Transcendental numbers powered by Cantor's infinities 

regarding the .99999999... =1.000000 Look, just because you look like Lex Luthor doesn't mean you need to undermine the fabric of reality. Bald people can be nice too, just look at Captain Picard.

or Ghandi :)

what course do you teach?

At the moment I am teaching calculus, linear algebra and a unit called the Nature and Beauty of mathematics in which I do whatever I like (lots of fun stuff :)

Ghandi if that's how you pronounce it, but its actually Gandhi (give some load on the "dhi" part as if you were trying to speak "thee" without long ee and in low pitch)

I laughed out loud at that, too.

So ... can we have a proof where between each "major" step of the proof there are an infinite number of "minor" steps of the proof? (where, of course, the "minor" steps are themselves divided by infinitely many "very minor steps" and so on?) or, as we say is Frankfurt: huh??

Lmao I rewound and relistened to this part. Then I saw this comment.

Mission accomplished :)

me 1.999999999999999

He's married :)

@@davelowinger7056 it should be me 1.99999.... If you don't give the dots, its doesn't reach 2 😁

@@sadkritx6200 Well not as much as Joe siu

Me 1,9999999999999999999…

My thoughts, exactly. I actually tried to pressure my high school Maths teachers to tell me, *_WHY_* a/0 is such a taboo, and not just infinity; which they failed to comply. So, after a while, I figured: ”Screw it!”, and went to figure it out, myself; coming up with basically the same explanation Mathologer gives us: Since division is really just reversed multiplication, think about multiplying some number x with 0, and getting: x*0 = a ≠ 0. Well, that’s just impossible; so, we can’t have gotten to a situation, in the first place, where we could divide some non-0-number a with 0; and therefore, a/0 is nonsense. That was so hard for my ”teachers” to explain, yet a high schooler could figure that out 😑.

Nice isn't it :)

Well, if you are interested in the books that I've written have a look here www.qedcat.com/books.html and if you are interested in all sorts of other things that I've been doing pre-RU-vid check out this website: www.qedcat.com :)

OMG! This is gold. Thank you!

This is great! Sorry, I'm arriving late to the party...

Glad the videos work so well for you :)

:)

Friend, can you replicate your Pi Measurement to using a ruler with higher precision? When you did that, did your "rational pi" value change? If so, how can you explain that a _constant_ sometimes has one value, and sometimes another? Cheers

Someone here would not miss out a minute of pi day

@Slimzie Maygen *finite polynomial with algebraic coefficients. An infinite polynomial can be shown to converge to pi.

@Slimzie Maygen not a mathematical voice here, but if pi was wrong already at the third decimal digit, most of our modern buildings would have crumbled , and even some professional works of carpentry. Not to mention nanotechnology, molecular & cellular biology, and other microcosmic fields where the "traditional" pi is applied just fine up to many digits.

exactly :)

alkankondo89 or you could say that if the transidentals were coutable, then you could just enumerate by taking a algebraric number, then a trancedental, then a algebraric and so on. or what?

Mathologer how do you prove that real numbers are only algebraic or transcendental? (the basis of this proof)

I guess the transcedantal numbers are specificly defined as the set of all real numbers that are not algebraic?

why pull out a theorem from set theory? Just count the elements using some of the tricks described in the video!

WarpRulez Good insight! I would just add two clarifications. 1. The alphabet must be at most countably infinite, which is true here since you can take the alphabet to be finite, and 2. An algebraic number is not uniquely determined by the polynomial of which it is a root, as there are multiple roots. However, this is not a big problem. Just order the roots in some well-defined way, say by dictionary ordering on their coordinates, and then append the number of the root at the end of the polynomial.

he did not consider 2/2, he skipped it, he took 2/1

"some algebraic numbers can not be expressed in terms of +/-*sqrt" This is actually something very deep and the proof that such algebraic numbers exist is the solution to another very old problem. A bit of a holy grail for somebody like myself who is into coming up with good explanations of complicated material. Pretty high on my list of things to do :)

It's a first degree polygnomal. Just as we all are a first degree polyhuman (or polyAI for the bots out there, but those could actually be higher degrees too).

