An equivalence relation T on $\mathbb{N}$ is defined for all $x,y\in\mathbb{N}$ by
$$xRy \to x=2^ny\;\;\;\text{or}\;\;\; y=2^nx \;\;\; \text{for some non-nagtive integer}\;\;\; n$$
Write down the equivalence class [1] using any set notation.
MMy attempt
Since $1R\frac{1}{2^n}$, but $y$ must be a natural number so we can rule this out.
Since $1R2^n$, by $y=2^nx$, the answer must be
$$[1]=\{2^n, \;\;\;\text{for some non-negative integer}\;\;\; n\}$$
However, my answer shows $$[1]=\{2^n|n\in\mathbb{Z}, n\geq 0\}$$
I would like to clarify, do my answer mean the same thing? I was thinking more of, since the $n$ is fixed at the beginning by the word "for some", and hence, there is only 1 element in the equivalence class.
But it seems the answer is saying that there can be infinitely many elements, like $(2,4,8,16,...)$
The way the equivalence relation is defined, given natural numbers $x$ and $y$, they are equivalent if and only if such an $n$ exists, but $n$ can vary between pairs in the same equivalence class.
The two answers you gave are not equivalent. Or rather I should say that the first answer is not well defined. The second one is correct.
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