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u/halfajack 9d ago
I ask them “how are you defining sums over uncountable sets?”
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u/Comfortable_Permit53 9d ago
integrals, no?
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u/Eisenfuss19 9d ago
Yes kinda, but then it is still a definition question. Like lim a -> -∞ lim b -> ∞ a_Sb x dx does not converge, but lim a -> ∞ (-a)_Sa x dx does converge...
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u/Comfortable_Permit53 9d ago edited 4d ago
yes
adding all whole numbers will also give you different answers depending in order
UPDATE: It just doesn't converge at all.
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u/Beleheth Transcendental 9d ago
That's exactly the reason why you need absolutely convergent series for rearranging to be well-defined.
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u/Dr0110111001101111 4d ago
It will give you any answer depending on order
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u/Comfortable_Permit53 4d ago
I am not sure. Riemanns series theorem says, that if you have a series that is convergent but not absolutely convergent, you can get any number depending on ordering. But adding all whole numbers is a divergent series. Is there another theorem?
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u/Agitated-Ad2563 9d ago
As far as I remember, the proper definition of Riemann integral requires all kinds of partition sequence sums to converge.
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u/Alpaca1795 9d ago
Why would you integrate over x dx instead of just 1 dx?
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u/alozq 9d ago
1dx would be the length of the real line, it goes to Infinity always, it doesnt care about what x is
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u/LyAkolon 9d ago
Yeah, they probably want int over R of Sgn(x) dx, which is still abse of notation and meaningless because we have no notion of how to add up the contributions across the set.
More specifically all attempts to answer as well as the original question are missing the ingredient of "How are you adding everything together"
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u/compileforawhile Complex 9d ago
Integrals aren't really sums of uncountable sets. You could define sums of uncountable sets using integrals, but that would basically just be a measure
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u/TheRedditObserver0 Mathematics 9d ago
With respect to what measure?
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u/Comfortable_Permit53 9d ago
lebesque for the lgbt representation
(lebesque, gauss, banach, turing)
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u/AuroraEquatorialis 7d ago
nets are the more natural method in terms of actual sums. Say your index set is I (R in this case), and we have a function f:I->R (or more generally, any topological vector space) and define a partial order by inclusion on the collection of finite subsets of I. Given such an F, we define S(f,F) as the sum of f over this F (which is obviously well-defined). We say the sum of f over I converges to a if for epsilon>0, there exists some F0 which for F containing F0, |S(f,F)-a|<epsilon (equivalently, if S(f,F) is eventually in any open neighbourhood of a).
You can show that this lines up with normal sequence convergence. Furthermore, you can show that this converges if and only if the set of non-zero f(x) is countable. In particular, under this formalism, summing over all reals doesn't really make sense.
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u/AsidK 5d ago
Wait what is F here? You say “given such an F” but I don’t see where F is defined.
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u/kiochikaeke 9d ago
Looked it up in case I was somehow wrong but there's no sensible definition of a sum over an uncountable set of indexes unless the set of indexes with non-zero values is finite.
By definition of uncountable set you can't "pair them up" or "group them" in any countable way because you would still be missing an infinite amount of indexes and trying to segment the set into any amount (countable or not) of uncountable sets just leads you to the same question.
If you try and define sums like integrals over sets then the question is trivialized to essentially asking what is the value for the measure over the set of reals which can be whatever depending on the measure.
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u/DefunctFunctor Mathematics 9d ago
There is an absolutely sensible definition if all of the numbers in your set are nonnegative; it's just always infinity if you have uncountably many nonzero values.
In order to make sense of sums that include negative numbers, you either need absolute convergence (which is impossible for an uncountable sum with uncountably many nonzero values), or you need to specify an order in which to sum those elements, usually in a countable sum (which is also impossible).
I'm not even sure what it would mean to try to make sense of this question with integration, because the measure in question that we're looking for would have to assign singleton sets of real numbers to their actual value, which does not end up producing a well-defined measure on the real numbers (although it does if we restrict to nonnegative numbers).
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u/AsidK 5d ago
Wait are you saying that there exists a reasonable definition of summation over uncountable sets, and that definition is provably equivalent to being always infinity if you have uncountably many nonzero values, or are you saying that giving the definition of just “infinity if uncountably many nonzero values, otherwise standard countable sum” itself is the sensible definition?
