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.
right, but how did Fourriere come up with the initial equation for x? it didn't just pop out of thin air thanks to Fourriere's geniality, no. there was a thought process behind that, that led him to deliberately choose precisely this definition for x. The motivation was to analyse the difference between "e" itself (as a sum of all terms of its Taylor/Maclaurin series) and the partial sum of the same series, up to b-th term. then you scale up the difference by multiplying it with b!. and that's what should have been said explicitly in the video, imo: Why are you doing that? why are you multiplying by "b!"? It is to make both, the fraction a/b and the partial sum, integers. The partial sum is an integer bcs you're summing for n=0..b, so b > than all n, and b!=1*2*3*..*b, so b! is divisible by every integer smaller than b => every term of the partial sum is an integer. - so that's why he deliberately chose x to be specifically THAT formula. bcs it makes it easy for him to prove that x is an integer. the second part, x < 1, comes from the fact that factorials grow so quickly and i actually like how the video treats that part.
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
I like another fun and easy proof of that result: Let for any positive integer n be In=int(x^n×e^x dx)[0;1] By using integration by part, we can easily proove by induction that for all n there is some integers (a;b) such as In=ae+b We can also see that lim In=0 Let suppose that e is rational, so e=p/q with p and q being positive integers So for a positive integer n, In=a(p/q)+b=(ap+bq)/q as for all n In>0 we have ap+bq>0 so ap+bq>=1 (because ap+bq is an integer) Meaning for all n, In>1/q wich is absurd, because Lim In=0 So e is irrationnal.
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.
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.”_
Small inaccuracy at 3:15: 1/b doesn't have to be STRICTLY less than 1. It could be equal to 1. It make no difference in the proof (there already was a strict inequality in the chain). If 1/b=1 then b=1, that is, "e" would have to be an integer. It is well-known that 2
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.
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!
This proof isn't really by contradiction, it's easy to repair it so that it becomes a direct proof. Outline: consider the funtion f(b,x) = b!(x-\sum_{n=0}^b 1/n!) for b natural and x real numbers. Step 1: show that, if x=a/c is rational with c\b!, then f(b,a/c) is an integer. This implies, in particular, that, if x is rational, then there exists some b such that f(b,x) is an integer. Step 2: show that f(b,e)\in (0,1) for all b. The assertion then follows simply by contraposition.
using a contrapositive doesn't actually "fix" the use of contradiction imo, i think its part of how "not" is used in the definition of irrational that means all such proofs will necessarily be non-constructive
@@mmmmmmmmmmmmm You're right about such proofs not being constructive because of the negation, but that's not the point: the issue I'm raising is that, in this theorem and many others, you can logically arrive at the conclusion directly from the definitions and assumptions, whithout needing a roundabout way such as a "reductio ad absurdum" argument. (Other examples of this problem are common proofs of Euclid's infinte prime numbers theorem and Cantor's uncountable real numbers theorem that you find in textbooks and online videos.) I suspect that, because negation is so difficult to process in our minds, many people have a bias of defaulting to contradiction arguments whenever a negation appears somewhere, and thus you see many "false" proofs by contradiction that are confusing and totally unnecessary. To illustrate this point, I suggest a little exercise: write down the definition of finite set ("there is a natural number and a bijective function..."), then apply the negation operator to it, and really do the symbolic manipulations it requires to eventually arrive at the full-blown definition of infinte set. Finally, use this definition to write a direct proof of Euclid's theorem on infinte primes. It's quite mentally demanding...
Beautiful. I had to pause on all the summation manipulations before understanding them, and I'm going to have to watch a few more times to get the rest. (I'm about 70% on board with the inequality at 2:53.) Thanks for the concise, quality explanation.
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
So if you assume _e_ is rational, you can prove there is an integer greater than 0 and less than 1. And you can prove other things like 2 = 6, or Abraham Lincoln was a carrot.
if u assume e is rational, u find there there needs to be an integer between 0 and 1 since there is no integer between 0 and 1, we know the initial assumption that e is rational is false therefore e has to be the opposite of rational, which is irrational please think for 2 seconds before u comment
I was a math tutor for 5 years and ive gone about 2 years without actively tutoring the subject or learning it. Gotta say, its an attractive subject but some of this definitely went over my head. I need to sit down and do this by hand to understand it better
This is proof is by Joseph Fourier, and for me it is one of the proofs that I find not too difficult to follow, as compared to proving pi for example. Please do more videos on more famous proofs!
