Conway's "calculus" proof of the irrationality of the square root of 2 (and more).

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This video presents a way of proving the irrationality of the square root of 2 that makes no reference to primes, factorizations, or fractions being in lowest terms. I heard this from John H. Conway several years back, and it seemed too interesting to let go. (The fact that the square root of 2 is irrational comes just a few minutes in---but the rest of the video shows how far you can go with this idea!)

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4 окт 2024

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Комментарии : 114

@MohamedSayed-bk7fq 4 месяца назад

I've seen many proofs of the irrationality of root 2, but this one is just phenomenal, It's so simple yet it has such an outstanding outcome that it feels like anyone (even without calculus! although calculus does help) could've come up with it. And of course conway was the one to share it, Conway was such a great mathematician. may he rest in peace.

@frankswenton3177 4 месяца назад

Thanks, he was certainly one of a kind!

@derendohoda3891 4 месяца назад

My favorite nitpick of irrationality of sqrt(2) is that the proofs only show that if the square root of 2 exists then it isn't rational. But by leveraging limits it's easier for my nitpicking mind to grant that the existence of irrationals as a prior given. Super cool proof.

@martinepstein9826 4 месяца назад

That's a fair point. To show that sqrt(2) exists in the first place you need a bit of real analysis. For example, you can let c be the least upper bound of the set of real numbers x such that x^2 x^2 is continuous, so since 1^2 < 2 < 2^2 we know x^2 = 2 has a solution by the intermediate value theorem.

@xatnu 4 месяца назад

@@martinepstein9826I wish I knew more about p-adic norms but I think there is a totally different concept of sqrt(2) in those number systems, so it is a very fair critique.

@drakebaxter957 4 месяца назад

Someone send this to Terrence Howard

@ceramicsky14 4 месяца назад

He still wouldn’t agree because he thinks 1x1 = 2

@mke344 4 месяца назад

He wouldn't understand a single thing here

@danielbriggs991 4 месяца назад

Noo yoo he has to agree at 5:24

@Mathematica702 2 месяца назад

He would say a bunch of gibberish, but couldn’t do as much as differentiate a linear function.

@abstractbill 4 месяца назад

This is a really neat proof, thanks for taking the time to make the video!

@frankswenton3177 4 месяца назад

Thanks, it was just an offhand comment years ago, but it stuck with me, and I figured it deserved to be out there!

@Demki 4 месяца назад

Essentially what we've shown (in the general case with monic polynomials) is that the only integral elements of Q over Z are already in Z. namely that Z is integrally closed, which is quite an important property of Z (to some people)

@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ 4 месяца назад

What does "integral" here mean? The closest thing I know of in this context is an integral ring: a*b=0 => a=0 or b=0, but that doesn't fit what you're saying

@frogkabobs 4 месяца назад

@@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذFrom the Wikipedia article for “Integral element”: In commutative algebra, an element b of a commutative ring B is said to be integral over a subring A of B if b is a root of some monic polynomial over A. In this case we take B to be Q and A to be Z. Being an integral element is a rather valuable property in algebraic number theory.

@academyofuselessideas 4 месяца назад

John Conway doesn't cease to amaze me! So many cool results in so many different areas of math! (Matt Baker has a blog with some of other Conway's results... and yet they only scratch the surface of what Conway did) Conway had such an incredible insight for mathematics! This proof deserves more attention. Thanks for making this video!

@thbb1 4 месяца назад

my favorite leveraging the fundamental theorem of arithmetic: if a and b are integers and (a/b)^2 = 2, then I can decompose a and b as unique products of powers of prime numbers: a = 2^a1 * 3^a2 * 5^a3 ... and b = 2^b1 * 3^b2 * 5^b3... Then by the first equation: a^2 = 2 * b^2 and developing: 2^(2*a1) * 3^(2*a2) * 5^(2*a3) ... = 2^(2b1+1) * 3^(2b2) * 5^(2b3)... because of the unicity of prime decomposition, we must have a2=b2, a3=b3... AND, the catch : 2*a1 = 2*b1 + 1. But 2*a1 is even and 2*b1 + 1 is odd while both a1 and b1 are integers. This is impossible. QED. The nice thing about this proof is that it generalizes to all roots of numbers that are not integral powers of an integer, and even to all roots of rationals that are not integral powers of a rational: the rational nth roots are "quite rare" among rationals.