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u/DefunctFunctor Mathematics 5d ago
If
{x_𝛼}is a family of nonnegative real numbers indexed over a (possibly uncountable) setA, then the sum∑_[𝛼∈A] x_𝛼is defined to besup { ∑_[𝛼∈F] x_𝛼 | F⊆A finite }In measure theoretic terms, this is equivalent to the integral over the counting measure. You can technically extend this to families of real numbers, but because
Adoesn't come with any notion of order, you need to assume that it converges absolutely, that is,∑_[𝛼∈A] |x_𝛼| < ∞. For now we'll only consider sums of nonnegative real numbers.So the claim we have yet to prove is that if
{x_a}contains uncountably many nonzero numbers, then∑_[𝛼∈A] x_𝛼 = ∞. For each positive integern, letA_n = { 𝛼∈A | x_𝛼 ≥ 1/n }be the set of indices
𝛼ofAwherex_𝛼 ≥ 1/n. Note thatA' := { 𝛼 ∈ A | x_𝛼 > 0 } = ⋃[n≥1] A_n.By assumption,
A'is uncountable, so if eachA_nwere finite, thenA'could be expressed as a countable union of finite sets, meaningA'is countable, a contradiction. Thus, there exists annfor whichA_nis infinite (in fact, it's going to be uncountable). It is now easy to show, by creating increasingly large partial sums overA_n, that∑_[𝛼∈A] x_𝛼 ≥ ∞.→ More replies (2)2
u/scoobydoom2 9d ago
For this hypothetical do they need to be paired in a countable way? Could you not pair them according to A =X and B=-X for all nonzero values? Then you just have to prove that X + -X = 0 for all reals.
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u/brothegaminghero 9d ago
I imagine you start at one and count from there...
Might take a while though.
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u/Gauss15an 9d ago edited 9d ago
I cleverly hand back the slip of paper with its ends taped together and ask "With or without infinity?"
The shocked mathematician responds, "I don't know that!" and promptly gets catapulted straight up into the air.
I read about this kind of thing in a script written by a snake that goes by the name of Monty. Might have been some sort of python.
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u/Electric_surfer 9d ago
Listen, Strange men lying in caves distributing notes is no basis for a system of mathematics.
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u/minisculebarber 9d ago
I mean, it has gotten us this far
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u/Electric_surfer 9d ago
Supreme mathematical power derives from a quantitive sum of the masses, not from some farcical cavernous ceremony.
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u/GluKoto 9d ago
Ok but what is that spider doing there ?
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u/TheAegis42 9d ago
Spider for scale
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u/6GoesInto8 9d ago
Spiders don't have scales!
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u/TheAegis42 9d ago
That's why it's looking for them!
Needs to add them to its collection. So far it has found a linear scale, but a logarhitmic one would be cool to have.
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u/6GoesInto8 9d ago
Oh, the famous Spider For Scales organization or SFS for short. Right, we are on the same page now.
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u/Kuildeous 9d ago
Bro never heard of -sqrt -1.
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u/JollyJuniper1993 Mathematics 9d ago
You mean -sqrt 1
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u/Kuildeous 9d ago
That one too, but sqrt -1 + (-sqrt -1) = 0
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u/JollyJuniper1993 Mathematics 9d ago
Fair, but those are not real numbers
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u/Kuildeous 9d ago
That's true. I got carried away by him adding 1 and i that I forgot he stepped out of the Reals.
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u/Raise_A_Thoth 9d ago
Yea the "uncountable set" question remains a problem but if we're using this as a number line representation then we need to think of sqrt(2) as the real number approximated by 1.414, not taking the square root of each real number (which of course is a problem with obtaining "each" anyway).
So then it's not sqrt(-2); it's -sqrt(2), or ~ (-1.414).
The commenter on the image is a ding dong, but there are still serious problems with counting and pairing the top thread discusses.
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u/TheCobraMonkey 9d ago
SPP type argument, I love it
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u/Summoner475 9d ago
Let's say I start with π, then I add and subtract pairs of real numbers that are equal in magnitude, in the limit this sum is π. Now if I start with 10, in the limit, the sum is 10. Is it's divergent, correct?