Sir can you make videos on conic sections including ellipse , parabola and hyperbola including its applications and also it's book for self study. Please sir.
3:00 I thought a^1 + a^2 + a^3 + ... = -1+ (1/(1-a)), because the summation is supposed to start at k=0 to use the expression 1/(1-a). For example, 1/2 + 1/4 + 1/8 + ... = -1 + (1/(1 - 0.5)) = -1 + 2 = 1. This seems like a big problem in your steps, but if that's the case, then we should have an easy route to the proof: 0 < -1 + 1/b, thus 1 < 1/b, thus b < 1, however, no such integer was chosen for b as b was supposed to be 1, 2, 3, ..., or etc.
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.
I wouldn't say 'the most beautiful' proof when you begin by defining a 'magical' weird expression as the one for x, which seems taken out of a black box. It would help to explain the intuition or logic behind such definition.
I retook a discrete maths exam a few days ago, and I saw that I answered correctly on almost all questions (it’s only 8 questions to begin with and a 60% to pass), but the last one stumped me. It was basically just to prove that a set P_n with a certain relation was a partial order and after that draw a Hasse diagram of the relations for P_3. My problem though was that I was going fine with the partial order proof, but when I got to the antisymmetry part I thought it was extremely trivial and too easy (because it would have resulted in concluding that we must have reflexivity, which had already been shown) , so I doubted myself whether I had misremembered, and that it is actually symmetry that’s a part of partially ordered sets (for this proof, that would have also been trivial but less). Suffice to say, I changed the proof and went onwards….
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.
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
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.
An issue with maniam I believe. Maybe he forgot to exit text or he called it and it didn’t exit properly. Otherwise, it’s extremely small to even see anyways
Here's a proof I found myself when I was in highschool (it was surely well know but i was proud at the time to figure out something by myself) Let for any integer n define In=int(x^n×e^x dx)[0;1] By using integration by part, we can easily proove by contradiction that: For any n, In=ae+b with a and b beings integers (but not constants). We also see that for any n, In>0 and that lim In=0 We suppose that e is rational, so we can write e=p/q with p and q being positives integers. So for any n, In=ap/q +b=(ap+bq)/q>0 so ap+bq>0 so ap+bq>=1 (because it is an integer) So for any n, In>1/q wich is absurd because lim In=0. So e is irrationnal.
Very fun video! Just wondering, before the video I tried it with e= a/b(assuming a is natural and b and integer, assuming you can't simplify a/b) and then did this: ln(e) = ln(a/b) 1 = ln(a) - ln(b) ln(b) + 1 = ln(a) a = e^(ln(b) + 1) a = e * e^(ln(b)) a = e*b if e is a whole number: you can simplify the fraction, which goes against assumption if e isn't a whole number: b is an integer--> e*b isn't an integer --> a isn't natural this seems easier than what he showed in the video. Is this proof faulty, or is the proof in the video just better for some reason? edit: minor spelling error Edit 2: Thanks for the replies! Kind strangers helped me figure out that 1. What I did in like a bajillion steps is a 2 step process 2. This does not disprove e being rational at all Conclusion: my proof went absolutely nowhere. I would delete this out of shame but this is a reminder that sometimes it's ok to be wrong. Also there's no such thing as a not stupid question in math. Every question is stupid. But you still have to ask those questions
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.
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)
@@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
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.
@@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.
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!
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.
