Irrational Root 2!

In this video, we introduce and prove the "Carpets Theorem" and then utilize the theorem to prove that both the square root of two and the square root of three are irrational by a visual infinite descent argument.
If you like this video, consider subscribing to the channel or consider buying me a coffee: www.buymeacoff.... Thanks!
This animation is based on an argument from Stanley Tennebaum. If you are interested, I recommend this article from John Conway and Joseph Shipman about the irrationality of square root of two (and others):
dev.mccme.ru/~m...
#irrationalnumbers #realnumbers​ #manim​ #math​ #mtbos​ ​ #animation​ #theorem​​ #visualproof​ #proof​ #iteachmath #mathematics #irrational #carpetstheorem #proofbycontradiction #root2 #algebra #infinitedescent
To learn more about animating with manim, check out:
manim.community

[@YouTube_username_not_found]

Infinite descent!

[@MathVisualProofs]

😀

[@attackoramic8361]

just like my life

[@O_79]

@@attackoramic8361xd

[@no-one-1]

@@attackoramic8361😀

[@hmkl6813]

First time its actually infinite

[@piecesofmathematics]

Beautiful proof! Thank you!

[@jan-timolobner]

I finally understood one of your videos. I might not understand every video, but I am determined to learn, so thank you for your hard work on these awesome videos.

[@MathVisualProofs]

Happy to hear that! Keep it up. It takes time and I do rely on other facts

[@alegoncalves472]

Beautiful to hear that!!! Hope to be like that as well mate 🤗🤗🤗🤗 greetings from Venezuela!!!

[@GiganFTW]

@@alegoncalves472shouldnt you be hunting for dogs in the street instead of trying to learn math from youtube?

[@alegoncalves472]

@@GiganFTW Math must be understood in order to hack the city mate 😁😂😂😂

[@Emily-fm7pt]

I appreciate the elegance of the proof! Especially the last part I feel is more intuitive than the usual “you can always divide by two,” because in this case it’s clear that you’ll eventually just run out of positive integers to work with, which causes the contradiction

[@Fire_Axus]

your feelings are irrational

[@aymangani5416]

@@Fire_Axusyour mass commenting is irrational… do something better with ur time lol

[@nargacugalover]

Luckily 2! is still 2. So the title is still right in a way

[@orisphera]

So r/unexpectedfactorial didn't change the meaning in this case

[@nargacugalover]

@@orisphera yep

[@ahmadmneimneh]

​@@orisphera r/ihavereddit

[@orisphera]

@@ahmadmneimneh r/ihaveihavereddit

[@ahmadmneimneh]

@@orisphera I shall summon the cursed subreddit, r/ihaveihaveihavereddit

[@holyek7892]

These videos make me so happy and mad. Where were they when I needed them in geometry and calculus? They make everything so clear and comprehensible.

[@Fire_Axus]

your feelings are irrational

[@Ninja20704]

I really like these geometric proofs because I was only ever taught algebraic proofs for proving this such as the classic proof, or other things like the FTA or rational zero theorem. Thank you for teaching me new things

[@maxborn7400]

before algebra, people specifically used geometric proofs. That's why it took a while before people accepted negative solutions to a quadratic formula, because it seemed "nonsense" to accept x = -2 for x^2 = 4.

[@dragansantrac4011]

Wow. One of the best proofs of irrationality of sqrt(2)

[@Fire_Axus]

no

[@bd8037]

@@Fire_Axusyes one of the best visual proofs. You are wrong

[@FLS96]

Nice proof, I've haven't seen a visual one of this before! If someone's still not getting it: If √2 were rational, there would have to exist (finite) integers a and b so that a/b = √2. a/b = √2, square both sides a²/b² = 2, multiply by b² a² = 2b² So proving the irrationality of √2 is equivalent to proving there is no square of an integer, that is double the square of another integer. The proof by contradiction in the video assumes the opposite. There are no finite a and b to satisfy these equations, so √2 has to be irrational. Correct me if I'm wrong, thanks!

[@hidude1354]

awesome! love how the well ordering axiom can prop up in fun ways

[@matthewbay1978]

I love that proof, proving root two is irrational is part of my favorite proof which proves you can raise an irrational number to an irrational number and get a rational number.

[@AndrewBalm]

Simple and intuitive. Amazing!

[@MathFromAlphaToOmega]

If you go in the other direction with increasing pairs, you'll get better and better approximations to sqrt(2). Those should correspond to solutions to Pell's equation a^2-2b^2=±1.

[@jessehammer123]

Well, as long as your original pair satisfies the Pell equation. If you start with, say, (8,5) instead of (7,5), the ratio tends to phi instead of sqrt(2).