@Niv0505 4 месяца назад

I feel like this is the normal proof but a little more rigorous lol because it actually mentioned the TTOA that makes it work

@sabotagedgamerz 4 месяца назад

I loved this proof. It is a really ingenious way of proving this result, and a great use of just a little bit of calculus.

@ighao6032 4 месяца назад

Great proof! For some reason I cannot explain, I've always felt that the standard proof for the irrationality of sqrt(2) is kinda boring. This one is neat and generalizes in a simple, logical way. Beautiful

@임한준-d7v 4 месяца назад

Maybe we can use pigeonhole principle instead of calculus? From 4:35 0 < (sqrt(2)-1)^n = (p_n / q) < 1 for any n with p_n integer So 0 < p_n < q and p_n integer for any natural number n We can define function {natural number} -> {1,2,3,...,q-1}; n -> p_n Clearly this can not be injection. (Pigeonhole principle) So there exists i < j natural number s.t. p_i = p_j thus (sqrt(2)-1)^i = (p_i / q) = (p_j / q) = (sqrt(2)-1)^j hence, (sqrt(2)-1)^k = 1 for some k ( = j - i > 0) This contradicts the fact that sqrt(2)-1 is less than 1. Maybe this is also a valid proof? Also maybe this proof can be applied to the algebraic integers.

@frankswenton3177 4 месяца назад

Indeed, this would substitue for that spot! Once you've had the idea to subtract the nearest integer and take powers, there are apparently multiple ways to get the rationality to give you something. Would definitely work for the algebraic integers as well, nice!

@frankswenton3177 4 месяца назад

Oh, though one little thing! We can't eliminate 0, either, so you'd have to say that the difference is 0 or 1...but it still gets what you need! :)

@임한준-d7v 4 месяца назад

Oh, I forgot to thank you for presenting such a wonderful proof about the irrationality of √2 without involving prime numbers or lowest terms. I always found the method of 'assuming any rational number can be represented in lowest terms and then showing that √2 cannot be expressed in lowest terms' a bit awkward. It felt like I was just assuming and contradicting the same argument. But I couldn't think of the proof that does not involve the lowest term by myself. This new proof (it's new to me) really resolved my curiosity.

@임한준-d7v 4 месяца назад

@@frankswenton3177 You're right... Clearly I have mistakes there not considering 0... Anyway thanks for your reply and videos!

@frankswenton3177 4 месяца назад

@@임한준-d7vNo problem at all, great idea and thanks for watching!

@Misteribel 4 месяца назад

Perhaps the best explanation of Conway's algorithm I've seen. Your channel deserves more subs! You've got my vote :). (nit: I'd have mentioned as well that on each iteration, the next numerator is the previous numerator plus denominator)

@frankswenton3177 4 месяца назад

Yep---I was between discussing the estimates and not, didn't put a lot of thought into a more complete discussion there! :)

@matron9936 4 месяца назад

What a nice argument. It seems to also show that an of degree d integral over the integers real number x is rational if and only if it is integer, i.e. that the ring of integers of the rationals are the integers, as if x can be written as p/q, then Z[x]\{0} is contained in Z/q^{d-1}\{0} yet latter has positive distance from zero and former doesn't due to picking the closest integer a to x and having powers of a-x, which lie in Z[x]\{0}, converge to 0.