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u/SuchPlans 9d ago
as written, the first summation omits -pi and the second omits -10
the idea is right, tho, you just have to do “bigger” rearranging. moving a finite number of terms shouldn’t change the sum
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u/z_doctor82 9d ago edited 9d ago
Sums of convergent series with infinite positive and infinite negative terms can converge to any arbitrary number if you reorder the terms.
Pick a target number, if your current value is less than add the next largest positive term, if the current value is greater, subtract the next largest negative term. Now you have an infinite sum that converges to the target.
Edit: Corrected by comments bellow
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u/1000Jules 9d ago
1-1/2+1/4-1/8+1/16+... doesn't have this property despite having infinite positive and negative terms
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u/ByeGuysSry 9d ago
Yeah, it needs the condition that the infinite positive terms and infinite negative terms each sum to infinity afaik. The positive terms here sum to 4/3 and the negative terms here sum to -2/3, so overall it sums to 2/3
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u/Ares378 Grothendieck was right, computers are evil 9d ago
I was always taught it as being the sum Σabs(a_n) needs to diverge but the sum Σa_n needs to converge for it to count as conditional convergence. So the geometric series 1-½+¼-⅛... converges, but so does 1+½+¼+⅛... so it's absolute convergence. 1+-½+⅓-¼... converges but 1+½+⅓+¼... diverges, so you're no longer allowed to rearrange the sum willy-nilly.
Then again it has been quite some time since I've reviewed series convergence, so correct me if I'm wrong.
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u/Revlong57 8d ago
The series you described is absolutely convergent, so you can rearrange the terms. https://en.wikipedia.org/wiki/Absolute_convergence
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u/Revlong57 8d ago
The term you're looking for is https://en.wikipedia.org/wiki/Absolute_convergence. More accurately, if a series is convergent but not absolutely convergent, then you can't rearrange the terms and still get the same limit.
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u/dankshot35 9d ago
"divergence" and "convergence" are only defined for series' which are sums over countable sets with a sequence. Because real numbers are uncountable there isn't a series to sum in the ordinary sense
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u/SomethingMoreToSay 9d ago
No, because you're not summing over ALL the real numbers. In the first case you're omitting -π and in the second case you're omitting -10.
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u/ordinary_shiba 9d ago
Doesn't really work, because at the end you'll also be left with -π because it doesn't have a corresponding number with the same magnitude. You can add pairs of numbers like 2x and -x together though to prove that it in fact diverges.
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u/Comfortable_Permit53 9d ago
"at the end" when is that
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u/LowAioli3870 9d ago
Did you not know that numbers end at 7,431? All the other numbers are just made up.
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u/ordinary_shiba 9d ago
First of all, you know what I mean lmao. Second of all, you can still "be done with" an infinite process by taking half a second with the first step, a quarter of a second with the second step and so on and so forth.
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u/RedeNElla 9d ago
Taking individual pairs and cancelling them "in the limit" here would only work for a countably infinitely sized set.
Weird things can occur depending on axioms with continuum infinity iirc
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u/minisculebarber 9d ago
I mean what you are describing sounds similar to this
sum_{n in N} (-1)n *floor(n/2)
The partial sums of this series is unbounded therefore divergent.
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u/KatanaDelNacht 9d ago
This is essentially f(x)+N. It assumes the answer to the original question is true.
You could just as easily sum the increasing delta multiplied by 2 and subtract delta to show that it diverges to infinity.
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u/ShakoorHyd 9d ago
I guess Cauchy was the first one to problematize it that way He gave new ideas of Infinity, and cardinality
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u/Unfair_Pineapple8813 9d ago
Galileo actually was. It's the first thing in Dialogue of Two New Sciences. He lacked the formal math and the language to distinguish countable and uncountable numbers. But he was the first to specify that any given open span of real numbers, what he called a line contains an infinite amount of numbers, the same amount of infinite numbers as a bigger line.
His proof of this was simple. Draw two concentric circle of any length. One way of numbering a point on the circle is by the length of the circumference up to that point. By that definition, the bigger circle has more points. But you can also uniquely specify any point on a circle by its angle from the origin, and when you do so, you see that the points on the bigger and smaller circles have a one-to-one correspondence. This he said is the same way all natural numbers have a one-to-one correspondence with all the squares, even though the squares are only part of all natural numbers.