[@MathFromAlphaToOmega]

@@jessehammer123 You won't always get solutions to Pell's equation, but the ratio should always approach sqrt(2). Assuming my calculations are right, (8,5) becomes (18,13), and then one more iteration gives (44,31). The ratio of those is pretty close to sqrt(2).

[@zytr0x108]

You're right! I wrote a Julia program to try it out: a = b = 1 sr2 = sqrt(2) while (a/b != sr2) a = a + 2*b b = a - b println("$a / $b = $(a/b)") end On my machine, the loop exits after 21 iterations when it gets to the maximum precision a Float64 can hold.

[@tamaz88]

This would have been a perfect video to be made into a loop, but looks like we didn’t get to see that

[@ThreadedNail]

Or, did you?

[@tamaz88]

@@ThreadedNail VSauce reference ❓❓❓

[@canyoupoop]

​And as always, thanks for watching​@@tamaz88

[@ThreadedNail]

@@tamaz88 its a math loop. So the reference fit.

[@jakobr_]

I feel like there’s a missing step where it’s explained that b is between a and a/2 so that the small carpets can overlap like that

[@bobh6728]

If the two smaller didn’t overlap, then a^2 wouldn’t equal 2b^2, because a^2 would just be the 2b^2 plus the uncovered area. So a^2 would be larger then 2b^2. So pointing that out missing.

[@Fire_Axus]

your feelings are irrational

[@tweep978]

This is the best explanation of this I've ever seen

[@orisphera]

I prefer the following instead of infinite descent: Take the smallest counterexample. By the step, there's a smaller one. This is a contradiction. You can note that this is essentially a special case of a reasoning that under a slightly different formulation is known as transfinite induction. It's similar to induction, but more powerful. It's also proven differently. Normal induction is proven from the definition of natural numbers (and 0) as ones reachable by the operations. (By “reachable”, I mean present in every set closed under them.) Transfinite induction is essentially the well-orderedness (existence of the smallest in every subset) of the natural numbers (or whatever set you're using), although for N it may be easier formulated as the normal induction over prefixes

[@noahblack914]

Took me a minute to figure out why a^2=2b^2 means √2 is rational. If √2 is rational, then there exists integers a, b where a/b = √2. Multiply both sides by b and then square them.

[@babusseus1105]

Cool proof!

[@Marcus-y1m]

This demonstration is a genius move

[@chixenlegjo]

Nice proof, but I’ll propose my personal favorite: The exponents in the prime factorization of a perfect square are always even. When you multiply this perfect square by a prime number, the exponent in that prime factor becomes odd. This means that you cannot multiply a perfect square by any prime number to get a perfect square. Therefore, the square root of any prime number is irrational.

[@MathVisualProofs]

A good one for sure. Hard to do it visually though :)

[@lenskihe]

Nice one!

[@SivaSiva-k5t]

All is well 🎉🎉🎉 ❤

[@ValkyRiver]

By the way, Mathologer also made a video with a similar trick on root3, root5, and root6

[@bluebow9347]

Magnificent!

[@narfharder]

For a long time something vaguely bothered me about this particular proof by contradiction, now I've finally put my finger on it. Thanks for the epiphany! (Surely this has been thought of before, I'm just an armchair amateur who wouldn't even know what book to read about it.) Consider this is really a proof that √2 is not rational. So, either "irrational" is merely defined as not rational, meaning we have only proven a contradiction by contradiction; or, if "irrational" has a positive definition, then we must first prove that "rational" and "irrational" are in fact mutually exclusive. Now I wonder if these two ways of defining "irrational" are themselves mutually exclusive - so, what obvious thing have I missed?

[@Ninja20704]

Irrational are only really definable as “not rational” so that automatically makes them mutually exclusive. There is no positive definition of irrationals because there is no specific form or anything that all irrational numbers fit/satisfy.

[@Emily-fm7pt]

The definition of rationality that’s most often applied is “a number is rational if it can be written as a ratio of two coprime (irreducible) integers,” and thus irrational is anything that cannot be represented under that system. Though, an important thing you start to realize as you go further into advanced math is that most big concepts have a bunch of different (provably) equivalent definitions, and it’s really just about picking which one is most convenient at the time. In this case, most proofs are based on showing that any ratio will be infinitely reducible, causing a contradiction, so this definition is favorable compared to something like a more set theory focused definition. If you are interested in this stuff, it’s never too late to just start learning math! Anyone with knowledge from Algebra 2 read most undergrad level math textbooks tbh

[@narfharder]

Well, I found the "book" I needed to fill my knowledge gap. en.wikipedia.org/wiki/Square_root_of_2#Constructive_proof gives a proof that √2 _is a number_ that cannot be rational, while all the proofs by contradiction only prove that _if it is a number,_ it cannot be a rational one - leaving it undecided whether √2∈ℝ.