@gqip 4 месяца назад

Great video! Another viewpoint: limits and the topology of the reals must be part of number theory! As another comment mentions, this is a proof that Z is integrally closed. A similar result, namely that C is algebraically closed, also uses analysis in a crucial way, despite being a purely “algebraic” statement.

@frankswenton3177 4 месяца назад

Agreed to a great extent---though the transcendentals in R and C are the only inherently [metric] topological bits, which is where the topology is forced to come into play. All that Pure Algebra can see are the algebraics. :)

@myboatforacar 4 месяца назад

So sad we lost Conway to the pandemic, this is great :( Would have loved to see that algebra with logarithms you mention at 1:46 though... been a long time since I've done math and I'm rusty :)

@frankswenton3177 4 месяца назад

Agreed, 2020 was a bad year. :( The "logs" bit goes essentially like this: we first have to suppose that the distance z = | sqrt 2 - 1 | is not zero (if it is, we're done!). Then for any target size T (e.g., 1/q), we can solve z^n < T by taking the log of both sides: log(z^n) < log T, so n log z < log T, i.e., n > (log T)/(log z). So any power n bigger than this will work (careful, dividing by log z switches the < to > because log z id negative!). Easier to see on paper than in a text comment, but that's the outline!

@myboatforacar 4 месяца назад

@@frankswenton3177 above and beyond. Absolutely fantastic :)

@afm4711 4 месяца назад

This is indeed a very old result packaged in a new form. The approximating fractions are the continued fraction approximants. The fact that they are really good approximations goes back to Liouville. And it is very easy to prove that a number is rational precisely when the continued fraction expansion is finite. Lambert has used this to prove that pi is irrational.

@frankswenton3177 4 месяца назад

Not a single result in this is claimed to be new...the curiosity is in the method of proving them, which is (to me) interesting and, apparently, little-known.

@afm4711 4 месяца назад

@@frankswenton3177 as to „little known“: my engineering students all learn it in first year linear algebra. An a student just gave a presentation on Lambert‘s proof this morning in a seminar.

@frankswenton3177 4 месяца назад

I think there is a miscommunication here. As I said, none---zero---of the _results_ here are claimed to be new. The point is about the cute argument of subtracting the nearest integer and taking powers, that's all.

@Nakshatrasengupta 3 месяца назад

This is pretty cool!

@proxagonal5954 4 месяца назад

Bravo, John Conway

@thewoodland7599 4 месяца назад

That is a very good video. I learned a lot of interesting facts! Thank you for making this video.

@JL-zl6ot 4 месяца назад

incredible video. thanks for the clear explanation and love this pace!

@Czeckie 3 месяца назад

this is pretty neat way of showing that integers are integrally closed. im so rusty concerning algebraic number theory, but im sure this is not the usual way

@cauchym9883 4 месяца назад

Cool! This is a really nice proof!

@Patrickoliveirajf 3 месяца назад

I wonder how we can applie this method to the decomposition of rational matrices into a product of two interger matrices

@jamesknapp64 4 месяца назад

interesting concept. I think its a good exercise for student why this "doesn't" work with rational numbers. i.e. why you can't apply it to say 5/4 and say its irrational.

@frankswenton3177 4 месяца назад

Certainly!

@TheEternalVortex42 3 месяца назад

Another generalization of this is: consider any positive real x. If there is a sequence of distinct rationals p_n/q_n that approximate x arbitrarily well, then x is irrational. In the video you produce such a sequence by taking powers. But this can be applied to show that non-algebraic numbers are irrational as well.

@mudkip_074 3 месяца назад

I'm not sure I understand your comment, the sequence (1/2)+(1/n) (so p_n=n+2, q_n=2n)positive integer n approximates 1/2 arbitrarily well. Maybe you mean that if in any such sequence the denominators have to grow without bound, then the 'target number' is irrational (as otherwise you could keep using p/q itself and be as close as was needed). And for the second part, the rational number p/q is the root of the polynomial qx-p (though admittedly, this isn't monic), so any non-algebraic number is irrational. I assume I'm misinterpreting something here.

