As an engineer who studied calculus and applied math, I still feel that real mathematicians live in another, fascinating universe.

That's how you solve a difficult problem in mathematics. By reducing it to two difficult problems. In case you wonder what to do with two difficult problems in mathematics, you obviously reduce them to four difficult problems.

@TykoBrian7

6 жыл бұрын

wiadroman REDUCE??

@Haannibal777

5 жыл бұрын

Are the number of difficult problems rational or irrational?

@TheDavidlloydjones

5 жыл бұрын

"Burn down the house and it reduces to the previous problem."

@Arycke

5 жыл бұрын

I mean we have done this before in the history of mathematics so you're sarcastic statement has truth of it in some sense ironically 😊😂 I see what you mean though, they posed this video in a strange way. However, solving other problems related to a problem/conjecture can oftentimes provide insight into the original. (See Birch-Swinnerton-Dyer conjecture and Fermat's Last Theorem and the numerous difficult problems posed from/in relation to, serendipitous or not, that that have actually gotten us closer to proving or disproving the BSD conjecture). So yeah. 😘

@beyondyourperspective8420

5 жыл бұрын

@@Haannibal777Or "Is the number of difficult problems countable or uncountable?"

Please use a monospace font next time you want to compare numbers like at 3:57

@mx2000

6 жыл бұрын

Yes, please.

@The_hungry_vegans

5 жыл бұрын

What?

@error.418

5 жыл бұрын

@@The_hungry_vegans Google: monospace font

@StefanReich

4 жыл бұрын

You have to forgive them... they're not so good with numbers

KZread channels like this and PBS ScaceTime makes life better. There need to be a PBS Chemistry series though, for organic chemistry.

This is my first try so please be gentle: 1. We assume there is a list which contains all numbers produced by exchanging digits of e and pi. 2. We write a non terminating binary sequence corresponding to each element of the list by following rule: If we dont exchange a digit we write zero, if we do we write 1. e.g. 0.111000... means we exchanged first three digits. 3. Now produce a new binary sequence which differs the digits of these binary sequence atleast at one position.(The way we show irrationals are uncountable) 4. This new sequence gives the encoding of a exchange scheme which was not in our original list.

Make a binary digit to represent which decimal place you change. So 0.11001101... will change the digits 1,2,5,6,8,... after the decimal place, if you make a computable list(for instance 0.1, 0.01, 0.001,...), you will eventually reach all* solutions, and then you can show via Cantor Diagonalization that they are in fact uncountable.

@Noah-fn5jq

7 жыл бұрын

dang... this was my back up proof. Well played sir.

@KaiKunstmann

7 жыл бұрын

I don't get it. How is "counting from 1 up" the same as "uncountable". I really don't get it.

@Noah-fn5jq

7 жыл бұрын

Kai Kunstmann The diagonolization proof is a proof by contradiction. It shows that no matter how you make the list, at least one element will always be left off because it's not possible.

@trulyUnAssuming

7 жыл бұрын

You have countable digits but the set of those numbers is uncountable. Proof: Given that every real number can be written in arbitrary base notation and you allow every binary string from 0 to 1 your set includes all the real numbers from 0 to 1 and that set is uncountable. EDIT: And the proof that you can write every rational number in an arbitrary base b with b>1. Be a in R, then floor(a):=max{n in N: n=a} It is kind of obvious that ceil(a)-floor(a)

@KaiKunstmann

7 жыл бұрын

The diagonolization proof only proves that the infinite list of real numbers is not finitely complete, because you can always find another number, but that is what you started with in the first place: an infinite list. In a sense, the proof shows, that your assumption of an infinite list being "complete" is wrong.

I am so excited that this channel is pulling from the entire internet community and look into the discussions on their videos. Well done!

e*pi*0 = 0. Do I win?

@jackcau1845

5 жыл бұрын

I think not, because you used 0

@douwehuysmans5959

5 жыл бұрын

You proved that e*pi*0 is rational

@filyb

5 жыл бұрын

*WHOA* now you can cancel out 0 from both sides and you are left with: e*pi = 1

@jackcau1845

5 жыл бұрын

@@filyb nope because to do that you have to divide by zero from both sides. And you cant do that

@filyb

5 жыл бұрын

@@jackcau1845 woa really i totaly didnt know!

Here's a proof for why there are an uncountable number of ways to swap the digits around: For each digit that is different between e and pi, we have only two choices: swap or don't swap. Therefore, we can label a given permutation by listing whether or not we swap each digit, skipping any matching digits. If we use a 0 to represent doing nothing and a 1 to represent swapping the digits, then each permutation is labeled by a unique infinite list of 0's and 1's. Now, if we take this to be the decimal part of a real number, we can see that we have a mapping of possible permutations to all real numbers between 0 and 1 (written in binary). Since we already know that the real numbers in an interval are uncountable the permutations must also be uncountable. (In fact, some of the possible permutations are being mapped to the same real number due to ambiguity with infinitly repeating 1's, eg. 0.0111... = 0.1. However, the important point is that every real number is mapped to at least one permutation.) Alternatively, you could apply Cantor's diagonal proof directly to the original list of 0's and 1's. If you try to create an infinite list containing all possible permutations, you can always find a number not on the list by making sure that the 1st digit does not match the 1st number, the 2nd digit does not match the 2nd number, and so on. This directly proves that the permutations cannot be listed completely, which means they are uncountable by definition.

@danielhall271

7 жыл бұрын

+

@KaiKunstmann

7 жыл бұрын

Dude, you can order and list all the strings of 0's and 1's, which is the same as counting. For example: 0.1, 0.01, 0.11, 0.001, 0.101, 0.011, 0.111, 0.0001, 0.1001, 0.0101, 0.1101, 0.0011, 0.1011, 0.0111, 0.1111, 0.00001, …

@KaiKunstmann

7 жыл бұрын

By the way, the same goes for decimals: 0.1, 0.2, 0.3, …, 0.9, 0.01, 0.11, 0.21, 0.31, …, 0.91, 0.02, 0.12, 0.22, 0.32, …, …, 0.99, 0.001, … and so forth. The 479th number will be 0.974 and the 4483218799699478700th number will be 0.0078749969978123844.

