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May
24
comment An example of a group such that $G \cong G \times G$
Oh. It's displayed wrong in the mobile site
May
24
comment An example of a group such that $G \cong G \times G$
I don't understand.. If H is finite for example it's not true
May
9
comment Homomorphisms and Groups proof
@danny $f(b^{2})=f(b\cdot b)=f(b)\cdot f(b)=f^{2}(b)$ . Now use induction
May
3
comment Do you know Legendre's conjecture ? Has it been proved?
Can't you just take epsilon to be small enough such that the difference between the expression with the exponent of 2 and the one with exponent of 2 plus epsilon is less then 1? If the claim holds for every positive epsilon this reasoning should show it holds for epsilon equals zero
May
2
comment Why is the zero polynomial not assigned a degree?
Increasing in what variable ?
May
2
comment Why is the zero polynomial not assigned a degree?
what about $+\infty$ ?
May
1
comment Find the splitting field of $x^6-2x^3-1$ over $\mathbb{Q}$.
Splitting field over what base field ?
May
1
comment Evaluating $\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!…n!)^2} \right)$
Did you try to cancel out some of the terms ?
Apr
25
comment Is a pattern proof?
No, this [post][1] contains many examples [1]: math.stackexchange.com/questions/514/…
Apr
25
comment Prove that this set is open
@drhab - I think you should keep it, it can be interesting for future readers as it is a confusing matter
Apr
25
comment Prove that this set is open
@drhab - It is prove! If $X=\Pi_{i=1}^{n}\mathbb{R}$ one can consider two topologies on $X$ - The product topology and the topology induced by the metric on $\mathbb{R}^{p}$. I proved that $A$ was open without relying on the fact that the two topologies on $X$ are the same since I assumed the OP does not know that
Apr
24
comment Topology, maps, continuity
@smits - then $f^{-1}(U)$ is closed. This condition is equivalent for the continuity of $f$
Apr
23
comment Subgroups of every order dividing the order of the group imply the group is abelian?
Thank you for your answer, Can you please rephrase or explain the theorem you are taking about "if divides the order of a group with a prime, then the group has a subgroup of order " it seems to be missing a word or two and it sounds like Cauchy theorem..
Apr
23
comment If $f^2$ integrable, then $|f|$ is integrable?
Don't you need to know that $|f|$ (or $f$) is integrate to begin with ? The space of which this inequality is deduced is the space of integrable functions $f$
Apr
23
comment Subgroups of every order dividing the order of the group imply the group is abelian?
@Timbuc - I see your point. But this seems to answer only (2), can you also give an example for (1) ? thanks!
Apr
23
comment Groups with “few” subgroups
@mathmandan - Very interesting, I'll be sure to read it. thanks!
Apr
23
comment Groups with “few” subgroups
@mathmandan - How do you know such subgroups of $T$ exist ?
Apr
23
comment Groups with “few” subgroups
How do you know that $P$ exist ?
Apr
20
comment Finding all possible pairs of positive integer values
Because I can assume without loss of generality that x>=y (otherwise y>=x and we change their names ). I have used this in the first equation, the LHS is positive and by that assumption the RHS is also positive
Apr
19
comment Proofs shorter than the statement of the theorem
Does "this is a direct consequence of the previous theorem" counts? :-P