Consider the relation on the set of all real numbers $\mathbb R$, defined by $a\sim b$ if, and only if, there exists an integer $k$ so that $a = 10^k b$. Prove or disprove: this is an equivalence relation.
I have no idea what it's talking about, anyone understand this ?!
The question is asking you to prove that a relation, $\sim$, is an equivalence relation. A binary relation is technically a subset of the cartesian product of a set on itself. Although the words may seem foreign, you've been using relations your whole life. Think of it as a way of "comparing" things. For example, $5<7$, we say that $5$ is "less than" $7$, and with respect to the relation "$<$" we would say that $5$ is related to $7$.
(Note that we don't require under normal circumstances that it be symmetric. We can have $5<7$ but $7\not <5$)
The specific relation given to us in this problem is $a\sim b$ iff there exists some integer $k$ such that $a=10^k b$. For example, $37\sim 3700$ since $37 = 10^{-2}\cdot 3700$. Also $890001\sim 890.001$ since $890001=890.001\cdot 10^3$.
To prove something is an equivalence relation you must show the following three things:
Reflexivity: Suppose $a\in\mathbb{R}$. Does there happen to be an integer $k$ such that $a=10^k a$?
Symmetry: Suppose that $a\sim b$. So, that means that $a=10^k b$ for some integer $k$. Does there happen to be a good choice for $l$ so that $b=10^l a$?
Transitivity: Suppose that $a\sim b$ and $b\sim c$. So that means there is some integer $k$ and $l$ such that $a=10^k b$ and $b=10^l c$. Does there happen to be a good choice for $m$ that makes it so that $a=10^m c$?