A particular relation $\star$ is defined on $\mathbb{N}$ by $x\star y$ iff $\exists k\in\mathbb{Z}$, $y=5^k x$. Is $\star$ an anti-symmetric relation?

Question

Its asked for other things too like symmetry, etc. but I'm confused on how to go about anti-symmetry, if I pick $k = 0$ then $x = y$ and $y = x$, is this enough to say yes? On the other hand I can pick $y = 5$ and $x = 1$, clearly $x$ relates $y$ and $y$ relates $x$ but $k = 1$ and $k = -1$ but $x$ does not equal $y$ and is thus not anti symmetric. So in summary I feel like it can go both ways and I'm not sure how to proceed.

Answer 1

The relation $\star$ is antisymmetric if and only if $$a\mathrel{\star}b \text{ and } b\mathrel{\star}a \implies a=b$$

So let $a,b$ be two integers so that $a\mathrel{\star}b$ and $b\mathrel{\star}a$. There exists $k,k'\in\mathbb{Z}$ so that $a = 5^{k}b$ and $b=5^{k'}a$.

So, you have $a=5^k b = 5^k(5^{k'}a) = 5^{k+k'}a$.

First case, $a=0$, then $b = 5^{k'}a = 0$ and $a=b$.

Second case, $a\neq 0$, then $5^{k+k'}=1$, i.e. $k+k'=0 \iff k=-k'$.

From this point, to prove that $\star$ is antisymmetric, we would need to prove that $k=k'=0$ (and $a=b$).

However, we can immediately build a counterexample with $a=10$ and $b=2$ : $a= 5 \times 2 = 5b$ and $b = \frac{10}{5} = 5^{-1}a$. We have $10\mathrel{\star}2$, $2\mathrel{\star}10$, and yet $2\neq 10$. So $\star$ is indeed, not antisymmetric.