@joshyman221

7 жыл бұрын

Kai Kunstmann this argument does not work because you will never list all the irrational numbers.. I.e numbers with an infinite non repeating decimal string (very informal) e.g where does e-2 and pi-3 come up? A way to prove these are uncountable is to assume they can all be listed, then by flipping the ith digit from 0 to 1 or 1 to 0 in the ith number on our countable list we create a new number still in the set but not on our list, contradicting assumption of countability.

@KaiKunstmann

7 жыл бұрын

By the same argument, you will never list all the natural numbers, when trying to list the natural numbers. In particular you will never list integers with an infinite expansion, like "…333333" or "…62951413". So, either the natural numbers are as "uncountable" as the reals, or infinitely expanding real numbers are "countable" like their integer counter part. Either way, they must both be of the same magnitude.

Use binal base. You can exchange ones and zeros of both. Goodbye

@TheDavidlloydjones

5 жыл бұрын

Hell, Rafeio, express them in base pi and base e. Then they're both 1

@JuniperHatesTwitterlikeHandles

5 жыл бұрын

Assuming you mean binary, the ones and zeroes won't line up perfectly, and there will be many points where both digits are 1 or both digits are 0, so you won;t be able to turn them into 1.0000 repeating and 1.111111 repeating if that's what you mean.

@Leidl.Michael

3 жыл бұрын

I paused the video at 0:50 and came up with the same answer as rafcio pranks. As a math student i have to mention "exchanging digits" is not specified enough in this video.

@tolkienfan1972

3 жыл бұрын

@@Leidl.Michael that just means there are (at least) two problems here. But one of them is obviously uninteresting. The other is fascinating

first episode that i truly fully understand. great job to PBS.

Oh, man. Just found this channel and I'm loving it!! Looking forward to seeing more math!! Can you guys touch on a bit of stochastic calculus?? Or maybe at least a bit of topology, say, compact spaces??

Find out next time on DbZ

@pronounjow

7 жыл бұрын

Matt would like this. lol

@soyokou.2810

5 жыл бұрын

Db zubstatution

wait, THE Terry Tao?

@guest_informant

7 жыл бұрын

No. A different one. Someone added two mathematicians and it came to Terry Tao. Strange coincidence.

@georgedosidis5734

7 жыл бұрын

Yeah, look at the original Math Overflow post, it's in the description. He's very active on the site.

@nsnick199

7 жыл бұрын

Are you sure you're not thinking of it backwards and Terry Tao didn't Banach-Tarski himself into 2 copies?

@nobodypi320

7 жыл бұрын

This is so funny because I know his son IRL. Before I knew how awesome Terry is.

@xXxBladeStormxXx

7 жыл бұрын

+NobodyPI Oh, that's cool! How old is his son?

This is a seriously awesome show. Thanks for the knowledge!

Very nicely explained as always, Kelsey! BTW, any chance you might make a video about the Futurama Theorem?

Did someone forget a coffee cup on your shoulder?

@DragonGod1012

7 жыл бұрын

LOOOL

@forcumj

7 жыл бұрын

It's a tattoo :) it's been discussed.

@atomicvomit8780

7 жыл бұрын

ifo Why the hell would anyone get a tatoo of a coffee cup stain? KABAAAAM

@Zahlenteufel1

7 жыл бұрын

definitely blood magic!

@forcumj

7 жыл бұрын

she talks about it in a QnA i don't have the link for or proof of lol

I'm no mathematician, but these videos are interesting. Kelsey, please return!!

@aniksamiurrahman6365

4 жыл бұрын

She's a college professor now. www.olin.edu/faculty/profile/kelsey-houston-edwards

You can encode a particular switching of the digits as a list of natural numbers indicating which positions the switched digits are in. But there are uncountably many subsets of the natural numbers.

cool episode, as always! and also... Schrodinger's Equation as wall paper!

14.6 actually​ has two repeating decimal representations: 14.6000~ and 14.5999~

@flyerfan8

4 жыл бұрын

@themistlink 14.59999... does not converge to 14.6 they are equivalent. 999999.... equals 1.0 therefore 14.6*1.0 = 14.6*.9999..... therefore 14.6 = 14.599999....

@petersullivan5702

4 жыл бұрын

TheMistLink it’s equivalent because there doesn’t exist a number in between

@stydras3380

3 жыл бұрын

@TheMistLink You are half right. The number 14.5999... is defined to be the limit of the sequence 14.59, 14.599, 14.5999, ... and this limit is exactly 14.6. So by definition 14.5999...=14.6.

@charlieangkor8649

3 жыл бұрын

Someone gives you two real numbers and you need infinite amount of time to compare if they are even identical.

Just started watching but I wonder if this idea is going to come up. *At most one of e+pi and e*pi is rational* Clearly, p(x) = (x-e)*(x-pi) is a polynomial whose roots are e and pi, so its coefficients cannot all be rational, by the definition of transcendental numbers. Expanding that expression, we get (x-e)*(x-pi) = x^2 - (e+pi)*x + e*pi This means that 1, -(e+pi), e*pi cannot all be rational. If all the coefficients were rational, we would have found a polynomial with rational coefficients that had e and pi as roots. In other words, at most one of e+pi and e*pi is rational. (We know that they cannot both be rational, so that's the most we can say). (From Ask Dr Math in 2001(!) mathforum.org/library/drmath/view/51617.html)

@Czeckie

7 жыл бұрын

well it didn't, but it is cool nevertheless

@ComputerNerd98234616

7 жыл бұрын

I don't see why it is not possible for both (e+pi) and (e*pi) to both be irrational. yes there must me at least 1 irrational coefficient but there is no reason to think that there is exactly 1 coefficient. plus you can probably directly prove that both are irrational.

@__-nt2wh

7 жыл бұрын

Daniel Flores He said *at most* one is rational, aka, didn't assume one is rational, just that both can't be.

@ComputerNerd98234616

7 жыл бұрын

flashdrive ahh OK I hadn't read that right. although it might be a neat challenge to show that both are irrational

@alxjones

7 жыл бұрын

It's more than a neat challenge, it's an open question in mathematics.

great explanation!

