some relation $R$ is defined on $\mathbb{R}$ such that $xRy \iff x = 7^{k}y,$ for some $k\in \mathbb{Z}$. Prove that $R$ is an equivalence relation some relation $R$ is defined on $\mathbb{R}$ such that $xRy \ \iff  \ x = 7^{k}y,$ for some $k\in \mathbb{Z}$. Prove that $R$ is an equivalence relation
I'm confused with proving that it is symmetric and transitive.
 A: If $x\sim y$ and $y\sim z$ then we have $x=7^ky$ and $y=7^nz$ so $$x = 7^k(7^nz) = 7^{k+n}z$$ so it is transitive.
Also $x\sim y$ then we have $x=7^ky$ so $y =7^{-k}x$ so it is also a symmetric.
A: Symmetric: If $x\sim y$ then there exists an integer $k$ so that $x=7^ky$  Then $y=7^{-k}x$ So $y\sim x$. 
Transitive: Suppose $x\sim y$ and $y\sim z$ so that there exist integers $k,m$ such that $x=7^ky$ and $y=7^mz$, from which it follows that $x=7^{k+m}z$. Thus $x\sim z$
Reflexive: For every $x$, we have $x\sim x$ because $x=7^0x$. 
So this is an equivalence relation. 
A: This holds for $\,G = 7^{\large \Bbb Z}$ and also for any set with the same algebraic (group) structure (closed under products & inverses). The proof is  more conceptual (and just as easy) when done this way.
Suppose that $\,G\subset \Bbb R\,$ satisfies $\,\color{#0a0}1\in G\,$ and $\,g,h\in G\,\Rightarrow\, \color{#08f}{gh}\in G,\,$ and $\, \color{#c00}{g^{-1}}\in G$ 
Then $\ x\approx y \!\overset{\rm def}\iff\! x = g\, y\ \ {\rm for\ some}\ \ g\in G\ $ is an equivalence relation, $ $  since
$\qquad\ \ \ \  \approx\,$ is $\ \ \rm\color{#0a0}{reflexive}\,\ \ $ by $\  x = \color{#0a0}1x\,\Rightarrow\, x\approx x$
$\qquad\ \ \ \  \approx\,$ is $\,\rm\color{#c00}{symmetric}\,$ by $\ x\approx y\,\Rightarrow\, x = g y\,\Rightarrow\, y = \color{#c00}{g^{-1}}x\,\Rightarrow\,y\approx x$
$\qquad\ \ \ \ \approx\,$ is $\ \rm\color{#08f}{transitive}\,\ $  by $\,x\approx y\approx z\,\Rightarrow\, y = hz,\,  x = g y = g(hz) = (\color{#08f}{gh})z\,\Rightarrow\, x\approx z$
Remark $\,\ G x = \{ gx\ :\ g\in G\}\ $ is called the $G$-orbit of $x.\ $ It is a basic concept in group theory. 
A quick way to recognize such group structure is by the subgroup test, i.e. a nonempty $G\subset H$ of a group $H$ forms a group $\iff$ it is closed under division, i.e. $\, g,h\in g\,\Rightarrow g/h = gh^{-1}\in G,\,$ which is clear for $\, G= 7^{\Large \Bbb Z} $
 since $\, 7^{\large j}/7^{\large k} = 7^{\large j-k}\!\in 7^{\large \Bbb Z}$ since integers are closed under subtraction. In fact this subgroup test is implicitly used in a complementary form since grade school in inferences like below
$\qquad$ 
