Do we even need the real numbers??

We explore the possibility of a calculus only over the rational numbers. This leads us to the important role of the axiom of completeness in the important theorems from first semester calculus.
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[@a-manthegeneral]

I guess the real numbers were the friends we made along the way

[@nan9180]

i love you

[@welperooni7401]

real real numbers

[@ImaginaryMdA]

This is such a dumb comment, and I absolutely love it so much. XD

[@mohammedjawahri5726]

They always ask "do we need the real numbers" but never "how are the real numbers" :,(

[@emanuellandeholm5657]

The math version of "what are you doing step bro" vs. "how are you doing step bro".

[@alejandroromansanchez5368]

Lol

[@tomkerruish2982]

They're keeping it real. I'll show myself out now.

[@ketzunet]

The real numbers are too dense to notice, anyways.

[@muckchorris9745]

There is a nice approach via Cauchy-Sequences and a nice Isomorphism. Maybe you already know the state there is only one Field that is nice ordered and complete. From there you know what you mean when you say "Real numbers" you mean that that is unique (except isomorphisms).

[@barutjeh]

For the first example, you could use x^2 < 2 instead of x < sqrt(2); that way the nay-sayers have nothing to complain about. (You could include a 'or x < 0' if you want to keep your function exactly the same.

[@MrRyanroberson1]

all x such that x

[@ahmedouerfelli4709]

This comment is underrated.

[@Lodeous]

If he used x < 0 then it wouldn't be continuous because you could find negative rational numbers arbitrarily close to 0 and you wouldn't get a two sided limit

[@MrRyanroberson1]

@@Lodeous you're misinterpreting since the "or" part takes the union of negative rationals with the already given set of rationals between -sqrt(2) and +sqrt(2) from the inequality x^2 < 2

[@pierrebaillargeon9531]

Reformulating sqrt(2) differently does not make it anymore part of Q. Plus, defining a function using a number fro a domain outside of the domain of the function can always make you cheat. For example, if you use an infinitesimal, then you can define a function f(x) on R as: 1 if x < 2 + infinitesimal, 0 if x > 2 + infinitesimal. This gives you a continuous function on R -> R even though there is a discontinuity. So the argument has nothing to do with Q and everything with placing a step transition on a point not in the domain.

[@SlidellRobotics]

The original function can be defined without referencing anything irrational: 1 if x^2 < 2 or x

[@toast_recon]

What an excellent video. I've always been curious where the rationals aren't good enough. They always seemed plenty abundant to me, but wow, they're missing some key features we take for granted in the reals.

[@MyOneFiftiethOfADollar]

It may have been a real analysis book by Rudin that defined real numbers as Cauchy sequences of RATIONAL numbers. I think I actually understood that at some point earlier in life, but always wondered how by hand we could do even the standard arithmetic operations on a pair of “unreal” I mean real numbers!

[@elKuhnTucker]

You gotta love that chalk screaming at 2:50

[@oussamanhairech5178]

cool

[@Alians0108]

cool

[@Nithesh2002]

kzbin.info/www/bejne/fZnEqKqOmLami9U Walter Lewin has a video on how it's done :3

[@JansthcirlU]

Oh, I didn't realize I was watching an NJ Wildeberger video ;)

[@samisiddiqi5411]

Based.

[@atonaltensor]

When I first saw the title of this video, I thought you were going to use the fact that the rationals are totally disconnected and zero-dimensional. You should do a sequel on calculus on the irrationals. The topology of the irrationals is homeomorphic to the Baire Space (N^N), not to be confused with spaces which satisfy the baire category theorems. I’ve heard descriptive set theorists refer to N^N as “digital calculus”. It might mean it’s more amenable to calculus-like theorems.

[@arnaudparan1419]

I already knew all of that and I am glad that I have had great math teachers on such topics, but that video is great. I am glad there are people here doing youtube videos trying to give others mathematical intuitions and I hope it serves many :)

[@valeriobertoncello1809]

2:49 oh God that is soo satisfying

[@alnitaka]

Also I note that there is a 3-dimensional field over Q, by adjoining roots of x^3-3x+1=0, but it is well known that there is no 3-dimensional field over R.

[@DeanCalhoun]

cal-Q-lus!

[@GRBtutorials]

No, cal-ℚ-lus.

[@hybmnzz2658]

Underrated

[@sanjursan]

All that in under half an hour. Amazing. Super intro to Real Analysis, or maybe mini-course.

[@sowoul_]

how many integrals i gotta do to get dat big arms??

[@tomkerruish2982]

Moar

[@myxail0]

he climbs

[@soyokou.2810]

2389 integrals

[@benheideveld4617]

Zero is a natural number because there can be zero cookies left in the cookie jar.