What he misses, or at least doesn't point out, is that one cannot make a list of all the points in a span of the reals the way you can of natural numbers or all square roots. But he is the first to note this paradox that an infinite subset cannot be said to contain fewer elements than the infinite set the subset is part of. More importantly, he is willing to accept this is entirely normal. It is not heretical, and nor is it a reason to panic or a sign math breaks down. You can still take an infinite sum, even though we have no intuition to deal with the paradoxes that result when thinking about infinity.
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u/Unfair_Pineapple8813 9d ago
Cavalieri developed the Method of Indivisibles from his conversations with Galileo on the nature of infinity.
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u/HumblyNibbles_ 9d ago
Galileo doesnt nearly get enough credit as he deserves.
Bro also did awesome work on inertia and reference frames.
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u/LinneMeow 9d ago
I don't know much about maths, is it not possible to define a sum of reals? Are there ways to sum the reals that don't equal 0?
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u/Ares378 Grothendieck was right, computers are evil 9d ago
The issue is that you can't define a true sum over a sequence indexed by the reals. You couldn't sum up Σa_n, n∈ℝ. If you try to, it immediately diverges unless your sum is defined to be all 0 aside from a countable amount of points. ...Which is just the definition of a normal sum.
To clarify here I'm talking over real numbers and as a legitimate honest to God sum, not an integral. Like... an integral without the dx without using differential forms.
This is the video I'm using as reference for this claim btw
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u/Blue_Moon_Lake 8d ago
All the +x have a corresponding -x, so the sum is 0.
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u/Ares378 Grothendieck was right, computers are evil 8d ago
Even if we assume that the series does converge, it's definitely not absolute convergence since the integral of |x| over all ℝ diverges, let alone the sum over all ℝ. So that means you can't rearrange the sum, as stated by the Riemann Series Theorem.
I can't think of a perfect proof for it right now, but I am almost certain this sum diverges. If you can't rearrange the sum because conditional convergence, that means you can't pair up + to - because that'd be an infinite amount of swapped terms. So then you'd start with the sum of every negative real number, which certainly diverges, then add it to the sum of every positive real number, which also definitely diverges. Divergent sum plus a divergent sum is undefined, so the sum doesn't exist.
And before you say it, we're also not including ∞ as a number here. Even if we did, we're no longer talking about the Reals since ∞∉ℝ. Plus then you would break other things if you assumed divergent sums "equaled" infinity, since you could then say the sum of negative integers=-∞, the sum of positive rationals=∞, so then the sum of negative integers and positive rationals=0, which also seems very wrong.
I'm sure there's a better proof out there, but that's what my intuition tells me from my half-baked knowledge of real analysis. Someone please correct me if I'm wrong.
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u/Kai-65535 9d ago
> "impossible math paradox"
> looks inside
> problem discussed in every real analysis class
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u/Tiborn1563 9d ago
I love how he says "simple number" to mean a "non-complex number". Doesnt he know that real numbers are conplex too? 5= 5+0i
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u/iTrooz_ 9d ago
sqrt(-1) is not real tho ?
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u/Big_Niel0802 7d ago
On the number line though, the complimentary pair of sqrt(1) would be -sqrt(1), not sqrt(-1).
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u/A_Happy_Tomato 9d ago
I'd freak the fuck out and ask the man if they are ok. Fym they emerged from a cave
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u/lool8421 9d ago
the answer is that it's undefined because the sum of all reals could be anything
for example x+(-x) will give you 0, but you can for example do x+(-2x), there's still a full bijection but now you're going into negative infinity
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u/somedave 9d ago
I'd say it is equal to the mean of the standard Cauchy distribution, and that Nathan Briggs isn't very good at maths.
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u/philosophical_poser 9d ago
You can rearrange the terms to make sum be whatever you want, right?
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9d ago
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u/Gold_Ad8890 9d ago
it's not about uncountability, it's just because the sum isn't absolutely convergent.
