the most viral "false" equation

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[@MichaelPennMath]

For more information about openings at Mat-X go to www.matx.com/michaelpenn

[@kubilayaytemiz7274]

At 14:52, multiplying with s+1/s+1 for s = -1 gives us 0/0 which is indeterminate.

[@adb012]

Yes, but that step would still be valid for s=-2. Does it mean that the sum of all squared integers is zero? Legit?

[@kubilayaytemiz7274]

@@adb012 To obtain a similar equation for s = -2, we have to multiply with s+2/s+2 which is indeterminate also. To validate myself, you can see at 23:09, after doing I.B.P, we have Gamma(s+3). Only way to get Gamma(s+3) from Gamma(s+2) is multiplying the integral with s+2/s+2. And this will be the case for every s. Your conclusion is not valid I'm afraid.

[@DeJay7]

My genuine reaction when the sum of all the integer squares is a "trivial" result and very obviously equal to 0.

[@vascomanteigas9433]

zeta(-2)=0. The best formula is to use Bernoulli Numbers. For all integers n>0, then zeta(-n+1) = B_(n+1)/n

[@spiderjerusalem4009]

@@vascomanteigas9433 Any book/articles to learn about bernoulli/euler numbers?

[@MarcinSzyniszewski]

Proof: it's trivial. 👍

[@goodplacetostop2973]

25:58 Ah, when Mathologer got mad at Numberphile… Those were the days

[@NotBroihon]

And rightfully so.

[@l.w.paradis2108]

@@NotBroihon 💯

[@Alan-zf2tt]

Em hmm I don't know and think context is important. In Math approach infinity goes on to infinitely many infinities In science approach quanta do things discretely yet there comes a place where quantum mechanics needs -1/12 to conform to observed behaviors. Each has its own context and I think those contexts need to be explored and explained in a way that satisfies both math and science parrticularly quantum observables

[@NotBroihon]

@@Alan-zf2tt that has absolutely nothing to do with what Mathologer rightfully criticized Numberphile for.

[@Alan-zf2tt]

@@NotBroihon I think what I am trying to say is that it is science and not a war. It deserves explanation and we can agree it seems controversial - for at least ten years

[@giovanninuno7963]

Swapped summation and integral without proving convergence?

[@jimallysonnevado3973]

i think this step is barely not allowed😂

[@user-jc2lz6jb2e]

It's been 10 years. I was in college taking calc 2 when that video dropped. I remember a classmate bringing up how it was "proved" that 1+2+3+... = -1/12, and our professor was just confused.

[@kazedcat]

Yeah Terence Tao has proven it. To be very precise he has proven that the sum of all natural numbers can be -1/12 if you extend the definition of partial sum into a weighted partial sum.

[@l.w.paradis2108]

@kazedcat In other words, if you redefine the operation of summation itself. Well, then.

[@kazedcat]

@@l.w.paradis2108 Partial sum is axiomatic so there is no proof that piecewise summation is the only valid way to do infinite sum. This is similar to the fifth postulate and rejecting it leads to Non Euclidean geometry.

[@l.w.paradis2108]

@@kazedcat That's fine, but it's not the same thing. Why is it that in geometry, we never confuse Euclidean with Lobachevskyan, etc.?

[@kazedcat]

@@l.w.paradis2108 Because you don't deal with geometry in your everyday life. When was the time you had to compute the area of a parking lot to buy groceries. Only Google Maps developers have to deal with the difference between a flat map and a spherical map.

[@mikederleres4300]

2 sign mistakes, which nicely cancel out: At 16:20, on du At 18:00, on the integral

[@nHans]

1:03 - Nice, I learnt a new mathematical term today: *_barely not allowed._* 🤣

[@ericbischoff9444]

5:19 the odd terms sum starts at 3, but on left side of = it starts at 1. I think n should start at 0 for odd terms. The error is canceled at 6:20.

[@williamperez-hernandez3968]

He used an incorrect expression for the odd numbers. Using 2n-1, allows to begin with n=1. Thus the sum over all naturals is recovered.

[@aryakaranjkar939]

Yeah, or instead it should be 2n-1 instead of 2n+1 .

[@FranzBiscuit]

Nice! Another way to look at it is as the indefinite integral of the generating function of the sequence which produces the sum-of-integers, ie. (n(n + 1))/2 = (n^2)/2 + n/2, which is of course -1/12.

