Pi is irrational (π∉ℚ)

Рет қаралды 58,217

Күн бұрын

We give a simple proof of the irrationality of pi.
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Пікірлер: 219

@sword7163 4 жыл бұрын

this proof is just tremendously, extremely mind blowing .

@malawigw 4 жыл бұрын

I knew about this proof and that 1 page article before, but I could not get this far when it comes to filling in the details. This video helped a lot!

@timurpryadilin8830 4 жыл бұрын

Please return the check-squares next to the tools you use. They are really awesome in the way you fill them with a clack!

@manucitomx 4 жыл бұрын

Timur Pryadilin I second this

@blazedinfernape886 4 жыл бұрын

I knew sinx was coming but it took like 20mins for it to come lol. Such a hard proof!

@goodplacetostop2973 4 жыл бұрын

23:18

@darkseid856 4 жыл бұрын

Noice

@user-mt9ux2di6u 3 жыл бұрын

Always so helpful!

@horseman684 3 жыл бұрын

so unfair how did you get that account name

@djbj1993 3 жыл бұрын

At 17:00 looks to me that there are a few mistakes on the exponents in about 4 different places. The k-th derivative of x^n is not (n!/k!)x^(n-k) in the case where k>n, which is the second sum. The same problem happens in the second factor of the first sum.

@demenion3521 4 жыл бұрын

the presentation of the proof was very good, but I think that the proof really involves way too many seemingly arbitrary choices and estimations, which makes the proof not quite understandable

@easymathematik 4 жыл бұрын

This is the big problem. This proof doesnt include Niven´s idea behind the choices which makes this proof very very hard to understand to 100 %. I dare to say that only Niven understand this proof to 100 %. All other steps fall from heaven and where they come from only Niven knows exactly.

@bttfish 4 жыл бұрын

@@easymathematik As Gauss described this style as “no self-respecting architect leaves the scaffolding in place after completing the building”

@joeg579 4 жыл бұрын

@VeryEvilPettingZoo this is well put. the rough work and logistics towards figuring out the "seemingly arbitrary choices" seen in this video are executed in the same way one would construct an epsilon-delta proof. these proof techniques are skills that high-level mathematicians have learned to use by working backwards.

@hybmnzz2658 3 жыл бұрын

If you understand every step in the way you have understood the proof. We are proving Pi's irrationality over here, and one that was not discovered until 1947. Hours of trial and error, blind guessing, experimenting with functions, etc.

@yunoewig3095 3 жыл бұрын

@@hybmnzz2658 Pi's irrationality was known since the XVIIIth century

@bobzarnke1706 3 жыл бұрын

I thought that there was an error because, when k > n, x^(n-k) would have a negative exponent. (udic01 notes this is a problem when x = 0.) But then I realized that x^(n-k) results from the k-th derivative of x^n; and it's 0 when k > n. So, in the second summation (for k = n to m) for p^(m)(x), all except the first term are 0. (The first m-n terms of the first summation are also 0 for a similar reason.) This doesn't affect the proof, but it seems noteworthy.

@pausesmaths3086 27 күн бұрын

It doesn't affect the proof but I call it a mistake anyway.

@gio5969 2 жыл бұрын

Quite true. I had a discussion with Pi the other day. Completely irrational.

@MathTutor1 2 жыл бұрын

I remember I learned this proof as an extra challenge problem when in high school. Proof of π being irrational is much longer than that of e. These two along with the irrationality of √2 are must to know for math majors if you planning to go in that direction.

@suramanujan 3 жыл бұрын

Wow!! This is an extremely helpful video at least for me!! Thank you for this one. 👍👍

@tomatrix7525 3 жыл бұрын

This one was really hard. Not calculation wise, but alot of those p(x) functions seemed to come from nowhere

@tracyh5751 3 жыл бұрын

I first saw this proof in my transcendental number theory class. You can use these same ideas to prove that some interesting numbers are transcendental. The main idea is to use algebraic number theory to get bounds on integrals, and then assuming your number is algebraic contradicts the bounds that you constructed. These kinds of proofs are really cool, but unfortunately it requires coming up with weird functions like the ones that we see in this proof. The ad hoc nature of the proofs makes them difficult to generalize, so we can go decades before we get new proofs using this technique (and historically this is what happened).