[@clivegoodman16]

Whether 0 is considered a natural number is a matter of opinion. However, although the mathematics are somewhat simplified if we start at 0, historically we started at 1. There have been notations for numbers which include the positive integers but not zero.. I believe that 0 was introduced by the Indian mathematicians who were responsible for our so called "Arabic" number system. I believe these Indian mathematicians also introduced the idea of negative numbers.

[@clivegoodman16]

Note. By Indian I do NOT mean Native American.

[@eli0damon]

Your counterexample to the Mean Value Theorem is also a counterexample to the Extreme Value Theorem.

[@herrk.2339]

and vice versa

[@ghislainleonel7291]

This is interesting. Great content

[@verbumtech]

Not using epsilon-delta definition "since we're aproaching this at the level of Calculus 1". My teacher of Calculus 1 who made me to decorate this definition: "Pathetic"

[@senhueichen3062]

Very meaningful talk!!!!!!!!!

[@CamEron-nj5qy]

This would lead to a great debate with NJ WIldberger

[@Entropize1]

Sure. You can define Calculus over anything, without limits even, thanks to Algebraic Geometry. The results don't have to be pretty in all cases, just consistent.

[@biohoo22]

* Solves global warming * "Okay, great."

[@ruanramon1]

*That’s a good way to stop *

[@edusoto91]

Note that the first step function can be defined without use of real numbers (or sqrt 2). Take f(x) = 1 if x

[@atimholt]

I don't have the “blackboard” symbols for types of numbers down in my head. I half expected this to be about using quaternions instead, lol.

[@sea34101]

It was interesting, thanks. On a second thought, some of the issues you raised disappear if we use uniform continuity instead, ie a function which is uniformly continuous over ℝ is also uniformly continuous over ℚ. Also if f is uniformly continuous over ℚ and for every point in ℝ f has a right limit and a left limit then f is uniformly continuous over ℝ (I added the right and left limit hypothesis to exclude pathological functions like indicator of ℚ).

[@swenji9113]

Exactly ! If we restrict to compact intervals, which does not change anything for most of the things we do, then there is a 1-to-1 correspondance between continuous fonctions on R and uniformly continuous fonctions on Q. From that perspective, real numbers can be seen as a tool to describe a global property (uniform continuity of a fonction on Q) locally (by saying that the extension of this function to R, provided it exists, is continuous at every point). I love how it reminds me of non-standard analysis, where you add infinitesimals to the real numbers in order to control very small variations locally. In this field one of the most basic theorem is that a function is uniformly continuous if and only if it is continuous not only at every real but also at any hyperreal point.

[@Abhisruta]

Sir, kudos for such videos. One suggestion: Please increase the volume little bit. It was great in some previous videos. Thank You.

[@zoltanposfai3451]

02:00 If we are pedantic, we already have a problem here, as you are using the > relation defined over (R;R) to define a function Q->Q. If you use "> over Q", then you can't use sqrt(2) (or any other irrational number). If you use "> over R", then you are in a catch-22. The same issue arises if you start digging into the definition of lim over Q, which, for the simplicity of this specific video, you "just started using".

[@MuffinsAPlenty]

Why can't we use the real number system to prove things about the rational number system? It's like using the complex number system to prove things about the real number system. Someone could reasonably accept the real numbers as a totally valid number system and still ask the question of whether we could do calculus over the rationals. Nevertheless, a lot of people ask this question after being exposed to Norman Wildberger, so they will really be asking the question: if real numbers are _not allowed_ can you still do calculus with the rational numbers? So I understand where your fear of using reals within these arguments comes from. Even under this more stringent setting, both sorts of issues you bring up are completely avoidable in a rather simple way. For the function, f : Q → Q, we can defined it as f(x) = 1 if x < 0 or x^2 < 2 f(x) = 0 if x^2 > 2 One can easily define limits within the context of rational numbers by simply taking ε and δ to be rational numbers, rather than real numbers. You can even do this when defining limits in the set of real numbers too, because of how rational numbers are dense in the real numbers.

[@zoltanposfai3451]

@@MuffinsAPlenty I didn't say there is no way around it. x^2

[@MuffinsAPlenty]

​@@zoltanposfai3451 I agree that we have to be conscious of where things could go wrong. I also think that it's okay to ask the viewer to consider why those things are okay, without explicitly mentioning them in the video. Plus this can also lead to some good discussions in the comments :) But yes, if someone asks, we should be prepared to work with them on answering the question.