[@fable4315]

I think the sketchy step is 10:50. Swapping an infinite summation and integration is probably only under some assumptions possible, for example I think one condition should be that the sum under the integral converges?

[@__christopher__]

In the partial integration step: x^s is0 for x=0 only if s>0. Which precludes not only inserting s=-1, but even taking the limit of s to -1.

[@CoreyKatouli]

Excellent video as always

[@gerryiles3925]

When you wrote out the result of the first integration by parts step at 18:03 you missed the negative sign but then the error got undone in the second integration by parts...

[@petterituovinem8412]

Is the sketchy step changing the order of summation and integration

[@janeknowakowski5732]

Yes, i think so

[@vincentbutton5926]

I missed that happening at about 10m20s in. Thanks for pointing it out.

[@davidcroft95]

Not quite sure if it's a problem (or one of the problems, maybe), but surely the problem is in the next board, while solving the "zeta integral": he multiply (and divided) by s+1, which is zero in this setup

[@jagatiello6900]

​@@davidcroft95 At 14:47 he's exploiting the fact that gamma(s+1) has a simple pole at s=-1, which is why he then says the (s+1) in the denominator is absorbed by using the functional equation for gamma.

[@davidcroft95]

@@jagatiello6900 he still divided by zero

[@coreyyanofsky]

although other functions that "look like" 1 + 2 + 3 + ... when analytically continued can "assign" arbitrary values to the series, the value -1/12 is special because in quantum physics when the series 1 + 2 + 3 + ... occurs only the value -1/12 gives answers that agree with observation there has been recent work on how regularization of divergent sums and integrals in quantum theory works; i won't pretend to understand sentences such as "...regulators that allow the infinite series of natural numbers to converge towards -1/12 are intimately connected to the preservation of gauge invariance at one loop in a wide class of non-abelian gauge theories coupled to an arbitrary number of Dirac fermions" but if this topic is of interest to you, dear reader, you can find more information in the arXiv preprint "Smoothed asymptotics: from number theory to QFT" by Antonio Padilla and Robert G. C. Smith

[@Noam_.Menashe]

My favourite way to "prove" this is with derivatives of geometric sums. You start with 1/(1+x)=sum (-x)^n, after derivation it becomes 1/(1+x)^2=sum n*(-x)^n-1, putting in x=1 we get 1/4=sum n*(-1)^n-1 and finally by isolating even terms and dividing by two we get -1/12=sum n I liked the multiplying and dividing by s+1, then setting s=-1. If the integral converges it converges.

[@galveston8929]

how do you justify exchanging the limit and the infinite sum?

[@MichaelRothwell1]

​@@galveston8929 by the limit I imagine you mean the limit as x→1, given that the sum is only true for |x|

[@rrr00bb1]

i like doing it this way too. I use recursion so that you use nothing more than high-school math. it really only requires taking the definition of "a=b" more literally; like a computer does. With recursion, you can avoid all infinite sums. This is good, because you don't end up arguing about what you can and cannot do with infinite sums. kzbin.info/www/bejne/oHvJn55urcSLnaM

[@bloodyadaku]

When you did integration by parts you skipped out on a negative, but you skipped out on the negative twice so it ended up cancelling out.

[@user-mw6tq2nq2e]

When you used the formula for geom. series, the geom. series only works if the absolute value of the common ratio is strictly less than 1, though it will not be if x = 0.

[@l.w.paradis2108]

NOW I know why ChatGPT is suddenly making mistakes in simple mathematics, and refusing to show its work! (Seriously, there's already a major paper on it.)

[@s4623]

4:57 wouldn't 2n+1 starting at n=1 gives you 3? Where did the leading 1 term go? (Should have been 2n-1)

[@gp-ht7ug]

At 02:36 the formula is the sum of alternating signs?

[@Kratos_legends]

Wooowww😮 you are the best teacher that I have seen in my whole life ❤❤ your skill of explanation is so great 👍

[@rdubeau]

At 11:40 you substitute t=nx but accept infinity as your upper limit of integration, same as before the sub. I can see this maybe being “barely not allowed.”

[@speedbird8326]

OMG i could have sworn @22:51 he was going to say "And that's a good place to stop" but he kept going !!!!!!