@James_Moton 3 жыл бұрын

If you haven't already, will you please prove that (m choose k-1) plus (m choose k) is equal to (m+1 choose k)?

@joeistead 3 жыл бұрын

Expand the sum, then factor out the common factor, so you have a common factor multiplied by the sum of two fractions. Combine those two fractions into one fraction by adding them. Then you have the common factor multiplied by one fraction. Now multiply the common factor by that fraction and you'll get (m+1)! / [k! * (m+1-k)!] which is the desired RHS. Send me a private message if you get stuck.

@johnpaterson6112 2 жыл бұрын

Consider the coefficient of x^k in the expansion of (x+1)^(m+1), and compare with the expansion of ((x+1)^m)*(x+1), which will contain the two coefficients of x^k you need. You can do it mentally! Of course that will not work for you unless you proved the choose function independently of the binomial expansion, and then used that to prove the binomial expansion directly, rather than by induction.

@mathismind 4 жыл бұрын

Nice video, ty!

@IbrAhMath 4 жыл бұрын

I was so impatient to finally see the part in the proof where you utilise properties of π and not “numbers” in general. That’s always the point I say to myself: “Oh this is how we are bringing π in there...”

@tomatrix7525 3 жыл бұрын

Yep, the problem with these crazy proofs is that there is so much work before you bring pi in so it not intuitive at all

@gavasiarobinssson5108 3 жыл бұрын

@@tomatrix7525 That is also the beauty of it.

@zhuolovesmath7483 2 жыл бұрын

KEEP GOING Michael!!

@0priest086 4 жыл бұрын

wow it is Ivan Niven's proof! It is very interesting that only basic calculus knowledge is needed.

@xshortguy 4 жыл бұрын

This seems like it would be an entry into a contest of ridiculous ways to prove pi is irrational.

@johnvandenberg8883 3 жыл бұрын

17:07 what about x^(n-k) for x=0 and n-k = -1, ... , n-m (i.e. n

@dr.rahulgupta7573 4 жыл бұрын

Excellent presentation of the topics in a beautiful manner. Thanks a lot.DrRahul Rohtak.India

@shambosaha9727 3 жыл бұрын

14:33 Won't it be (n-m+k+1)? Also at 15:43 where is the m choose k term?

@mrmathcambodia2451 3 жыл бұрын

I like this solution so much ,Good explen in this exam.

@udic01 4 жыл бұрын

17:07 when evaluating the second part of P(m)(0), what is 0^(n-k) where k is bigger than n?!

@udic01 4 жыл бұрын

I watched it again and realized that the formula that he wrote is not correct. The Kth derivative of x^n where K is bigger than n is 0. for example (d3) of x^2 is 0.

@yunoewig3095 4 жыл бұрын

@@udic01 he's a bit sloppy, that's all

@ryderpham5464 4 жыл бұрын

udic01 so both terms go to zero?

@yunoewig3095 4 жыл бұрын

@@ryderpham5464 there's no limit here, some terms are zero, that's all

@udic01 3 жыл бұрын

@@angelmendez-rivera351 i know about the convention of n choose k ( i myself answered someone else's question about it. My problem with the formula is the fact that you have 0 and then multiplied by 0^(negative ) which is undefined. Like i wrote above, We know that the Kth derivative of x^n is n(n-1)...x^(n-k) when k is smaller or equal to n. But evaluating it at 0 and saying that because for all other values of x it is 0 therefore it's value for x=0 is also 0 is incorrect. Examine the tunction f(x)=0/x. For all x!=0 it is 0 of course. But for x=0 it is undefined. All i am saying is that michael shouldn't have written it that way.

@thomasborgsmidt9801 3 жыл бұрын

For the first time: a) I was able to follow (sort of) the argument. b) That n! beats exp(n) given n large enough. Sounds reasonable; but news to me. c) The generalised product rule. Just wonder what it looks like for a general number of functions. d) Good proof, as the contradiction is in the squeeze theorem. e) I lack rules for when you can swap integral and summations around.

@niallloughran5149 2 жыл бұрын

Since integrals are linear, the integral of a finite sum of functions becomes the finite sum of the integrals of the functions. For infinite sums things get a bit trickier.