You didn't count the gnome who wears it ;)

Sci Twi he's not a gnome, he's an elf. stop being a fantacist

Well, not least because the truly uncountable set is made out of undefinable numbers... but mostly because even writing down or generating all the rationals between 0 and 1 would take an infinite time - countable or not.

Yup!

Hey I'll be a math friend. Math is cool, and I need a math friend

@@jetison333 S A M E

I've always been a little wary of describable/indescribable numbers. Something about the definition seems too lax. It makes me worry that the definitions may allow for some nasty self-reference. Do you know if this is not a problem? Edit: Sorry, that was a sloppy comment. Allow me to be clearer. Here's why I'm concerned about describable numbers: Consider the positive integers. Then there are positive integers which can be described in twenty or fewer words in English. We will make an argument that actually _all_ positive integers can be described in twenty or fewer words in English. Suppose the set of positive integers which cannot be described in twenty or fewer words in English is nonempty. Then since the set of positive integers is well-ordered, there is a smallest element in this set. Therefore, there is a unique "smallest positive integer which cannot be described in twenty or fewer words in English", which is a description in twenty or fewer words in English. This is a contradiction. Hence, every positive integer can be described in twenty or fewer words in English. But if there are only finitely many words in English, then there are only finitely many twenty-or-fewer-word descriptions in English. Therefore, it is impossible for all infinitely many positive integers to be describable in twenty or fewer words in English. What gives? The thing that gives is self-reference. Describability/indescribability is itself describable. The definition of describable is actually too loose in that it reasonably admits self-reference. But this example isn't exactly the same as what you're talking about in terms of describable numbers. So I wonder if perhaps the describable numbers avoid this sort of issue of self-reference.

MuffinsAPlenty There are subtleties to it, when you look at it formally you have to remember that the concept of something being definable needs to be formalized within the language of the model of set theory you are talking about. If you try to talk about something like “the set of definable sets” then you can run into problems where the concept of definable isn’t itself definable within the language of set theory, it’s a second order concept above it. In other words if you ask the question “is the set of sets that aren’t definable sets included in itself?” you encounter a paradox. There are formal treatments of this though. As I recall there is a decent Wikipedia page on Definable Sets that outlines the different ways to talk about definability of numbers. In fact there are some counterintuitive results that if I recall right prove that if you are talking about definability in terms of the ability to define a set theory model then you can show that there must exist models of set theory for any given real that can define that real.

Thank you very much for the information! I'll definitely look into that.

@@Bodyknock All this really rings a loud ”Russell’s Paradox” -bell, in my brain. 😅

I got it from here www.zazzle.com.au/polygnomial_t_shirt-235678195975837274 :)

Sadly it's false advertising, there's only one gnome so it's a mognomial :'( Clever shirt though

Adel D well x = 1 is also technically a polynomial so... but it would be even better if the gnome was above the x (x to the power of gnome). Yes i am petty

@@KosteonLink *Monognomial.

Skytern... he is sliding to 2/1

So would I 🤩.

This is a clever thought. Can someone please explain what is wrong with it?!

You correctly identify the issue with your argument. None of the real numbers with infinitely many decimal places are on the list. And no, just because the list is infinite does not imply that they are on it. Another way to think about countability is this way: A set is countable if you can order it in such a way that, starting from the first element, you can count up to any element you want in a finite amount of time. If you think about it in terms of lists, this is the same as saying that each individual element is in a finite position on the list. For example, you can do this with the integers: 0, 1, −1, 2, −2, etc. Every integer n is in a finite position. 0 is the 1st number on the list. If n is a positive integer, then it is the (2n)th number on the list. If n is a negative integer, it is the (2|n|+1)st number on the list. So we see that every integer appears in a finite position. The same thing can be done with the rational numbers. Look at the spiral given in this video. Every symbol in the grid will be reached in a finite number of steps. You can see this by taking any symbol of the form n/m where n and m are integers. Now, consider |n|+|m| = k. Notice that −k/0 is the last symbol you will ever reach where the sum of the absolute value of the numerator and denominator add up to k. Notice that −k/0 is inside a square of side length 2k+1. There are (2k+1)^2 numbers in this square, and every number n/m with |n|+|m| = k appears inside of it. Thus, for any rational number n/m, we know it must appear before the (2(|n|+|m|)+1)^2-th position on the list. Therefore, every rational number appears in a finite position on the list. As you correctly pointed out about your own listing attempt, none of the real numbers with infinitely many digits appear in a finite position on the list. So the listing technique you gave does not work.

