Show $xRy$ iff $\exists k \in \Bbb Z$ such that $x=(10^k)y\in\Bbb Z$ is an equivalence relation.

Question

Define a relation $R$ by $xRy$ iff $\exists k \in \Bbb Z$ such that $x=(10^k)y\in\Bbb Z$. Prove R is an equivalence Relation

I know how to show it is symmetric but can't figure out how to write out the reflexive and transitive portion

Answer 1

$R$ is reflexive if $xRx$ for any $x$ in your set (which you haven't specified). Now $xRx$ if and only if $x = 10^k x$ for some $k\in\Bbb{Z}.$ Take $k = 0$ to see that $x = 10^0 x = x,$ so $R$ is reflexive.

$R$ is transitive if $xRy$ and $yRz$ imply $xRz.$ We need to show that $x = 10^k z$ for some $k\in\Bbb{Z},$ and we know that $x = 10^r y$ and $y = 10^s z$ for some $r,s\in\Bbb{Z}.$ Thus, $$ x = 10^r y = 10^r(10^s z) = 10^{r+s}z. $$ Since $r$ and $s$ are integers, $r + s$ is as well, and this shows $xRz.$

Answer 2

Keep in mind $k$ can be $0$ or negative. Then just do algebraic manipulation.

Reflexive. $x = 1*x = 10^0x$ so $x Rx$ ($0$ is an integer).

Symmetric. $x R y\implies x = 10^k y$ for some integer $k\implies y=10^{-k}x$ so $y R x$ (If $k$ is an integer so is $-k$)

Transitive $xRy$ and $yR z\implies $ there are $k, j$ so that $x = 10^k y$ and $y = 10^jz$ so $x= 10^k = 10^k*10^j z = 10^{k+j}z \implies xR z$ (If $k,j$ are integers so is $k+j$).