[@dneary]

@MichaelPennMath I saw you in Nina Williams's China Beach documentary! I know you from math videos, but apparently my son's know you more from climbing stuff!

[@KipIngram]

6:27 - Whoa; hold it. Your sums from 1 to infinity were for (2*n+1) and (2*n). That starts at 2. But you replaced it with a sum that starts at 1.

[@ruffifuffler8711]

So, you sail a boat past all the natural #s, using all the previous #s to pick up the next one, but, at some point the boat sinks, and reaches equilibrium at -12 feet? Just goes to show you that most numbers aren't worth a dam when their daid.!

[@vascomanteigas9433]

The reasoning was to use the Fermi-Dirac integral and apply integration by parts to find the result.

[@djfahed3002]

15:24 Once you multiplied and divided by s+1, you assumed that s!=-1 thus you lost the privelege of replacing s by -1

[@Mohanchous]

1. Why the Riemann zeta function? Couldn't other functions be used to deruve a value for 1+2+3+...? If so, would other functions imply the same -1/12 evaluation? 2. Is there a number system, a variant or extension of the reals in which 1+2+3+... is convergent?

[@debtanaysarkar9744]

Answer to ur first question, there's another function which was devised by Ramanujan himself to prove this result, and guess what, that uses complex analysis 💀

[@MuffinsAPlenty]

Great questions! 1. Good question! Because your concerns are valid. We can develop other series-defined functions which "evaluate" to 1+2+3+4+... outside of their region of convergence, and which analytically continue to assign values _other than_ -1/12 to input which "evaluates" to the series 1+2+3+4+... [See appendix B of this comment.] The only examples I've seen, however, are pretty contrived and aren't useful. So the question then becomes how can we justify that the Riemann-Zeta function is the best choice to use. 2. No. And this is because convergence of the sequence of partial sums leads to a _stable, linear_ summation method. And it is impossible for 1+2+3+4+... to be assigned a value under any _stable, linear_ summation method. [See appendix A of this comment.] *APPENDIX A* A "summation method" is a function S with inputs being (some) infinite series and outputs being numbers. A summation method S assigns a value to a series ∑aᵢ if S(∑aᵢ) is defined. A summation method is considered to be _linear_ if S is a linear function: 1. If S(∑aᵢ) and S(∑bᵢ) are defined, then S(∑(aᵢ+bᵢ)) is defined and has the value S(∑aᵢ)+S(∑bᵢ) (essentially, term-wise addition of adding series is consistent with adding the sums of the series). 2. If S(∑aᵢ) is defined an c is a constant, then S(∑caᵢ) is defined and has the value cS(∑aᵢ) (essentially, distribution/factoring of constants over an infinite series is consistent with multiplying the sum of the series by that constant). A summation method is considered to be _stable_ if the following holds: If S(a₁+a₂+a₃+...) is defined and a₀ is any number, then S(a₀+a₁+a₂+a₃+...) is defined and has a value of a₀+S(a₁+a₂+a₃+...). (essentially, you can add a term to the beginning of the series, and the sum changes by adding the same amount). Again, one can prove that convergence is both stable and linear. Let's now show that 1+2+3+4+... cannot be assigned a value under any stable, linear summation method. We will do this by contradiction. We will assume 1+2+3+4+... can be assigned a value under a stable, linear summation method, and we will arrive at the conclusion that 0 = 1. Assume S = 1 + 2 + 3 + 4 + ... Then using stability, we can add a 0 term at the beginning, which will make the new series have the value 0+S = S. S = 0 + 1 + 2 + 3 + ... Using linearity, we can term-wise subtract the bottom series from the top series, and it will result in a series having the value S-S = 0. 0 = 1 + 1 + 1 + 1 + ... Using stability, we can add a 1 term at the beginning, which will make the new series have a value of 1+0 = 1. 1 = 1 + 1 + 1 + 1 + ... However, this new series is identical to the previous series. So we have 0 = 1 + 1 + 1 + 1 + ... = 1 or 0 = 1. So we have arrived at a contradiction. This shows that 1 + 2 + 3 + 4 + ... cannot be assigned a value under any stable, linear summation method. And since convergence of the sequence of partial sums is always a stable linear summation method, 1 + 2 + 3 + 4 + ... cannot have a sum using convergence, no matter what "number system" one uses. *APPENDIX B* I can't take credit for this example. I obtained it from the Math Stack Exchange post entitled "Uniqueness of analytic continuation to assign sum values" asked on January 14, 2018. Let C be any complex number, let h(x) = (1−x) − C(1−x)^3, and consider the function: g(x) = 1/(1−x)^2 − h(x)/(1−x)^3. Note that 1/(1−x)^2 = d/dx[1/(1−x)] = d/dx[1+x+x^2+x^3+...] = 1+2x+3x^2+4x^3+..., and this converges absolutely on (−1,1) Additionally, −h(x)/(1−x)^3 = −h(x)/2 * d/dx[1/(1−x)^2] = −h(x)/2 * [2+6x+12x^2+20x^3+...] = −h(x) * [1+3x+6x^2+10x^3+...], and this converges absolutely on (−1,1). Adding together, we get (1−h(x)) + (2−3h(x))x + (3−6h(x))x^2 + (4−10h(x))x^3 + ... = 1(1−2h(x)/2) + 2(1−3h(x)/2)x + 3(1−4h(x)/2)x^2 + 4(1−5h(x)/2)x^3 + ... = the sum of n (1−(n+1)h(x)/2) x^(n−1) from n = 1 to infinity. Let's take f(x) = ∑n∙(1−(n+1)h(x)/2)∙x^(n−1), which is defined on (−1,1). Notice that h(1) = 0, so f(1) = ∑n∙(1)∙1^(n−1) = ∑n = 1 + 2 + 3 + 4 + ... So we have found a function f(x) so that f(1) is 1 + 2 + 3 + 4 + ... Now, we have already shown that on (−1,1), we have f(x) = g(x). On this same interval, g(x) = 1/(1−x)^2 − h(x)/(1−x)^3 = 1/(1−x)^2 − [(1−x) − C(1−x)^3]/(1−x)^3 = 1/(1−x)^2 − (1−x)/(1−x)^3 + C(1−x)^3/(1−x)^3 = 1/(1−x)^2 − 1/(1−x)^2 + C = C The constant function C is an analytic function; hence, j(x) = C is the analytic continuation of f(x). Therefore, we associate 1+2+3+4+... = f(1) with j(1) = C through this particular analytic continuation. But recall that C was _any_ complex number. Hence, we can use analytic continuation to associate the series 1+2+3+4+... with _any complex number we want._ The function f(x) here isn't particularly important, and there's no reason we should expect to use f(x) to assign a value to the series 1+2+3+4+... in any given context, but it shows that there should be more than merely saying "analytic continuation". One should justify why the Riemann zeta function is the particular function we want to use. And there are good reasons to believe that the Riemann zeta function is a good choice, but I believe people should do a better job of explaining that.