And of course, even the computable numbers are countable, since we can list the programs that compute them. In fact even the *definable but noncomputable* numbers are, since we can list their definitions. So really, even though we assume uncountable sets of numbers to exist, it's literally impossible to give an example a specific number that isn't in a cleaner, countable subset.

+MrCheeze A nicely written wikipedia article on the topic: en.wikipedia.org/wiki/Definable_real_number

anon8109 You can go a step even further and consider the set of undefinable numbers. It's fairly clear that the set of possible definitions is countable. Imagine that you have them written out into a document, which is scanned into a computer. Then the resulting file will be a string of 1s and 0s, which can be trivially mapped to a unique real number between 0 and 1. This means the set of possible files is countable, which, if you make the reasonable assumption that there aren't any "magic documents" that are somehow both readable and yet cannot be represented by any form of digital photograph, means that the set of possible definitions is also countable. Of course, narrowing down which documents are valid definitions and which are gibberish is impossible - just think about one with really bad handwriting that no one can decide for certain if that digit is a 1 or a 7 - but we don't need to. If all the documents, including the gibberish ones, are countable, then any particular subset you claim to be the valid definition ones will be countable as well. Anyway, these undefinable numbers are interesting in that it's completely impossible to ever give an example of one - to be able to do so would imply you have some way of defining that very number. They are completely untouchable by any mathematical formulation. And yet, you can produce them at will. For instance, let's say you roll a die over and over, writing down each roll as a digit. The number you end up with as you continue rolling forever will be undefinable. The key is that there's no way some other person could independently produce the exact same number. Their dice will roll different numbers, and if the see the first 10 rolls you made, they can't determine what the 11th should be unless they see you roll it too. Now here's a real poser: Is it possible to have a number that is definable but not computable? That is, something that can be proven to exist and be unique, but with a value so convoluted that no computation can ever approximate its value. Actually, I can think of one. Consider the Halting Problem, which states that it's impossible to produce an algorithm that determines whether a given program will terminate. Now suppose we define an integer to be the number of programs of a given "size" that terminate. This number is unique for a given program "size", since every program will either terminate or it won't. However, figuring out such a number in general for sufficiently long programs would require solving the Halting Problem! Now, you might argue that although the halting program is unsolvable in general, specific programs can be proven to never terminate, so how do you know that for a given "size" you have any programs that can't be reasoned about? The simple solution to this is to produce a number by somehow combining all the termination counts. Maybe set the nth digit to the first digit of the termination count of programs of "size" n. Then to calculate this number, you must solve the Halting Problem in general for all programs.

+MrCheeze What do you mean by "cleaner"?

Steve: Well, every real number is - I assume - in _some_ countably infinite set. So by "cleaner" I just mean that every number we can talk about or access is _also_ in a countably infinite set with a fairly simple definition, e.g. the set of definable numbers.

That proof will be a LOT more accessible than the one for pi :)

I'll put it on my list of things to ponder :)

You're missing one detail: that the infinite sum 0.9+0.09+0.009+... converges to 1 (in the sense that the limit of the sequence of partial sums 0.9, 0.99, 0.999, ... is 1) is *precisely* what it means that the number 0.999... is equal to 1.

It seems that I forgot to add "s'il vous plaît" xD even If I don't think it will change the rate of replies...

There are no known counterexamples, and this is conjectured to be true, though I don't think there's a proof for all such decimal expansions.

Drew Duncan, Thank you for you answer, Do you have any links on research papers in this topic ? How do you know it is conjectured to be true ?

arxiv.org/abs/0908.4034

No. Some would be algebraic.

yes, a simple back and forth counting argument is all that's needed. This is what Mathologer expected to find in the comments, but I think you're the first to see it.