This is really cool. I'm very interested in how to reserch/investigate math. Sitting in a classroom, learning about mechanics is much less cool than thinking abstractly like this.

@TheManxLoiner

7 жыл бұрын

Thinking abstractly about physics is just as cool as thinking about abstract mathematics (and I'm a pure mathematician, too!). Check out Feynman's Lectures on Physics, which is a series of physics textbooks now available online for free. In it, Feynman proves, for example, how to derive an expression for gravitational potential energy from the principle that you cannot energy.

My proof: If pi and e differ by infinitely many digits then you can label every digit that's different among them by d_1,d_2,... starting at the first digit that they differ by and going further. You can then assign 0 or 1 to every number d_n such that you if you switch the digit d_n then you assign 1 to it and if you don't then assign 0. You then have countable many choices for 0 or 1 and every string of 0s and 1s corresponds to exactly one number that's on the list and it is known that {0,1}^ℵ0=continuum q.e.d.

@Macieks300

7 жыл бұрын

correction: I should've written |{0,1}^ℕ|=continuum or 2^ℵ_0=continuum

@Noah-fn5jq

7 жыл бұрын

You beat me too it... I guess now I'm going to have to write a modified diagonilization proof :P

@luketurner314

7 жыл бұрын

My thoughts exactly. Even trying to create a number not on the list fails because for digit n there are 2 choices (0 or 1), half of the list match and half does not; for digit n + 1, each half of the list that matches (and does not) for digit n matches for digit n + 1 and half does not match for digit n + 1. This pattern never ends. Another way to express this is for any two digit swappings there are 4 ways to swap, represented in binary as {0,0}, {1,0}, {0,1}, and {1,1}. All 4 of these will already be on the list, and therefore not be able to make a new number not on the list. Q. E. D.

@Noah-fn5jq

7 жыл бұрын

Luke Turner I don't think the "4 expressions" are necessary. If you only need one one rational number, then only one digit needs to be considered.

@ffffhhgg4104

4 жыл бұрын

You beat me to it.

If I lived in a room like yours I also would become crazy. Buy a chair, tv-set and a plant.

@error.418

5 жыл бұрын

A TV set does not make your life better.

Unless I'm misunderstanding, in a normal number the sequence 888 should occur more often than 123. This is because you can have a sequence like 8888 where you get 888 twice in four digits.

@nielsen425

3 жыл бұрын

I like your insight here.

Let A be the set of numbers obtainable by swapping digits between pi and e, and assume that the digits of pi and e differ infinitely often. We construct a function from P(N) to A as follows: Let B be an element in P(N). For each n in B, swap the digits at the nth differing decimal place. By construction, this is an injection, and A has cardinality at least 2^aleph0. In fact, the cardinality is exactly 2^aleph0, since A is a subset of R.

Also interesting is the question if e and pi added, subtracted, multiplied, etc. can lead to a rational number; I think that even e+pi and e*pi are not proven to be irrational

@finnaginfrost6297

7 жыл бұрын

I presume so (at least for addition, I haven't proved it for multiplication), because if they were rational then e+pi could be written as (a/b)+(c/d), or ((ad + bc)/(bd), and anything in the form ((ad + bc)/(bd) can't contain enough information in its repeating decimal to create infinitely many pairs of place values for pi and e- eventually, as all permutations of ways to split ((ad + bc)/(bd)'s repeating place values are used up ('

@danclaydon6588

7 жыл бұрын

I think you mean irrational at the end, since π+e is clearly real. But you are right, this is still an open problem.

@raphaelmillion

7 жыл бұрын

dan claydon Thank you, I didn't realize

@finnaginfrost6297

7 жыл бұрын

Thank you from me as well, I'd made the same mistake

@danclaydon6588

7 жыл бұрын

Finnagin Frost It doesn't follow that if a+b is rational then a and b are rational, as you seem to imply. For example, (1-π) and π are both irrational, but add to 1.

Terry Tao says that if pi and e are normal numbers, then you can't make either rational by switching digits (in base 10). Perhaps an amended question would be: is this still the case for any base b, still assuming that pi and e are normal numbers?

@guillaumelagueyte1019

7 жыл бұрын

I think we don't know that yet, as it is not yet known whether or not the Champernowne constant is normal in any base that's not 10

@MrCheeze

7 жыл бұрын

jledragon I'm fairly sure it would be true for base >= 3, but have absolutely no idea about base 2.

@nickharland6473

5 жыл бұрын

If e and pi are actually normal numbers (which is conjectured), then it wouldn't matter which base

I LOVE PBS INFINITE SERIES!!!!!!!!!!! Oops got a little too excited there >

The t-shirt is funny. I cracked up, and then couldn't really even think how to explain why it's so funny to anyone else. Some of the thoughts that went through my mind were "that's funny because the math would all totally fall apart and because only mathematicians would understand that and for some reason I don't really understand, things are funny (and kewl!) if they're only understood by an inside group" and then "but epsilon always has to be greater than zero, so it must be tired of being so restricted. free epsilon, yeah! *let* epsilon be less than zero for a change! go epsilon!" :) Really, I just know that I WANT THAT SHIRT!!!

Do exchanged digits have to be in the same position (i.e. you must exchange the nth digit in pi with the nth in e)? If not, then the solution is trivial... replace all of the 0s in pi with any non-zero in e. the result is rational.

@Bodyknock

7 жыл бұрын

In fact the numbers don't even need to be normal if you can exchange digits from anywhere in the expansion to anywhere else in the other expansion. Since pi, for instance, is irrational there must exist a digit than appears an infinite number of times. Let's say for the sake of argument that the digit 7 appears an infinite number of times (maybe other digits do, maybe they don't, it only matters that 7 does). Then swap the first instance of 7 in pi with the first digit of e, the second instance of 7 in pi with the second digit of e, and so on, resulting in one number that is all 7s and hence rational. The only difference with normal numbers is that you can know in advance that any digit you pick will appear an infinite number of times. But even if the irrational number isn't normal there is still guaranteed to be at least one number that will work above if you can swap digits to any place you want.

I made a new number, Pi + Li = 10 3.141592 + 6.858408 = 10 So that's how to make pi = 10

This video deserves more views.

