@shaihorowitz does that look right or am I supposed to replace the term inside the paranthesis?

whats the bottom term in the bottom right fraction

$8(1-\xi)^{\frac{5}{2}}$

hi @RandomVariable how are you? :)

is that /xi? as in ${\frac{1}{xi}}$ or am i missing something

$\xi$ as in the variable in the Taylor remainder

ah ok hold on gotta read through everything

semiclassacal is prob doing something like that=

alright

Hello, I have a problem about arithmetic...

@Semiclassical save me help steve

@DanielCortild shoot

Just ask; don't ask to ask.

its ok to ask about asking to ask tho

but then is it ok to ask to ask to ask

where does the regression stop?

Ok, so I have to prove that there is an infinity of $a$'s satisfying this property: If, for all $(x,y)$, $a$ divides $(x+y)^5-x^5-y^5$, then $a$ divides $(x+y)^7-x^7-y^7$ as well.

I've seen that it is the same as $(x+y)^7-(x^7+y^7)$... But from there

hmmm

I think it has to do with binomial coefficients

It is proposed as a arithmetic problem, so it is probaley more number theory oriented....

while I wait for @Semiclassical to answer, anyone in the mood for a discrete math question?

I'm more in mood for someone helping me... I am really stuck...

@DanielCortild note when y=-x then $(x+y)^5-x^5-y^5$=$(x+y)^7-x^7-y^7$

@shaihorowitz Are you sure?

everything is zero then

so of course

wolframalpha.com/input/…

@r9m Hello. I haven't seen you around here for a couple months. Have you been very busy?

any points of these solutons will yeild infinite a but thats the easiest

@shaihorowitz hi =) how are you?

i'm alright how are you null

i feel like nothing :(

@shaihorowitz Ohh yea, but it is if for all (x,y) a divides the first, then prove that it also divides the second...

nothings better then the best feeling ever

@RandomVariable ya .. coursework workload etc ..

oh right hold on @DanielCortild

@shaihor you have misunderstood the question, you have to show that the number of $a$ so that $\Bigg( a$ divides $(x+y)^5 -x^5-y^5$ $\Bigg)$ $\implies$ $\Bigg($ $a$ divides $(x+y)^7-x^7-y^7$ $\Bigg)$ is infintie

@shaihorowitz puns puns puns :D

How many bridge hands with a 4-3-3-3 distribution are there? I think I have to use binomial coefficients and the product rule, but I'm not quite sure

there are no conditions on the $x,y$

@s.harp Yes, you are right

just to write it out

Wait...

Are there infinitely many $a$ that divide $(x+y)^5 - x^5 - y^5$ for every $x$ and $y$?

Yes that is absolutely correct... But I can't see where it leads...

you can always divide out at $xy$ term, so you can assume $a$ doesnt divide $xy$ but divides the right term for the $t$ part (otherwise $a$ would divide both and you would be finished)

@SteamyRoot thats not the question

@SteamyRoot No, but there are infinitly many a's so if it divides the fist then it also divides the second and vice-verse

It was the original question? "If, for all $(x,y)$, $a$ divides..."

@shaihorowitz did we discuss encryption? (i was named saturated)

If you find infinitely many $a$ that do not satisfy the first condition, you're done

because $p \implies q$ is true if $p$ is false

@s.harp Yes, but that is the problem...

@DAniel I'm just writing out some thoughts, not pretending to solve it

@SteamyRoot That's not stupid...

Hello!

its possible i too have a problem names

@SteamyRoot finding infinitly many doesnt mean that it holds for all a, do you see why? (i have absolutly no clue about the task, just wanted to say that)

@r9m I'm always surprised when someone pings me in chat because it happens so infrequently.

@shaihorowitz Sorry, I don't understand

@Null I know, but the question is to prove there are infinitely many $a$'s satisfying the implication

@SteamyRoot ah ok

sorry @DanielCortild not to you, @Null its possible i too have a problem with names

@SteamyRoot if i partition N in even and odd, and show that for all odd something holds, ...

