If $n>1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.
If you know Bertrand's Postulate, then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k...
Мне интересно вот, что.
Формулы выписанные ниже есть решение в общем виде Диофантова уравнения Лежандра.
Они позволяют сразу по коэффициентам получить формулу описывающую их решения.
Решения будет в том случае когда хоть один корень есть целое число.
Когда ни один корень не будет целым, надо буд...
It looks like he's had it lifted and the doctors who did it meant a bit too well. Now there is not enough skin left to move any part of his face. Hahaha.
Basically, I have a question that reads as: Let $A$ be any infinite set and $B$ any countable set. Prove that $|A \cup B|$
I was able to create a nice bijective function between $A$ and $A \cup B$ ASSUMING that B was an infinite set. However, now if I take the case where $B$ is FINITE - I hit a few problems. In fact, I believe that if B is finite, equality does not hold. Any advice ?
I see ! Well, I'm going to try that tomorrow morning then. Thanks for your help @KarlKronenfeld. Appreciate it. Hopefully I smash it out tomorrow then :) .
As for the exercise, anyone interested can see Dummit. It is a known result that solvable quintics over $\Bbb Z[x]$ have galois groups as subgroups of $F_{20}$ and you need some trickseys to rule out the proper subgroups.
@KarlKronenfeld nah
@Karl Consider $\frac{\Xi}{\overline \Xi}$ such that $\bar{a}(t_{a_i})$ if a functor from $\Bbb {\hat O}$ to ...
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If you consider the duality mapping from a normed space to it's continuous dual and the duality mapping is shown to be an isometry, if V is a finite dimensional normed space does it follow that the duality mapping is also surjective?
Let $D$ be a bounded domain with piecewise smooth boundary. If $f(z)$ is an holomorphic function on $D$ that extends smoothly to $\partial D$, then $$ \int\limits_{\partial D} f(z)\,\mathrm{d}z = 0 \, . $$
If you consider the duality mapping from a normed space to it's continuous dual and the duality mapping is shown to be an isometry, if V is a finite dimensional normed space does it follow that the duality mapping is also surjective?
Here is the proof for my book, I thought they needed to show that there exists a continuous mapping / transformation from $1$ to $2$, otherwise it seems somewhat strange =(
I meant you could add lines into the circles like I did above, with the green and blue. These line integrals are the same but with opposite direction and hence canceling each other out
Then you are left with a simple connected curve with no singularities, plus the piecewise disks around each singularity
So you sort of go from $1$ to $2$ by using those green and blue line integrals, which vanishes.
Surely you know that $$\oint_C f(z) dz = \oint_{C'} f(z) dz$$ for C and C' being the boundaries of a torus in $\Bbb R$, upto continuous deformation. So that kinda is what you use in the name of Cauchy =P
@Sawarnik Cool, congratulations.
@Sawarnik I am free, at least today. Let's see what you have.
@BalarkaSen Wat, you are free! This is nice and easy I think: $x^5+y^5=2x^2y^2$. Prove that if $x$,$y$ are rational numbers then $1−xy$ is a perfect square.
Hello. I have a question. There seems to be a general idea where you can take a lie algebra / algebra and consider formal power series representations of the lie algebra (for example, exponentiating the lie algebra to a group, and the vertex operator algebra ) is there a general theory of taking series representations of an algebra? thank you
