Prove that $aRb$ if $a = 2^kb$ is an equivalence relation.

Question

Let $R$ be a relation on the set of integers given by $aRb$ if $a = 2^kb$, for some integer $k$. show that $R$ is an equivalence relation.

I don't understand how it will be equivalence. Is it the case that it should be reflexive, symmetric and transitive at least one value of $k$?

e.g. Relation will be reflexive for $k=0$

Could someone help me with this?

Answer 1

The equivalence relation is: $aRb$ if there exists some $k$ such that $a=2^kb$. This is what is meant by the phrase "some integer k".

To prove that $R$ is an equivalence relation, you are allowed to choose the $k$. So for example, $aRa$ holds since there exists a $k$, namely $k=0$, such that $a=2^kb$.

Answer 2

1) reflexive

$a=2^0\cdot a \Rightarrow aRa$

2) symmetric

if $a=2^kb$ then $b=2^{-k}a,$ where $k;-k \in \mathbb Z$

3) transitive

if $a=2^kb$ and $b=2^mc$ then $a=2^{k+m}c$ where $k;m, k+m \in \mathbb Z$