Yes, sadly common sense is not that common :)

why is it named common sense if it isn't common ?

ah fellow Lovecraft fan :)

I think only set theorists truly care about Chaitin constants and such things you're talking about, and very few understand what set theorists are producing today. No disrespect to Cantor and his contemporaries, yes, these ideas are crazy cool but concrete math is super hard already, but I also can build stuff with it. I can program with it. I can control hardware to work for me. Sadly, most practitioners of the newer stuff, like category theorists, tend to take similar risks when they don't learn from the mistakes of the mathematicians of the past.

The definition you propose makes sense, despite the halting problem. It could be that what you propose is equivalent to real numbers in a constructive axiomitization of set theory, although I'm not an expert in logic...

That's great. What I do to create the English subtitles is to simply take the autogenerated English subtitles that RU-vid provides and correct them. The same is possible for other languages. Have a look at this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-b9cKgwnFIAw.html So when you are on the page that allows you to add translated subtitles you simply first press the "autotranslate" button and then correct what the program generates. Anyway, it would really be great to have French subtitles. Oh, one more thing, recently RU-vid also added the option of adding customised titles and descriptions to videos. This means that if such a customised title is available in French for example, the youtube search engine will show you those titles by default if your accounts preferred language is set to French :)

Oh, well, when I tried to auto translate, sometime it was ok and sometimes it was completly messed up, I'll do my best then ^^

The Math Edumacator One of the important properties of a "measure" is what is called "countable additivity", i.e. if you have a countable collection of pairwise disjoint measurable sets (i.e. no overlap between any two sets), then you can find the measure of the union by adding up the measures of the sets. This does not apply to uncountable collections, however. Which is why you can't union all the singleton points on the line together and say the line has measure zero.

Ultimate wisdom or knowledge? Try breath meditation. 😉

Well, if nothing goes wrong three weeks from now :)

Don't jinx it please :) Love your stuff!

So many worthwhile things to do and at the same time so little time to do them all. Anyway one thing I am puzzling over at the moment is how to prove in a calculus-free way that pi is irrational. Even that is remarkably tricky. Eventually doable I think but very tricky :)

Have you looked at how Spivak did it in his calculus text?

No, the original Liouville constant is in decimals. However, interpreting this number as being written using other bases, e.g. 2 also yields transcendental numbers :)

Sure, it's just a lot more tedious than the 2d spiral. E.g. in 3d systematically inspect all integer triples (points with integer coordinates) via expanding cubes of integer diameter. Within every cube order the finitely many integer triples contained in it in lexicographic order, etc.

Gospel numbering? What is it? BTW - I didn't pick up the point where Burkard said that all algebraic numbers can be written out using basic operations and roots, and at the moment I cannot listen again to 20 minutes of video, but either you misunderstood or he misspoke. Algebraic numbers are those corresponding to the solutions of polynomial equations with rational coefficients. However, not all those solutions can be represented with roots and basic operations: for example, the only real root to x^5-x+1 cannot be represented with radicals, but it is still algebraic (by definition!). [edit: replaced x^x with x^5 in the above eq - sorry for the typo]

sorry I meant Godel numbering but autocorrect got the better of me. I'll watch to see what was said and report back. Also I don't think x^x can be in a polynomial. But maybe I'm not understanding something here. edit: on rewatch I misspoke. I got the implication backwards. He said "all such expressions are algebraic" not "all algebraic numbers can be written as such" But I'll rewrite my earlier way and still use Godel numberings. Take the Godel numbering of every polynomial and pair it with a number representing the Nth real root (ordering roots by

This must be one of the funniest (and apt!) autocorrect entries I have read in a while! Gospel's first incompleteness theorem: any Gospel contains propositions whose truth cannot be established from within the Gospel. Gospel's second incompleteness theorem: any sufficiently complex Gospel cannot prove its own consistency. Yep, I think the Gödel numbering argument works, but it's a hell of a lot more complicated (or at least lengthy) to build the Gödel numbering than using a spiral + diagonal argument for building the bijection. x^x is not algebraic (unless you restrict it to rational x), since non-rational powers are non-algebraic (by the Gelfond-Schneider theorem)