But what about other properties of e. In standard calc classes e is introduced as a limit of (1+1/n)^n. So if proof that your e is solution of this limit is complex, you just moved the complexity to another part…
Similarily e^2 is also irrational. Same proof write e^2 = exp(2) = sum 2^n / n! except multiply the sum by (2b)! Now we have that e cannot be a root of polynimial bx^2 - a One more step in this direction and we have that e cannot be a root of ANY polynomial i.e. e is trancedental
It is extra funny, as the standard definitions, like e.g. lim (1+1/n)^n, all contain fractions....in contrast, e.g., to pi (while of course there are infinite fraction representations for pi)
These irrationality proofs are interesting, but I just have practically zero interest in "why 1+1 = 2." I'm perfectly willing to just say "because that's how we've defined our numbers." The symbol 2 represents that number which we get when we add the number we call 1 to itself. Full stop - that's good enough for me. That gets you all the integers. I think the first interesting stuff might appear when we started justifying our positional notation system for representing numbers larger than our base, but the DIGITS I will just take as "defined givens." Actually I wouldn't say the 1+1=2 thing the way I did above. I'd just say that the symbol 2 is defined as representing the "successor of 1." 1 is the successor of 0, and so on up to 9. Actually we could just define 10 as a SINGLE SYMBOL to be the successor of 9, so that reasoning gets us all the integers. The interesting stuff would have to do with how those >9 symbols related to positional notation and all. That is, "Why is 10 the SENSIBLE symbol to use for the successor of 9?" So, the video that would interest me would be one that covered the "math theoretic" aspects of justifying positional notation - I just bet there's some interesting stuff there.
what i don't get is how the sum of rational numbers(the taylor series of e) leads to a number that is irrational, like is'nt a rational plus a rational supposed to be rational?
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 + …
Actually, if you shift this proof by a constant (6), you will find that x is the integer between six and seven, namely thrembo. Hence by counter-contradiction, e is rational, most definitely.
Wouldn't it work to just say that e is rational bc you will get to 1 divided infinity so it's a number divided by anumber with infinity numbers and a irational number is tehnically a number that has rationality of 2 numbers with infinite digits
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
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.
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.
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
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?"
This might be a dumb question, but why do we need to establish that upper bound at the end? Why not just stop at the truncated factorial 1/n(n-1)… since this is not an integer already?
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
It's beautiful, yes, and easy to understand, but I hate that it all starts with that definition of x. I don't like it when a step depends on a certain genius, that makes it so that most people can't deduce that proof without the "epiphany" of that definition of x.
Infinity is what makes e irrational. Look carefully at the expansion definitions of e, and we simply have the sum of many many rational numbers which, by law, must give a rational number. Until it doesn't.
Since Absolute values make negative numbers into positive ones. (Ex: |-3| = 3) and i MIGHT be a negative number, then what if we take the Absolute value of i?
1:46 Can someone explain to me how that happened? How did the limit change from n=0 to n=b to n=b+1 to n=infinity? I haven't encountered it yet. It's intriguing though.
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.
There are integers greater than zero in less than one. They’re not really talked about much I know lots of them like for example, quantum physics that actually number if we convert into our real number we get 54/99 So yeah, you just forgot about them
Is the sum part in 1:24 really an INTEGER? Sum of b! over n! from just 0 to 3 wouldn't seemingly be an integer.. I mean N seemingly will be even sometimes, so 3/even-number makes it a non-integer.
0:39 As a part of your proof by contradiction, you assume it will lead to a contradiction. This sounds like confirmation bias. Imagine how e would feel that you’re gaslighting it into thinking it’s irrational.
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
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.
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)
I don't like these styles of proofs when group theory makes a proof like this trivial. I have fought many mathematicians over this! FIGHT ME! What does group theory say? Oversimplified: two elements of a group when combined via a binary operation always yield another member of said group. It should be this simple to prove/disprove group membership, and it is.
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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.
right, but how did Fourriere come up with the initial equation for x? it didn't just pop out of thin air thanks to Fourriere's geniality, no. there was a thought process behind that, that led him to deliberately choose precisely this definition for x. The motivation was to analyse the difference between "e" itself (as a sum of all terms of its Taylor/Maclaurin series) and the partial sum of the same series, up to b-th term. then you scale up the difference by multiplying it with b!. and that's what should have been said explicitly in the video, imo: Why are you doing that? why are you multiplying by "b!"? It is to make both, the fraction a/b and the partial sum, integers. The partial sum is an integer bcs you're summing for n=0..b, so b > than all n, and b!=1*2*3*..*b, so b! is divisible by every integer smaller than b => every term of the partial sum is an integer. - so that's why he deliberately chose x to be specifically THAT formula. bcs it makes it easy for him to prove that x is an integer. the second part, x < 1, comes from the fact that factorials grow so quickly and i actually like how the video treats that part.
