/sci/ - Science & Math

i don't even know what's being asked
any help?

>>4428294
its about numbrs yuo take from sets and than you join numbers and get moar numbers

no

>>4428294
Yeah, the "disjoing union of three sets..." confuses me too..

>>4428336
disjoint means the intersection of each pair is empty, i trust that you understand what union and three means

>>4428294
You define a function, which maps a real number to a set of integers like this

http://www.wolframalpha.com/input/?i=Table[Floor[2.819*n]%2C{n%2C1%2C100}

(here for \alpha = 2.819)

do this 3 times for any \alpha, \beta and \gamma

And you have to show that you can't construct N disjointly out of these three.

>>4428346
bitch looks plastic as fuck who even likes that.

>>4428362
I don't. Anyway, back to topic

>>4428362
>who even likes that
every nonfaggot likes that

>>4428467
No.

tip: it's almost all about the number used to create the integer 1.

Alright, let's go.

Since the other two sets are nonempty and the disjoint union contains 1, define your smallest parameter alpha such that.
<span class="math"> \alpha = 1 + \frac{a}{A} | A > a \neq 0 s.t. a, A \epsilon \mathbb{Z} [/spoiler]
and define
<span class="math">
n \equiv \lceil \frac{A}{a} \rceil
[/spoiler]
which forces
<span class="math">
\beta \equiv n + \frac{b}{B} | B > b \neq 0 \\
m \equiv \frac{B}{b}
[/spoiler]
where <span class="math"> \gamma [/spoiler] is then forced to be equal to m.

Further notes: n, m > 1 (I think this fact solves the problem, but I can't think of how to state it)

Anybody who finds this little doodle useful is welcome to solve the problem with it.

>>4429023

<span class="math"> \beta \equiv n + \frac{b}{B} | B > b \neq 0 [/spoiler]
<span class="math"> m \equiv \frac{B}{b} [/spoiler]
is the missing bit.

Apparently line breaks aren't fair game

oh look, the /sci circlejerk of the day.

something to do with infinite prime numbers?

Just to make sure: it's essentially asking to show that there's at least one positive integer <span class="math">k[/spoiler] that is a member of more than one of <span class="math">S(\alpha)[/spoiler], <span class="math">S(\beta)[/spoiler], and <span class="math">S(\gamma)[/spoiler].

>>4429120
or isn't a member of any of them

>>4429120
Or isn't in any of them.

>>4429142
>>4429330
Duh, right. Forgot about that.

Well I have no idea where to begin.

>>4429336
Well this is all I have so far.

a,b,y>1, otherwise we get repetition. It's also clear the integer parts of a,b,y must be different, so the floor isn't the same. assign a<b<y.
1<a<2, as 1 must be in the set, although all of the above is trivial. I've been mucking about trying to find an expression for elements not contained in S(a), with no luck.

how about thinking of the problem with cartesian graphs, where a, b, and y are slopes. the only way for a disjoint union to happen is if all three lines are parallel, ie: a = b = y

>>4429455
Given a,b,y>1 and 1<a<2, you've pretty much gotten it. (Also, defining here that a<b<y)
Assume a=1+10^-n for some number n.
The set S(a) will contain all of {1,2,3...} until the number Ceil[10^n], which will be skipped. The numbers skipped will always be x*10^xn+x-1 for some integer x. (For example, if a=1.1, S(a) contains 9 and 11, but not 10. Then, 21 will skipped, then 32, etc.)
In order for S(b) to be disjoint from S(a), but for S(b) to still be part of the set {1,2,3...} would be if 10^n<=b<10^n+1. But for S(b) to not repeat parts of S(a), S(b) must fulfill Floor(b*2)=2*Floor(b)+1, which W|A claims has no real solution. The same is obviously true of S(y).

Kinda just wrote that out to see if it was going somewhere, seems like an answer at least.

You can derive a contradiction showing that the partitioning of the positive reals leads to non disjoint sets.

>>4428763
You're a faggot.
Case dismissed.

>>4431381
No.

Given S(a), S(b), S(c)

in S(a) there exists a member such that a*n is not relatively prime to b or c. (that is, n is a multiple of b or c)

Therefore no triad of sets can be disjoint, ever

weak proof, yes, but that's basically why

>>4431399

(continued because i forgot something)

because a*n is a multiple of c (lets say c),

there exists a number x such that x* c = a * n.... and that x is a natural number, and so it is in the set S(c) and therefore S(c) upsidedownU S(a) is not null

>>4431405
I was thinking on similar grounds.
I'm sure on this question 1-2 points would be given for a weak proof.

So does /sci/ ever solve these?
How hard are they? Undergrad or mathematician with PhD tier?

>>4431695
>PhD tier
They are im principle quite solvable. But they are not easy and sometimes a little number theory is very helpful. In any case, these problems are nothing like research level mathematical problems, alone because there is more to math than number theory and results in set theory and these problems are almost every time about these subject. It's so that the problems can be stated so that you can understand the question without having to study mathematics.
If one takes a look at mathematicans, then it seems every second is into algebraic topology, which has nothing to do with these questions.

>>4431695
Undergrad.

So necessarily in order to contain 1 and not 0, 1<alpha<2. It can be known that for any three rational numbers alpha, beta and gamma there will be some n for each that sets them to be equal thus causing their greatest lesser integer to be equal. For irrational numbers I guess you could take a sandwich theorem approach. Each irrational number is arbitrarily close to a rational number, which means that they will reduce to the same number for an arbitrarily high n, allowing you to essentially attach each irrational number to a rational number. I know it's not the best or most formal proof, but there's my take on it.

>>4431965
correction, 1 <= alpha < 2. Obviously the idea still applies, or worst case scenario the special case of alpha = 1 would be easy to prove.