The Most Beautiful Proof 

I absolutely ADORE this proof. Can’t believe I could follow, and even spent time making sure of parts where you go too fast for me. However, the real genius is in picking x. Where does that come from, what prompts it? Honestly, when I try to understand that, I feel very stupid again. Any insights? Thank you.

​​@@212ntruesdale It's not genius, let alone "real genius". Genius is over 160 IQ.

It makes sense, though, since irrational is defined, pretty much, as "not rational". So, proofs of irrationality are, more or less, proofs of "not rationality". And the most natural way to tackle something of this nature is contradiction!

I don't know how you could prove a number can't be written as a fraction without proof by contradiction. We know rules about fractions and what is needed to be considered a fraction but what rules are there for irrational numbers? There aren't really any consustent ones so you kind of have to show it by just showing it's not a fraction.

Pretty much having an easy to define property and proving something doesn't have that property will always be contradiction. When we assume not, it gives us the power to use that property's easy definition.

Technically it’s a proof by negation not contradiction

Quote from Lara Alcock’s book “How to Think About Analysis”: _“Proofs by contradiction pop up a lot in work with irrational numbers, precisely because it is hard to work with irrationals directly. Effectively the thinking goes, ‘I know this number is going to be irrational, but rationals are easier to work with so let’s suppose it’s rational and show that something goes wrong’. This is exactly how proof by contradiction works.”_

ever*

@@FanitoFlaze thx

Meth*

@@triangle2517Bro

Reading and writing are up there too, even just having a standardized alphabet.

Now explain your rationale.

e is e-rational-rational

e is e-rational-rational-rational

I is I-rational

why do you think it is simple?

@@Fire_Axus A way to say “I’m so smart.”

@@Fire_Axus It’s definitely hard. But I may have found a mistake. Why is 1/b < 1? Since b is a positive integer, b could be 1, in which case, 1/b

@@212ntruesdale We have 0 < x < 1/b, so even if b were 1, we would get 0 < x < 1 which has no integer solutions.

@@jarige4489 No, actually. If b=1, then x 1, not just a + integer, like he said.

But I think that can be proved that 2

It's trivial that b can't be 1, because e isn't a integer.

@@CrimsonFlameRTR Yeah, that's what I said.

You're right about b=1, but there's already a strict inequality in the line (just to the right of the blue text) so even after correcting the mistake you pointed out, we still get x < 1.

@@olaf7441 That's what I said.

Sorry, I think I replied to the wrong comment after reading another one which pointed out the b=1 case but didn't mention the other strict inequality like you did!

Thank you very much 👏👏. I tried to understand what x was but the video is really not exhaustive and clear. I went to the comments hoping to find something and there you are😁 thank you

Who is Fourriere? When and where did he publish this proof?

​@@diegogamba2601 It's Fourier, the name of mathematician. Heard of Fourier Transforms? He did a lot on integration.

Since n > b ( the summation starts at b+1 ), of course that (b+1)(b+2)...n > (b+1)(b+1...(b+1) Or written more compact: (b+1)(b+2)...n > (b+1)^(n-b) (The power is n-b because n is supposed to be b + the number elements and so the number of elements in this product is (n - (b+1) +1), (the +1 because we count the inital element too) Think of how, how many number are in the sequence 3,4,5,...,17? Well it's 17 - 3 + 1 ( 17-3 = 14, but the doesn't take account for the inital 3 itself in the sequence since differences count distance between a number to another, not from a number to a number and so we add the +1) And since we are talking about reciprocals we invert the sign. That's the inequality that Brit shows

YES. Thanks, QG!@@quantumgaming9180

I recall e is the easiest of the bunch outside of roots, and pi has some funky integral with an otherwise similar argument.

found the engineer!

g = 10, sqrt(g) ≈ 3, pi = 3

It would be easier come up with the idea if we think of the entire infinite series and multiplying by b! first. e = 1 + 1/1! + 1/2! + 1/3! + … + 1/b! + 1/(b+1)! + … a/b*b! = b! + b!/1! + b!/2! + b!/3! + … + b!/b! + b!/(b+1)! + … a(b-1)! = b! + b!/1! + b!/2! + b!/3! + … + b!/b! + b!/(b+1)! + … The LHS is an integer. On the RHS, all the terms up to and including b!/b! are intergers, meaning the remaining terms must add to an integer. So we just call that whole thing x and investigate it. Hope this is more intuitive to think about.

Math is usually like that. Some miraculous choice for x appears in the proof and the rest is easy. Too be honest I really hate it too that people don't give a reasons for things like this, but eh, that's that. But when you are given a reason for something a proof, or even better find out thag hidden reason yourself, it sure feels fulfilling

@@quantumgaming9180 It can’t be random. Definitely intuition. Some people just know. We call them geniuses.