@RandomVariable well I haven't been frequent on chat for some time .. thought I'd say hi! ;)

@Null I know all that.

mmh

then i really dont understand the question haha

You have to prove that there are infinitely many $a$ such that $p \implies q$

So I'm just saying: if you prove there are infinitely many $a$ for which $p$ is not true, you're done

@SteamyRoot ah... because ex falso, got it

no, $p(a)$ can be false for every odd $a$ but for every even $a$ $p(a)$ can imply $q(a)$

besides, if $p(a)$ is false for every $a$, then $p(a)\implies q(a)$ for every $a$ tautologically

... @s.harp But how do I solve it then?

so basicly: if i show that for all $a\in$ Primes p isnt true, i showed that for infinitly many a, p->q is true. (because there are infinitly many primes)

@Null Yes, but how is that proven?

not with some easy trick spares you from looking at the actual details of the question

@steamyRot nice proof but probably not what the teacher wanted

MYEA... I am seaching an elementary proof...

@DanielCortild ex falso quod libet. (out of wrong i can conclude anything)

@Null... Yes, but that isn't the whole proof...

A simple example: if a<1 then a>1, for all a>2

@Null Well that is true, but such a will never exist...

Well it is true becuase such a will never exist...

there are certainly a>2 (in R)

Yes, but none of them are smaller than 1...

exactly!

But how does this help?

in any intervall of R there are infinitly many numbers (call them a)

so show that you found an intervall for that p isnt true

and as steamy said, then you are done

So I just have to prove that there are some a's where it isn't true and that's it? But how is this then proven?

no

not only "some" a's, you got to show that an intervall of numbers doesnt satisfy p

these confusions will not help you

... Are you sure that there isn't a more elementary proof?

a>the xy stuff, will certainly not devide it

and you can prove very easily that there exist infinitly many of those

Hmmm... I reallt don'y follow you...

$\forall a>(x+y)^5-x^5-y^5: $ a devides the stuff$\Rightarrow$ a devides the other stuff

But a doesn't divide the stuff if it's bigger...

exactly

therefore p is untrue

therefore the statement is true

I have a question about C-vector spaces

I still havn't understud the answer...

ok so here is a different proof consider the line x=-y as i stated earlier this simipliefies the equation down to 0=0 but it still is an infinite line i.e infinite pairs (x,y) next. note that 0/a=0 so 0 is divisible by any integer therefore there are infinite such a, therefore If, for all $(x,y)$, $a$ divides $(x+y)^5-x^5-y^5$, then $a$ divides $(x+y)^7-x^7-y^7$ as well Qed.

Ok, so I have to prove that there is an infinity of $a$'s satisfying this property: If, for all $(x,y)$, $a$ divides $(x+y)^5-x^5-y^5$, then $a$ divides $(x+y)^7-x^7-y^7$ as w

@null we good?

as long as a not equal 0

@shaihorowitz well, i just read somewhere that 0 has no divisors in rings, but i can follow your argument

i'm looking at the wiki of divisors and it seems allowed

it depends how you define "divisor"

truth

i think your proof is more what the teacher expected tho

You are not proving the statement

@Null $2 \cdot 3 = 0$ in $\Bbb Z/6\Bbb Z$

right i proved infinite (x,y) not all x,y

you have to show that there are infinetely many numbers $a$, so that whenever $a$ divides $(x+y)^5-x^5-y^5$ for some $x,y\in\mathbb N$ then it must also divides $(x+y)^5-x^5-y^5$

the numbers $x$ and $y$ are not fix

@Fargle mmh, so 0/3=2 ? which doesnt make much intuitive sense

your proof is so much simpler @s.harp and actually proves

@Null I wouldn't say that--it makes no sense to talk about division in that ring precisely because it has zero divisors

I'm not proving anything, nobody has proven anything

you proved an infinite false p in p implies q...

