Convergence
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Hello. So today in class, we talked a bit about <g> as the set of all integer powers of g. Made enough sense. Then we did some examples in \mathbb{Z} / 7\mathbb{Z}, and I got a bit lost. I think this is more due to the fact that perhaps I don't quite grasp \bmod{n} as an equivalence class. So I know that the elements of \mathbb{Z} / 7\mathbb{Z} are \{ [0], [1], [2], [3], [4], [5], [6] \}, but we were looking for <g> where g = [5]. The idea was we were showing that \mathbb{Z} / 7\mathbb{Z} is cyclic. Since we were "brute-forcing" it so to speak, we wrote down elements in a certain order, which I have written down as \{ [0], [5], [3], [1], [6], [4], [2] \} and then it cycled back to [0].
So I suppose I'm having trouble doing computations in \mathbb{Z} / 7\mathbb{Z}, or \mathbb{Z} / n\mathbb{Z} in general. I mean, I know that if a \equiv b \bmod{n} then n \mid a-b, but in this case I'm not sure where to take it. Is it:
5^0 = 1 \equiv something \bmod{7}, then 5^1 = 5 \equiv something \bmod{7} etc?
Sorry if my question(s) is a bit hard to read; perhaps I didn't explain it as best I could. But any help would be appreciated.
So I suppose I'm having trouble doing computations in \mathbb{Z} / 7\mathbb{Z}, or \mathbb{Z} / n\mathbb{Z} in general. I mean, I know that if a \equiv b \bmod{n} then n \mid a-b, but in this case I'm not sure where to take it. Is it:
5^0 = 1 \equiv something \bmod{7}, then 5^1 = 5 \equiv something \bmod{7} etc?
Sorry if my question(s) is a bit hard to read; perhaps I didn't explain it as best I could. But any help would be appreciated.