[@IanKjos]

Turn it around and say that a function is continuous precisely when the intermediate value theorem holds in the limit coming from both directions. E.G. for Y=X^2-2, you can get Y as close to zero as you like from either positive or negative direction with rational choice of X; you need not necessarily have a rational root to accomplish this. And then you can talk about Dedekind.

[@Djorgal]

Damn, drawing that doted line at 2:48, that's some skill with the chalk. I am a maths teacher myself, and I can't do that.

[@almightysapling]

You can teach yourself in like... 5 minutes. It's really not difficult you just gotta angle the chalk at the board the right way, more like pushing the chalk into the board than dragging it along the board

[@kimulvik4184]

Walter Lewin popularized it. There is a video out of him doing it several times. There are also some tutorials available

[@richardfarrer5616]

Interesting. So far as I can see, the Intermediate Value Theorem does hold over algebraic numbers however, so long as functions are ratios of polynomials. I'm not sure about the Mean Value theorem or Extreme Value Theorem but intuitively I think that they are also valid, given that you can't have a similar counterexample

[@conorbrennan5838]

Could you do a video on the set of constructible real numbers and then maybe calculus on the constructible reals?

[@conorbrennan5838]

@Ron Maimon I'm not sure? I know constructible reals are numbers we can describe with an algorithm, so all numbers we can describe as the limit of a sequence and we also know how far we need to go in the sequence to get within any error. So this includes pi and e and all these irrational real numbers we know of. But we can prove non constructible numbers exist , which is wierd because we can never really write them down in a sense? Yea the set of all constructible numbers is countable but you don't have completeness so you won't get all of the theorems we get using calculus with the reals? at least that's what I thought anyway

[@conorbrennan5838]

@Ron Maimon Ahhhh sorry yes computable real numbers is what I meant , been awhile since I thought of these things!

[@omipi4798]

Excelent question.

[@kumoyuki]

This is actually really important for Machine Learning applications. Really any Numerical Methods course should be teaching this because when we actually calculate (by hand or computer) we use rationals.

[@adamnevraumont4027]

If you want an approach that gives you usable calculus on computers, look at construtive analysis. I find it gives great intuition on what is possible, and you get most of calculus with slight changes.

[@benjaminbrady2385]

Funny that a piecewise discontinuous function of constants has a derivative of 0 everywhere with this which would make the antiderivative of 0 the set of all piecewise constant functions. This seems strongly reminiscent of a formal argument I read before that the antiderivative of a function (which it refered to as a mapping via a set) is the set of all possible functions which have the derivative of the integrand (namely, the antiderivatives plus any constant which is where the +C from integration comes from). I wonder if this hints at a more "fundamental" structure of mathematics beyond calculus where calculus-y things return these sets (which yet again is hinted at by the fact that Lebesque integration is this on piecewise sets anyway)

[@itays7774]

8:01 Yes! Finally! *0 is not a natural number!*

[@jesusalej1]

There are many proofs that confirm it

[@jesusalej1]

0!=1, a^0=1, etc... are deffinitions besides deffinitions in numbers, naturals and integers. 0 is a number that adds nothing but eats all.

[@itays7774]

@@jesusalej1 first of all, i think that the set of the natural numbers is a definition, so how could it be proven that 0 is a natural number? I'd be glad if you could provide a link to one such proof because im genuinely curios. In the way i see it, the natural nubers ate used to count things, but you can't count 0 things, it is meaningless in the real world That us my rational of why i don't consider 0 as a natural number

[@jesusalej1]

kzbin.info/www/bejne/m3yZnqarh9aNi5o... that is a bprp video that proves those things I told you before. In the naturals set are opperations such as sum, product. The integers are deffined by 0... given a, there is an opposite -a such that a + (-a) = 0... but 0 cannot be even a rational number, bc there is no number in Q wich results from division by 0.

[@schweinmachtbree1013]

@@itays7774 I have a different opinion: I would call the numbers which are used to count things "the counting numbers", which would be 1,2,3,... . On the other hand, I think of the natural numbers as the (whole) numbers which occur in nature, which are 0,1,2,3,... (e.g. when a wolf kills a herd of sheep, there are 5 sheep, then 4 sheep, ..., and then 0 sheep). Of course it is a matter of convention and personal taste whether 0 is a natural number - the most common definition of the natural numbers, the Peano axioms, defines the natural numbers as beginning with zero, but the Peano axioms can be easily modified so that 1 is the first natural number.

[@mxpxorsist]

The problems pointed out in the video already disappear if you consider the algebraic numbers and algebraic functions. So you'd have to cook up a natural example where a transcendental number pops out e.g. integral of 1/x is ln(x). Are there any nice examples that arent integration and arent trying to take an arbitrary limit that doesnt exist?

