# 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.

• For symmetric, note that $k$ can be negative. – wjm Jul 8 '19 at 13:56
• For transitive, note that $7^a7^b=7^{a+b}$. – wjm Jul 8 '19 at 13:57

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.

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.

This holds not only for $$\,G = 7^{\large \Bbb Z}\,$$ but also for any set with the same algebraic (group) structure (closed under products & inverses) and the proof is more conceptual (and just as simple) 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 \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

$$\qquad$$