Mathematics desk
< August 2	<< Jul \| August \| Sep >>	Current desk >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

August 3

Natural numbers to modulus

Is it true that for every set of n distinct natural numbers, there exists a natural number $m\leq 2n^{2}$ , that modulo which all of these n numbers are in the range $[1,m/2]$ ?David Frid (talk) 15:40, 3 August 2018 (UTC)[reply]

Sure, just use

m = 2

(or even 1, if you're allowing that). All the numbers in your set are definitely going to be 0 or 1 mod 2. –Deacon Vorbis (carbon • videos) 15:49, 3 August 2018 (UTC)[reply]

Ummm... It's kinda edge case, so I edited my question to handle this. David Frid (talk) 16:23, 3 August 2018 (UTC)[reply]

You changed

[0,m/2]

to

[1,m/2]

. Now it's false for all n. Pick n numbers which are divisible by all numbers up to

2n^{2}

, e.g.

k\times (2n^{2})!

for k = 1..n. For any

m\leq 2n^{2}

, all the modulos will be 0. PrimeHunter (talk) 17:24, 3 August 2018 (UTC)[reply]

The answer to any nontrivial version of this will be "no": for fixed n, take one of the numbers to be congruent to -1 modulo 2, 3, ..., 2n^2 and this particular number will always be in the wrong half. Moreover, the same construction (using the Chinese remainder theorem) will work for any subsets as long as the collection of moduli is determined in advance. --JBL (talk) 20:17, 3 August 2018 (UTC)[reply]

Assuming also that each of these n numbers is $\leq 2^{n}$ , is the statement correct? (I ask this because both your answers assumed otherwise) David Frid (talk) 06:01, 4 August 2018 (UTC)[reply]

There are lots of counter examples for 5 numbers, e.g. {1, 4, 7, 13, 26}. PrimeHunter (talk) 10:58, 4 August 2018 (UTC)[reply]

Thank you! David Frid (talk) 08:09, 5 August 2018 (UTC)[reply]

Zero mass

Sorry back again.

Not understanding why something can't have negative mass...could some explain in basic terms? Chromagnum — Preceding unsigned comment added by 165.16.75.66 (talk • contribs) 22:15, 3 August 2018 (UTC)[reply]

Not really a math question. In the future, WP:RD/Science would be a better place to ask.

But as long as you're here, see principle of equivalence. If you had an object with negative gravitational mass, then that means that in a gravitational field, it should accelerate in the "wrong" direction. But in a situation in, say, an elevator, which is supposed to look equivalent locally — how is that going to work, exactly? Why should the elevator accelerating upwards make the negative mass accelerate upwards inside it? --Trovatore (talk) 22:24, 3 August 2018 (UTC)[reply]

Actually, now that I think about it, maybe that's not a problem. If you have an object with negative gravitational mass in the Earth's gravitational field, the force on it due to gravity will be upwards (away from the center of the Earth). But if it also has negative inertial mass, which why wouldn't it, then the upwards force will make it accelerate downwards, just like it's supposed to. --Trovatore (talk) 22:59, 3 August 2018 (UTC)[reply]

Cool wrong desk my bad. Does that apply with anti matter. Is there any math that may suggest an answer to the problem.... — Preceding unsigned comment added by 165.16.75.66 (talk • contribs) 22:41, 3 August 2018 (UTC)[reply]

Anti-matter has positive mass. Turns out we do have an article on negative mass. I haven't read it, but you might like to. --Trovatore (talk) 22:45, 3 August 2018 (UTC)[reply]

Mass is by definition positive:

m={\sqrt {p^{\mu }p_{\mu }}}

- positive square root from the square of 4D momentum. It is either positive or purely imaginary (in case of tachyons). Ruslik_Zero 20:06, 5 August 2018 (UTC)[reply]

Hmm — this seems like a relatively modern "definition". Is it really a definition, or just a true relationship, as far as we know? It seems to me that gravitational mass and inertial mass are not equal "by definition", but rather as an empirical fact, so they can't both be intensionally the same as the 4-momentum one.

To put it another way — we know how we expect a body with negative mass to behave differently from objects of positive mass. For example, if you have a body N of negative mass interacting gravitationally with a body P of positive mass, then N will accelerate towards P, but P will accelerate away from N.

Can that be reconciled somehow with the 4-momentum formulation? --Trovatore (talk) 20:53, 5 August 2018 (UTC)[reply]

Mass is always defined as a positive value even in GR. Gravitational mass is a red herring here as the gravity is produced by energy-momentum, not by mass. In short, there is only one mass in GR and it is always non-negative. Ruslik_Zero 20:14, 6 August 2018 (UTC)[reply]

Umm — I feel you have not really addressed the heart of the issue here. Do you dispute the content of the negative mass article, where it talks about negative mass being possible if the stress component of the stress–energy tensor is greater than the mass density? --Trovatore (talk) 22:07, 6 August 2018 (UTC)[reply]

The same article explains why it can not exist in reality. And as I said above it is more question of definitions then substance. It is always possible to define mass as purely positive value. What really matters is how energy and momentum are defined. Ruslik_Zero 20:12, 7 August 2018 (UTC)[reply]

Well, it does not seem to be so categorical on "it cannot exist in reality". It does say that physicists tend to find it "preposterous" (in one quote), but that is not quite the same thing.

I am not persuaded that it is just a matter of definition. The runaway acceleration of a body of negative mass chasing one of positive mass seems like a thing that could potentially be observed, so if indeed there is no negative mass, this appears to be a synthetic rather than analytic fact. --Trovatore (talk) 21:07, 7 August 2018 (UTC)[reply]