Activity for Snoopyâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #286985 |
Post edited: |
— | over 1 year ago |
| Edit | Post #286985 |
Post edited: |
— | over 1 year ago |
| Edit | Post #286985 | Initial revision | — | over 1 year ago |
| Question | — |
Given two angles of a triangle, finding an angle formed by a median > Problem: Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a median) of $\triangle ABC$. Show that $\angle MBC=\angle BAC$. > Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a median of $\triangle ABC$.... (more) |
— | over 1 year ago |
| Comment | Post #286957 |
Thank you! Do you have a reference for the result in the second paragraph? (more) |
— | over 1 year ago |
| Edit | Post #286956 | Initial revision | — | over 1 year ago |
| Question | — |
If $\mathbf{R}$ is thought of as a vector space over $\mathbf{Q}$, what is its dimension? It is known that $\mathbf{R}$, as a vector space over the field of real numbers, has the dimension $1$. I know that $\mathbf{Q}$ is also a field. Question: If $\mathbf{R}$ is thought of as a vector space over $\mathbf{Q}$, what is its dimension? (more) |
— | over 1 year ago |
| Comment | Post #285036 |
Here is a meta complain by a former Math SE mod: https://math.meta.stackexchange.com/q/28168. (more) |
— | almost 2 years ago |
| Comment | Post #285036 |
@#8046 That's a long story. Basically, on Math SE, there is a small group of people who dominate the "deletion queue" of the site in a chat room. They have a very strong opinions on what kind of questions should *not* be allowed. Whenever they think the posts are not "good questions" using *their* st... (more) |
— | almost 2 years ago |
| Comment | Post #285036 |
"What draws newcomers in?" Many good contributors suffer from (a lot of) their valuable highly upvoted answers being deleted frequently by a clique in Math Stack Exchange. Some of them have stopped doing any further contributions on that site. Those are some potential users of codidact. (more) |
— | almost 2 years ago |
| Edit | Post #286788 |
Post edited: |
— | almost 2 years ago |
| Edit | Post #286788 | Initial revision | — | almost 2 years ago |
| Question | — |
The gcd of powers in a gcd domain Question: $\def\gcd{\operatorname{gcd}}$Let $R$ be a gcd domain. Does it always hold that $\gcd(x^m,y^m)=\gcd(x,y)^m$? Context. If the ring is a Bezout domain, then we can apply this method. However, a gcd domain may not be a Bezout domain. I don't know how I can go on. (more) |
— | almost 2 years ago |
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