[@stephenbeck7222]

The first continuity example remains a problem for algebraic numbers if a transcendental number is chosen as the boundary (I suppose the issue then is how do you finitely define a particular trans number with reference only to algebraic numbers).

[@tomkerruish2982]

2^sqrt(2) is transcendental, though you may consider that to be taking a limit. If you confine yourself to rational functions of rational powers, then the algebraic numbers work fine. However, as you point out, you can't integrate.

[@danconcep]

I would say that those theorems come from the topology being connected for R, and not connected in Q.

[@thephysicistcuber175]

Non completeness of Q also causes problems.

[@chhabisarkar9057]

Hey mate , what's topology ?

[@hydraslair4723]

Most generally, the usual theorems of single variable calculus come from the connectedness but also the compactness of the underlying set. For the real numbers, both these properties are implied by the fact that they form a linear continuum: for all x,y, there is a z such that x

[@tomkerruish2982]

@@hydraslair4723 R is also noncompact, but it is locally compact and sigma-compact. (For those completely mystified by obfuscatory jargon, these properties lead to nice results on R such as "Every real-valued function continuous on a closed finite interval achieves its maximum and minimum values on that interval" and "Every bounded infinite sequence has a convergent subsequence.")

[@alexfan3816]

Q is even totally disconnected. Every subspace of it with more than one element is disconnected

[@sugarfrosted2005]

My favorite continuous functions are point discontinuous.

[@arturjorge5607]

in the second theorem you gave an example that contradicts what you said in the begining, that all functions are a polinomial

[@jabunapg1387]

This is great

[@llchan]

I wish my high school math lessons would be like this.

[@willnewman9783]

Just because those properties do not hold does not mean this is useless. Algebraic geometry can be done over any field, in particular the rational numbers, and your definition of rational derivatives of rational functions is equivalent to what is used there. But in algebraic geometry, the order of a field (if it has one) is not considered, so one does not talk about things like open intervals in that setting. What algebraic geometry can do is define things like tangent spaces, dimension, and differential forms all in the context of the rational numbers

[@MrAlRats]

In theoretical physics, there are clues emerging to indicate that spacetime is an emergent phenomena. One hypothesis that I find quite compelling is what's called the Causal Sets program (en.wikipedia.org/wiki/Causal_sets). Do you think that the Algebraic Geometry definition of things like tangent spaces, dimension and differential forms in the context of rational numbers or algebraic numbers has some relevance to this line of research?

[@iabervon]

I'm curious as to how hard it is to say things about whether limits converge when working in the rationals. You can obviously work in the reals and then note that the limit is rational, but I think a lot of tests for convergence aren't natively going to tell you whether your sequence is headed for a hole.

[@countiblis1867]

What does work is to replace the reals by a lattice with lattice spacing epsilon. One can do discrete calculus that is well established in the literature, which is not the same as real calculus, results of computations can differ from the real calculus equivalents by a number of order epsilon. So, one can then take the limit of epsilon to zero at the end of any computations and reproduce the conventional calculus result. So, the set of real numbers is analogous to the infinitesimals that were used in the early days of calculus. To get rid of those, on needed to introduce a limit procedure. Similarly one can get rid of the reals by introducing an even more elaborate limit procedure.

[@Pika250]

Then again, we did define the real numbers as Dedekind cuts on the rational numbers, which in turn were defined from ordered triples of natural numbers...which themselves came about purely through first order logic.

[@aggelosgkekas3113]

This is very interesting. I was thinking something similar some months ago and found out that the ε, δ definitions make sense in Q, but with some similar counterexamples of the classic real analysis theorems I really appreciated the definition of the reals.

[@jordanrozum]

Interesting stuff. Is it possible to "rescue" everything by replacing statements like "has" or "exists", etc. by "can be approximated by" and by using a slightly different definition for "continuous". Maybe one using Cauchy sequences? I haven't thought about this out carefully, but something like this might work: f:Q->Q is continuous on [a,b] iff |f(x_n)-f(y_n)|->0 for all pairs x_n and y_n of (rational) Cauchy sequences bounded between a and b with |x_n-y_n|->0. You could say f is continuous at a point c if there is a an interval I=[c-a,c+a] (a>0) such that f is continuous on every subinterval of I containing c. Admittedly, this is sort of cheating because you can construct R as equivalence classes of Cauchy sequences, but at least we never have to admit that sqrt(2) actually exists :) Because some functions can be discontinuous on [a,b] but continuous at every point in [a,b], some "point-wise" results might need to be tweaked a bit. I feel like this sort of rescue should be possible formally, because we do it informally all the time -- we can only ever express the value of an irrational number as an algorithm to approximate it (i.e., as a Cauchy sequence of rational numbers). Maybe people who specialize in numerical methods would have some additional insights into rational (and even finite) versions of calculus?

