dashdart
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 Apr8 revised Mathematical Career Advice to a young 16 year wannabe mathematician removed personal information. minor edits otherwise Apr8 revised A Young Math Enthusiast's Fear retracted personal information, and minor edits Mar19 awarded Notable Question Mar16 revised Assume that $1a_1+2a_2+\cdots+na_n=1$, where the $a_j$ are real numbers. Formatting Mar16 suggested approved edit on Assume that $1a_1+2a_2+\cdots+na_n=1$, where the $a_j$ are real numbers. Dec10 awarded Caucus Oct28 comment finding numbers to make both sides equal @bof Sorry! I meant the largest. Silly error! Oct28 comment finding numbers to make both sides equal Either by solving $1000=x(x+1)(x+2)$ or by brute-force, you'll find out that $45$ is the smallest $x$ that works. So, anything below $45$ is fair game. Oct17 comment Solving an exercise in Pinter's Abstract Algebra The set you describe is called the normalizer of $H$ in $G$. Check out proofwiki.org/wiki/Normalizer_is_Subgroup Sep30 awarded Explainer Sep12 awarded Popular Question Sep11 awarded Notable Question Jul24 awarded Good Question Jul2 awarded Curious May28 awarded Yearling May24 awarded Good Answer May2 comment Piecewise bijection $f: \Bbb R \to (\Bbb R$ \ $\{1\})$ @user129120 Wow, My bad! For some reason, I thought you were asking for a function $f:\mathbb R \setminus \{1\}\to \mathbb R$. However, Since the function I proposed is a bijection, its inverse is a function from $\mathbb R \to \mathbb R \setminus \{1\}$. Kind of a cheat but still! May1 comment Piecewise bijection $f: \Bbb R \to (\Bbb R$ \ $\{1\})$ Is there a particular reason why you would need a piece-wise function? Then you could use $\ f:x\to \frac 1{1-x}$. Feb26 revised Prove that $e ^ π$ > $π ^ e$. Formatting Feb26 suggested approved edit on Prove that $e ^ π$ > $π ^ e$.