Prove that if $a \mid b$, then $(2^a-1) \mid(2^b-1)$.
So, I've said the following:
- Let $a \mid b$.
- $ \implies b=am$
- Assume $(2^a-1) \mid (2^b-1)$
- $ \implies (2^b-1)=(2^a-1)x$
But I can't figure out what to do with any of this from here forward.
Prove that if $a \mid b$, then $(2^a-1) \mid(2^b-1)$.
So, I've said the following:
But I can't figure out what to do with any of this from here forward.
I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...
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