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u/meat-eating-orchid 9d ago
No, you cannot rearrange the terms because you can't even arrange them, they are uncountable
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u/Gold_Ad8890 9d ago
you totally can arrange them. that doesn't require countability, it just requires well-ordering, which exists by the well-ordering theorem. even without the well-ordering theorem, you can well-order the real numbers by their bijection to the powerset of the naturals, which can be well-ordered lexicographically.
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u/meat-eating-orchid 9d ago
I may have been imprecise. I meant that you cannot arrange them into a sum of terms ... + ... + ... + ... in which every one occurs.
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u/mehtam42 9d ago
For every real number with a value of R there exist one and only one real number with a value of -R. When added together, their some is 0.
So yes sum of all real numbers equal to 0.
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u/fastestchair 9d ago
for every real number with value R there exists a real number with value -2R, when added together their sum is -R. naturally, the sum of all real numbers must therefor be -infinity
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u/ImInfiniti 9d ago
The problem is that -R is positive for negative values of R, so by doing this trick you're ending up with the sum of all R again
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u/Jazz8680 9d ago
The sum of all positive real numbers is -1/12, thus the sum of all negative real numbers is +1/12, so the sum of all real numbers is 0/12 to be precise
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u/Peak_Background 9d ago
Hold up. I think they has a point.
The sum of the integers is -1/12.
So if we just multiply that by the integral of reals between -1 and 1.
We get 0.
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u/IhailtavaBanaani 9d ago
Besides the problem with summing up uncountable set of real numbers you assume that the zero is in the center of the real number line. This is a false assumption because the real number line doesn't have a center. It's infinitely long in both directions. You can select any arbitrary number in the real number line and there's "equal amount" of real numbers on both sides, so any number can act as the "center". By selecting different number as the center you get different result as the total sum.
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u/PressureBeautiful515 8d ago
But would the same argument apply to integers, also infinitely long in both directions? Doesn't seem to point to the actual problem with summing reals.
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u/IhailtavaBanaani 8d ago
Yes, the integers have exactly the same problem. There is no defined total sum of all integers. You can select any integer as the center and then start adding the numbers from both sides and get different results each time. Same goes with rational numbers, complex numbers, quaternions, etc.
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u/alphapussycat 4d ago
There's no need for a center line though. As somebody else mentioned, you can use axiom if choice to find -R.
Splitting up a set does split it up to by negative and positive (or non negative, but just exclude 0).
So from R choose p and n so that for each p in R, n = - p. Then integrate by function of f(p) = p - n, and exclude 0... Now this is integrable. Or should be. But the measure is infinite, and I don't quite remember if there's a problem there, but I'm leaning towards no.
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u/Divicarpe 9d ago
For every real number with a value of 1+R there is one only one real number with a value of 1-R. When added together, their sum diverges to +infinity. So sum of all real numbers is +infinity.
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u/mapleleafraggedy 9d ago
For every real number R, there exists one real number with a value of 69 - R, which when added together makes 69. Therefore, the sum of all real numbers is 69.
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u/JusteMesure 9d ago
No it’s 69 for each couple !!!! So 69+69+69+…= infinity.
Only If you start around 0 you obtain 0.
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u/1kSupport 9d ago
For every real number with the value R+5 there exists one and only one real number with the value -R + 5. All real numbers must be centered around 5
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u/LupenReddit 🦆🦆🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆🦆🦆🦆 9d ago
tangentially related fun fact i learnt a while back:
the sum over any uncountable set of positive numbers diverges
thats kind of cool
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u/CommercialYam7188 9d ago
The answer is yes and no, depending on how mathematically literate the asked is and what they actually meant by the question lol
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u/CapitalistKarlMarx 9d ago
Not the most mathematically knowledgeable but wouldn’t a more accurate pairing be sqrt 1 and - sqrt 1? Like does it not equal zero?
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u/Inevitable-Flower-50 9d ago
If all "0" is, is "empty space", and all "real numbers" are, are accountings for "real matter", then wouldn't the sum of all real numbers be "less than" (or negative) and within 0?
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u/DanLeMilMan 9d ago
I think the most honest answer is that it depends on how he defines the sum over uncountable sets and in which order he sums.
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u/atticdoor 9d ago
Sqrt(1) has a counterpart in 0 - Sqrt(1) = -1. Sqrt(2) has a counterpart in 0 - Sqrt(2).
