Karl Kronenfeld
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 Mar 21 awarded Yearling Mar 5 answered formulating English sentences to logic Nov 21 comment Need help on Linear algebra diagonalization The problem you're experiencing is based on the fact that your eigenvectors are not linearly independent: indeed 2v1+v2=v3. Also, they all belong to the eigenspace for $\lambda=-3$, you still haven't found an eigenvector for $\lambda=5$. Nov 9 comment If D is an Integral Domain and has finite characteristic p, prove p is prime. Cleaner: $0=p^ja^j=(pa)^j$ implies $pa=0$. Aug 6 comment Exercise 2.35 in Lee's introduction to topological manifolds (It's a hobby of mine to come up with overly cumbersome, cute answers) Of course, there's a one-liner. Aug 6 answered Exercise 2.35 in Lee's introduction to topological manifolds Aug 3 answered Homomorphism from $\mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z}$ has non-trivial kernel: elementary argument Jul 21 comment a question about analysis, how to find the largest cardinality in the following examples I cannot see how you have resolved the question. Uncountable sets may have larger or if $\neg$ CH smaller cardinality than $|\mathbb R|$, so you haven't shown that B, for instance, is right or wrong. Jul 20 comment Proving that $\langle \omega+1, \leq \rangle$ and $\langle \omega + \omega^*, \leq \rangle$ are not elementarily equivalent @Nagase Yes. See what you can do to simplify it now. Jul 20 comment Proving that $\langle \omega+1, \leq \rangle$ and $\langle \omega + \omega^*, \leq \rangle$ are not elementarily equivalent @Nagase Even assuming you intended $\le$ in place of $<$, it is incorrect since you have not ensured $z\ne y$. Jul 13 comment Composite Function Equality @Jason The a,b nonzero condition should have a negligible effect. I suspect a counterexample still exists, even given those conditions on $g$. Jul 10 comment Set Theory (Example of Set) Are you capable of labeling them (with distinct labels) somehow? Jul 8 revised Composite Function Equality Polished my answer a little. Jul 8 comment Composite Function Equality Yes, +1 for the statement and proof of the (correct) inequality. Jul 8 comment Composite Function Equality I don't think you can delete accepted answers. An edit explaining the error is typical, though I am no expert on MSE etiquette. Jul 8 comment Composite Function Equality Yes, $f(x)=x$ may have multiple solutions--just have $g$ map all of them to $x^{\ast}$. Jul 8 answered Composite Function Equality Jul 8 comment For signature $s= \{ f\}, n_f=1$, show that if for structures $M, N$, $f^N, f^M$ have only one orbit, of infinite order, then $M,N$ are isomorphic It becomes obviously right, as the guess the OP made would be a valid isomorphism. Jul 8 comment For signature $s= \{ f\}, n_f=1$, show that if for structures $M, N$, $f^N, f^M$ have only one orbit, of infinite order, then $M,N$ are isomorphic @DavidC.Ullrich Well, the first thing OP "didn't tell us" was which way to read an ambiguous, poorly written statement. I suspect the first revision is the problem as stated. There are always going to be conventions/assumptions that appear earlier in the text which an inexperienced mathematician will assume to be standard. The invertibility of a given interpretation of $f$ seems to be one of those, albeit a very odd convention! Jul 8 revised Injective map between power series ring added 13 characters in body