Cantor's diagonal argument is used to prove that the infinite swaps of digits of pi and e are *c* in cardinality. Suppose to the contrary that they were countably infinite: let A and B be the numbers obtained after swapping digits away from pi and e respectively, and consider the list of As. Change the digit representing 10^(-n) of An to its B-counterpart. Do this for each n. This new number is still an A, but it wasn't in the list

What if we do not insist that the digits be swapped from the same digit place? In this case, we can arrange that both numbers become rational after the swapping. See plus.google.com/u/0/+JoelDavidHamkins1/posts/ND1qdRmw63N

@pbsinfiniteseries

7 жыл бұрын

That's a neat follow up question! Thanks Joel!

The previous T-shirt was hilarious. All who disagree are robots with no soul. Clearly.

that line about a normal number having any digit or sequence of digits being equally likely gave me flash backs of my probability and statistics course and a formula that had something to do (or was compared to) a hypercube or something.

i absolutely love this channel!!!!!!!!!!!

Why wouldn't you use a monospace font for this?!

Said i to pi, "Be rational!" Replied pi to i, "Get real!"

This (great) video got me thinking whether this problem is basis dependent or not. It sounds to me like a natural and possibly trivial question, but I couldn't figure it out thus far. Also, how about algebraic numbers? Is there a digit exchange between any algebraic numbers that would yield a rational number? Or is there a digit exchange between transcendental that would yield a rational number (is this problem equivalent to the transcendental->rational?)? Anyway, keep up the excellent work. PBS Infinite Series definitely diverges from most KZread "content".

another great video

pi-pi+e-e

It's uncountable because it essentially boils down to a power set of the natural numbers right? I'm not crazy?

@TheManxLoiner

7 жыл бұрын

Slightly crazy - you should say `*the* power set of the natural numbers' instead of `a power set of the natural numbers'.

Is that a Full Metal Alchemist Ouroboros tattoo on your (left?) shoulder?? Whether it is, or it isn’t, sooooo cool!!

Without actually solving the _modified_ question as now-pertains to digits, one can give a method that might lead to a solution: TO WIT: write e,π in radix-f(n) such that the nth-digit is what you want to make the swapping process produce a rational-'pthpthpthpthpthpth' (that's a 'razzberry' from Bedazzled or its remake)...so yes, tentatively, 'we' can....

((pi/e) x (e/pi))/8=0.125 lol!!!! XD

@ComputerNerd98234616

7 жыл бұрын

ossiebird0 I don't think you understand the question being posed here. yes the product of 2 irrational numbers can be rational. e*(e^-1)=1

@ossiebird0

7 жыл бұрын

Daniel Flores, the subtly was lost on you, I was being flippant hence, lol!!!! and XD emoticon. However I did answer the title question.

@FlamingAnimation

7 жыл бұрын

Wow!!! (pi*e)/(pi*e) is 1, who would've guessed?

@oussamawahbi4976

7 жыл бұрын

ossiebird0 this is not the point of the video

@ossiebird0

7 жыл бұрын

Oh really? :‑o

Wait...What about this... We know that pi and e are irrational. We also know that every number appears at least once in the decimal expansion of pi - and - as such, we can also conclude that the digit "0" appears infinitely many times in that expansion ( as there are infinitely many numbers of form 10^n). As such, we could list the digits that appear in the decimal expansion of pi as being - for instance- the first "0" in the expansion, the second '"0" and so on. As we are able to list them, they're countably infinite, just as the digits of pi or e, which can also be listed as first digit of e, second digit of e and so on. Since both "spaces" have the same dimension, we could create a bijection between the "0" digits that appear in the decimal expansion of pi and the digits in e - that is, shift the first digit of e with the first zero of pi, the second digit of e with the second zero in the decimal expansion of pi to infinity. This way we could create, in place of e, any non-repeating rational number smaller than ten ( assuming we cannot shift the position of the decimal point in e or pi), just by simply building the initial significant digits of the desired number by some makeshift rule, then applying the "0"-shifting described above to all the remaining digits of e. Does this make any sense? If so, then the answer to the problem is "Yes, we can".

@guillaumelagueyte1019

7 жыл бұрын

We don't know if there are infinitely many 0s. If pi and e were normal numbers, there would be infinitely many, but it's not proved.

@joaomarcoscorreiamarques4899

7 жыл бұрын

I see! Thank you for your answer! Yet, since pi is irrational, we must have at least 2 numbers that appear infinitely often in its expansion ( otherwise at some point that number would start repeating itself over and over and pi would be rational). This way, the logic can be applied to any of the (at least) 2 infinitely repeating numbers to generate a rational number, no ?

@guillaumelagueyte1019

7 жыл бұрын

Good point, I didn't push the reasonning further but you're right that we could do that I guess

Since for each decimal place in pi and e you can choose either to swap or not to swap the digits, the number of possible options is equivalent to 2^aleph-0, or the power set of countable infinity. Cantor's diagonal argument shows that the power set of countable infinity is uncountable, so the number of possible swap options is uncountable.

I still don't like questions in mathematics that depend on how we represent the objects, rather than on itrinsic properties of them, independent of the representation we choose. This one in particular has some flavour of that, although it actually manages to avoid it.

Do not combone Pi and e because it will explode destroying everything

@sulflix

5 жыл бұрын

pi+e=pie

@anticorncob6

5 жыл бұрын

Ulsk!

** She meant *tau* and e

@pco246

7 жыл бұрын

Hmm... However the decimal expansion she wrote is the one for τ/2. Interesting

@FractalsByX

7 жыл бұрын

I see what you did there...

@pco246

7 жыл бұрын

Tau is not really about convenience in formulas (actually it is a little bit, but not totally). It just makes more sense to use 2pi as the circle constant because the defining characteristic of a circle is its radius, not its diametre ;) Plus, radians are just messed up if you use pi, whereas if you use tau they make a lot more sense.

@PlayTheMind

7 жыл бұрын

PCreeper394 : You're officially my best friend in this comment section.