It is slightly more intuitive to come up with it with if you just looked at the infinite series of e first then multiplied by b! e = 1+1/1!+1/2!+1/3!+…+1/b!+1/(b+1)!+… a/b*b! = b!+b!/1!+b!/2!+b!/3!+…+b!/b!+b!/(b+1)!+… a*(b-1)! = b!+b!/1!+b!/2!+b!/3!+…+b!/b!+b!/(b+1)!+… The LHS is an integer. On the RHS, all the terms up to an including the b!/b! term are integers. Thus we know all the remaining terms must add to an integer. So we just call that x and then investigate it. I hope this helps.

One way of thinking about it is that the rationals are "spread out". If you pick the rationals which have denominator at most y, then the gap between any 2 of them is at least 1/y^2. If both have denominator of exactly y, they have a gap of some multiple of 1/y. The trick here involves finding a rational, namely sum( b!/n! for n=0 to b)/b! that is extremely close to e. Well it's more an infinite sequence of rationals that rapidly converges to e that you want. Namely sum( b!/n! for n=0 to b)/b! for b=1 to infinity. The x is just multiplying a rational of fixed denominator by that denominator. If you have some other number that can be approximated by rationals Really well, then it's also irrational. ie q=sum(10^(-q!) for q=1 to infinity) is also irrational.

a¹ + a² + a³ + ... = -1 + (1/(1-a)) if you reduce at the same denominator you will have a¹ + a² + a³ + ... = a / (1-a)

There’s no issue. Your sum is correct, but note that -1 + 1/(1 - a) = a / (1 - a). Plug in a = 1/(b + 1) and you get exactly what’s in the video. Where do you think the error is?

@@Deathranger999 I must have misread the denominator in the 4th expression from the left as 1 instead of 1/(b+1), then immediately looked down and did calculations while ignoring the rest of the video.

Yup. That would be a long-form video to say the least. :) E.g., math.colorado.edu/~rohi1040/expository/eistranscendental.pdf

The sum formula is a_1/(1 - q) where a_1 is the first term in the geometric series. The first term may be equal to 1 but it doesn't have to be. The infinite sum starts with n = b + 1 b!/(b + 1)! = b!/(b!*(b + 1)) = 1/(b + 1) So 1/(b + 1) is the first term in the series. Edit: If you want, you can start with S_n = a_1 * (q^n - 1)/(q - 1) which is the sum formula we use for finite geometric series. When the quotient is between -1 and 1 and n goes to infinity, the numerator goes to -1. So you get -a_1/(q - 1). Reverse the minus signs in both the numerator and denominator and you end up with a_1/(1 - q).

@@UmarAli-tq8pl I see what happened here. When we learned about the geometric series, we were told that it always starts at n=0, so your a_1 was always raised to the 0th power, so it was always 1. We didn't get to see what happens with n=1, we simply were supposed to say "Oh, that doesn't apply to the geometric series, we have to add the 0th term and then subtract it afterwards to make it work". But thanks a lot for your text, I just learned something!

There was already a strict inequality in that line, so saying 1/b

If b is one then a=e

​@@SteveThePsteroh yeah, thats true. even so, saying 1/b < 1 is kind of an unnecessary jump in reasoning that doesnt really align with what he was saying out loud. maybe im just nitpicking at this point

@@Memzys Because we already know that e is not an integer, therefore b >1.

your feelings are irrational

Well, I suppose if you didn't love contradictory proofs you wouldn't have made that comment, so that tracks.

b must be an integer. Since e is positive, we can assume WLOG that both a and b are positive. If b=1, then e=a which we know isn’t possible since e isn’t an integer. Thus, we know b>1

True, the final step of the proof would be proving that if b=1, then e would be an integer, and showing that e is bounded between 2 and 3, and thus not a possible integer.

ty guys now i understand

​@@Ninja20704You would have to give a separate proof of the fact that e isn't an integer im the case of b=1 I.e. that e is bounded by 2 and 3

@@quantumgaming9180 do we really need to though? The value of e is already known at least up to a couple decimal places so we already know it isn’t an integer.

Hey I could totally be wrong but I think the error in your proof is the line that says if e isn’t a whole number and b is an integer then e*b isn’t an integer. Let’s say e is 0.5 which isn’t a whole number and b is 4. 0.5*4 is still an integer. I could be misunderstanding the wording though.

@@lukeforestieri6322 That is true. Yea that could be the error. Thank you!

