maths

[Anonymous No.5860221]

/sci/entist gentlemen, I invented a potentially harder problem when trying to solve an easier one.

Let <span class="math"> p_1, ... p_k [/spoiler] be distinct primes. Are there rational numbers <span class="math"> a_1, ... a_k [/spoiler] (besides the obvious case of them being zero) so that

<span class="math"> a_1 \sqrt{p_1} + ... a_k \sqrt{p_k} = 0 [/spoiler]

It's obviously impossible when <span class="math"> k \leq 4 [/spoiler]. But how about the rest?

Besides that, bitching about maths thread, I guess.

[Anonymous No.5860270]

Why is it impossible when k \leq 4?

[Anonymous No.5860281]

>>5860270

Because (a+b+c+d)(a+b-c-d) = (a+b)^2 - (c+d)^2. You can reduce 4 roots down to 2 roots. Then put the roots on one side, and square the whole thing again, to make it become 1 root. Can't be rational.

[Anonymous No.5860290]

>>5860221
The standard negotionary between those sequence doesnt converge on an infallible notation, the distribution of p is indicative of being a constant subset of a ... tesingfuckingdicksman do you even know what you're trying to describe here OP?! ... ffs.. 1\=\k can be better standardized as 0.0.∞|->∞^0.(i) / sqrt a^p.a...(p) ... I really have no idea of the standard of 0 you're shooting for, what you're describing there is effectively an operator of... ffs to ms paint!

Hokay!
>this is pi in infinitive functions on what you would call from that operator sin boundries
>say hello to it, it's very small and shy
The rest requires a bit of demonstration! I'm not familiar with your notation for these kinds of things

[Anonymous No.5860304]

>>5860290
I don't know what you're smoking but lemme try it again

<span class="math"> a_1 \sqrt{p_1} + a_2 \sqrt{p_2} + ... + a_k \sqrt{p_k} = 0 [/spoiler]

[Anonymous No.5860312]

>>5860304
the numbers being an expression of 0.00... (letters)000... correct? just in effect I dont really care about what you think of my limits notation

[Anonymous No.5860318]

>>5860312
Are you sure you have the right thread? Because I don't even know what you're talking about.

[Anonymous No.5860320]

>>5860312
the letters meeting eventually a convergant no? A to Z so to speak

srry I've found alot of math has alot of problems... kinda like steam compared to electricity :)

(we still use steam to make electricity btw, lets hope we can change that in atleast math)

[Anonymous No.5860326]

>>5860318
Well would you like to see my side of the field?

[Anonymous No.5860394]

>>5860290

[Anonymous No.5860401]

>>5860326
>my life
.over

[Anonymous No.5860428]

Bump.

It's okay, /sci/, I'm used to being disappointed.

[Anonymous No.5860448]

>>5860428
It's a very hard problem m8, probably warrants a deep investigation in analytics number theory.

[Anonymous No.5860556]

Well I can tell you that it doesn't work for arbitrary <span class="math">k[/spoiler].

If it works for <span class="math">k_0[/spoiler], then
<span class="math">a_1\sqrt{p_1} + \cdots + a_k\sqrt{p_k} = 0[/spoiler].
I can already tell you it doesn't work for <span class="math">k_0 + 1[/spoiler], though.
<span class="math">a_1\sqrt{p_1} + \cdots + a_k\sqrt{p_k} + a_{k + 1}\sqrt{p_{k + 1}} = 0 + a_{k + 1}\sqrt{p_{k + 1}}[/spoiler]

The only way for this to equal 0, is if <span class="math">a_{k + 1}[/spoiler] is 0, which you already said is not the case.

[Anonymous No.5860566]

>>5860556
Wait, I just realized I might have misunderstood the problem, in which case this is absolute horseshit.

[Anonymous No.5860617]

I'm pretty sure this would follow trivially from basic Galois/field extension theory by considering the field extension by each of those square roots. I don't know Galois theory quite well enough to give the details.

[Anonymous No.5860628]

Lrn2 galois m8s. This forms a basis for the extension of Q containing these square roots, so only the trivial solution works.

[Anonymous No.5860635]

>>5860617
Following up, I think you could show that the degree of the extension is 2^k by giving automorphisms sending
<div class="math">\sqrt{p_1},\dots,\sqrt{p_k} \to \pm \sqrt{p_1} \dots, \pm \sqrt{p_k}</div>

>>5860628
I'm just not sure if that's circular or not. Is there no nontrivial solution because they are a basis, or is it a basis because there is no nontrivial solution?