@pco246

7 жыл бұрын

Yay! :D

For roots of rationals I like the simplest proof: list the prime factors of the numerator and denominator, strike out duplicates, you've got a fraction in lowest terms. Squaring simply duplicates the lists. So if you raise a rational to any power and get denominator one, i.e. an integer, you started with one.

very nice.Please make video about curve shorting flow or ricci flow.

Is this something to do with e^i*pi=1?

@troger147

7 жыл бұрын

Kyrator -1 I believe.

@Kyrator88

7 жыл бұрын

Yeah I think it's -1, I was thinking of e^i*pi +1 = 0

@troger147

7 жыл бұрын

Kyrator Haha that's good. I do something I call FYI on my Facebook page and for pi day I did one on that equation.

@Pfhorrest

7 жыл бұрын

e^iτ=1

@guest_informant

7 жыл бұрын

e^i*pi + 1 = 0

Is there any potential future in which I know what the rest of that tattoo looks like :[

They way I think about the hypothetical list of possible numbers formed by swapping digits to be uncountable, is that you can form a bijection with the set of infinite strings of 1's and 0's. When building your swapped digit number, put a 1 as the first digit in the infinite string if the original number was pi, and a 0 otherwise. Then, go through each digit that differs between e and pi, and if it was swapped, add a 1 to the string, if it was no swapped add a 0. This mapping is surjective because given any infinite string of 1's and 0's, you can find a pattern of swapping digits to give that string. It is injective because if any two strings are the same, then the same digits must've been swapped, leading to the same original numbers. This set of infinite strings of 1's and 0's is fairly easy to see is uncountable being in a simple bijection with the closed interval [0,1] by just writing every element of [0,1] in binary. Thus, the set of swapped digit numbers is uncountable

Nice I'm doing transcendental number theory for my current work :)

@11:45 Lol, my profile name is based on combining the words "circuit" and "neutrinos"

@healthystrongmuslim

7 жыл бұрын

Circuitrinos ok.

@CircuitrinosOfficial

7 жыл бұрын

I was the first comment response at the end of the video. I was amused by how she tried to pronounce my profile name.

big fan of the show, but I think you talk too much and dwell too long on the basics, like what pi and e are, and what irrational numbers are. most of the audience is already engrained in mathematics, and would know what these terms are.

@connorp3030

7 жыл бұрын

Ah, I think some people would be lost if she didn't, and that it probably only adds like a minute to the length. I personally like the "explain it to a child" approach.

@EvelynDayless

7 жыл бұрын

Presumably as the series goes on she can point to older videos to explain simpler topics that have already been discussed, but if she never goes over them the vast majority of youtubers who stumble onto these will have no idea what she's talking about.

@Eric_Pham

7 жыл бұрын

Noah Faso-Formoso I didn't knew what e was

@vaibhavjain3998

7 жыл бұрын

no...not all...😅

@Mastermindyoung14

7 жыл бұрын

You're nuts. I could listen to her read cereal ingedients.

It's possible to correlate every rational number between zero and one to a switching of digits, by converting it into binary, and switching every one while leaving every zero. It would also work in reverse alowing labeling of every switching with a real number. we can then use cantor's diagonal argument or a similar one to proof the uncountability of the set.

I googled "transcendental numbers" and went down a wonderful rabbit hole.

I did giggle at let epsilon < 0 shirt lol

Here's a way to see that the list of all pairs of numbers from swapping digits is uncountable. For each digit write a 0 if you don't switch them, and a 1 if you do. Then you can interpret that as a the decimal portion of a number in binary (between 0 and 1 inclusive). Thus, there is a bijection between the set of all possible pairs and the real numbers between 0 and 1, which is uncountable.

I love this video.

The proof of the uncountability of this "list" is, I believe, straightforward. I'll sketch it out for anyone interested in seeing it for themselves if they aren't confident enough to give it a shot-- though I recommend trying for a little while. You're usually more clever than you give yourself credit for. Say you make a list of hypothetical numbers created by switching digits between pi and e. You have a first number, a second, and so on. Now, construct your own new number that is *still* a combination of pi and e by looking at this list and choosing the value of pi or e in the ones place that is different than the value in the first number in the set. For example, if your first number in the list starts with 2, start yours with 3. If it starts with 3, start yours with 2. The second digit of your new number will be inspired by the second digit in the second number of your list. If the second number of the list has 7 as its second digit, your new number should have 1 as its second digit. Vice versa if it's got 1 as its second digit. If you run to a position in your number where the n-th place matches between pi and e, you can't elect to change that one so you look further in the digits of this number until you are in a place that you *can* change. For example, when you're figuring out what the 8th digit of your number is, you see that both pi and e have 8 as their 8th digits, so you have no choice but to choose 8. Then, while you still look at the 8th number, look at its 9th digit and choose the digit of pi or e that doesn't match the 9th digit of that number. Every time you happen to be in a spot where the digits of pi or e match, it's guaranteed that if you look further down in their digits, it won't be matching at all. It's important to note that you are not allowed to skip a number on your list while making up your new number. Your new number has to have a little inspiration for each of its digit choices from each of the numbers in your list as you continue your number's construction. This new number, because it was inspired by changing the digits of each of the numbers of your list, cannot be a member of your list. If it were the m-th number and you compared your new number to the actual m-th number, you'll find that your number has the be different from the m-th number by at least one digit because of how it was constructed. Because your number is guaranteed to be a combination of pi and e, but it has no place in the list, the list can never be complete or all-encompassing, so no list can exist and therefore the amount of these constructed numbers is uncountable.