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.
I like another fun and easy proof of that result:
Let for any positive integer n be
In=int(x^n×e^x dx)[0;1]
By using integration by part, we can easily proove by induction that for all n there is some integers (a;b) such as In=ae+b
We can also see that lim In=0
Let suppose that e is rational, so e=p/q with p and q being positive integers
So for a positive integer n, In=a(p/q)+b=(ap+bq)/q as for all n In>0 we have ap+bq>0 so ap+bq>=1 (because ap+bq is an integer)
Meaning for all n, In>1/q wich is absurd, because Lim In=0
So e is irrationnal.
I swear contradiction is in every proof of irrational numbers. I swear.
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.”_
Math is the best thing that humanity has ever accomplished.
ever*
@@FanitoFlaze thx
Meth*
@@triangle2517Bro
Reading and writing are up there too, even just having a standardized alphabet.
e is e-rational
Now explain your rationale.
e is e-rational-rational
e is e-rational-rational-rational
I is I-rational
Small inaccuracy at 3:15: 1/b doesn't have to be STRICTLY less than 1. It could be equal to 1. It make no difference in the proof (there already was a strict inequality in the chain). If 1/b=1 then b=1, that is, "e" would have to be an integer. It is well-known 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!
3:23 technically you haven't ruled out b=1 at this stage so that last < should be
But I think that can be proved that 2
This proof isn't really by contradiction, it's easy to repair it so that it becomes a direct proof.
Outline: consider the funtion f(b,x) = b!(x-\sum_{n=0}^b 1/n!) for b natural and x real numbers. Step 1: show that, if x=a/c is rational with c\b!, then f(b,a/c) is an integer. This implies, in particular, that, if x is rational, then there exists some b such that f(b,x) is an integer. Step 2: show that f(b,e)\in (0,1) for all b. The assertion then follows simply by contraposition.
:D
using a contrapositive doesn't actually "fix" the use of contradiction imo, i think its part of how "not" is used in the definition of irrational that means all such proofs will necessarily be non-constructive
@@mmmmmmmmmmmmm You're right about such proofs not being constructive because of the negation, but that's not the point: the issue I'm raising is that, in this theorem and many others, you can logically arrive at the conclusion directly from the definitions and assumptions, whithout needing a roundabout way such as a "reductio ad absurdum" argument.
(Other examples of this problem are common proofs of Euclid's infinte prime numbers theorem and Cantor's uncountable real numbers theorem that you find in textbooks and online videos.)
I suspect that, because negation is so difficult to process in our minds, many people have a bias of defaulting to contradiction arguments whenever a negation appears somewhere, and thus you see many "false" proofs by contradiction that are confusing and totally unnecessary.
To illustrate this point, I suggest a little exercise: write down the definition of finite set ("there is a natural number and a bijective function..."), then apply the negation operator to it, and really do the symbolic manipulations it requires to eventually arrive at the full-blown definition of infinte set. Finally, use this definition to write a direct proof of Euclid's theorem on infinte primes. It's quite mentally demanding...
I've honestly never seen a proof that e is irrational before, and now I'm surprised that the proof is so simple.
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.
Beautiful. I had to pause on all the summation manipulations before understanding them, and I'm going to have to watch a few more times to get the rest. (I'm about 70% on board with the inequality at 2:53.) Thanks for the concise, quality explanation.
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
So if you assume _e_ is rational, you can prove there is an integer greater than 0 and less than 1. And you can prove other things like 2 = 6, or Abraham Lincoln was a carrot.
if u assume e is rational, u find there there needs to be an integer between 0 and 1
since there is no integer between 0 and 1, we know the initial assumption that e is rational is false
therefore e has to be the opposite of rational, which is irrational
please think for 2 seconds before u comment
@vergilfan6818 What was wrong with his comment?
But what if we assume there IS an integer between 0 and 1?
Now prove e is transcendental.