[@SocratesAlexander]

Do we have a subset of real #s which we can safely do calculus? (With algebraic #s etc.?)

[@mtaur4113]

Outside of the rational numbers, the irrational numbers are man's best friend. Inside of the rational numbers, it's too dark to read.

[@robshaw2639]

Let's not forget that not only do we need the reals, but combinatorial game theory (Winning Ways -- Berlekamp/Conway/Guy) shows that we need the infinitesmals as well ;-) Michael -- would you do a non-standard analysis vid?

[@s4623]

Question: How does it work differently if we are defining similar ideas over polynomial functions within the algebraic numbers? Because the algebraic numbers is equinumerous as the rationals and yet none of the problems listed in the video seems to be an issue as long as the terms are defined using algebraic numbers.

Change sqrt(2) to pi in the example?

[@MagicGonads]

I think for the "rational derivative" and the other "rational limit"s you need to do the limit over all sequences phi: N -> Q, that converge to the limiting point, not just h = 1/n like "prove the image over all convergent paths converge" kind of thing so lim f(x) x -> a = lim f(phi(n)) n -> inf for all phi: N -> Q where lim phi(n) = a n -> inf (and if there are any that don't equal that means the limit of f(x) as x approaches a in the rationals does not exist) you might even need phi: Q -> Q rather than N -> Q like how paths in multivariate calculus are R -> R^n, but I'm not sure It would be simpler maybe to just use the metric-convergence space open ball definition instead (e.g. epsilon-delta)

[@ernestomamedaliev4253]

Working in Q, I think we will alwayd have troubles with pointwise statements since it is not topologically closed. Afterall, we can understand that R is the closure of Q, making all the stuff work.

[@kobethebeefinmathworld953]

For the troubling example, I wonder if we can find the Lebesgue integral using measure theory

[@jakesecondname2462]

What about just removing the non-computable numbers, i.e. just working with computable numbers?

[@tim57243]

There is a "Computable Analysis" book by Weihrauch on my reading list that I didn't get to yet. Specker sequences are an interesting counterexample well described in Wikipedia.

[@MCLooyverse]

As a type-theory novice: nope! (answering the title question). What do we need uncomputable numbers for? Of course, we *do* need more than the rationals. I have no idea where to get the calculable(?) trancendentals (e, π, ...) from though, yet.

[@hjdbr1094]

This is awesome! I've wondered for a while now about rational and real numbers, and I have a question. You can define the natural numbers based on Peano's axioms, the integers assuming there's always an predecessor (and dropping the axiom that states there's no number whose successor is 0) and the rational numbers by assuming that for every integer k=/=0, there's a number k^-1 such that k*k^-1=1. Then, making this set closed under addition and multiplication, you have Q. However, how do you expand Q to become R the same way you expand N to Z, or Z to Q? I've only been able to see definitions of R that have nothing to do with Peano's axioms, and the only "expansion" of Q to R I've seen is by defining the irrational numbers as a convergent limit of a rational sequence, such as 1, 1.4, 1.41, 1.414, ..., and assuming its limit also belongs to R. I personally find this definition a bit vague, since you'd have to prove such a sequence converges and that it converges to the desired number (e.g., x such that x²=2) for EVERY IRRATIONAL NUMBER. Is there a better way to expand Q to R, or if this is the only way, is there an alternative so it becomes a bit more intuitive how all these properties always hold in the limit? (Also, one might argue that, in the same way Z is an extension of N based on addition and Q based on multiplication, R might be based on exponentiation. However, exponentiation leads to complex numbers, and that's not what we're looking for, is it?)

[@davidfenoll2332]

The rational sequences used to define the real numbers are Cauchy sequences, in which consecutive terms get arbitrarily close and that are bounded. If a Cauchy sequence doesn´t converge in Q, then you define an irrational number from it. In some sense a Cauchy sequence is the closest you can get to the idea of convergence in R without explicitly referencing the value of the limit, which only makes sense once your space is complete.

[@schweinmachtbree1013]

en.wikipedia.org/wiki/Construction_of_the_real_numbers

[@davidfenoll2332]

@Peter J. I prefer them too, since the concept extends naturally to arbitrary banach spaces.

[@mazona9985]

A detail, but still: At 14:20, the contradiction here is that \sqrt(2) would be a rational number (which it obviously isn't). It is, contrary to what was said in the video, not a contradiction that c is in the rational numbers (in fact, c is rational ab initio).

[@cerwe8861]

Press harder when you do the dottet lines, then it sounds greater.