For the challenge question: Consider the infinitely many places where e and pi differ (assuming that is indeed the case, I mean) and label them with the natural numbers. i.e., the first place they differ is labeled 0, the next is labeled 1, etc. Any swapping of digits of pi and e is uniquely specified by some subset of the naturals (specifically, for a given subset of natural numbers, you swap only the digits with those numbers as labels). E.g., the subset {2,4,6,8, ...} means swap the digits labeled with an even number (and don't swap any others). Note that because only differing digits are labelled, we are skipping over all of the matching digits and never bother to involve them in a swapping. Crucially, every such swapping results in different pairs of numbers,* since we are only labelling and considering digits that differ. For example, if only the digits labeled 4, 8, 15, 16, 23, and 42 were swapped (i.e., the relevant subset is {4,8,15,16,23,42}), any different swapping would necessarily either leave at least one of the digits unchanged or change some digit that the original swapping didn't. Note again that this only works because "swapping" is the same as "changing" when we're ignoring all the matching digits. So, there is a bijection between the subsets of the naturals and the relevant swappings of the digits of pi and e. The powerset of the naturals (the set of all subsets of the naturals) has the uncountable cardinality 2^(aleph_0), and Cantor's diaganol argument shows this is bigger than countable infinity (aleph-zero). By the very definition of equinumerosity (two sets are of the same size iff there exists a bijection between then), this means that there are uncountably many ways to exchange the digits. As shown in the preceding paragraph, each swapping results in different pairs of numbers. Therefore, there are uncountably many numbers you get by swapping the digits of pi and e (assuming that they differ in infinitely many places)---2^(aleph_0), to be precise. * Note: now that I think about it, I realized I missed a small gap---I'm only considering two list entries to be "seperate" if they come from the same number. For example, the subset {} produces the pair (pi, e) (no digits were swapped), while the subset {0,1,2,3, ...} produces the pair (e, pi) (every digit was swapped). This doesn't affect the argument, though---if you take the list and remove all numbers came from e, you're left with numbers that *are* all different from one another, so the bijection exists and the argument applies. Removing a bunch of elements from a list can't make it bigger, so the original list must also be uncountable. ____________________________________ Fascinating video as usual! A couple questions: * I've read about some stuff tangentially related to this before; it's not known whether or not e+pi, e-pi, pi^e, and other such combinations are irrational. However, I've also read that e^pi is known to be irrational (and transcendental too, if I remember correctly). Is there any reason why this is easier to prove than the others? * The digit swapping is base specific; does anything different/interesting happen in other bases? * Intuitively, it seems like things like pi-e *should* be irrational and transcendental---most numbers are, in some sense---but proofs of transcendence are elusive. After all, pi and e aren't random, no matter how their digits may appear---just because something occurs with 100% certainty for two randomly selected real numbers doesn't mean it holds for the very precisely defined e and pi. There are numbers, though, that are defined by there lack of definability---I've read that only countably many real numbers can be uniquely specified by a 1st-order predicate. I have some kind of vague sense that these two ideas are connected---do the so-called "undefinable" reals which make up most of the real line have any connection to these properties (like summing to a rational number) that only very special cases of numbers should have? I know this is a vague question (if it can even be considered that), but it just seems like there's something I'm missing here: there are these properties that should only hold for very special cases of numbers (e.g., being non-normal (abnormality?)) and fail for the rest, and there are these numbers that can't be defined with any first order predicate; it seems like these numbers shouldn't have what it takes to be one of those special numbers with those properties, you know? Or maybe my intuition is conpletely off here. If you've managed to make it this far into my ramblings, could you help me out here? Thanks :)

Two remarks: 1) Let n at least one. The probability that with n randomly picked number you can form one (or more) rational number by swapping digit is 0. 2) The answer might depends on the basis i.e there are two numbers such that you can form a rational number by swapping digit in basis 2, but you cannot in basis 4. Of course e and pi are not randomly chosen number (and even if there were a zero probability might still happen), however it illustrate why it is reasonable to conjecture that swapping digit of pi and e will not allow us to construct a rational number. proof of 1) To be clear we randomly uniformly pick n-tuple of number of [0,1]^n (i.e with the usual measure). Let b greater than one be a basis. Let q be a sequence of digit (in {0,...,b-1}), let k be a natural number The probability that a randomly picked number have the same k-th digit than q is 1/b, so the negation of this has probability (b-1)/b. Thus the probability that among n randomly picked number none of them share the same k-th digit than q is ((b-1)/b)^n. Therefore the probability that among n randomly picked number at least one of them have the same k-th digit than q is p=1-((b-1)/b)^n. Note that p

The digit strings 123 and 888 are *not* equally likely to appear in pi or e. This arises because sometimes there are more than 3 consecutive eights, so e.g. the substring 8888 counts as 2 occurrences, whereas no 4 digit strings contain 123 twice.

The resulting list in case of infinitely many differences between e and pi is uncountable because for each digit we have to decide between SWAP and NO SWAP. That's two possibilities. The number is 2^x, where x is the number of choices we make. That x is countable infinity, 2^x is the size of its power set, which always has greater size than the set it is based on.

Here’s my take on the challenge problem. Consider all possible infinite sequences of binary digits, such as 01010101010…, 010001001001…, 11111111111…, 0000000…, etc. As Cantor first showed in 1891, there is no bijection from these sequences to the integers because given sequences S1, S2, …, SN, … we can construct D such that S1(1) ≠ D(1), S2(2) ≠ D(2), …, SN(n) ≠ D(n), … simply by negating the digit with Boolean logic (if it’s 1, we choose 0, if it’s 0 we choose 1). Thus our D is not on the list yet we have run out of integers, thus there is no bijection. That said, there is a bijection from the infinite sequences of binary digits to possible digits swaps between π and e. The nth digit in each binary sequence can represent whether we swapped the nth differing digit of π and e, with 0 meaning no swap and 1 meaning swap. By this mapping, every possible swapping corresponds to a single infinite binary sequence, and every infinite binary sequence corresponds to a single swapping. Thus, there is a bijection between the two, which implies there isn’t a bijection from swappings to the integers, and the swappings are uncountable.

Great video! Is there a chance we could get a video about differential geometry?

@kostasch5686

7 жыл бұрын

totalnastoka I knew you d like to see some curvature going on

it is very cool......could you please explain about Fourier transform and Laplace transform concepts?

Thanks for the information, nice video! Please make a video about the "game of life" by bitstorm there are fun maths there.