@jansmotlacha1077 3 жыл бұрын

As I understand the proof, almost all the steps would hold for arbitrary rational. The only exception is on the fourth line from bottom on the table (if I consider the last 15 seconds) - it states that int(p(x)sin(x)) = P(0) + P(pi) which is integer. At the same time, this proves that int(p(x)sin(x)) cannot be integer for arbitrary rational (because if it would be integer, it would contradict the last inequality as well).

@TwilightBrawl59 4 жыл бұрын

I would ask a question about pi, but the answer will never end.

@mr.mirror1213 4 жыл бұрын

pUn

@emanuellandeholm5657 4 жыл бұрын

Oh but it will! In base pi.

@sword7163 4 жыл бұрын

@@emanuellandeholm5657 if you combine the euclidean algorithm and the completeness of R, real analysis will end.

@xriccardo1831 4 жыл бұрын

@@sword7163 how?

@sword7163 4 жыл бұрын

@@xriccardo1831 very badly

@anon6514 Жыл бұрын

I love this proof. The version of the proof I read first didn't have that nice extension to to the product rule you proved in the first half. If the assumption that pi is rational were true... p(x) would be positive in the interval (0,pi) but have roots at x=0,x=pi -- just like the sin function. in fact all of its derivatives would evaluate to an integer at x=0, x=pi -- just like the sin function it would be symmetrical about x=pi/2, i.e. p(x) = p(pi-x) -- just like the sin function It's almost like this proof works because the assumption translates to: "what if the sin function were a polynomial?"

@elliottmanley5182 2 жыл бұрын

Do you have a vid proving pi is transcendental?

@txikitofandango 3 жыл бұрын

Around 16:40, if n-k is always

@kimi20F1c 3 жыл бұрын

It took me like 20 minutes to figure out what's going on there: When he puts in 0, the only case this sum isn't zero anyways, is when n=k. Since when n

@mikesteele5935 Жыл бұрын

It's more fun to prove the Leibnitz rule by (1) first noting it is obvious except or the coefficients.(2) to identify the coefficients we can use any f and g. This step you can do by picking eponentials \exp(ax) and \exp(bx). Then set x=0 and remember the binomial theorem. Voilà.

@ogasdiaz 7 ай бұрын

I really like that this proof never even describes pi. It only requires that sin of pi is an integer.

@1_1bman 3 жыл бұрын

17:36 I don't understand this step. What makes (m - k)! a multiple of (n!)?

@jacoboribilik3253 3 жыл бұрын

If you multiply (m-k)! by all the integers between (m-k)and n you get n!.

@victoralfonsovicentebenite2414 3 жыл бұрын

A really nice proof! The next step for the next video.... Pi is a trascendent number :D

@eamon_concannon 2 жыл бұрын

17:00 This part is an integer for x=0 by first noting that nCk = 0 for k>n, so this second series collapses to a single term by putting k=n. The x term becomes 1. The power of a-bx becomes 2n-m which is non negative as 2n >=m. We eventually get nC(m-n)*(m-n)!*(-b)^(m-n)*a^(2n-m) which is a non-zero integer since 0 0. To show the first series is an integer for x=pi, one can write k!(m-k)!/n! = (1/(nCk))*(m-k)!/(n-k)! so we get for first part of first series, (nCk)(nC(m-k))*k!(m-k)!/n! = (nCk)(nC(m-k))*(1/(nCk))*(m-k)!/(n-k)! = (nC(m-k))*(m-k)!/(n-k)! which is an integer as m>=n. The second part of the first series for x=pi is(-b)^(m-n)*a^(n-k)*0^(n-m+k). The powers m-n, n-k are non-negative, The power n-m+k is non negative if m-k

@barrankobama4840 2 жыл бұрын

I'm perplexed by the nonchalant use of 1/(-1)! = 0 I would have preferred a sum for k between 1 and m and them we could adjust the cases for 0 and m+1 using bin(m, 0) = 1 = bin(m+1, 0) and bin(m, m) = 1 = bin(m+1, m+1).

@cicik57 3 жыл бұрын

emm.... i do not get, why the second x ^n-k is not 0 and entire second part is not 0 the

@yashvardhan6521 3 жыл бұрын

10:46 U must be Fermat's descendant

@brianjesusdiaz 4 жыл бұрын

I had feeling you were going to use Niven's proof! What a 1-page wonder!

@michellauzon4640 2 жыл бұрын

the induction base can be m = 0.