[@tokajileo5928]

please can u explain in simple terms what is the so called Higgs branch and Coulomb branch ? thanks

[@cognology604]

Couldn't we say something like: We don't have the IVT, but for some arbitrary value predicted to exist by the IVT, we can find an arbitrarily close approximation to that value? This isn't the real IVT but could it be good enough to be useful anyway?

[@dusaprukiyathan1613]

I thought this was gonna be about non-real numbers. And my immediate thought was "unless you have something BETTER than i, you kinda do, since they're very closely tied together."

[@eiseks3410]

Since the function is from Q->Q you cannot define the limit, so it is not continuous

[@jadonjones4590]

What if you were to use the algebraic closure of Q and restriction it to rational functions

[@ramziabbyad8816]

I suspect that a lot of the discussions can be saved by restricting ourselves to cases where everything has a finite quality to it. That is we should have boundedness in the range. Or just restrict yourself to finite degree polynomials over rationale what you've got there is some kind of infinite degree polynomial.

[@donaastor]

also integral from 1 to 2 of 1/x doesn't exist in rational calculus. right?

[@user-A168]

Good

[@peterhall6656]

Welcome to 19th century analysis. If you did analysis in the UK around1900 there was a textbook by T J I'A Bromwich "An Introduction to the Theory of Infinite Series" which is full of problems based on the works of Dedekind, Pringsheim, Dirichlet, Cesaro, Gauss ,Jacobi, Mellin etc etc . Hardy oversighted the textbook which was described as being designed for "elementary analysis". It is useful to compare that standard to today's standards. As a general proposition all the problems were at least at Tripos level and many were much, much harder.

[@pietervandenakker419]

I would like to see the same on the field of describable numbers (real numbers that can be uniquely defined in a finite number of words). Still has gaps, but you can't make them explicit :)

[@schweinmachtbree1013]

by "word" do you mean in the everyday sense, or in the sense of some formal language (e.g. first-order logic)?

[@pietervandenakker419]

@@schweinmachtbree1013 I think it doesn't matter (as long as your language is strong enough).

[@IanKjos]

Descriptions are countable, so it has the cardinality of the integers. The corresponding "troubling examples" would follow from the incompleteness theorem. (Aside: real numbers are imaginary.)

[@Raynover]

Impressive stuff! How about a video on that last bit, about the Axiom Of Completeness? Also a correction on 16:10 the function is continuous everywhere. The points -sqrt(2) and sqrt(2) are not in the domain of the function, so there is no point asking if it is continuous there.

[@PubicGore]

That's not a correction. That is exactly what he said, the function is continuous on all of Q.

[@azhakabad4229]

Hello sir, Do you have any pdf of book to learn descrete mathematics?

[@schweinmachtbree1013]

a lot of books are available online for free as PDFs. I imagine there is a reference-request thread on math.stackexchange.com for discrete maths.

[@PASHKULI]

numbers are like elementary "particles" in physics - we can divide them and group them - but in fact there is always a unknown realm of elements (numbers) that we have not considered yet - as we have not encountered the problem in the "real" world when humanity discovers the inner layers of atoms and even beyond - the layers of the fine matter (spiritual energy), then we might have a better picture of what is going on with all the functions in the real domain as well as with some of the complex domain

[@Jim-be8sj]

At least it would simplify measure theory to work on a set like the rationals. :)

[@nikitakipriyanov7260]

9:02 no need to excuse, a and b appear symmetrically, so both a^k b^(m-k) or a^(m-k) b^k have to be same thing. This fact relates to the symmetry in the binomial coefficients. What if we use a set of all algebraic numbers with zero imaginary part instead of rational? (Note algebraic numbers defined as polynomial roots could be complex.) That will also seem to be full of transcendental holes. But, sqrt(2) and so on will be there, the trick which allowed you to disprove intermediate value theorem and extreme value theorem wouldn't work anymore. Actually it seems much of "well known math" is done in this set, and then results were expanded to the complete continous set of real numbers.

[@nikitakipriyanov7260]

@@angelmendez-rivera351 how can I choose transcendental number in a algebraic setting? Theorems in the video all failed because polynomyal with rational coefficients could have irrational roots. This even leads to the definition of an algebraic number: that is, any root of any rational poly-al. But, a poly-al with algebraic coefficients also could only have algebraic roots. To prove this you can put a transcendental into poly-al, and since none of poly-al operations are transcendental, nothing could remove transcendency (with the exception to mul by 0 or raise to the power of 0), your number would stay transcendental. In rational setting we had a restriction on powers, only to allowing integer powers in the poly. Now we can lift that restriction and have rational powers. Still, such "poly-al" would only have algebraic roots. While there will still be holes, those holes shouldn't be that wide and deep like in rational setting. Apart from this, which functions I can permit and which statements will be retained from "real" calculus?