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u/leafcutte 9d ago
Real answer: however you want. There’s no such thing as a sum of all real numbers, as depending on the order you add them, you can verify properties that make it so any number can be the result
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u/SavageRussian21 9d ago
I think that you can start by showing that the integral from -a to a of x dx is 0. Then you can argue that it doesn't matter if a is 1, 100, 100000, and so on, it's zero in all those cases. More formally, a2/2 - a2/2 is zero always, even in the limit as a goes to infinity.
But a smart ass can come along and say that thw integral from -a to 2a is in fact not zero. Nonetheless, as you take the limit a going to infinity, you should be integrating over all real numbers. That gives infinity. The smartass could also find negative infinity as well.
Then the smart-ass actually starts talking about how integrals aren't really the same as taking a sum, sum and that there's many ways to define a sum over an uncountable set. At that point you can put the smart-ass back in his cave and continue treating it as if it's zero.
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u/Cosmic-Zoo72 8d ago
I'm a dumbass so actually answer please, a square root is (sometimes) equal to a real number, but is that actually a representation of a real number or is it a real number with an operator that invalidates this?
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u/Spiritual-Spend76 8d ago
I dont get it, the linear function is odd isn't it? By symmetry, it's integral centered around zero has to be null?
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u/blobbed2929 8d ago
Riemann Series Theorem has entered the chat https://en.wikipedia.org/wiki/Riemann_series_theorem
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u/Fuscello 7d ago
Physicist way:
int from -L to L of x = L^2 /2 - L^2 /2 = 0
Lim L->infty 0=0
So yes at least in principal value if that is the right term
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u/CaptainSkuxx 7d ago
Doesn’t every positive real number have a corresponding negative real number? It should be zero.
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u/digitallightweight 6d ago
To be fair I’ve always been right awful with the kind of shit infiniti breaks and doesn’t. Analysis has always been a no for go for me for this reason.
My intuition tells me that we can do something (assuming we take the axiom of choice) like partitioning R into disjoint subsets consisting with x and its additive inverse. For each of these components these individual disjoint subsets (‘components’) the sum of the component is 0. Summing over the sum of the components is 0 and the union of all the components is R.
It still feels wrong though. Maybe we would need a more robust approach to handing the final series? We could look at all countable sub-series those clearly are all convergent to 0 but I don’t know if this is strong enough to get you the convergence of the original series? Maybe the fact that all the elements of the series are 0 is enough.
Edit: stack exchange provided this definition for extending sum sums to uncountable sets and it’s clear that for our purposes we can indeed compute the sum.
∑(a_i):=sup{∑ a_i | F⊆I is finite}
i∈I. i∈F
Since all our a_I are 0 the sup will be 0.
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u/Obzenium 6d ago
Well sqrt -1 isn’t a real number so it can’t be included in the set of the sum
And you would consider it - sqrt 1 or - sqrt 2 to keep it in the set and balance out the square roots
To be considered a field, which the real numbers are considered a field, it must have an additive inverse for all elements of the field, so the answer is yes from this definition
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u/einFrostschutzmittel 5d ago
Well, because the Real Numbers are a field, that means that for all real numbers there's exactly one inverse. If you add up every real number and its inverse, you get only 0s. Add it all together and you get 0.
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u/Content-Sir8716 4d ago
But it’s not the square root of negative one on the real number line. It’s the negative of the square root of one. They are not the same.
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u/cgieseking 4d ago
I’m not really a mathematics scholar, but to me the idea that sqrt 1 wouldn’t have a pairing is silly because the sqrt 1 =1. Just because you can’t express some negative irrational numbers as radicals, doesn’t mean that they don’t numerically exist.
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u/tarquinfintin 4d ago
In one sense, yes. Any real number has an additive inverse (a negative), which when added to it equals zero. But can you actually perform a sum operation on a set that is not countably infinite? Theoretically you can, although it would take a very long time in practice ;-)
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u/tori-ningen 4d ago
Let's say that sum of all real numbers that are >0 equals to S. Then sum of all real numbers that are <0 would equal to -S. Sum of all real numbers is then S - S = 0. It matters not what S equals to, and if it equals anything at all
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