...aaand I had a typo too!!! I meant x^5-x+1, not x^x... Sorry! (Fixed in the post above as well, for the benefit of posterity).

dlevi67 no worries. We both make mistakes. Everyone does. As for the complication of Godel numberings. Yes they're complicated when constructed explicitly but frequently I make no use of the actual construction. Just that there is one. Also I'm a computer scientist so encoding things as natural numbers is kind of second nature for me to think about. This is why Godel numberings are my go-to for things like this. But yea I agree this would make an awful explicit construction.

Just two remarks: 1. The solutions to Ax^2+Bx+C=0 are in the first list if and only if B^2-4AC is a square. But more importantly 2. I really only insisted on avoiding duplication to make the exposition cleaner, but the fact of the matter is that you actually don't have to worry about this. The main thing is that the list contains all the numbers you want to list (rational or algebraic in the case of this video). It does not matter at all how many times they pop up for diagonalization to spit out a number outside the list. So what this means is that for the quadratic list you simply list ALL the real roots, the same for the cubic etc. :)

the problem is no matter how hard you try, you can never reach exactly pi due to the accuracy constraints this universe gives you Or as someone else put it (and I dont remember the exactl calculation or number, sorry) calculating pi after a certain amount of digits is most likely a very useless task "for the real world", since you can only use the whole universe and the smallest particles and arrange them so you can get a circle with x digits exact to pi itself After that x'th digit (assuming you dont find infinitely much space or infinitely small particles) its impossible to improve that "accuracy"

There are numbers and then there are various representations of numbers. Transcendence is a property of the numbers and does not have anything to do with the different representations of numbers for example in terms of different bases. Considering bases other than natural number ones is possible but I don't think it gives any more insight into any of the things that we are talking about in this video :)

called it :P

I put a pdf version here: www.qedcat.com/misc/trans.pdf :)

Mathologer thank you very much. I love pi and his transcendental brother

But the next one, 2/1 *is* 2.

Pi is not 22/7, pi is irrational (and transcendental). 22/7 is an approximation of pi correct to 2 digits after the decimal point.

+Abhishek Bhattacharjee I understand your reasoning (and therefore your confusion): since, between any two consecutive whole numbers, an (even) infinite amount of rationals exists, should Q not contain way more numbers than N? Looks reasonable, right? With infinite sets, all kinds of counter-intuitive things "can happen". Think of the following reasoning. Consider the following set P(n,n+1) = { x ϵ Q | n = 0. It seems "obvious" that P(0,1), P(1,2), ..., P(n,n+1) all contain "the exact same number" of elements, since they are just shifted copies (by a whole number) of one another. Together they form the set of non-negative rational numbers. If you represent rational numbers as lattice points (with whole numbers as coordinates) as done in the video, then P(0,1) is the set of all such lattice points above the line y=x. The set of non-negative rational numbers can be represented by the set of lattice points of the 1st quadrant. (I consider only lattice points with non-negative coordinates, since together they already represent the whole set of non-negative rationals; a bit different from what is shown in the video) Does it not seem plausible (viewed in this way) that P(0,1) contains half of the number of rationals as the set of all non-negative rationals? But we just "saw" that the set of all non-negative rationals can be viewed as the union of an infinite number of "copies" of P(0,1) ... Let us consider the following two sets of rational numbers: P1 = { x ϵ Q | 0 < x < 1 } and P2 = { x ϵ Q | x > 1 } "Clearly" set P2 is way larger than P1 (same reasoning about copies as above). But if we define the following function f: f : P1 → P2, f(x) = 1/x, then f is a bijection between P1 and P2, meaning that every element of P1 is related to a unique element of P2 and vice versa. This "clearly" demonstrates that P1 and P2 must contain "the same number of elements" ... Mathematicians have struggled with these "contradictions" too. They have settled for (and are happy with) the notion of "cardinality", that coincides with "the number of elements" in the case of finite sets. (see e.g. en.wikipedia.org/wiki/Cardinality for more details) Hope this helps a bit ...