this post was submitted on 06 Jan 2024
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I considered deleting the post, but this seems more cowardly than just admitting I was wrong. But TIL something!

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[–] WilloftheWest@feddit.uk 64 points 10 months ago (7 children)

This kind of thread is why I duck out of casual maths discussions as a maths PhD.

The two sets have the same value, that is the value of both sets is unbounded. The set of 100s approaches that value 100 times quicker than the set of singles. Completely intuitive to someone who’s taken first year undergraduate logic and calculus courses. Completely unintuitive to the lay person, and 100 lay people can come up with 100 different wrong conclusions based on incorrect rationalisations of the statement.

I’ve made an effort to just not participate since back when people were arguing Rick and Morty infinite universe bollocks. “Infinite universes means there are an infinite number of identical universes” really boils my blood.

It’s why I just say “algebra” when asked what I do. Even explaining my research (representation theory) to a tangentially related person, like a mathematical physicist, just ends in tedious non-discussion based on an incorrect framing of my work through their lens of understanding.

[–] balderdash9@lemmy.zip 34 points 10 months ago (2 children)

For what it's worth, people actually taking the time to explain helped me see the error in my reasoning.

[–] WilloftheWest@feddit.uk 21 points 10 months ago (1 children)

There’s no problem at all with not understanding something, and I’d go so far as to say it’s virtuous to seek understanding. I’m talking about a certain phenomenon that is overrepresented in STEM discussions, of untrained people (who’ve probably took some internet IQ test) thinking they can hash out the subject as a function of raw brainpower. The ceiling for “natural talent” alone is actually incredibly low in any technical subject.

There’s nothing wrong with meming on a subject you’re not familiar with, in fact it’s often really funny. It’s the armchair experts in the thread trying to “umm actually…” the memer when their “experience” is a YouTube video at best.

[–] MonkeMischief@lemmy.today 8 points 10 months ago

It’s the armchair experts in the thread trying to “umm actually…” the memer when their “experience” is a YouTube video at best.

And don't you worry, that YouTuber with sketchy credibility and high production values has got an exclusive course just for you! Ugh. Lol

[–] ook_the_librarian@lemmy.world 0 points 10 months ago

I don't know why you see it as an error. It's the format of the meme. The guy in the middle is right, the guy on the left is wrong. That's just how this meme works. But the punchline in this meme format is the the guy on the right agrees with the wrong guy in an unexpected way. I'm with the guy on the right and no appeals to Schröder–Bernstein theorem is going to change my mind.

[–] intensely_human@lemm.ee 9 points 10 months ago

Yeah I sell cabinets and sometimes people are like “How much would a 24 inch cabinet cost?”

It could cost anything!

Then there are customers like “It’s the same if I just order them online right?” and I say “I wouldn’t recommend it. There’s a lot of little details to figure out and our systems can be error probe anyway…” then a month later I’m dealing with an angry customer who ordered their stuff online and is now mad at me for stuff going wrong.

[–] Skates@feddit.nl 6 points 10 months ago (3 children)

The two sets have the same value, that is the value of both sets is unbounded. The set of 100s approaches that value 100 times quicker than the set of singles.

Hey. Sorry, I'm not at all a mathematician, so this is fascinating to me. Doesn't this mean that, once the two sets have reached their value, the set of 100 dolar bills will weigh 100 times less (since both bills weigh the same, and there are 100 times fewer of one set than the other)?

If so, how does it reconcile with the fact that there should be the same number bills in the sets, therefore the same weight?

[–] WilloftheWest@feddit.uk 20 points 10 months ago* (last edited 10 months ago) (1 children)

I like this comment. It reads like a mathematician making a fun troll based on comparing rates of convergence (well, divergence considering the sets are unbounded). If you’re not a mathematician, it’s actually a really insightful comment.

So the value of the two sets isn’t some inherent characteristic of the two sets. It is a function which we apply to the sets. Both sets are a collection of bills. To the set of singles we assign one value function: “let the value of this set be $1 times the number of bills in this set.” To the set of hundreds we assign a second value function: “let the value of this set be $100 times the number of bills in this set.”

