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Jan
25
comment Equation of a plane through line and point
Assuming I did the computations correctly, I get a constant term of 101 rather than 201. You seem to be doing the right things. Assuming your line of intersection is presented correctly, any problems should be computational.
Nov
29
answered Precalculus math question natural logs
Oct
29
awarded  Yearling
Sep
22
comment What is the purpose of the Codomain?
@Thomas - Right. Equality requires same domain, same codomain, and same image. Perhaps my parenthetical muddled rather than clarified what I was saying.
Sep
21
comment What is the purpose of the Codomain?
We can approach this question from the definition of equality of functions. Two functions are equal if they have the same domain, the same codomain, and agree on their common domain (that is, they have the same image).
Aug
27
accepted What is the injective envelope of a product of abelian groups?
Aug
27
comment What is the injective envelope of a product of abelian groups?
Never mind, I was being dense.
Aug
27
comment What is the injective envelope of a product of abelian groups?
This seems to work. One question, and I may just be being dense here: Why is the group $E$ divisible?
Aug
26
revised What is the injective envelope of a product of abelian groups?
Further clarification of question.
Aug
26
revised What is the injective envelope of a product of abelian groups?
clarifying the desired form of the answer
Aug
26
asked What is the injective envelope of a product of abelian groups?
Aug
4
comment Why is it that if one NP complete problem is solved, all NP problems are solved?
I believe the term NP complete was chosen because all NP problems can be realized as instances of the NP complete problem.
May
4
comment For a polynomial $f\in K[x]$, when is there a constant $c\in K$ such that $f+c$ is irreducible?
Suppose $K=\mathbb{C}$. What are the irreducibles in this case?
Apr
15
comment Subgroups of $ \mathbb{Z}_n$ (integers mod $n$)
Following up on what @quid said, $m$ and $n$ will generate the same subgroup of $\mathbb{Z}_{18}$ if and only if $\gcd(m,18)=\gcd(n,18)$, which you may verify by listing all the cyclic subgroups of $\mathbb{Z}_{18}$. You may restrict each of $m$, $n$ to be in $\mathbb{Z}_{18}$ of course.
Feb
9
comment Non abelian subgroup of a abelian group.
You can have abelian subgroups in a non-abelian group, but not non-abelian subgroups in an abelian group.
Jan
18
comment is complex number under absolute value a group?
Is the number 0 allowed? If so, what is its inverse?
Dec
29
comment Why rationalize the denominator?
@Ahaan - +1 for the answer. Nonetheless, a good teacher, at any level, should have no trouble realizing that your three radical expressions represent the same number. However, having said that, and being a US resident, maybe not.
Dec
29
awarded  Nice Answer
Dec
22
answered Removable discontinuity question
Oct
29
awarded  Yearling