I think you've seem why your proof doesn't chain togheter from the previous comments. But I also want to say that you forgot that the asssumption is that e is rational, not an integer. Also you could've just multiply by b to get from e =a/b ==> a = eb. ( don't worry, I know the feeling of doing unnecessary steps in my calculations and proofs as well)

@@quantumgaming9180 LOL. This sums up how useless the proof is. I was onto nothing

@@pokerpoking3207 nice try though

Only when you are working with finite sums. When you allow infinite sums, you can sometimes get irrational values. Consider, for example, that pi = 4(1/1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ...), or four times the alternating sum of odd rational numbers between 0 and 1 inclusive, (-1)^k/(2k+1) from k=0 to infinity. Even though each individual, finite operation never produces any thing that isn't rational, the whole collection produces something that is irrational (indeed, it produces something _transcendental,_ which is even further away from rational numbers!) Essentially, the trick is that once you add infinity to the mix, a LOT of rules stop working correctly. Finite sums are always commutative and associative, but infinite sums are NOT always so. The problem is, these properties are only defined over finite operations, and when we try to extend their definition to infinite sets, we have to use some form of "limit" argument, and limits only work when you have certain properties like continuity. Infinite sums often do not have continuity, and thus some algebraic properties break down when you try to apply them to non-finite sums. (However, if the series in question is _absolutely_ convergent, then there's no problem. But the sum above is not absolutely convergent; the sum of the odd rational numbers between 0 and 1 diverges because the harmonic series diverges. You can see a similar effect with stuff like the alternating harmonic series, S=(-1)^(n+1)/n from n=0 to infinity, which equals ln(2), an irrational number. The harmonic series diverges, but the _alternating_ harmonic series converges. Since the alternating harmonic series does not have absolute convergence, we can't apply many of the nice algebraic properties of addition, like the associative property or the commutative property. Rationality is not preserved for (some of) these sums that are not absolutely convergent.

You're right. Rational + rational = rational ( in thr finite context). But when you talk about infinity all rules need to be rechecked. You've seen enough numerical examples from the previous comment, but herenis my favorite geometrical example: Take a square and a circle inside it that touches all 4 sides of the square. Fold the corners of the square just so the corners touch the circle. Repeat this process to the new smaller corners. In the limit you should have folded the square exactly like the circle! ( no corners, no nothing, pure curved) Here is an interesting question, what happend to the lenght of the sides of the square? Each time we fold the corners the total lenght is still 4 * L But you're telling me that the limit is the same as the circle so it's circumference should be 2*pi*(L/2) = pi*L ?! What?! The lenght stays the same every time so it's 4L, 4L, 4L, ... but the limit is pi*L??? (Informal answer to why is this happening if you want to know: the limit of the lenght of a curve doesn't necessarly mean is the same as the lenght of the limit of a curve) Crazy interesting stuff. If you want to know more, the subject is called Measure theory

Long story short: Every real number can be written as an infinite sum of rational numbers. In fact, this is exactly how our “decimal” system operates! For example: 3.14159… = 3 + 1/10 + 4/100 + 1/1,000 + 5/10,000 + 9/100,000 + …

@@quantumgaming9180I still remember the day I first learned about non-measurable sets, my mind was absolutely blown 🤯

Thank you, everyone, this has helped me understand irrationality a bit better!

but there are no integers between six and seven like there are no integer between 0 and 1

@@undecorateur There exists exactly one unique integer between 6 and 7 and its the integer known by mathematicians as thrembo.

@@undecorateurIt’s a joke from somewhere (possibly a TV show? Maybe just online? I don’t remember). Anyway, just ignore and move on.

@@Deathranger999 yes , i was really confused before It is a thing from youtube

this is a bad video to comment this on

@@Fire_Axuswhy?

it's really a gniess proof, and a tuff one too, quite a marble of human ingenuity

It is pretty obvious that b cannot be 1, because that would imply e=a, but we know very well that e isn’t an integer, so we can eliminate that possibility.

That's right. We would need to prove that e isn't an integer. I.e. thaf is bounded by 2 and 3

I don't quite understand what you mean by this? What do you mean by a irrational number is a number that has rationality of 2 numbers and has infinite digits?

@@quantumgaming9180 tehnically that is bc they have infinite nonrepeting digits so it should be a infinity long number divided by 10 to the power of infinity

Quite unrigorous but I see what you mean. Also can you explain exactly what you meant in your whole comment cuz I still don't understand

I'm unsure myself what he means, though I like the point about an irrational number being expressed as an infinitely large number of non repeating digits divided by an infinitely large power of 10

Counterexample to your reasoning: 1 = 1/2 + 1/4 + 1/8 + ... You can get to rational and even whole numbers even when terms in series approach infinity.