Now, if we compare the value restricted to two finite subsets (set within a set) of the same size, the subset of hundreds is valued at 100 times the subset of singles.

Comparing the infinite set of bills with the infinite set of 100s, there is no such difference in values. Since the two sets have unbounded size (i.e. if we pick any number N no matter how large, the size of these sets is larger) then naturally, any positive value function applied to these sets yields an unbounded number, no mater how large the value function is on the hundreds “I decide by fiat that a hundred dollar bill is worth $1million” and how small the value function is on the singles “I decide by fiat that a single is worth one millionth of a cent.”

In overly simplified (and only slightly wrong) terms, it’s because the sizes of the sets are so incalculably large compared to any positive value function, that these numbers just get absorbed by the larger number without perceivably changing anything.

The weight question is actually really good. You’ve essentially stumbled upon a comparison tool which is comparing the rates of convergence. As I said previously, comparing the value of two finite subsets of bills of the same size, we see that the value of the subset of hundreds is 100 times that of the subset of singles. This is a repeatable phenomenon no matter what size of finite set we choose. By making a long list of set sizes and values “one single is worth $1, 2 singles are worth $2,…” we can define a series which we can actually use for comparison reasons. Note that the next term in the series of hundreds always increases at a rate of 100 times that of the series of singles. Using analysis techniques, we conclude that the set of hundreds is approaching its (unbounded) limit at 100 times the rate of the singles.

The reason we cannot make such comparisons for the unbounded sets is that they’re unbounded. What is the weight of an unbounded number of hundreds? What is the weight of an unbounded number of collections of 100x singles?

[–] Passerby6497@lemmy.world 4 points 10 months ago (2 children)

Is it possible for infinite numbers to be larger than others? Or are all infinite numbers equal?

[–] WilloftheWest@feddit.uk 5 points 10 months ago

Yes, there are infinities of larger magnitude. It’s not a simple intuitive comparison though. One might think “well there are twice as many whole numbers as even whole numbers, so the set of whole numbers is larger.” In fact they are the same size.

Two most commonly used in mathematics are countably infinite and uncountably infinite. A set is countably infinite if we can establish a one to one correspondence between the set of natural numbers (counting numbers) and that set. Examples are all whole numbers (divide by 2 if the natural number is even, add 1, divide by 2, and multiply by -1 if it’s odd) and rational numbers (this is more involved, basically you can get 2 copies of the natural numbers, associate each pair (a,b) to a rational number a/b then draw a snaking line through all the numbers to establish a correspondence with the natural numbers).

Uncountably infinite sets are just that, uncountable. It’s impossible to devise a logical and consistent way of saying “this is the first number in the set, this is the second,…) and somehow counting every single number in the set. The main example that someone would know is the real numbers, which contain all rational numbers and all irrational numbers including numbers such as e, π, Φ etc. which are not rational numbers but can either be described as solutions to rational algebraic equations (“what are the solutions to “x^2 - 2 = 0”) or as the limits of rational sequences.

Interestingly, the rational numbers are a dense subset within the real numbers. There’s some mathsy mumbo jumbo behind this statement, but a simplistic (and insufficient) argument is: pick 2 real numbers, then there exists a rational number between those two numbers. Still, despite the fact that the rationals are infinite, and dense within the reals, if it was possible to somehow place all the real numbers on a huge dartboard where every molecule of the dartboard is a number, then throwing a dart there is a 0% chance to hit a rational number and a 100% chance to hit an irrational number. This relies on more sophisticated maths techniques for measuring sets, but essentially the rationals are like a layer of inconsequential dust covering the real line.

[–] kogasa@programming.dev 3 points 10 months ago

There are different things which could be called "infinite numbers." The one discussed in the other reply is "cardinal numbers" or "cardinalities," which are "the sizes of sets." This is the one that's typically meant when it's claimed that "some infinities are bigger than others," because e.g. the set of natural numbers is smaller (in the sense of cardinality) than the set of real numbers.

