroup$ – DonAntonio Sep 2 '16 at 21:32
  • $\begingroup$ @DonAntonio I'm guessing this was a typo on the part of the OP. $\endgroup$ – 211792 Sep 2 '16 at 21:35
  • $\begingroup$ yes was a typo error, just fixed. thanks to everyone! $\endgroup$ – Alfonse Sep 2 '16 at 21:59
  • $\begingroup$ @Alfonse, I think you will also want to change $c=2^k a$ to $c=2^p a$ ? $\endgroup$ – user326210 Sep 2 '16 at 22:19
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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 $a = 2^\ell b$.)

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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.

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  • $\begingroup$ I totally forgot to think to "negative subset" of integers, thanks a lot! $\endgroup$ – Alfonse Sep 2 '16 at 21:55

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