I was a math tutor for 5 years and ive gone about 2 years without actively tutoring the subject or learning it. Gotta say, its an attractive subject but some of this definitely went over my head. I need to sit down and do this by hand to understand it better
This is proof is by Joseph Fourier, and for me it is one of the proofs that I find not too difficult to follow, as compared to proving pi for example.
Please do more videos on more famous proofs!
I recall e is the easiest of the bunch outside of roots, and pi has some funky integral with an otherwise similar argument.
I wouldn't exactly call it beautiful, since you have to invent a magic formula out of nowhere to accomplish it.
Okay, now prove it's transcendental. (I'll wait).
Yup. That would be a long-form video to say the least. :) E.g., math.colorado.edu/~rohi1040/expository/eistranscendental.pdf
Aujourd’hui il pleut beaucoup à Bordeaux. Mais cette démonstration m’a ouvert l’esprit et remonter le moral. Un grand merci
Sir can you make videos on conic sections including ellipse , parabola and hyperbola including its applications and also it's book for self study. Please sir.
3:00 I thought a^1 + a^2 + a^3 + ... = -1+ (1/(1-a)), because the summation is supposed to start at k=0 to use the expression 1/(1-a). For example, 1/2 + 1/4 + 1/8 + ... = -1 + (1/(1 - 0.5)) = -1 + 2 = 1. This seems like a big problem in your steps, but if that's the case, then we should have an easy route to the proof:
0 < -1 + 1/b, thus 1 < 1/b, thus b < 1, however, no such integer was chosen for b as b was supposed to be 1, 2, 3, ..., or etc.
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.
I wouldn't say 'the most beautiful' proof when you begin by defining a 'magical' weird expression as the one for x, which seems taken out of a black box. It would help to explain the intuition or logic behind such definition.
wdym? e=3=pi=sqrt(g)
found the engineer!
g = 10, sqrt(g) ≈ 3, pi = 3
Brit, he math guy.
I retook a discrete maths exam a few days ago, and I saw that I answered correctly on almost all questions (it’s only 8 questions to begin with and a 60% to pass), but the last one stumped me. It was basically just to prove that a set P_n with a certain relation was a partial order and after that draw a Hasse diagram of the relations for P_3. My problem though was that I was going fine with the partial order proof, but when I got to the antisymmetry part I thought it was extremely trivial and too easy (because it would have resulted in concluding that we must have reflexivity, which had already been shown) , so I doubted myself whether I had misremembered, and that it is actually symmetry that’s a part of partially ordered sets (for this proof, that would have also been trivial but less). Suffice to say, I changed the proof and went onwards….
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.
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.
Is there a single program that all these math youtubers use to illustrate these videos so well?
Jazyk programmirovannija «Python 3»
Why is there a very small written "e=a/b" in the middle of the screen starting at 1:00 minute? 😂
An issue with maniam I believe. Maybe he forgot to exit text or he called it and it didn’t exit properly. Otherwise, it’s extremely small to even see anyways
Here's a proof I found myself when I was in highschool (it was surely well know but i was proud at the time to figure out something by myself)
Let for any integer n define In=int(x^n×e^x dx)[0;1]
By using integration by part, we can easily proove by contradiction that:
For any n, In=ae+b with a and b beings integers (but not constants).
We also see that for any n, In>0 and that lim In=0
We suppose that e is rational, so we can write e=p/q with p and q being positives integers.
So for any n, In=ap/q +b=(ap+bq)/q>0 so ap+bq>0 so ap+bq>=1 (because it is an integer)
So for any n, In>1/q wich is absurd because lim In=0.
So e is irrationnal.
I like how direct the video was!
This looks pretty simple to follow, but the clever part is coming up with the equation for x out of thin air.
I am in love with this proof. Do you have something similar for the number "pi"?
Very fun video! Just wondering, before the video I tried it with e= a/b(assuming a is natural and b and integer, assuming you can't simplify a/b) and then did this:
ln(e) = ln(a/b)
1 = ln(a) - ln(b)
ln(b) + 1 = ln(a)
a = e^(ln(b) + 1)
a = e * e^(ln(b))
a = e*b
if e is a whole number: you can simplify the fraction, which goes against assumption
if e isn't a whole number: b is an integer--> e*b isn't an integer --> a isn't natural
this seems easier than what he showed in the video. Is this proof faulty, or is the proof in the video just better for some reason?
edit: minor spelling error
Edit 2: Thanks for the replies! Kind strangers helped me figure out that 1. What I did in like a bajillion steps is a 2 step process
2. This does not disprove e being rational at all
Conclusion: my proof went absolutely nowhere.