Yes you're right. Brit forgot the case where b = 1. But that is easy to solve since in this case e =a/b ==> e = a, whereas we have to prove that e is not an integer to conclude the proof. You just need to provr that e is bounded by 2 and 3 and you're done

The first sum is from 0 to infinity and you subtract the second sum which is from 0 to b. The initial common part cancels hence only the part from b+1 to infinity remains. At 1:46 the top left is just updated with the result arrived at in the middle.

@@Jester01 Ohh, so that's how it is. I didn't think of that. It was just a simple numerical operation. Thank you very much. Guess I will remember it forever.

Cause you still have to sum it, it could maybe converge to some integer (although it doesn’t you can’t be sure of it) and the easiest way to go forward is to find that 1 bounds x

@@rpfp4838 oh i overlooked the sum symbol completely, thx!

I doubt it

-infinity would be a solution if we were working in R U {-inf, +inf} but in just plain R there isn't a solution

​@@quantumgaming9180I don't think -inf would be a solution

@@Eye-vp5de oh, actually now that I think about it yeah

Yeah, I was gonna say, I didn't see why it became a strict inequality, because if it weren't strict, b=1 is valid. All other results are of course invalid.

Because when you look at the original

i differ

I would go no further than calling it a nice proof.

Yes : e = e² / e a = e² and b= e are both irational

Because the sum goes over all n from 0 to b.

Where?

When we define x at 0:35 we are assuming that b is greater than n from the summation from n = 0 up to b. If it weren't then the summation wouldn't been defined in the first place and neither would be x. Question for you: "How do we know that b > 0 as to make the summation well defined in the first place?"

@@quantumgaming9180because that’s what we assumed in the first place for contradiction

You can always imagine Pi as 3 + 0.1 + 0.04 +... etc

|i|=1 i isn't a negative number

Thanks!

I watched a video on that just yesterday 💀

I mean maybe but the article doesn't say how large SCP-033 is, it's redacted, but clearly just a single digit since the black boxes are one character wide, but it's unlikely to be between 0 and 1, since there's 8 other options

@@thevalarauka101 i remember somewhere it talking about how it’s an integer between 5 and 6, but the idea still stands.

Good one!

because the sum of n isn't really -1/12 ( -1/12 is ζ(-1) where ζ is the riemann zeta function, but that doesn't necessarily mean that 1+2+3..=-1/12) 1+2+3+...≠-1/12 if you use classic definition of infinite series, partial sums, limits... (it is true if you use Ramanujan summation , but Ramanujan summation isn't really a summation if you look at the definition)

How is it not an integer? By the definition of a factorial it has to be.

@@ngc-fo5te A factorial OVER a factorial isn't.

@@ngc-fo5te Wait.. it might always be an integer because all factorials after 1 are even numbers, so maybe that could work.

@@PhrontDoorwell, b>=n, so b!/n! Is infact an integer

​@@PhrontDoor b>=n so b!/n! is an integer For example 6!/3! = (6×5×4×3×2×1)/(3×2×1) = 6×5×4 = 120

Since we assumed that e = a/b with a and b integers and since e is not an integer b cannot be equal to 1 else we will have e = a, which would mean that e is an integer, which is not the case.

there is no “constructive” proof that e is irrational 💀💀 like what would u even construct

n! means n factorial, which means you take the integer n and keep multiplying it by smaller and smaller integers until you reach 1. So 4! is equal to 4x3x2x1 which is 24.

@@EdKolis ok thanks

gamma... isn't.... a number.... It's a function.... unless you're talking about the euler-mascheroni constant, which is irrational

@@kemcolian2001 gamma has not yet been proven rational or irrational yet. It is still an open problem

​@@kemcolian2001euler mascheroni is believed to be irrational, not yet proven

@@quantumgaming9180nope, it's been proven. there's multiple papers on this.

@@Ninja20704It has been proven actually. just look up "euler-mascheroni constant irrational" and you should find a couple papers.

No it’s not. If you’re speaking to the mistaken strict inequality at the end, that doesn’t actually matter since there’s a strict inequality earlier in the chain. So there’s a minor error in the proof, but it doesn’t require any extra work to handle.

Try it yourself. Find an infinite summation form of Pi and try to find a contradiction. Although I've never seend this done before

Factorial is always an integer Factorial of a bigger number divided by a factorial of smaller number is always an integer (can be proven trivially from the definition of a factorial as a product)

@@Eye-vp5de Thank you.

This proves indeed that e is not rational the assumption made at the beginning is that e = a/b we found a contradiction so this assumption is false so e ≠ a/b

if a number is a rational number then this number can be written as a quotient of two INTEGERS a and b must be integers but e isn't an integer so we cannot chose e as a and 1 as b