@sujalsagtani6868 4 жыл бұрын

Please make a video on Find all x such that x+1 is a perfect square and 2x+1 is also o perfect square

@sujalsagtani6868 4 жыл бұрын

Also on Find x such that x+1 and 5x+1 is also a perfect square

@azhakabad4229 4 жыл бұрын

Epic title famous all over mathematics!

@jesusmauro3114 2 жыл бұрын

Such a beautiful proof.

@jhoes_inf 4 жыл бұрын

Great!

@AbuMaxime 2 жыл бұрын

That was epic!

@all462 4 жыл бұрын

Are you gonna participate in #MegaFavNumbers

@javierlazaro7951 2 ай бұрын

Wait, why is the expression always less than 1. Checking case n=1 we have a^3/b^2 but since a/b is greater than 2 at least and something greater than 1 squared is bigger then the expression is false

@psychSage 3 жыл бұрын

22:49 plus, not times

@vardaandua3585 4 жыл бұрын

Oh i am very happy, I had requested this one thank u so much

@easymathematik 4 жыл бұрын

For me irrationality proofs are very fascinating because of 2 reasons. 1) the definition. number r is irrational : r is not rational so there is a "not". And r being "rational" is defined via: there exist integer p integer ,q natural > 0 s. t. r = p/q. So irrational : for all ratios p/q : r != p/q And what is the big problem of this definiton? How you can check all ratios? In other words: The definiton is not constructive. It doesn´t tell you: "use this method to show the irrationality" So you have to be very creative and smart and find something what can help you. 2) specific definition of the number (here pi) The second problem is, that every number has specific behaviour. But how to use it? A "problem" of Pi is, that there is no "nice" definition in some sense. Should I work with the geometric def? u = d*pi? Or should I work with the analytic def? Pi = 2 * smallest positive zero of cosine? Or should I work with some crazy series? Or should I work with some crazy integral expression? Everybody with little calc knowledge can follow this steps. But one thing is very hard to understand. Why this chosen f(x) should help? What is the deep idea behind this choice? Why all these steps show the way to the goal? This makes this proof marvelous and everytime I see it I am impressed.

@easymathematik 4 жыл бұрын

Compare for example this proof with the proof of irrationality from sqrt(2). The proof "sqrt(2) is not rational" is very easy and all steps are more or less natural. There is no deeper understanding.

@jimallysonnevado3973 4 жыл бұрын

Why we only claim that the derivative of p at 0 and pi is interger? Why can’t we claim that it is 0 since x^(n-k) is 0 if x=0 and (a-bx)^n-m+k=0 if x=pi. Also is division by zero a problem here since fo k>n n-k is negative and 0^negative is undefined?

@udic01 4 жыл бұрын

I also wondered about x^(n-k) and commented before I saw your comment. I watched it again and realized that the formula that he wrote is not correct. the Kth derivative of x^n where K is bigger than n is 0. for example (d3) of x^2 is 0.

@yunoewig3095 4 жыл бұрын

it's algebraic shorthand, the derivative of a polynomial will actually be 0 if the order of the derivative is greater than the degree of the polynomial, so no division by 0 actually occurs.

@yunoewig3095 4 жыл бұрын

@ That's a complicated way of thinking. Really, he should have been more careful in the way he wrote those expressions, but it's forgivable, we can see what he actually meant to say.

@tylerantony7399 3 жыл бұрын

I don't think removable singularities need come up here. In the second term, note that n - m + k is always positive (this follows since k >= n, and m

@yunoewig3095 3 жыл бұрын

@VeryEvilPettingZoo Thank you for referencing me

@user-A168 4 жыл бұрын

Good

@featherton3381 4 ай бұрын

Interesting. Doesn't this also imply that the inverse sine of any non-zero rational number is irrational?

@p0gr 3 жыл бұрын

this "tool" is not necessary at all. you simply have to know that a monomial of order n needs n derivatives to become constant and gains a factor of n! on the way. thats trivial.

@asok5597 4 жыл бұрын

7:00 why both are 0?

@udic01 4 жыл бұрын

when you have m choose k and k is either negative or bigger than m than it is defined as 0. In how many ways you can choose -1 objects out of m? 0 ways. The same goes for m+1 objects out of m.

@asok5597 4 жыл бұрын

@@udic01 thnks. I was thinking that as a defined form. But i didnt know if it was right.