[@sigmaoctantis5083]

@@narfharder I think you are making this more complicated than necessary: all common proofs of "the irrationality of √2" show that there is no rational number a/b whose square equals 2. They don't need any assumptions about whether √2 _is_ a number. This is a different story: since √2 naturally occurs as the length of a diagonal in a unit square, it should be a number, and therefore we are well-advised to extend the rational numbers to the reals or at least the algebraic numbers.

[@stephenliao63]

simplify sqrt to r a/b = r(2) for some a,b inZ+, gcd(a,b)=1 a=r(2)b aa=2bb lhs has factor 2 but there are two a so 4 | a Thus 4 | 2bb and this imply 2 | bb Which means 2 is factor of b So gcd(a,b) does not equal to one. Therefore, no such coprime integer pair (a,b) exists I think this is the same to the video but in different way of recursively decreasing a and b If we use the well ordered property of integer and state (a,b) is the least element satisfying the equation then we do not have to state the recursive part

[@TannerJ07]

It would be cool to show it for perfect squares so the difference is visible

[@erberlon]

It might have been helpful to explicit that such an infinite sequence cannot exist because the positive integers are a finite going down.

[@kexcz8276]

Wow, cool, yet I know I will forgot that lin like next 10 mins... 😂

[@tellusorbit]

There is a far simpler proof of the square root of two being irrational I used to demonstrate to my students which didn't involve either geometry or algebra. It was an indirect proof that used simple elementary school arithmetic. My students loved it.

[@iyraspusjfzifzocyoyxyoxyoxoy]

thanks, Vsauce!

[@MdTanvir-eh6so]

I love mathematics ❤❤

[@jimnewton4534]

why is (2b-a)

[@KDev-fq9iq]

Since a>b , say a=b+x Implies 2b -a = 2b-b-x= b-x Seemingly ,a > 2b-a Now , let's say a-b =b in that case a^2 = 4b^2 { a=2b} which is not our assumption as we considered a^2= 2b^2 Therefore , b has to be >1/2 of a Implying that a-b

[@dazai826]

b^2 < a^2 (talking only in natural numbers ) so b

[@rohitjohn6180]

What if instead of two, we try to prove sqrt(4) is irrational? Does it still work?

[@MathVisualProofs]

Can’t do it. Because 4 1x1 squares actually fit perfectly in a 2x2 square. No overlap so no carpets theorem infinite descent.

[@rohitjohn6180]

I see thanks

[@tdubmorris5757]

You forgot to explain how we get a^2 = 2b^2 which is confusing, I had to look it up √2 = a/b 2 = a^2/b^2 (power 2 both sides) 2b^2 = a^2 (multiply by b^2)

[@DanDart]

I'm interested to know if there's a proof not involving contradiction, or if that can't exist. I'm category theorying this real abstract stuff.

[@MathVisualProofs]

They all have to use contradiction I think-though some are better at hiding it :)

[@evdokimovm]

But (2b - a)^2 is not equals to 2(a - b)^2 if we open brackets. (2b - a)^2 = a^2 - 4ab + 4b^2 and 2(a - b)^2 = 2a^2 - 4ab + 2b^2

[@MathVisualProofs]

But remember that our assumption is that a^2=2b^2

[@evdokimovm]

​@@MathVisualProofs Oh, right! I forgot that. Now we get `2b^2 - 4ab + 4b^2` and `4b^2 - 4ab + 2b^2`. Sorry for another perhaps silly question, but does not the fact that they are equal mean that the sqrt(2) is rational?

[@JordanBeagle]

It's squares all the way down!

[@joaoneves4150]

You could've included this in the video: Sqrt(2) = a/b

[@MathVisualProofs]

Yes. Probably should have. It’s in the longer one linked.

[@jesusthroughmary]

a^2 = 2 * b^2 is an equivalent statement

[@99.googolplex.percent]

Everything means something. That's why this guy says "this means" so much.

[@slayeryt637]

Proof by contradiction :)

[@JustinChan-u3r]

How did you get the first equation (a^2=2b^2)?

[@MathVisualProofs]

if root(2) = a/b, then root(2)*b = a, so 2b^2=a^2.

[@iyraspusjfzifzocyoyxyoxyoxoy]

fractal imagery...

[@roberteospeedwagon3708]

why does this work for actual squares?

[@platonicgeometryportal5567]

You are showing a geometric sequence with an arithmetic proof?

[@David-bh7hs]

But it doesn’t? How can I suppose it?

[@lkytmryan]

I was just thinking That.