@Fargle ah so, "no zero divisors" is actually a sound definition?

no, I just remakred that $p$ being false makes $p\implies q$ a true statement

@Null Yup, that's what an integral domain is.

@Fargle and division somehow requires a field, if not some "wierdo" devision is meant

(i mean integral domain)

(but the step to a field is smal^^)

@Null Well, there are integral domains where division isn't okay also

one small step for man kind one inoperable leap from logic

But it turns out that every finite integral domain is a field by Wedderburn's little theorem

devision is basicly: multiplying by the inverse

?

Right, and you only get that for sure if you're working in a division ring (a field minus possibly commutativity of multiplication)

i think the question opener has already left :(

@NaCl hi :) its me, saturated, are you better now? how's your pc problem?

What are examples of topological vector space that are not normed vector space ?

@Null the pc problem will not solve until I buy a new one (new mainboard would cost 300, and it's very likely the main board)

My health got better

How are you?

@Astyx I would guess you could use a metric that is not translation-invariant?

@NaCl my pc has costed 200, what kind of beast you run? im fine, thanks!

@SteamyRoot So some kind of a norm that does not have the 1-homogeneity axiom ?

@Astyx Space of real functions with pointwise convergence.

I may have forgotten what the 1-homogeneity axiom is... heh

Or rather, the product topology.

Just take, say, $\mathbb{R}^2$, and define the metric as $d(x,y) = \|x\| + \|y\|$

@Null Dell XPS17, I was 13 and thought it would be cool, but it wasn't. The fan is loud as sonic boom, battery life is low as a simple dipole and the case melted after I played Minecraft back in the days, so yeah. I'm running an overpriced junk-pc

Ok thanks @BalarkaSen @SteamyRoot

I wonder how a proof for that could look like

@Steamy It's still isomorphic to $\Bbb R^2$ with the standard norm though

@NaCl got to watch tht calmly^^

hello guys, I was just wondering

what does determine (if so)

what?

^

hello all, I attempted to answer my own question, could someone check it over?

Determinists determine.

whethere one complex number is greater than another

Nothing, $\Bbb C$ is not an ordered set

there is no standard ordering on the complex numbers, and there is none compatible with the arithmetic operations like there is for real numbers

You can order lexicographically, but there is no conventional order relation

cause I was to determine is that possible to use rule of 3 sequences if sequence consists of complex number

"rule of 3 sequences" what?

@BalarkaSen why is that?

many convergence tests for infinite series of complex numbers use absolute values of terms, which will be real numbers.

theory of 3 sequences*

what does "theory of 3 sequences" mean?

aahh

squeeze theorem

he means this I think en.wikipedia.org/wiki/Cross-multiplication

sorry, I was translating from polish

you can use squeeze theorem for the absolute values of complex numbers

carefully

@SteamyRoot Any two equidimensional topological vector spaces are isomorphic, as long as they are finite dimensional.

'carefully' disposes a devil

hello, can somone tel me how to start the proof of $I'$ is continuous , from where i mmut take the sequence :math.stackexchange.com/questions/2013871/…

Actually maybe I need Hausdorff down there somewhere.

ask vague question, get vague answer

ohh, you mathematics

lol

are so vague

:|

unfortunately nobody understands what you are asking @Dartek12

such vague, much meta

im just doing fun

For a vector bundle $p \colon E \to B$, how are orientations of $p$, $E$ and $B$ related? Is there a canonical orientation of $p$ when $B$ and $E$ are oriented?

@abenthy What do you mean by orientation of a map?

An orientation of a vector bundle is family of orientations of each fiber that is compatible in the sense that around each $b \in B$, there is a local trivialization which is fiberwise orientation-preserving.

I am aware of what orientation of a vector bundle means.

Oh, by orientation of $E$, did you mean of the total space?

yeah

Neat question.

Hmmm... I still don't understand the proof, and I am searching an elementary proof! Like just using some of the basic theorems and things like that... If anyone could help... Here is the problem: Show that there is an infinity of $a$'s so that if $a$ divides $(x+y)^5-x^5-y^5$ then it must divide $(x+y)^7-x^7-y^7$... Anyone?