@meryemnour9996 3 жыл бұрын

Please someone clears it up for me ,i don'tget it when he said m choose m+1 is equal to zero ,isn't it supposed to be indefined?

@hybmnzz2658 3 жыл бұрын

Its a matter of extending the definition. You can imagine Pascals triangle has imaginary zeroes outside. It is fine because many theorems still work with this definition.

@meryemnour9996 3 жыл бұрын

@@hybmnzz2658 thank you so much.....but"m choose m+1"=m!/(m+1)!(-1!) And by using gamma function you can poove that (-1)! Is undifined ,the same thing happens when k=0 in the first part of the sum.

@meryemnour9996 3 жыл бұрын

@@hybmnzz2658 do you have sources that allow me check for what you are saying?

@stefanstojkovic8712 4 жыл бұрын

Next step : prove e+pi is irrational ;)

@valeriobertoncello1809 3 жыл бұрын

e is irrational, any prime p is rational and p*i is a rational complex number. Thus e+pi is the sum of an irrational number and a rational number, therefore it is an irrational number qed

@minh9545 3 жыл бұрын

@@valeriobertoncello1809 i think by "pi", the person means 3.14159...

@TechToppers 3 жыл бұрын

@@valeriobertoncello1809 😅 I think you are confused...😂

@dinocoder1281 3 жыл бұрын

@@minh9545 r/woosh

@mygoogleaccount4404 3 жыл бұрын

@@dinocoder1281 stop bullying him he thought he could help

@sword7163 4 жыл бұрын

could you review some proofs from the book entitled :"Problem-Solving and Selected Topics in Number Theory", you can find a pdf version in the website libgen.is. It contains beautiful proofs and methods. Also, here is a nice brain teaser: prove that the fractional part of sqrt(4n^2+n) is less than 1/4

@sword7163 4 жыл бұрын

@VeryEvilPettingZoo a simpler formulation would be: 4n^2 =< 4n^2+n =< 4n^2+n+1/16=(2n+1/4)^2 which directly implies two things: -that the integer part of sqrt(4n^2+n) is 2n and that 0=b and x ay+bx a direct proof would be pretty. let's exchange problems !!

@sword7163 4 жыл бұрын

@VeryEvilPettingZoo yeah i wanted to write x>y. check out some of the resources here: artofproblemsolving.com/wiki/index.php/Olympiad_books and this pack of resources: www.dropbox.com/sh/pwfeve60hdbpgqt/AACJ_JNddxclbpn41p9ebePxa?dl=0

@trumplostlol3007 2 жыл бұрын

The real line is a strange animal. If I change the scale and label pi as 1 and 2 pi as 2 and so on. Then it is a rational number. LOL So, whether a number is rational or irrational actually is scale dependent. I can even make sqrt(2) a rational number and 2 an irrational number. Relabel sqrt(2) as 1. Then sqrt (2) becomes a rational number and 2 become an irrational number cause sqrt ( sqrt(2)^2 + sqrt (2)^2) now has a value of sqrt(2) even though it was 2 in the old scale. LOL

@CglravgHRjsksgS 2 жыл бұрын

We know that e:=n-->∞ lim[(1+1/n)^n], but which limit of a sequence gives us π? Please i have been searching that for 2 years... If this channel found it, it would be perfect!

@roronoazoro-bc6ev 4 жыл бұрын

A great proof

@azhakabad4229 4 жыл бұрын

I think he's gonna start his merchandise!

@malawigw 4 жыл бұрын

has been available for quite some time now.

@jimiwills 3 жыл бұрын

Nice!

@heliceaberlin 4 жыл бұрын

but how this Pi which cannot be rational is connected to Pi defined as the defined as the ratio of a circle'a circumference to its diameter? The connection must be made when the trigonometric functions appear, but it is not obvious (to me)

@pbj4184 4 жыл бұрын

The connection is due to the definition of radians. One radian is the angle subtended at the center by an arc whose length is same as the radius. Since we defined π is circumference by diameter, when we take a full circle as the arc, the arc length is the circumference which is 2πr due to our definition of π. So, a full rotation is 2π radian because arc length is r*theta _That_ is how angles and π and arc lengths are related to each other :)

@yunoewig3095 4 жыл бұрын

when evaluating the integral, sin(pi) = 1, cos(pi) = 0

@heliceaberlin 4 жыл бұрын

@@yunoewig3095 yes, of course! Thanks

@yunoewig3095 3 жыл бұрын

@@angelmendez-rivera351 You are right. I just wanted to point out that the fact that a/b=pi comes up when evaluating the integral. I got the values wrong because I messed up pi and pi/2.