@joshwhite522 4 месяца назад

Great video!

@ryantamburrino3289 4 месяца назад

Love it! Thank you for sharing!

@xatnu 4 месяца назад

A small note: this proof relies on there being no integers between zero and one. This is an easy fix, and probably most easily shown by considering the powers of such an integer and using the Well-Ordering Principle. This is still a very elementary proof however!

@frankswenton3177 4 месяца назад

With the integers being ordered and consecutive integers being one unit apart, we don't have to work quite so hard for that, but it's a good thing to be concerned about!

@xatnu 4 месяца назад

@@frankswenton3177 I'm not so sure it is so trivial from an elementary standpoint! Who says numbers fit on a line? Who is to say a huge pair of positive and negative integers don't sum to a number in that so so important gap? That would be such a pain in the neck! Besides, the WOP is not very hard I think. It is a one or two-liner, and it is crucial to the proof in the video... What can I say, I like my axioms!

@xatnu 4 месяца назад

@@frankswenton3177 My thoughts arose mainly because prime factorisation relies on the WOP (at least in proving the GCD algorithm terminates)... I felt it had to sneak back in elsewhere! Otherwise, all of set theory might crumble and I won't stand for it 😁

@frankswenton3177 4 месяца назад

@@xatnuOh, I have nothing against WOP or even prime factorization. :) My thought is that because the integers are ordered, once you have consecutive integers 0 and 1 (say), every other integer is either greater than one or less than zero, so we can deduce there are none between. It seems like the construction of the integers from the natural numbers should allow an inductive proof of all of that, but I honestly haven't thought it through in detail!

@frankswenton3177 4 месяца назад

Come to think of it, induction is equivalent to the WOP for integers anyway, so with all of the suppressed induction, we're on the same page!

@vivan1310 4 месяца назад

can you please give the proof by induction for fact 2?

@frankswenton3177 4 месяца назад

Here goes for square roots (will be a little messy in a comment!): call z = sqrt(N) - A for convenience (the first example is N = 2, A = 1). We want to show that for n >= 1, z^n = k sqrt(N) + l for some integers k,l. Base case [n = 1]: take k = 1 and l = -1, then k sqrt(N) + l = sqrt(N) - 1 = z^1, done. Inductive step [true for n ==> true for n+1]: Suppose z^n = k sqrt(N) + l. Then z^(n+1) = z^n • z = (k sqrt(N) + l)(sqrt(N) - A). FOILing out, this gives kN - kA sqrt(N) + l sqrt(N) - lA = [-kA+l] sqrt(N) + [kN - lA], which is again an integer [-kA+l] times sqrt(N) plus an integer [kN - lA]. (These would be your "new" k and l.) Might be easier to follow copied onto paper, but that's how the induction could go!

@JOSHUVASRINATH 4 месяца назад

U can also use Z[√2] which is same

@frankswenton3177 4 месяца назад

There are lots of ways to prove this---the point is the method (using no algebra and not specific to 2).

@twelvethirteenyo 4 месяца назад

When you expand (rad(2)-1)^n=0 into k•rad(2)+j, and you haven't established the irrationality of rad(2) yet, how do you know that integer j doesn't contain additional multiples of rad(2) which may confound the proof?

@frankswenton3177 4 месяца назад

By definition of the square root, every even power comes out to an integer, and thus every odd power comes out to an integer times root 2 (an even power times one more root 2)...does that help? With that in mind, we can inductively FOIL stuff or use the binomial theorem and be assured that k root 2 plus j is all that we'll see.

@Bayerwaldler 4 месяца назад

We have to invoke an additional axiom which is not part of number theory though, the Archimedian property. But never mind, it’s a cool proof.

@deepdockproletarianarchive4539 4 месяца назад

Would this be considered algebraic number theory?