I'm trying to understand the Thom isomorphism theorem for compact supported cohomology. Its interesting that if $p \colon E \to B$ is a rank $m$ vector bundle with $E$,$B$ oriented and $n = \dim(E)$, then one can use poincaré duality to form the composition $$H_c^{n-m-k}(B) \xrightarrow{\cong} H_k(B) \xrightarrow{\cong} H_k(E) \xleftarrow{\cong} H_c^{n-k}(E)$$

My question is: Is this the same as the Thom isomorphism if $p$ is (appropriately) oriented?

Can someone help me understand this process: take the definition of MVT $\frac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}} = f'(x_c)$ for some $c \in [x_i - x_{i-1}]$ when you shrink the interval such that $\Delta x_i \to dx$ does this theorem preserve itself? does the limit of $x_i - x_{i-1}$ go to 0, for $\Delta x_i \to dx$? I'm not sure if this is a meaningful question.

But, if the limit does go to 0, the theorem no longer makes sense. Yet, we are able to use the theorem in finding the length of a curve by transforming: [the square root of the sum of its squared components] to [the sum of the magnitudes of its derivative multiplied by infinitesimal subintervals].

@arctictern I'm still confused about matrix algebras. So I can prove that $R=\mathbb{M}_n(\mathbb{C})$ is simple. Since $R$ is a $\mathbb{C}$-algebra, any left $R$-module is a $\mathbb{C}$-vector space. Now I don't see the connection between the dimension of this vector space over $\mathbb{C}$ and the fact that $R$ is simple...

@BalarkaSen You are here more often than I am, who can I ask for number theory?

Could someone look for my problem please??

Is (-a/-b)^1/2 a rational if 0<a , 0<b ?

is there some truth in the saying "if you cant explain it to a [six year old], you dont understand it" (replace six year old with whatever)

no

@Krijn Depends on what kind of number theory you want answered. I think @arctictern knows some.

Why? @s.harp

@ffahim i was talking to @null, but look at $a=2,b=1$

@Null have you ever tried explaining anything to a six years old ?

no, some concepts are too technical or rely on too many facts that need to be established before

@s.harp well, can u answer my one ?

More seriously, I believe this saying only means that you only understand something fully when you can always answer the question "why ?" a six years old would keep on asking

I did...

@BalarkaSen Can you please look at my question?

@ffahim e.g. a = 2 and b = 7 would that make a rational?

Yes @Dartek12

@arctictern, do you know a bit of class field theory?

Ohh no

Sorry

@ffahim then give coefficients p,q so that (2/7)^(1/2) = p/q

:/

My question :is (-a/-b)^1/2 real if 0<a, 0<b ?

@Astyx thats a NICE assumption!

@ffahim do you know when real roots exist?

does anyone know this limit? $\lim_{x\to 1}\left(\zeta(x)-\frac{1}{1-x}\right)=\gamma$ where gamma is the Euler-Mascheroni constant

Yes.. only if the inside one is positive... @s.harp

wolfram alpha says zeta(1+10^-6)-10^6=0.577 so that's how I made the conjecture

if a>0 and b>0, what is the sign of (-a)/(-b)?

Positive

so what about (-a/-b)^(1/2), can you say that there is a real root?

That means the roots are real ???

if you want to know whether x^(1/2) is real this is the question whether or not x has a real root

@s.harp there would be one real root yes, but did mean exponent 0.5 or root?

Same

Isn't it @Null

no

What!!

principal root doesnt equal to exponent 1/2

definition bs

What is that principal root? @Null

@Null How does it not ?

the principal root is always positive

And so is x^1/2 where defined

$a^{0.5}$ isnt always positive as complex numbers dont have positive/megative

So is $\cdot ^ {1/2}$

is $(-1)^{\frac{1}{2}}$ positive?

Of course they differ if you consider one over R and one over C...