@gaufqwi 4 жыл бұрын

Nice presentation as usual. Is it just me, or does this channel have a lot of ads? I think I had at least six interruptions.

@xriccardo1831 4 жыл бұрын

The ads are important because it's the only way he can earn money he deserves from his videos. Just remember that his videos are free to watch

@gaufqwi 4 жыл бұрын

@@xriccardo1831 Yes, I understand the revenue model on KZfaq. Note that I was commenting not on the presence of ads but the volume of ads. I estimate this channel shows me 3-5 times as many ads as any other channel I am subscribed to.

@xriccardo1831 4 жыл бұрын

@@gaufqwi i think that the amount of ads depends on two things 1) the lenght of the video and 2) the creator (they can probably decide to increase it, together with their partnership), but i'm not 100% sure

@gaufqwi 4 жыл бұрын

@@xriccardo1831 If he indeed has control over it I hope Michael will consider turning down the number of ads. An ad block every ten minutes or so is reasonable, but I'm getting ads within two minutes of each other. That's excessive.

@shambosaha9727 3 жыл бұрын

@@gaufqwi Well, compared to the tremendous amount of hard work he puts into his (almost) daily uploads, he has a very low number of subscribers. Which is why he needs more ads to get the ad revenue he needs. Together, we can change this if we spread information about him and help him grow. Unfortunately, the modern society looks down upon learning, considering it 'uncool'. But we can do our part. Besides, if you wish to be uninterrupted, just go to the end of the video and hit Replay. the ads will no longer interrupt as they will be considered watched.

@mohammadhosseinkheradmand1912 3 жыл бұрын

There is question in my mind Althogh this proof is so complicated,pi has some definitions.but this proof doesn't use any of those.so even if I use it for e,golden ratio or even an integer,I will reach the contradiction.isn't sth wrong here?

@dOncRiMe616 3 жыл бұрын

no as you don't get the same integral values as you are plugging in pi into sin/cos.

@illogicmath 4 жыл бұрын

Now please prove that pi is transcendental

@maypiatt3766 4 жыл бұрын

I always thought that the generalized product rule was called the leibniz rule

@yunoewig3095 4 жыл бұрын

Leibniz Rule is the 'simple' product rule.

@pbj4184 4 жыл бұрын

@@yunoewig3095 Then what is the generalized one?

@yunoewig3095 4 жыл бұрын

@@pbj4184 I don't think it has a special name, but, looking it up, it seems to be called the 'general Leibniz rule'.

@yunoewig3095 4 жыл бұрын

@@pbj4184 It isn't called that way, but personally I think Leibniz-Newton rule would be a fitting name, since it combines Leibniz product rule with Newton's binomial rule.

@pbj4184 4 жыл бұрын

en.m.wikipedia.org/wiki/General_Leibniz_rule

@bttfish 4 жыл бұрын

How did the mathematician come up with such complicated but actually working constructions of sums and functions?

@brianjesusdiaz 4 жыл бұрын

To be fair, Niven's ideas build from Cartwright's proof and, hence, Hermite's, which date within a century before Niven.

@wesleydeng71 4 жыл бұрын

Try something to see if it works. If not, try another. Sometimes something works out.

@bttfish 4 жыл бұрын

Wesley Deng There must be some underlying theory otherwise it is even impossible to get started

@FareSkwareGamesFSG 4 жыл бұрын

I have a math question I did some research about, but came to no satisfying answer. I was wondering if anyone would be able to tackle it: Are there 3 distinct right triangles, ΔΑ, ΔΒ, ΔC, with sides of positive integer length and equally sized hypotenuses such that the sum of the areas of two of the triangles equals the area of the third?

@komis5555 3 жыл бұрын

Даже не так . К меня электронный ставочек , совпадение большое .

@tcmxiyw 3 жыл бұрын

Ivan Niven was a master. His book “Irrational Numbers” is a joy. It includes proofs of the transcendence of both e and pi. He also published a paper on the transcendence of pi in the American Mathematical Monthly (< 3 pages). He has many interesting articles in the AMM.

@usernameisamyth 4 жыл бұрын

Nice proof.