[@MathFromAlphaToOmega]

One kind of strange thing about the sum is that if you use the function f(z)=0/1^z+1/2^z+2/3^z+3/4^z+.... and continue it to z=0, you get f(z)=zeta(z-1)-zeta(z), so f(0)=5/12. That suggests that 0+1+2+3+4+5+...=5/12, so these kinds of arguments are very sensitive to the exact function you use.

[@vascomanteigas9433]

Because this renormalization only are valid using the Riemann Zeta Function.

[@kazedcat]

There is another way to get -1/12 using weighted partial sum.

[@KipIngram]

I understand why this upsets people. Most people realize that they don't know "high math," but they do at least think they understand adding integers. Then this comes along and more or less says to them "you know NOTHING." The thing to understand here is that the -1/12 does NOT come from actually adding up these numbers. You can write an expression that is equivalent to this series when its parameters are set a certain way, but that expression itself isn't defined everywhere. You can then use analytic continuation to carry it from the region where it IS defined to "everywhere else," and it is THAT function that has the value -1/12 at a key point. It's not the same function - you've replaced it with a different one at that point. In some ways it's a "cheat," though in some ways it's perfectly valid and actually has physical applications.

[@vascomanteigas9433]

Abel-Plana formula goes brrrr... 😂

[@MasterHigure]

I would be pedantic and call 1+2+...=-1/12 an *equality* not an equation.

[@the_nuwarrior]

it can be proved from hankel integral over C

[@glenneric1]

To prove it is wrong I just go 'der'.

[@debguha8727]

At 17.19 x^s is 0 only if s>0. So we cannot put s=-1 in the last integral.