[@eli0damon]

Do you know if it's possible to define completeness in general without reference to the real numbers (i.e. without metric spaces and the Cauchy condition)?

[@thedoublehelix5661]

Topologically right?

[@willnewman9783]

Completeness is defined most generally on what is called a "uniform space." A uniform space is a topological space with extra structure. Metroc spaces and topological groups have canonical uniform structures

[@asklar]

But... how can you take the limit of something without an epsilon/delta?

[@benjfactor]

Are there theorems in calculus that apply to complex numbers that don’t apply for real numbers?

[@almightysapling]

Oh god yes. This is why "Complex analysis" is a thing. In particular you get that once differentiable implies infinitely differentiable, so in a sense the calculus over complex numbers is much much "nicer" than over the real line

[@youssefwahba6120]

Hey michael, i was wondering if you could make a video/series on the theorems and tricks in math olympiads?

[@JonathonV]

Michael does some of them, but look up Prof Omar Math; he has tons of those!

[@Dystisis]

3:10 you're using dotted lines because they have holes everywhere, but aren't the axes rational (2:38)? so then there wouldn't be holes

[@pierreabbat6157]

No need for two cases in (x²-2)². x1=(x0+2/x0)/2.

[@maciejkubera1536]

Finding the antiderivative(s) of function 1/x would be an interresting question in Q.

[@HeraldoS2]

About a month ago, I did a video introducing the axiom of completness (really just prove that sqrt(2) is a real number) if you are interested on it I can dm you the link...

[@chjxb]

How about possibility of a calculus only over algebraic numbers?

[@JamesLewis2]

That isn't possible either, precisely because the algebraic numbers aren't complete, but there aren't simple counterexamples to the IVT, MVT, or EVT as with a "rational-number calculus".

[@benheideveld4617]

Actually any rational definition of continuity over the rational numbers should take the following definition: lim{h->0+} f(x-h)-f(x+h)=0 (h and x in Q) And then the step-function and the function 1/(x²-2) are both discontinuous at SQRT(2).

[@willnewman9783]

But sqrt(2) is still not in the rational numbers, so it cannot be discontinuous there. The definition you gave is implied by the definition in the video

[@erikev]

I expected this video to be calculus defined over the complex numbers minus the real line.

[@tomholroyd7519]

I want to like this twice.

[@EAdano77]

Maybe the rationals aren't enough for good calculus, but you can use the p-adics instead of the reals.

[@schrodingerbracat2927]

Define x(n) ~= a ("x(n) can be arbitrarily close to a") iff given rational positive episilon, there exists N such that |x(n) - a| < episilon for all n >_ N then IVT should work if you just require f(x_n) ~= 0 instead of f(x) = 0 exactly.

[@MuffinsAPlenty]

It seems to me that you're reconstructing the reals here.

[@turdferguson3400]

Brilliant! Now I wonder: if you restrict to algebraic numbers then define calculus on those algebraic numbers only. Then the intermediate value theorem would hold true. Maybe the mean value theorem and extreme value theorem would also work? AS A BONUS the algebraic numbers have the same cardinality as the rational numbers. I wonder if you could make a video about that?

[@willnewman9783]

Well, the (real) algebraic numbers work for polynomials/rational functions, but not for all continuous functions. And I do believe the extreme value theorem and mean values theorem work on them for rational functions: the point c (in either theorem) must exist in the reals, and it will be the root of some polynomial with real algebraic coefficients (because the derivative will have rational coefficients) and so c must be real algebraic.

[@frankwilhoit]

It has just occurred to me that, whereas there is a real between any two rationals, there is also a rational between any two reals. What does that do to any of the proofs that the reals have a greater cardinality than the rationals?

[@schmetterling4477]

Just guessing, but there are a lot more reals between any two rationals than there are rationals between any two reals? More precisely, the cardinality of the reals between any two rationals is the same as the cardinality of the reals, while the cardinality of the rationals between any two reals is the cardinality of the natural numbers.

[@IanKjos]

It makes no difference. Variations of the diagonal argument show that |Q| = |N| and also that |R| > |N|, but one intuition is that there are no holes among the reals: absolutely every quantity between two real numbers is a real; the same is not true of rationals (or the algebraics, or even the recursively-enumerable numbers).

[@The_Green_Man_OAP]

I think you should redefine the derivative as f'(x) such that it is the difference between a secant line slope ∆f/h and an error function Q(x)/h, i e. "tangent of curve at x"=f'(x)= (f(x')-f(x))/h -Q(x)/h, or f(x')=f(x)+h.f'(x)+Q(x) , and don't use limits to zero, infinity or anything, just h=p/q , i.e. rational, so that q(f(x')-f(x)-Q(x))=p.f'(x), for some natural numbers p,q. This is John Gabriel's method, aka "New Calculus".