[@anisurrahaman4490]

But ....i think... the side length of a square may be irritationl...then it will hold

[@Cruizzerr]

Why can't such a list of decreasing positive integers exist?

[@JuliusDofarios]

Sorry, I do not understand. Never was Sqrt2 mentioned except the end and it was rushed. What are you proposing?

[@shilohmagic7173]

it's just a proof for why square root 2 is irrational, I think.

[@jesse_cole]

Quick, somebody explain this to Terrence Howard.

[@Rutherford_Sam]

I must've messed up somewhere. This feels right, but wrong.

[@BlackLegVinesmokeSanji]

It is actually right Root 2 is irshinal for this reson

[@john.john.johnny]

I truly wish I could understand this cuz I feel so close to understanding but anyway

[@MathVisualProofs]

Try the linked longer version. It’s slower and includes the carpets theorem.

[@brandon.m]

This is definitely a different proof for the sqrt(2) not being rational, but I would like to nitpick that you showed that if the sqrt(2) does exist, then it cannot be rational. You did not show sqrt(2) is irrational though since you did not show its existence.

[@deon6045]

So is 'a' the square root of 2, and 'b' is equal to 1? Without the context, it just feels like you threw a random ab equation at me.

[@MathVisualProofs]

I say “suppose that root 2 is rational so that there are integers a and b with … “ Root 2 is not an integer

[@deon6045]

​@@MathVisualProofs To be fair, I missed that, but I don't think you appreciate how little that explains to someone out of the know, and too out of practice to easily infer how you jump from assuming the root is rational, to the area of random squares. If I am taking a second guess... Are we dealing with squares, because both sides of the equation were squared so that we would have a solid 2, rather than 'a = b*(2^1/2)' ?

[@3_14pie]

turtles all the way down

[@JustRandomLights]

Too bad we don't have a perfect square that's exactly half of another perfect square, cause then we could just have their square roots as the numerator and denominator😞

[@suhnih4076]

😮

[@adamosburn754]

🤔🧐 A golden square…

[@nilsalmgren4492]

I think mosr people accept the fact that unless you are taking a square root of a squared number, the result is irrational.

[@mohammadyusufchaudhary7121]

Bro , √ 2 =

[@WilliamWizer-x3m]

this isn't a perfect proof. there's a similar, yet more complete, proof by requiring that the ratio a/b is in simple form. meaning a and b are coprimes. it can be proven that both must be even. which is a contradiction.

[@RunstarHomer]

What is missing from this proof?

[@hidude1354]

this is a perfectly fine proof. he hasn't formalized the ending but this is completely valid

[@GD-wg4yl]

My teacher lied...

[@nbecnbec]

I think it's more intuitive if you take a definition of rationality we are A and B are in lowest terms, similar to the classic proof

[@velmrok1660]

( (a-b) + (a-b) )² , you cant do 2(a-b)² xd

[@DrPillePalle]

Beautiful proof but you are going way too fast.

[@MathVisualProofs]

yes. hard to get in 60seconds. The linked video is longer and slower. I get to take my time there :)

[@rajulsomaiya]

There is a simpler way to prove

[@salerio61]

I don't like that as a proof

[@ytbvdshrtnr]

If √2 is rational it can be written as √2 = a/b This would make a² = 2b² a² would be odd if a is odd so a must be even. a² would be divisible by 4 if a is even so b² must be even, so b must be even. But if a and b are both even, a/b could be reduced to a fraction where one of them is odd, which wouldn't satisfy a² = 2b². Therefore, √2 can't be written as a/b.

[@Jivvi]

If 𝒂 and 𝒃 are the smallest integers for which this is true, then it follows that 𝒃 = 𝒂 - 𝒃 (which is true if 𝒂 = 2𝒃) and also that 𝒂 = 2𝒃 - 𝒂 (which is true if 𝒂 = 𝒃). All of this can only be true if 𝒂 = 𝒃 = 0, so the square root of 2 must be 0.

[@hidude1354]

how does it follow that b = a - b?

[@Jivvi]

@@hidude1354 If 𝒂 and 𝒃 are the smallest integers for which this is true, and it's also true for the integers (2𝒃 - 𝒂) and (𝒂 - 𝒃), those integers can't be _smaller_ than 𝒂 and 𝒃, because of the first premise, so they must be the same. It could also be possible that it's the other way around, so 𝒂 = 𝒂 - 𝒃, and 𝒃 = 2𝒃 - 𝒂. But the result is the same, because 𝒂 = 𝒂 - 𝒃 gives us 𝒃 = 0, and 𝒃 = 2𝒃 - 𝒂 gives us 𝒂 = 𝒃.

[@canyoupoop]

Damn.