I would delete this out of shame but this is a reminder that sometimes it's ok to be wrong. Also there's no such thing as a not stupid question in math. Every question is stupid. But you still have to ask those questions
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
The graphics are looking a lot better. Great stuff Brian
3:10 wouldnt it be more correct to say that 1/b
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.
@@phiefer3yes but it's just that is should have been said, rather than skipping directly to it. It's obviously quite easy to show that 2 < e < 3
One plus one equals two;
by definition.
Because that's how we define Two.
This has become one of my favourite mathematical proofs. It feels so satisfying and unexpected
3:17 i am sorry but when did we assume/prove that b>1 ?
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.
3:01 Are you sure that the numerator is 1/b+1 and not 1? I always thought a geometric series has the value 1/1-q (|q|
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!
I love contradictory proofs
your feelings are irrational
Well, I suppose if you didn't love contradictory proofs you wouldn't have made that comment, so that tracks.
Awesome demonstration
2:50 you didn’t justify why you switched from
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
This video is proof that video proofs are much easier to follow than paper proofs.
But what about other properties of e. In standard calc classes e is introduced as a limit of (1+1/n)^n. So if proof that your e is solution of this limit is complex, you just moved the complexity to another part…
Similarily e^2 is also irrational.
Same proof write
e^2 = exp(2) = sum 2^n / n!
except multiply the sum by (2b)!
Now we have that e cannot be a root of polynimial bx^2 - a
One more step in this direction and we have that e cannot be a root of ANY polynomial i.e. e is trancedental
That is superb
It is extra funny, as the standard definitions, like e.g. lim (1+1/n)^n, all contain fractions....in contrast, e.g., to pi (while of course there are infinite fraction representations for pi)
These irrationality proofs are interesting, but I just have practically zero interest in "why 1+1 = 2." I'm perfectly willing to just say "because that's how we've defined our numbers." The symbol 2 represents that number which we get when we add the number we call 1 to itself. Full stop - that's good enough for me. That gets you all the integers. I think the first interesting stuff might appear when we started justifying our positional notation system for representing numbers larger than our base, but the DIGITS I will just take as "defined givens."
Actually I wouldn't say the 1+1=2 thing the way I did above. I'd just say that the symbol 2 is defined as representing the "successor of 1." 1 is the successor of 0, and so on up to 9. Actually we could just define 10 as a SINGLE SYMBOL to be the successor of 9, so that reasoning gets us all the integers. The interesting stuff would have to do with how those >9 symbols related to positional notation and all. That is, "Why is 10 the SENSIBLE symbol to use for the successor of 9?"
So, the video that would interest me would be one that covered the "math theoretic" aspects of justifying positional notation - I just bet there's some interesting stuff there.
what i don't get is how the sum of rational numbers(the taylor series of e) leads to a number that is irrational, like is'nt a rational plus a rational supposed to be rational?
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!
Actually, if you shift this proof by a constant (6), you will find that x is the integer between six and
seven, namely thrembo. Hence by counter-contradiction, e is rational, most definitely.
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
Proof that e is rational: to be rational there needs to exist an a and b such that e = a/b: a = e, b = 1; e = e/1
Question: is this what you do for your job? I finished watching your actuary video.
i didnt get it at 3:32 why cant b just be 1? why does 1/b have to be less than 1 just because b is a positive integer? b could also be equal to 1
You can dismiss this hypothesis because x is strictely less than 1/b, so even if 1/b = 1, x < 1 and x > 0 meaning x is not an integer
Wouldn't it work to just say that e is rational bc you will get to 1 divided infinity so it's a number divided by anumber with infinity numbers and a irational number is tehnically a number that has rationality of 2 numbers with infinite digits
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.
if all we know about b is that it's a positive integer, then b can be 1, making 1/b rational.