Ordinal numbers are another. Whereas cardinals extend the notion of "how many" to the infinite scale, ordinals extend the notion of "sequence." Just like a natural number always has a successor, an ordinal does too. We bridge the gap to infinity by defining an ordinal as e.g. "the set of ordinals preceding it." So {} is the first one, called 0, and {{}} is the next one (1), and so on. The set of all finite ordinals (natural numbers) {{}, {{}}, ...} = {0, 1, 2, 3, ...} is an ordinal too, the first infinite one, called omega. And now clearly {omega} = omega + 1 is next.

Hyperreal numbers extend the real numbers rather than just the naturals, and their definition is a little more contrived. You can think of it as "the real numbers plus an infinite number omega," with reasonable definitions for addition and multiplication and such, so that e.g. 1/omega is an infinitesimal (greater than zero but smaller than any positive real number). In this context, omega + 1 or 2 * omega are greater than omega.

Surreal numbers are yet another, extending both the real and hyperreal numbers (so by default the answer is "yes" here too).

The extended real numbers are just "the real numbers plus two formal symbols, "infinity" and "negative infinity"." This lacks the rich algebraic structure of the hyperreals, but can be used to simplify expressions involving limits of real numbers. For example, in the extended reals, "infinity plus one is infinity" is a shorthand for the fact that "if a_n is a series approaching infinity as n -> infinity, then (a_n + 1) approaches infinity as n -> infinity." In this context, there are no "different kinds of infinity."

The list goes on, but generally, yes-- most things that are reasonably called "infinite numbers" have a concept of "larger infinities."

[–] elrik@lemmy.world 10 points 10 months ago* (last edited 10 months ago) (1 children)

once the two sets have reached their value

will weigh 100 times less

there should be the same number bills in the sets

The short answer is that none of these statements apply the way you think to infinite sets.

[–] intensely_human@lemm.ee 2 points 10 months ago

How would you even weigh an infinite number of dollar bills? You’d need an infinite number of assistants to load the dollar bills onto the scale for you, and months to actually do it!

[–] superbirra@lemmy.world 3 points 10 months ago

I see what you did here...

[–] volvoxvsmarla@lemm.ee 3 points 10 months ago (1 children)

So to paraphrase, the raging person in the middle is right? I'll take your answer no questions asked.

[–] WilloftheWest@feddit.uk 5 points 10 months ago

In short, yes.

[–] assa123@lemmy.world 2 points 10 months ago (1 children)

If we only consider the monetary value, both "briefs" have the same value. Otherwise if we incorporate utility theory with a concave bounded utility curve over the monetary value and factor in other terms such as ease of payments, or weight (of the drawn money) then the "worth" of the 100 dollar bills brief could be greater for some people. For me, the 1 dollar bills brief has more value since I'm considering a potential tax evasion prosecution. It would be very suspicious if I go around paying everything with 100 dollar bills, whereas there's a limit on my daily spending with the other brief (how many dollars I can count out of the brief and then handle to the other person).

[–] WilloftheWest@feddit.uk 2 points 10 months ago* (last edited 10 months ago)

I admit the only time I've encountered the word utility as an algebraist is when I had to TA Linear Optimisation & Game Theory; it was in the sections of notes for the M level course that wasn't examinable for the Bachelors students so I didn't bother reading it. My knowledge caps out at equilibria of mixed strategies. It's interesting to see that there's some rigorous way of codifying user preference. I'll have to read about it at some point.

[–] cows_are_underrated@feddit.de 1 points 10 months ago (1 children)

Correct me if I'm wrong, but isn't it that a simple statement(this is more worth than the other) can't be done, since it isn't stated how big the infinities are(as example if the 1$ infinity is 100 times bigger they are worth the same).

[–] WilloftheWest@feddit.uk 1 points 10 months ago* (last edited 10 months ago) (1 children)

Sorry if you've seen this already, as your comment has just come through. The two sets are the same size, this is clear. This is because they're both countably infinite. There isn't such a thing as different sizes of countably infinite sets. Logic that works for finite sets ("For any finite a and b, there are twice as many integers between a and b as there are even integers between a and b, thus the set of integers is twice the set of even integers") simply does not work for infinite sets ("The set of all integers has the same size as the set of all even integers").

So no, it isn't due to lack of knowledge, as we know logically that the two sets have the exact same size.

[–] cows_are_underrated@feddit.de 1 points 10 months ago

OK thanks for your explanation.