@elvidri 3 жыл бұрын

Let me bring a suggested problem on number theory: Let's say a number n is "twisted" if n is divisible by 4, n+1 is divisible by 5 and n+2 is divisible by 6. How many "twisted" numbers exist less than 2019? Thanks.

@MrSimmies 3 жыл бұрын

Fermat found a proof for this but there wasn't room in the margin of his notebook to show us it!!!!

@gaeb-hd4lf 3 жыл бұрын

I really wonder how do people come up with this kind of problem setups and strategies. Seems so random and out of the blue at first...

@TheSimCaptain 7 ай бұрын

Yep, there's no point in having a conversation with Pi. It's just too irrational.

@tobiasgorgen7592 3 жыл бұрын

That is the first prove I ever learned for pi being irrational. But I never saw the original paper. That is something else ^^

@Qoow8e1deDgikQ9m3ZG 3 жыл бұрын

crazy proof !!!!

@felipelopes3171 6 ай бұрын

I don't know what's the deal with these irrationality of pi proofs. If you know continued fractions it's very easy to prove. Even the simplest series for pi, the Leibniz series, can be converted into a continued fraction with Euler's formula, and when you write an infinite continued fraction involving rational numbers, that's it. You proved pi irrational.

@SlidellRobotics 3 жыл бұрын

C'mon, if you have lower case and upper case P as different functions, please make them look more different.

@tmpqtyutmpqty4733 3 жыл бұрын

naa

@mijmijrm 3 жыл бұрын

pi is irrational? .. i'd go further .. pi is irresponsible!

@samtcmu Жыл бұрын

This proof felt very rushed. I would love to see this proof done at a pace and depth of a typical video on this channel. Also, this proof is somewhat unsatisfying (though correct). I wish there was a more elementary proof. Perhaps something similar to the argument that e is irrational provided here: kzfaq.info/get/bejne/etVxlaRn177Wp3U.html

@route66math77 3 жыл бұрын

Ahhh very neat proof. Thanks for sharing!

@Nameless.Individual 4 жыл бұрын

We can also go by the contradiction that the trigonometric functions have an irrational value at 0 and pi, but that would nullify their geometric representations.

@JM-us3fr 3 жыл бұрын

This was the first proof of pi’s irrationality I ever saw, quite a while back. I’m still wondering what the motivation was behind each of these steps

@gavasiarobinssson5108 3 жыл бұрын

My guess is that the motivation came from the one and only place pi is used in the proof? At the end?

@skipii1234 4 жыл бұрын

I'm sad there are no backflips anymore 😭

@xriccardo1831 4 жыл бұрын

Same, i loved them

@twistedsector 4 жыл бұрын

asnwer=1

@zvonimirbolotinski652 3 жыл бұрын

Can you also prove that pi is transcendental?!

@anushrao882 4 жыл бұрын

Are you going to participate in #MegaFavNumbers ? I know this is copied but I genuinely want to know.

@mza3764 3 жыл бұрын

Pi is actually transcendantal ( not a solution to any natural polynomial ) look at von Lindemann and Liouville proofs

@ezequielangelucci1263 3 жыл бұрын

and? it doesnt mean it cant be irrational

@mza3764 3 жыл бұрын

@@ezequielangelucci1263 in english actually has the meaning of "in addition" too my friend Ezequiel and "in addition" is used to add new information to the reader/listener so by itself is not a questioning statement or an opinion.

@mza3764 3 жыл бұрын

@@ezequielangelucci1263 and in mathematics being transcendental, is like being ultra irrationnal which is irrational by extension.

@ezequielangelucci1263 3 жыл бұрын

@@mza3764 oh, well you a re right

@klementhajrullaj1222 2 жыл бұрын

Pi is equal cubes roots of thirty one! 😀😉 ...

@killermakd2015 4 жыл бұрын

how did he even think of proving like this? that to only in page......!!!!!!!!!!!!!!!!!!!!!

@IanKjos Жыл бұрын

Allegedly the ancient Greeks knew pi was not the ratio of any two integers, and they didn't have calculus as we know it. Is a non-calculus proof accessible to the general public?

@matiasgarciacasas558 9 ай бұрын

Did they really know that? It's hard to believe there's a geometric proof of it, which is all they did basically.

@ngc-fo5te 5 ай бұрын

There is no evidence they knew that.