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comment Every module over a division ring is free?
Have you tried going through your favorite proof that every vector space has a basis, and checking whether it will work for arbitrary division rings? If it does, then you've answered your question. If it doesn't, then pinpointing the step that fails will give you a crisper target for constructing a counterexample.
12h
comment Selection of distinct (positive factors) of 50
Hint: the factors come in pairs of two that each multiply to 50.
15h
revised If $\bf{R=\left\{(x,y): x\; is \; wife\; of\; y\right\}}.$ Then $\bf{R}$ is Transitive or not
edited body
15h
revised If $\bf{R=\left\{(x,y): x\; is \; wife\; of\; y\right\}}.$ Then $\bf{R}$ is Transitive or not
added 245 characters in body
15h
revised If $\bf{R=\left\{(x,y): x\; is \; wife\; of\; y\right\}}.$ Then $\bf{R}$ is Transitive or not
edited tags
15h
answered If $\bf{R=\left\{(x,y): x\; is \; wife\; of\; y\right\}}.$ Then $\bf{R}$ is Transitive or not
16h
revised Can one define alternative algebras?
typo
16h
revised Can one define alternative algebras?
added 737 characters in body
16h
answered Can one define alternative algebras?
17h
comment role of definitions in proofs
It would be easier to write an answer that addresses your misgivings if you could give a concrete example of a definition that you feel should be accompanied by an existence proof.
19h
comment How would I prove this strange combinatorial identity?
My gut suggests something about probabilities, possibly with some inclusion/exclusion.
21h
comment Pointclass of $\text{dom}(F)$ where $F:\omega^\omega\rightarrow\omega^\omega$ is partial recursive.
It seems your reasoning is right, then. Perhaps if you reveal which book/notes you have these definitions from, someone would be able to check if you have missed something pertinent in them.
21h
comment Pointclass of $\text{dom}(F)$ where $F:\omega^\omega\rightarrow\omega^\omega$ is partial recursive.
@喻良: The OP specifies that $\operatorname{dom}(G) = \operatorname{dom}(F)\times \omega$ (as it must logically be if $F(f)$ is always either total or undefined). Given that assumption, $\forall i\exists j \,G(f,i)=j$ is equivalent to $\exists i\exists j \,G(f,i)=j$, which is $\Sigma^0_1$.
23h
awarded  Good Answer
1d
comment Is there an “opposite” of a geodesic?
@IberêKuntz: I'm assuming that when nothing else is specified, the Levi-Civita connection (which is derived from the metric) is to be used.
1d
revised Is there a well-defined notion of measure zero on topological manifolds?
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1d
answered Is there a well-defined notion of measure zero on topological manifolds?
1d
awarded  Nice Answer
1d
answered Why are quadrants defined the way they are?
1d
awarded  geometry