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 Sep 3 answered Method of Least Squares-Why is it preferred? Aug 31 comment Intuitive significance open sets (and software for learning topology?) @MichaelGreinecker: limit points are not unique in arbitrary topologies. Aug 31 comment Intuitive significance open sets (and software for learning topology?) @rafiki: yes, many $\varepsilon$-$\delta$ definitions can be rewritten in terms of open and closed sets. Aug 31 comment Intuitive significance open sets (and software for learning topology?) @MichaelGreinecker: limit points are rather shaky in arbitrary topological spaces. Aug 31 comment Intuitive significance open sets (and software for learning topology?) @BabyDragon: also, metrically you usually define open sets first and then closed sets, since it is not quite as trivial to define closed sets in terms of "for any $x$ there is a ball of radius $\varepsilon$"-type sentences. Aug 31 comment Intuitive significance open sets (and software for learning topology?) @BabyDragon: you are right; it's just that some things take 2 fewer characters to define in terms of open sets. Aug 31 answered Intuitive significance open sets (and software for learning topology?) Aug 29 comment Proving $(x^s)^t=x^{st}$ In fact, there is a mistake in your induction. To be proven in the second step should be that $(x^s)^{t+1}=x^{s(t+1)}$ for all $s$. Aug 29 answered Show that $z^{-1} = \frac{\bar z}{|z|^2}$ Aug 29 comment Use a particular method to prove if $a^n \mid b^n$ then $a\mid b$ what do you mean by $(a,b)$? a representation of $a/b\in\mathbb{Q}$? or the gcd of $a$ and $b$? Aug 29 comment If $A+A^T$ is negative definite, then the eigenvalues of $A$ have negative real parts? Fabian: if I'd have a proof I'd give it. These facts can be concluded straight away. Of course the complicating factor here is that $A$'s eigenspaces are different from $A^T$'s. Aug 29 comment If $A+A^T$ is negative definite, then the eigenvalues of $A$ have negative real parts? $A+A^T$ is symmetric, thus diagonalizable, and since it is positive definite it is thus similar to a diagonal matrix with strictly positive eigenvalues. $A$'s eigenvalues are equal to $A^T$'s. Aug 27 comment Products of groups with no non-abelian quotients Nicky, that's a more interesting question but one I suppose is difficult to answer. Do all groups have nontrivial normal subgroups? Aug 27 comment Decomposition of Permutation Group Agreed, I'll leave this in place for reference. Aug 27 revised Decomposition of Permutation Group added 154 characters in body Aug 27 comment Existence of an Infinite Length Path You are right, I am a Mathematica noob and entered the integral incorrectly. Aug 27 comment Existence of an Infinite Length Path TonyK, your current arc length is bounded on $[0,1]$, according to mathematica. Aug 27 comment Existence of an Infinite Length Path Your $f$ is not a function, or not differentiable at $x=0$. Aug 27 comment An operator $T:\mathbb{R}^4\to \mathbb{R}^4$ such that $T$ has no (real) eigenvalues. Okay, you can. But then you need to prove this. More specifically, you would have to give an example or otherwise prove the existence of such $M$. Your answer is not very helpful like this. You translated the problem of finding $T$ to a much harder question of finding the right $M$. Aug 27 revised Decomposition of Permutation Group added 315 characters in body