(sort of on a tangent) How to get 3.111*.... using e and pi. Straight lines on a graph with co-ordinates, (-e, 0) to (10, e), (0, -e) to (e, 10) both cross each other at the co-ordinate ([e^2]/10, [e^2]/10), which occurs at the nearest approach to each other of the curves y = e^x and y = ln(x), while these straight lines also touch at a tangent to both curves y = e^x and y = ln(x). The algebra for the first line is y = (e/[e +10])x + ([e^2]/[e +10]) And the second line is y = ([e +10]/e) - e Plus a circle fits the co-ordinates (-e, 0), (10, e), (0, -e), (e, 10) as well..... *the centre of this circle is (x= 3.111, y = 3.111) Maybe you get 3.111111111111 ? - I don't know. I did this ages ago and didn't bother to do the co-ordinates to that accuracy (not realising the centre of the circle was at (x= 3.111, y = 3.111) The radius of this circle appears to be an irrational number (as it involves pythagarus using e), as well as the circumference of the circle (it is multiplying two irrational numbers.. the radius with pi). LOL - it also looks like an inappropriate crotch pic so I'm told (jus' saying) Also, (I don't think it does but) if you then convert the numbers of the axis into degrees and ploting y = arctan(e^tan[x]), then you get a straight line (approximately) y = x/2 + 45degrees, where x and y are in degrees... (I think it can actually only be expressed as an infinite series and doesn't form quite a straight line... If you map y = x/2 +45degrees onto a hemisphere, when viewed from above it very nearly fits a fibonacci curve... But I don't think y = arctan(e^tan[x]) quite matches a fibonacci curve using this method either - though it is very close [maybe it is the same and I got it wrong using may back of the envelope methods]) stuff Now I am going to eat some pie (e eat some pie pi - get it? sigh)

I knew my brain would hurt if I watched this video, and sure enough, it hurts trying to comprehend all of these infinities.

It is actually very easy to construct infinitely many normal numbers (in a fixed base). Once you have one such number (and there are several such constructions), changing finitely many digits in it produces another such number, so already you can construct countably infinitely many such numbers. You can also change infinitely many digits, if these digits have zero density (for example, all the n^2 digits), so you can build unaccountably many such numbers. I'm not sure what happens with numbers which are normal in every base.

It is easy to misunderstand this "swapping procedure", if you judge by all the comments to this video that got it wrong. 1. You can use a base of 2 for the e and pi numbers, simplifying to 0 and 2 as digits swapped. 2. Each swap taking place is defined to be between digits being at the n:th place (same place) of both numbers. 3. Talking about an actual "infinite number of decimals" is just a figure of speech, since real numbers don't allow an infinite number of parts, instead limits are used to work with the values of "infinite" decimal numbers. An actual infinite decimal is therefore not allowed, and would lead to contradictions. What you can do is to use "and so on" as a description of the infinite within the real number decimals, for example for the repeated patterns of "rational" real numbers repeated without end. 4. Philosophically, defining rational numbers as ratios or as decimal numbers with repeating endless part is not exactly pointing to the same mathematical object. The rational decimal number continues to be a real number with the same value as a rational number but with added real number properties.

With no rigor, let B be the interval (0, 1) represented in binary (we assume a unique representation for the rationals). so every number in B represents a unique exchange of digits between pi and e, and thus we got an injection from B to the set of exchanges. Therefore, the set of exchanges has uncountable many numbers.

You are born to Inspire

If I love math and also love this channel, named Infinity, does this create infinite love?

Answer to the challenge question. Suppose that the places at which the digits of e and pi differ be S. Since they differ at infinitely many places, cardinality of S is equal to the cardinality of the natural numbers (its countable). We can exchange digits at any of the places we want. This means we can take a subset of S and exchange digits at those places to get our two numbers. The set containing all the subsets of S (power set of S P(S)) tells us all the ways we exchange the digits. Since S is countable, P(S) has to uncountable (cardinality of the power set is always bigger than the cardinality of the original set). Therefore there are uncountably many pairs of numbers that are formed by exchanging the digits of e and pi.

@anhadvir

7 жыл бұрын

Suppose that the set of places at which the digits of e and pi differ is S.*

Just for info, e^(iπ)=-1 (a rational number). ^ stands for power i is rootover-1

@error.418

5 жыл бұрын

Instead of "rootover-1" most people will say "the square root of -1"

I think I can prove that, if π and e differ at infinitely many digits, the number of digit-swaps is uncountably infinite: Take the decimal expansions of π and e and drop every digit that matches in both of them; you now have two infinite (countable) sequences of digits which match at 0 locations. When generating a list of the swaps, you can choose to swap the digits at any position on this list, and the decision to swap at any location is independent- you can swap it if you damn well feel like it thankyouverymuch. It should be pretty straightforward that there are 2^n ways to swap digits- at any position, you can make one of two choices unconditionally; each time you make a decision, the number of outcomes doubles. According to Cantor's Theorem, the cardinality of a set's power set 2^S is greater than that of the original set; since the set we started with is countably infinite, and any number which is greater than the one countably infinite number is uncountable, the number of ways to swap is uncountable. QED. *_(This was written while I was exhausted; it may be incoherent, but I know for a fact the math is right)_*

Trivial general way to show existence of two irrationals whose digits can be exchanged to make a rational is to take e.g. 0.1111111..., and a number whose decimal differs from this in an infinite number of places. Then take a non-repeating subset of those places where it differs, and form the new number out of 0.1111.. and this new number. Naturally we can then reverse the exchange to get our rational 0.111111.... back.

My own shot at the challenge question: the different ways we can choose to exchange infinitely many digits of pi and e are a subset of P {N}, as we are basically creating a new set by deciding which of the countably infinite digits to exchange. In particular, it's the set of all the possible infinite sets contained in N. Now, we know that the cardinality of P {A} is greater than the cardinality of A whichever set A is, so P {N} is an uncountable infinity. Also, if we subtract a countably infinite number of elements from an uncountable infinity we still get an uncountable infinity. Thus, if the number of all the finite sets possible in N is countably infinite than we have proved our theory. Now my argument gets a lot complicated to explain with words but I'll try my best. We must show that the number of all of the finite sets found in N is countable. For that, I'm gonna use something similar to a 3D construction. Imagine to have a 3D orthogonal space. The X axis shows the minimum number of the desired set, the Y axis shows the number of elements in each of the desired subset and the Z axis shows the maximum number contained in that desired subset. In this way we have divided our space into a set of well specified points with well specified sets corresponding. For example, the point (1; 3; 4) represents the subsets (1;2; 4) and (1; 3; 4), the point (2; 3; 6) the subsets (2; 3; 6), (2; 4; 6), (2; 5; 6) and so on. Now, the number of these points is clearly countably finite as it's the sum of many countably finite sets (the Z directed straight lines of points) and while it's true that every point may contain more than one point, it would still be a finite number, and a countably infinite sum of finite numbers is itself countable infinity. I know this probably us extra confusing and certainly flawed in many ways, but at least I tried lol. Any response from the owner of the channel would be great!