[@donach9]

Γ(s + 1) = Γ(0), is undefined. Isn't that the dodgy step?

[@gp-ht7ug]

If I’m not wrong there is a new Numberphile video on this matter

[@thes7274473]

Two videos, actually. While watching the first one I was like "Yes, a redemption arc!" as they had a guest explaining analytic continuation. Then they posted another video where they doubled down and "proved" that the sum is actually equal to -1/12 by totally changing the problem.

[@l.w.paradis2108]

I hope not.

[@kazedcat]

​@@thes7274473It's not changing the problem but just changing the axiom of a partial sum. There is no solid reason why a partial sum needs to be the way it is. This is similar to changing the fifth postulate to derive Non Euclidean geometry there is no valid reason why the fifth postulate must only be true.

[@NotBroihon]

Probably Numberphile's worst video. Completely unmathematical and basically just clickbait.

[@l.w.paradis2108]

@@NotBroihon It was despicable.

[@theartisticactuary]

Must be swapping the order of summation and integration that's the problem. Everyone knows that you're only allowed to do stuff like that in physics. Do it in maths and you're always going to run into problems. This is why we have rules.

[@xaxuser5033]

i don't know why people are talking about multiplying by s+1/s+1 , or swapping integral and summation. The proof is perfectly fine and the value of zeta at -1 is indeed -1/12. The only problem is that the two expressions of zeta coincide at {s / s>1} and not in the real line. The first expression of zeta holds for s>1 , the other one holds for a bigger set, the two expressions coincide at a set which is not all the real line. And example of what i say : let f(x) = (x^2-1)/x-1 and g(x) x+1 f and g coincide in the real line except for x=1 where f is not even defined. We can't say then that f(1)=1 cz it doesn't even have sense, but we have still an equality nearly everywhere.

[@Alan-zf2tt]

I cannot help but wonder if Euler had troubles with his new math. Did notion of imaginary numbers unsettle people? It is rumored that Cantor had problems introducing notions of different infinities s0 maybe there is a thing happening there than needs solving to mathematical satisfaction? My own take is that there has to be a point, surface, event space where discrete math becomes continuous math, integer sums become integer integrals. And if it is a boundary is it a fuzzy boundary with epsilon like error bounds or a pointwise boundary like the circumscribing of a circle? In essence the three assumed bits/parts/partitions are: discrete, boundary, continuous seeming to correspond to quanta, boundary, continuity -shrug-

[@wolliwolfsen291]

3:38 The Zeta function converges for s>1, but you evaluate it later for s=-1. So 1+2+3+4+5+6+7+… can not be equal to -1/12, that‘s where the „proof“ fails

[@vladimirlinhart4486]

It is not equal in a sense of equivalence. It is an analytical continuation of dzéta function. Do not mix the result with the sum of the natural numbers.

[@wolliwolfsen291]

@@vladimirlinhart4486 of course zeta(-1)=-1/12 but the zeta(s) is defined by the dirichlet series (sum of 1/(n^s)) for Re(s)>1

[@coreyyanofsky]

of course this is true -- the series is divergent; however there is a certain well-defined sense in which this series is "equal to" -1/12 + ∞ you get there by (instead of looking at the limit of partial sums) smoothing the sum with a parameterized regulator function such that the terms of the (convergent) sum becomes closer and closer to those of the series 1 + 2 + 3 + ... as the parameter grows without bound and then doing asymptotic analysis on the parameter; the zeroth order term is -1/12 and the first order term diverges along with the parameter Terrence Tao wrote a blog post about this approach entitled "The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation"

[@vascomanteigas9433]

You need to derive the Analytical Continuation of Riemann Zeta Function. My favorite Proof is to apply the keyhole contour integral of Bose-Einstein integral and it os derived. Also it is possible to deduce a Laurent Series for Zeta valid for all complex Numbers except the pole itself at x=1. Zeta(x) = 1/(x-1) + sum from n=0 to Infinity of Stietjes(n) * (-1)^n * (x-1)^n / n!. The Stietjes constants are a generalizarion of Euler-Mascheroni constant.

[@djttv]

But, the sum of all positive integers does not equal -1/12

[@MichaelRothwell1]

That's why he always wrote "=" and not =. Only some physicists and intellectually dishonest mathematicians actually believe the sum in a literal fashion.

