The full question:
Having an equivalence relation $\sim$ on $\Bbb N$ defined by: $a \sim b$ meaning $a,b\in\Bbb N$ such that $a=b*10^k,$ for some $k\in\Bbb Z$, give a complete set of equivalence class representatives.
I am having trouble visualising these. I'm thinking you would need the whole set $\Bbb Z$. Does anyone have any ideas?
Help would be greatly appreciated! (this is my first post as well, so please say if anything is unclear)
On
Here's another answer that may be easier to "visualize". Your equivalence relation says that $a$ and $b$ are equivalent if you can get $a$ by appending zeroes to $b$ or deleting them from the end of $b$. That's the same as saying $a$ and $b$ start with the same sequence of digits before the trailing zeroes (if any) begin). Then the integers with no trailing zeroes represent each class just once.
Since $0$ is in a class by itself, it's the representative of its class.
Those are exactly the numbers in @dREaM 's correct answer.
The numbers that are not multiples of $10$ give us a complete set of representatives, and in fact we have exactly one representative for each class. Let $X$ be the set of all positive numbers not divisible by $10$
To see why it is a complete set of representatives take $n\in\mathbb N$, suppose $10^k$ is the largest power of $10$ dividing $n$ ($k$ might be $0$), notice $n\equiv \frac{n}{10}\equiv\dots\equiv \frac{n}{10^k}\in X$.
To see why it does not contain two elements in the same class notice if $a\equiv b$ then one of $a$ and $b$ is a multiple of $10$.
Edit: I assume $\mathbb N$ does not contain $0$, if we want $0\in\mathbb N$ just notice $0$ would be in a class by itself.