The Most Beautiful Proof
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- čas přidán 18. 02. 2024
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Are you fascinated by the enigmatic mathematical constant e? Ever wondered why it can't be written as a simple fraction? In this video, we'll dive into the elegant proof that demonstrates the irrational nature of e. Get ready to understand this mathematical marvel like never before!
#math #brithemathguy #mathematics
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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information. Viewers should always verify the information provided in this video by consulting other reliable sources.
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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.
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.
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!
e is e-rational
Now explain your rationale.
e is e-rational-rational
e is e-rational-rational-rational
I is I-rational
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.
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
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.
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
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
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
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.
Brit, he math guy.
The graphics are looking a lot better. Great stuff Brian
This has become one of my favourite mathematical proofs. It feels so satisfying and unexpected
I like how direct the video was!
Awesome demonstration
wdym? e=3=pi=sqrt(g)
found the engineer!
g = 10, sqrt(g) ≈ 3, pi = 3
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.
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.
Hello! Why in 2:51 you change that "less or equal to" into a strictly "less than"?
I am in love with this proof. Do you have something similar for the number "pi"?
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.
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!
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 ???
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.
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.
couldnt you do e/1, or 2e/2 or ae/a, where a is any complex number
Question: is this what you do for your job? I finished watching your actuary video.
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.
Amazing proof
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.
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.
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
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.
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)
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!
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
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.
Awesome
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.
Now prove e is transcendental.
I have just been listening to some Verdi and therefore would like to include music as one of mankind’s greatest achievements.
This video is proof that video proofs are much easier to follow than paper proofs.
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
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?"
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
So cool
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
Can you prove e is transcendental?
e keeps me up at night
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 really believe universe selects some people for its exploration . math enthusiasts are one of them !
this is a bad video to comment this on
@@Fire_Axuswhy?
Proving that e is transcendental (transalgebraic) is much harder.
Do the same with pi?
Try it yourself. Find an infinite summation form of Pi and try to find a contradiction. Although I've never seend this done before
Ok 0:12 and I understood why e is irrational, yeah it's super easy how could I not think of it.
wow
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
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
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
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
The Most Beautiful Proof I ever Seen
i differ
I would go no further than calling it a nice proof.
incomplete but beautiful
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
01:09 Why?
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.
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.
👍👍
What if b=1?
If b=1, then e=a, meaning e is an integer. But we know this isn’t possible because we know that e=2.718… which is definitely not an integer, so we can elimitae that possibility
@@Ninja20704 maybe so, but I still think the proof is incomplete if you don’t specify that
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’s easier (more beautiful?) to look at b/a=e^(-1).
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.
Is not it contradictional that sum of rational numbers is irrational?
It's not just irrational it's transcendental!
Good one!
WOW what a prooof
Taylor sum isn't definition, it's a property
It is certainly a proof. As for beautiful... Well that's a stretch
Pi is rational
22÷7=π
You technically assumed b>1 when saying 1/b
another way of being irrational
que tristeza...
Now... apply the same strategy to gamma. 😉 Clearly, we need more than contradiction to prove the irrationality of gamma, though, or it would have been accomplished already.
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.
sure buddy
a=e
b=1
This proves e is not an integer.. It doesn't prove that it isn't irrational.
First you have to prove that e is not an integer. Otherwise the proposed proof is obviously wrong.
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!
my friend :bro how u so intelligent ?
me who just copied ur method of proving random problems : i m born with natural powers
Where is beauty
e irrationality proofs tend to get very lengthy compared to this one. Although length doesn't imply beauty either. I think it's simplicity.
My favorite part about videos like this is going into the comments section and seeing all the people who clearly have no idea what they're talking about trying to "prove" the video wrong
a and b don't have to be positive for a/b to be rational.
Not necessarily, it just makes the proof easier for a video.
For cases where a or b were negative (or both of them), the proof only needs some redefining to still work.
When a is negative, the proof is actually the exact same (since the only place where a appears is in a(b-1)!, which is still an integer if a is negative.
