$$R=\left\{(f,g)\,\Bigg|\, \exists c\in\Bbb Z,\forall x\in\Bbb Z, \frac{f(x)}{g(c)}\le 2\right\}$$
I can show that this relation is reflexive by showing that $(f,f)$ is in $R$ and so $f(x)/f(c) \le 2$, for $c=x$
But i can't seem to find a way to show this relation is also symmetric and transitive.
Any help would be appreciated! Thanks.
As Brian pointed out, $R$ is not an equivalence relation because it is not reflexive.
To see why $R$ is not symmetric either, consider the constant functions where for all $n\in \mathbb{Z},\text{ }f(n)=1$ and $g(n)=3$. Then certainly we know that $(f,g) \in R$; for example, if we choose $c=7$, then for all $x \in \mathbb{Z}, \dfrac{f(x)}{g(c)}=\dfrac{f(x)}{g(7)}=\dfrac{1}{3}\le2$. Yet we have $(g,f) \notin R$ since for all $c\in \mathbb{Z}$, there exists an $x \in \mathbb{Z}$ (for example, $x=9$) such that $\dfrac{g(x)}{f(c)}=\dfrac{g(9)}{f(c)}=\dfrac{3}{1}>2$.
To see why $R$ is not transitive either, consider the constant functions where for all $n\in \mathbb{Z},\text{ }f(n)=4,\text{ }g(n)=2,$ and $h(n)=1$. Then since $\dfrac{4}{2}=\dfrac{2}{1}=2\le2$, we know by a similar argument from above that $(f,g),(g,h) \in R$. Yet we have $(f,h) \notin R$ since no matter what values of $c$ and $x$ are chosen, $\dfrac{f(x)}{h(c)}=\dfrac{4}{1}=4>2$.