[@The_Green_Man_OAP]

Here is one of John Gabriel's videos below. He has links to his New Calculus applets (which just use geometry, no limits) in the description. It's on his channel "New Calculus": "Were Newton and Leibniz wrong about discarding terms in h or setting them to 0?"

[@therealAQ]

I live for some adelic calculus

[@Protoex]

I have seen a particular problematic phenomena several times, but as far as I now, it has not been named or addressed properly. It appears on statistics as the problem of giving a reasonable & fair interpretation for a statistical number; And this problem is source for controversy all over the place. IMHO the problem arises from the gap between a pure mathematical object and our human cognition & intuition about it. I see the same problem here about the discussion of continuity over rationals and it's troubling results. Why epsilon-delta IS continuity? I would say all of this is a proof for the fragility of epsilon-delta argument and how It fails to fairly represent what we intuitively mean by continuity when we restrict ourselves on rational numbers. Or at least I would say this is a VALID interpretation. Should we find a new criterion for continuity that works for rational numbers as well? What is PRIMARY on math? Our intuitive understanding about things, or the default symbols we use to represent it?

[@jesusalej1]

We need them at all.

[@sophieward7225]

So, the short answer is "not really." Many of your arguments rely on the fact that sqrt(2) is irrational, so what if instead of asking if calculus works on the rationals, we ask if calculus works on the algebraic numbers?

[@tomkerruish2982]

As pointed out in another comment, you can't integrate. E.g., integrating 1/x leads to the natural logarithm and every positive algebraic number (other than 1) has a transcendental natural logarithm.

[@gnarlybonesful]

I don't understand the first example - if something satisfies the continuity definition why isn't it continuous? What is the failure other than "it doesn't look continuous when we graph it?"

[@schweinmachtbree1013]

he's saying that the definition of continuity doesn't work as we would like it to over the rational numbers (any "good" definition of continuity should make step functions discontinuous).

[@JB-ym4up]

Satisfied the definition when it was not supposed to, its like proof by contradiction. When you look at the graph you can see the discontinuity, but due to the lack of a real value at the discontinuity it is continuous along the reals.

[@gnarlybonesful]

@@schweinmachtbree1013 It feels arbitrary to me, if we didn't know real numbers existed why wouldn't it be continuous? The limit exists and it is equal to the function at that point.

[@gnarlybonesful]

@@schweinmachtbree1013 to put it another way, suppose we didn't know about the existence of the irrationals. Then that graph would have no "missing points" - what would be wrong with it bring continuous?

[@schweinmachtbree1013]

@@gnarlybonesful I think you mean "at all points" because by "that point" I assume you mean sqrt(2), which isn't in the domain. and even if we didn't know about the real numbers (in which case the example would still work, by changing "x < sqrt(2)" to "x < 0 or x^2 < 2", and "x > sqrt(2)" to "x > 0 and x^2 > 2"), we would still want the function to be continuous because step functions should be one of the prototypical examples of discontinuous functions; they "jump" (indeed I believe some people call them "jump functions"). so when a definition doesn't work as we would like it to, we had better change it.

[@alsatusmd1A13]

So we only need the real numbers if we care about finding derivatives or intermediate, mean or extreme values of arbitrary functions?

[@tomkerruish2982]

Or if you wish to integrate 1/x or many, many other functions.

[@schweinmachtbree1013]

@@tomkerruish2982 Indeed, integration is where we run into real issues. this video could easily become a series as there is a lot more which could be talked about.

[@almightysapling]

Well it's a bit silly. Whether anybody "needs" them or not is a lifelong philosophical debate, not something you can derive mathematically. But in this case, the question is really "can we make calculus work without them?" and yes, for calculus, those three particular tools are important.

[@MrAlRats]

@@almightysapling Why is it a lifelong philosophical debate? It's fairly easy to figure out what one needs at any given time or is likely to need in future. It just depends on the particular individual.

[@mtaur4113]

For the first example, you can find nested intervals of rationals for which f = +/-1, and the intervals are arbitrarily small. (e.g., the interval of positive rationals r such that 2-1/n < r^2 < 2 + 1/n). You could not do this in the interior of a real interval where a function is continuous. Overall idk how much this affects day-to-day math, but you definitely can't prove anything working with just rationals. And not proving stuff is sad....

[@gibson2623]

Isn't Q a subset of R....I would have assumed so, then, yes, we need R ;)