@ethanbottomley-mason8447 4 месяца назад

No, algebraic number theory is the study of number fields, I.e. finite extensions of Q (as well as some other fields like p-adic fields and related fields). In particular, of great interest are their rings of integers which encode very useful information about how to factor numbers if you allow more exotic factors. E.g. 6 = 2 * 3, but also 6 = (1+sqrt(-5))(1-sqrt(-5)). The contents of this video are more closely related to diophantine approximation.

@connorkearley7789 13 дней назад

Good video

@pauselab5569 Месяц назад

i still feel like the usual proof with gauss’s lemma is more intuitive, the result is much more algebraic than analytic

@frankswenton3177 Месяц назад

@@pauselab5569 Everyone's entitled to their opinion! The point of this proof is that it's elementary and analytic, though; most would consider these results to be algebraic, so algebraic proofs are rather expected.

@RandomPerson-nv3sm 4 месяца назад

I appreciate the video very much.

@frankswenton3177 4 месяца назад

Glad you enjoyed it!

@thechessplayer8328 4 месяца назад

Don’t need induction for fact #2. Just consider even and odd powers of root 2.

@frankswenton3177 4 месяца назад

Well, formally you need induction for all of that, as this is how.powers are defined---but all of it is obvious enough not to bother.

@thechessplayer8328 4 месяца назад

@@frankswenton3177 Formally you need induction for all proofs involving integers, rationals or reals as integers are defined by induction.

@thechessplayer8328 4 месяца назад

@@frankswenton3177 you need induction for everything involving integers, rationals, reals since natural numbers are defined through induction

@vnever9078 4 месяца назад

Great video bro

@aadityapratap007 4 месяца назад

Finally an easy one.

@ts4gv 4 месяца назад

that's pretty cool

@timwhite7127 4 месяца назад

I hope he didn't go to all that trouble on my account. I would have taken his word for it...

@danielc.martin 4 месяца назад

Great!

@양익서-g8j 4 месяца назад

개인적으론 아직도 귀류법이 왜 옳은지 확신이 없어요.말그대로

@glauberk9018 4 месяца назад

4 - 3 = 1 If root(2) and root(3) are rational (A/B)^4 - (C/D)^4 = 1 AD^4 - CB^4 = BD^4 But Fermat's Last Theorem was solved by Sir Andrew Wiles in 1993. Today we are certain that at least one of these numbers is irrational 😂

@tinafeyalien 4 месяца назад

Nice

@cantcommute 4 месяца назад

oh i know you from quora, been years lmao

@frankswenton3177 4 месяца назад

Ha! Yeah, I was pretty active a while back! Only pop in occasionally now. :)

@JOSHUVASRINATH 4 месяца назад

Nice but sound is not clear bro

@frankswenton3177 4 месяца назад

Hmm...volume up doesn't fix it? I usually leave the levels alone and figure people can turn it up to where they want it. I gave it a watch and it seemed ok, but if you can be more specific, maybe I can do something about it!

@frankswenton3177 4 месяца назад

@@JOSHUVASRINATH Thx---bizarre about the audio. I've checked it on two different phones and a computer with no problems. Is it just that the audio is soft, or something else?

@frankswenton3177 4 месяца назад

Have you tried a different device, by the way? I'm at a loss what might be going on here.

@JOSHUVASRINATH 4 месяца назад

Its too low

@JOSHUVASRINATH 4 месяца назад

It's working sorry for that

@bigbusiness3000 2 месяца назад

Where is the real world use for this information.

@youteubakount4449 4 месяца назад

I too can make wild statements about numbers that don't exist!

@__christopher__ 4 месяца назад

The proof in the video explicitly constructs a Cauchy sequence of rational numbers whose squares converge to 2 (the approximations mentioned at the end). By the construction of the real numbers from Cauchy sequences, the existence of the Cauchy sequence means the existence of the number it converges to.