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
Beautiful? Maybe. The most beautiful? Far from it.
why is b not equal to 1 ?
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
Can you prove e is transcendental?
2:43 why can we use the symbol "
1:12 why is b clearly greather than n?
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
Yeah I didn't understand this at all
This might be a dumb question, but why do we need to establish that upper bound at the end? Why not just stop at the truncated factorial 1/n(n-1)… since this is not an integer already?
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!
Amazing proof
Is e transcendental? Why?
You really got to be careful when you take things for granite.
it's really a gniess proof, and a tuff one too, quite a marble of human ingenuity
incomplete but beautiful
Hello! Why in 2:51 you change that "less or equal to" into a strictly "less than"?
It's beautiful, yes, and easy to understand, but I hate that it all starts with that definition of x. I don't like it when a step depends on a certain genius, that makes it so that most people can't deduce that proof without the "epiphany" of that definition of x.
e keeps me up at night
Wait… but why would rational numbers add together to form a irrational number?
Infinity is what makes e irrational. Look carefully at the expansion definitions of e, and we simply have the sum of many many rational numbers which, by law, must give a rational number. Until it doesn't.
Vy vsie na zapade takie tupyje?
Proving that e is transcendental (transalgebraic) is much harder.
Since Absolute values make negative numbers into positive ones. (Ex: |-3| = 3) and i MIGHT be a negative number, then what if we take the Absolute value of i?
|i|=1
i isn't a negative number
Thanks!
I have just been listening to some Verdi and therefore would like to include music as one of mankind’s greatest achievements.
Hi! What solution for 1^x=0?
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
1:46 Can someone explain to me how that happened?
How did the limit change from n=0 to n=b to n=b+1 to n=infinity?
I haven't encountered it yet. It's intriguing though.
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.
But can there be e=a/b with a and b being *irrational*? That is, can e be the quotient of two irrational numbers?
Yes : e = e² / e
a = e² and b= e are both irational
the pi + e
Proof it is transcendental is trivial using the transcendental nature of pi and euler's identity. Isn't proving it is irrational kind of redundant?
I'm not comfortable with proof by contradictions, I'm a constructivist
there is no “constructive” proof that e is irrational 💀💀 like what would u even construct
couldnt you do e/1, or 2e/2 or ae/a, where a is any complex number
So cool
There are integers greater than zero in less than one. They’re not really talked about much I know lots of them like for example, quantum physics that actually number if we convert into our real number we get 54/99 So yeah, you just forgot about them
Is the sum part in 1:24 really an INTEGER? Sum of b! over n! from just 0 to 3 wouldn't seemingly be an integer.. I mean N seemingly will be even sometimes, so 3/even-number makes it a non-integer.
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
0:39 As a part of your proof by contradiction, you assume it will lead to a contradiction. This sounds like confirmation bias. Imagine how e would feel that you’re gaslighting it into thinking it’s irrational.
Is there an 'intuitive demonstration' as to why an infinite sum of rational quantities gives an irrational result?
You can always imagine Pi as 3 + 0.1 + 0.04 +... etc
Now prove that e is transcendental.
I am still waiting.
This proof seems to be inspired by the irrationality measure of e. Or maybe the other way around?
3:21 clearly you haven’t heard of theta prime
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.
Not BriTheMathGuy, it is BrainyMathGuy
It's not just irrational it's transcendental!
Good one!
i love maths
The Most Beautiful Proof I ever Seen
i differ
I would go no further than calling it a nice proof.
Why are those numbers alarmed? Are they stupid?
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
You technically assumed b>1 when saying 1/b
Nice!
It’s easier (more beautiful?) to look at b/a=e^(-1).
Taylor sum isn't definition, it's a property
1:45 why is the result of the infinite sum of b!/n! an integer and greater then 0 if the infinite sum of n to infinity is -1/12 ???
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)
sure buddy
That little tiny e=a/b in the center is very distracting
I don't like these styles of proofs when group theory makes a proof like this trivial. I have fought many mathematicians over this! FIGHT ME!
What does group theory say? Oversimplified: two elements of a group when combined via a binary operation always yield another member of said group. It should be this simple to prove/disprove group membership, and it is.