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Jun
26
comment Number of monic irreducible polynomials over a finite field
Can you make use of the Möbius inversion formula?
Jun
25
awarded  Notable Question
Jun
24
awarded  Nice Answer
Jun
24
answered Subgroup generated by $G-H$ , $H$ is a subgroup of $G$
Jun
24
revised Proof that irrational exponents exist, and are unique?
Typos
Jun
23
answered Why can a multiplicative system not contain zero divisors?
Jun
23
comment Is there a direct proof of the following?
@StefanHansen ... provided $|I|\ge 2$. By the way could the OP please specify his definitions of "pairwise disjoint" and "disjoint"? - I suppose that the claim - as stated- - is wrong
Jun
23
comment Is there a direct proof of the following?
Things may look confusing especially because disjointness is inherently defined by negation ...
Jun
23
comment Counting nodes in a random tree
The thing is not as simple as your attempt suggests. For example, even if $n$ is prime and $n=n_1n_2$ cannot be fulfilled nontrivially and $\delta(n)=0$, say, we can have $N_2(n)>0$.
Jun
23
comment Prove the Sequence of Inequalities
If $n$ is twice a Mersenne prime, ...
Jun
23
comment What is the number of distinct subgroups of the automorphism group of $\mathbf{F}_{3^{100}}$?
Another interesting item might be: If $E/F$ is a Galois extension, what is the set of subgroups of $G(E/F)$ in (inclusion-reversing) bijection with?
Jun
22
comment Why is removing the negation worse than adding it?
It may depend on your opinion on whether $\neg\neg\phi$ is equivalent to $\phi$. And I find the judgement strange - either the rule is valid even when not used as last resort, or it is invalid especially when no other rules work.
Jun
22
answered Why is this relation irreflexive? And how can I prove it?
Jun
20
answered Is this polynomial irreducible in $\mathbb{Q}$?
Jun
20
comment Probability with coin toss
The problem statement is unrealistic insofar as politicions voting behaviour is not determined by them tossing a coin, but rather by otthers tossing many coins at them :)
Jun
20
comment Prove that this transformation is a reflection
Yes, that's fine with me. However, it is not even necessary to use a base and argue with matrices ...
Jun
19
answered Proof about isometries
Jun
19
revised Proof about isometries
added 62 characters in body
Jun
19
comment Vertices that create a convex quadrilateral
@Epic Alomost, but not quite. If you choose $AB$ and $BC$ out of $ABC\ldots Z$, say. there are several more solutions than just $ABCD$ and $ABCZ$ (which are also counted under $(AB,CD)$ and $(ZA, BC)$. There are also $ABCE, ABCF,\ldots , ABCY$.
Jun
19
comment Prove that this transformation is a reflection
@DanielFischer He says: unit vector