Nayuki Minase
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 Apr 16 comment Prove that $\lfloor \lfloor x / a\rfloor / b \rfloor = \lfloor x / (ab) \rfloor$ @MathGems: I looked at your proof, but I'm not convinced about why those bi-implications are true. I found the proof by Brian M. Scott to be better explained and easier for me to understand. Apr 13 comment Prove that $\lfloor \lfloor x / a\rfloor / b \rfloor = \lfloor x / (ab) \rfloor$ Wow André, that was a pretty terse hint. But with some work, I was able to expand it into a full solution. Thanks! Apr 13 comment Prove that $\lfloor \lfloor x / a\rfloor / b \rfloor = \lfloor x / (ab) \rfloor$ You're right. Though I feel embarrassed about asking a duplicate question, I'd like thank you very much! Apr 13 asked Prove that $\lfloor \lfloor x / a\rfloor / b \rfloor = \lfloor x / (ab) \rfloor$ Mar 3 awarded Commentator Mar 3 comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ I mean introduction of a new variable, whose role and meaning is unexplained. Mar 3 comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ I am extremely uncomfortable with how much you left implied, such as the instantiation of $K$, the references to GCD, and the very non-obvious invocation of Fermat's little theorem for $K$. Only after rereading your proof a few times on separate days did I start to believe that it plausibly leads to an answer at all. -1. Mar 3 comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ I didn't understand how your answer fits with my question. Please see the other answers as examples of proofs that I did understand. Mar 2 comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ Other than that, I'm sorry but I don't understand your argument at all. Mar 2 comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ Why are all your variables in uppercase? Feb 29 accepted Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ Feb 28 awarded Editor Feb 28 revised Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ Added better answer Feb 27 comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ Alternate explanation: $p$ divides $2^n-1$, so $2^n-1$ divided by $p$ leaves no remainder, thus $2^n-1 \equiv 0 \mod p$. Feb 27 comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ Ohh right, I understand now. I guess I'm not comfortable enough with modular arithmetic that such an elementary fact slipped past me, heh. Feb 27 comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ You said "$p$ divides $2^n-1$, so $2^n \equiv 1 \pmod p$". How does this step work? Feb 27 answered Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ Feb 27 asked Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ Aug 11 comment Is there an everywhere discontinuous increasing function? I admit in full honesty that I asked this question out of curiosity; it is not a homework problem for me as my calculus course did not go this far. Aug 11 awarded Student