[@joseoncrack]

@@MichaelRothwell1 Yes, judging from the current state of theoretical physics (/astrophysics), I would even dare say a bit more than "some physicists". 🙃

[@tenjan75]

Your April fools prank is 5 days late….

[@NarutoSSj6]

The annoying thing is that this divergent sum equals - 1/12 using all those sketchy methods but never anything else, like - 10 or whatever. If only one could find a way for it to equal somethings else than - 1/12 then i am happy to disregard it.

[@StephTBM4]

There is a way to (falsely) prove that the sum of all integers (let's call it S) equal -1/8 ... Regroup 2+3+4 =9 5+6+7 =18 8+9+10=27 ... So that S=1+9(1+2+3+...) Then S=1+9S so 8S=-1 and S=-1/8 !

[@StephTBM4]

So, continuing on false assumptions, it appears that -1/8=-1/12, thus one can "prove" that 0=1 😂😂 Never make computations on diverging series ...

[@StephTBM4]

kzbin.info/www/bejne/bHe3qICLgZtsq5Isi=n650LTzq_cy_a6Dr

[@vikramanbaburaj525]

While I appreciate your videos, I also find them to have little to no use in practical life.😅

[@mircoceccarelli6689]

1 + 2 + ... + n = n ( n + 1 ) ÷ 2 n € IN => n >/= 0 , n + 1 > 0 ( + )( + ) = ( + ) =/= ( - ) Ramanujan N.O.K. !!! 😊🤪👍👋

[@decaydjk8922]

Glad you're getting some sponsorship dollars out of the "AI" machine learning bubble anyway but it's on par with advertising for a crypto company

[@galveston8929]

Somehow sloppy math. You cannot change the order of a infinite sum an integral without proper justification. Let's call this "a lucky accident" unless there's a better exposition using Ramanujan sum and analytic continuity.

[@noahprentice751]

I think that was the "trick that's not allowed in one of the steps" he mentioned around 0:55.

[@galveston8929]

@@noahprentice751 it's more of "carelessness" than a "trick".

[@noahprentice751]

@@galveston8929 given the video was already 25 minutes long and not about moving limits across integrals, I think it's more likely that they intentionally decided not to delve into that detail, rather than mere carelessness. If you want a proof that goes into every detail, you can probably find a paper on the topic (or work through it yourself).

[@davidcroft95]

Yeah, that's the point of the video. Congrats 🎉👏🏻

[@galveston8929]

@@noahprentice751 There are plenty of examples where you may use a "trick" like this on a diverging sum and come up with an arbitrary finite value. Sufficient rigor must be carefully put into assigning a finite value to a diverging sum. KZbin is rife with clickbait videos about sum of all integers being -1/12, Mathologer's is the only one I've seen so far that treats the subject properly.

[@nightmareintegral5593]

copied from flammable math

[@randomguyontheinternet_69]

what are you talking about ? does he own this infinite series?

[@jellyfrancis]

​@@randomguyontheinternet_69yo 😂💥💀

[@nightmareintegral5593]

@@randomguyontheinternet_69 not but method is same XD

[@nightmareintegral5593]

@@randomguyontheinternet_69 that s why this is copy past

[@MuffinsAPlenty]

@@nightmareintegral5593 Have you accused flammablemaths of copying Riemann?

[@ignore_for_your_sanity]

A quirky startup for ASICs? Do they have the billions in capital of nvidia and the decades of experience required? Or is it simply a means to capitalize on corporate welfare and get bought out in a few years once TSMC is bought out by nvidia as a consequence of being the only company with ASML's EUV nanometer precision thanks to intel and amd price fixing and stagnating for the past 15 years? No, the answer this is all coming out samsung (heaven's mandate of SSD technology). I hecking love industrial (capitalist) society. It's so antiracist and inefficient. Also, cool vid. I can't wait for my hecking favorite reddit-tier channel (3b1b) to make a video twice this length to discuss why infinite series is abusive notation and technically youtube is a waste of time as all the academic minds in the world with light speed communication still can't solve if p=np. Education is a racket, and america is an organ plantation for the slavers.

[@restcure]

Welp - if they can make something that nvidia wants to buy them out for a goodly sum, then I say to them: good job of you getting yours, and enjoy your retitrment. Go, go, gadget capitalism.

[@patricius6378]

>"I hecking love industrial (captialist) society. It's so antiracist and inefficient..."