There are uncountably many ways to swap infinitely many digits of e and pi. Proof: If e and pi differ at uncountably many decimal places, the conclusion is trivial. Otherwise, they differ at countably many decimal places, which means that the set of these decimal places has the same cardinal number as N, the set of all natural numbers. Let M denote the set of indexed decimal places at which e and pi differ, i.e. if e and pi differ at the 27th decimal place, then M contains 27. It is clear that M is isomorphic to N. Consider now the following transformation: replace each decimal place that we decide to swap with 1 and each decimal place that we decide not to swap with 0 and let P denote the set of all such transformations. It is clear that P is isomorphic to 2^M, i.e. to 2^N, since we have only two choices for each decimal place. Consequently, P has the same cardinal number as 2^N, i.e. 2^(Aleph 0). Since R, the set of all real numbers, has cardinal number 2^(Aleph 0), this means that P is isomorphic to R. But we know that R is uncountable. Therefore, P is uncountable, which ends the proof.

When n is any normal #, e - n = irrational # and pi - q = irrational #. It follows that any sum computation of e and pi's digits would be irrational (given they are normal numbers) bc subtraction is just the inverse process of a sum.

Is there any matematical equation that relates pi, e, phi like Euler's equation? maybe also we can include sqrt(2)

I have an idea for a video on this channel. After watching a few videos about infinity on other channels (e.i. Vsauce, Vihart, Numberphile), I arrived and an interesting conundrum. If you take any finite number and raise it to the power of any other finite number, what you end up with is still a finite number. f^f=f But if you take a finite number that's greater than 1 and raise it to the power of the smallest infinity - countable infinity - you end up with a larger infinity - uncountable infinity f^c=u The conundrum arises when instead of an exponential function, we use a logarithmic function. log(f)c What is the solution to that function? What exponent must we apply to a finite number in order to arrive at the smallest, countable infinity? It can't be finite because you would end up with a finite number. But it can't be infinite either because you would end up with an uncountable infinity. Could it be that there is some number larger than all finite numbers yet smaller than countable infinity? I would love to see this conundrum addressed in a future video.

@MisterrLi

7 жыл бұрын

+Nathan Warford Your question is similar to asking what number is closest to 1 but not equal to one, in real or rational numbers. The answer is that there is no such number; if you name a candidate number you can always find a closer one by adding 1 and the candidate and divide by 2. So there can't really be a smallest infinite set, if you define infinite countable sets to change in value when you subtract a set from it (like with infinite "numerosities").

simple: there are countably infinite decisions to swap or not. that gives you the power set of the integers, which is well known to be uncountable.

ok, my answer to the challenge question: - take any real number between 0 and 1, and consider its binary expansion 0.b1b2b3b4b5b6... (if it ends, fill it up with zeroes, so e.g. 0.1=0.10000...) - take the first digit where e and pi differ. if b1=1, swap it, if b1=0 don't swap it. - take the second digit where e and pi differ, if b2=1, swap it, if b2=0 don't swap it. - go on like this, for all the infinitely many digits where e and pi differ. - by this method, for every real number between 0 and 1, we get a different pair of numbers, that results from swapping (infinitely or finitely many) digits of pi and e. - we know that the set of real numbers between 0 and 1 is uncountable. hence, the set of such pairs on this swapping-list must be uncountable as well.

The only combination of pi and e that I can recall that has a finite solution is indeed e^(pi)i=-1 but that's more of a theorem than a combination, so not sure it it meets the technical definition of "combine"

123 and 888 aren't equally likely to appear in a normal number because you can fit two 888s into a four digit long string, but you can't do this with 123. (ie '8888' contains two 888s)

*a rational number can be written as n/m where and n and m are integers, BUT also m≠0

Yes, and it's quite easy too. However, I suspect that there is a restriction that makes things more complicated. Probably, you are for example not allowed to exchange the 1st digit of pi with the 3rd digit of e; the restriction being that you can only exchange the i:th digit in pi with the i:th digit in e. I might have missed this restriction in the video but here's an easy way to prove that it's possible to exchange digits of e and pi such that one of them becomes rational without the restriction. Any real number x must have either infinitely many digits that are 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9 because every real number x has infinitely many digits (note that for instance 4 = 4.0000... which is infinitely many zeros) and there are only finitely many digits to choose from. Further note that any real number x has a countable infinite number of digits since it's easy to label all of the digits with natural numbers. Let's assume that the digit d occurs an infinite number of times in pi (which by the previous argument must be true for some digit d). Now, simply exchange the first digit in e with the first d in pi, the second digit in e with the second d in pi and so on. Since both the number of digits of e and number of d:s in pi are countably infinite, we can keep doing this until e has been turned into d.ddddd... wich obviously is repeating and hence rational.

Switch the first digit of pi with the first digit thats a three in e. Switch the 2nd digit of pi with the 2nd digit thats a three in e and so on. Then pi becomes 3.33333... doesn't this work or am i missing something?

@GarryDumblowski

5 жыл бұрын

Well, for one, that assumes that e is normal, and includes infinitely many 3s. But the bigger mistake is that you can only swap digits in the same place. You can't swap the first digit from the left in pi with the second digit from the left in e, for example, you can only swap the nth digit in pi with the nth digit in e.

let A be the set of decimal positions where pi and e differ. for every subset B of A, a unique irrational number can be made by exchanging the digits of pi and e at the positions of B and only B. the set of subsets of A is the powerset of A, and so the number of corresponding number made by exchanging the digits of pi and e at the positions of the subsets is also equal to the powerset of A. if A is infinite, the powerset of A is uncountable, so this completes the proof. qed