When b is negative, you can just put a - wherever b appears in the proof to make it possitive (so instead of writing b!(e- sum from 0 to b of 1/n!), you write (-b)!(e-sum from 0 to -b of 1/n!))
Considering all these cases unnecessarily complicate things, its safe to assume a and b are possitive. Yes, the proof is technically incomplete, but it's easier that way.
@@fertrex7340 Yes, necessarily. -5/2 is rational and a and b can't both be positive.
"For cases where a or b were negative (or both of them), the proof only needs some redefining to still work," is another way of saying the proof doesn't work.
My point is your definition of a rational number is incorrect. All the negative rational numbers are not rational by your definition.
All the positive rational numbers are not rational by your definition.
Now all the positive rational numbers can be expressed as a/b where a, b ∈ Q+ ∪ {0}, and b ≠ 0.
But all the positive rational numbers can be expressed as a/b where a, b ∈ Q- ∪ {0}, and b ≠ 0.
That's my only point.
@@markgraham2312 "your definiton of a rational number is incorrect". Where did I define a rational number. Dude don't assume things where they weren't said.
In a proof where you use variables, you have to consider all posibilities (these being, in this case, pos/pos, pos/neg, neg/pos and neg/neg). I am aware that neg/neg is exactly the same as pos/pos, but I never said those two where the only forms of a rational number, you assumed that. The thing is, in video proof neg/neg doesn't work, and that's why I said the proof needs rewriting.
@@fertrex7340 First, don't call me dude. I have multiple degrees in mathematics, and not from Tech High.
Second, my definition of a rational number is correct. How else would you show that -5/2 is rational?
For a number to be rational it must be expressed as a/b where a, b ∈ Z, not Z+ and b ≠ 0.
I didn't assume anything and I don't do crack.
I KNOW the definition of a rational number and there are an infinite number of rational numbers < 0. With your definition, you can't get a rational number < 0.
Deal with that.
@@markgraham2312 multiple degrees yet you still can't show where I defined rational numbers. Also, you claim "my definition of a natural number is correct", when did I said anything about how you define them? Unlike you, I'm not jumping to conclusions from a simple CZcams comment. Yet you keep yapping about "my 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.
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.
No! That is not true. There is no such thing as an "Irrational Number"
Here's the proof:
Consider:
d = e - 27182818284590452353602874713526624977572...(n digits)/10^(n-1)
It follows that the difference d between e and the ratio of two integers can be made arbitrarily small by choosing a sufficiently large value of n. In the limit, as n tends to infinity d tends to zero.
The fatal flaw in the video is the restriction of the integers a and b to a finite number of digits when e is transcendental. Without this restriction x = infinity * Zero which is indeterminate.
The infinity concept causes far more trouble than it is worth and should be abandoned.
Proof by contradiction aka Reductio ad Absurdum is a valuable technique when used without error.
This is a proof of negation, not a proof by contradiction.
You mean contrapositive? Like in P-> Q ~Q-> ~P ?
It is a contradiction. He proved that if e is rational, there has to exist aan integer between 0 and 1.
Gee, how can we prove that there is no such integer, anyway?
A proof by contradiction goes like this: to prove “A”, we assume “not A” and derive a contradiction. This is inherently nonconstructive because it implicitly makes use of double-negation elimination.
A proof of negation goes like this: to prove not A, we assume A and derive a contradiction. For example, to prove “e is not rational”, we assume “e is rational” and show that it leads to an absurdity. This is constructive, because this is the definition of “not”.
@@zeyonaut are you saying that we can't assume that not not A implies A? That's a weird kind of logic that works that way, wonder what kind of practical use it might have?
@@EdKolis Absolutely not, classical logic is important! I’m just trying to correct a misconception around the term “proof by contradiction”, because it has a meaning more specific than any proof that involves proving false.
You go much too fast in the "reindexation" at 2:53. It took me longer than your video to convince myself that it was correct. It's the only unclear point imo.
the most incoherent proof
consider x is complex we prove that e is a complex rational value :))
wot
Eh, sorry.
It is that I saw the same proof from Mathologer, otherwise I wasn't able to follow your proof...
:o
e = (2*e)/2
2e isn't an integer