Erdos' irrational numbers

[@tokahuke]

Neat! The generalized result can be used to prove that both e and pi are irrational, as corollaries. For e, it's just the classical proof. For pi, consider the expansion of sin(x) where x in Q. It is guaranteed to be irrational by the above prove. However, sin(pi/6) = 1/2 is rational. Therefore pi/6 cannot be rational and therefore pi is irrational.

[@tracyh5751]

3:58 Michael says (implicitly) that 0 is a natural number. Nature is healing.

[@user-ne1nw6hw2q]

I suspect that the "homework" fact that c is necessary less than a_N is not provable. In fact, we can just choose large enough prime to make this condition satisfied, but it wouldn't work for any p > c

[@Jack_Callcott_AU]

This is very helpful to me. I have been struggling to prove his homework exercise statement.👍

[@skylardeslypere9909]

The contradiction seemed random to me, so here's another way of seeing it. If we take any prime p greater than c, we conclude that p = 2. So the only prime greater than c is 2. But that's obviously not possible.

[@wesleydeng71]

10:25 If a_n+1>2, then 0

[@GiornoYoshikage]

This is the first thought I got after looking at example with `e`. If series sum is `b/c` then multiply by c! and look at series remainder from `N+1` to infinity where N is the greatest `n` such what a_n ≤ c. The remainder is in (0; 1) but (b/c)*c! is an integer. That's the contradiction

[@yoav613]

There should be a new channel: "michael penn's Homework solutions".

[@illumexhisoka6181]

Yeh they are often very hard

[@Macieks300]

13:50 p>c actually comes from the definition of p at 2:48. The homework exercise was to prove c

[@eauna0029]

Since p=p. If a_N were =N we'd have N

[@eauna0029]

Additionally, we do use this fact, when we say a_N!•b/c is an integer. This would actually work with just a_N>=c, which is always true even with p=c+1, so I imagine that was the true exercise (?)

[@Macieks300]

@@eauna0029 That's true that he uses it at 7:45, I forgot about that part. But your proof of c

[@icew0lf98]

@@eauna0029 we have that p=

[@nonameAccountable]

@@Macieks300 a_N can be equal to c and everything would work the same, so c

[@BlackIntegral]

This proof is too fast it feels sketchy. Like at 12:56, for example, the claim that e^x/p must be an integer makes no sense to me from your reasoning. It might be better to write a(N+1)!/a(N)!=p*q (q is and integer) and say that since e^x/(p*q) is an integer, e^x must be a multiples of p*q, so e^x/p must be an integer.

[@anshumanagrawal346]

You don't need to do everything he does, a lot of it is redundant, once you get the inequality for R, on the one hand aN! R must be an integer on the other hand it's strictly positive and less than e/a_(N+1) which is less than 1 since e

[@nathanbraswell2484]

This is also somewhat easy to to prove by just waving your hands and shouting analysis jargon. basically, the rational numbers arent complete, and you end up trying to take an infinite product to find a common denominator. factorial is just there to make sure everything is convergent- {a_n} is bounded below by {n}, so the series is bounded above by e.

[@racheline_nya]

4:24 this doesn't work on its own, you didn't prove that a_0

[@____victor]

what if a_0 > p? so N+1=0 and there's no a_N. (I'm guessing it's not possible as it would contradict c

[@illumexhisoka6181]

Just change the sequence by putting a new a_0 that's smaller than p You just add a rational number on both sides so the rationality or the irrationality of the number doesn't change

[@illumexhisoka6181]

Did you find out how to show that c

[@anshumanagrawal346]

​@@illumexhisoka6181I don't think it can be shown, since it need not be true, I think in the construction you have to take p to be large enough

[@eauna0029]

​@@illumexhisoka6181 the bound is strict (

[@nonameAccountable]

I don't think that would ever be the case since you can always find a p great than both a_0 and c.

[@weltkaiserendzeit2417]

Is there a way to prove that the same sum is also transcendental ?

[@beaver3393]

It's an interesting follow up question, but i doubt there's already a way to Show it considering how little we know about transcendentals.

[@ianmassa8572]

I would guess it is in fact transcendental, because of the factorial we have a irrational that can be well approximated by rationals

[@zh84]

In general, I don't think so. Proving transcendality is very complicated. Perhaps the approach used for Liouville numbers could help, but that's beyond my ability.

[@Convergant]

Proving something is transcendental is usually an absolute nightmare. I wouldn't bank on it.

[@trueriver1950]

If there is, then I bet it starts by assuming bwoc that there exists a polynomial with that sum as a solution. So what contradiction might that throw up?

[@goodplacetostop2973]

14:12

[@Calcprof]

So what is your Erdos number? Mine is 2.

[@TheEternalVortex42]

Mine is e

[@zh84]

∞

[@Convergant]

Undefined

[@eauna0029]

2 is nuts, i wish i were able to do papers and no less with only 2 degrees to Erdős

[@kokainum]

5:11 What if a_0 >= p ? Looks like separate case or it needs a proof it can't be the case. I guess it must be similar to the "homework" part, though I still need to find contradiction for that. 11:34 So integer divided by integer is always integer? I don't think it works that way. xD Maybe it is true in this case but I don't see you reasoning. R times (a_N)! is integer, got it, but R itself is an integer? It looks like insinuation that R is R times (a_N)! divided by (a_N)! so integer by integer but it only means R is rational. Idk if I'm missing something but it seems wrong reasoning. However you can show that e^x is integer (because a_(N+1)!/a_N! is integer) so e^x = 2 and maybe that can be used later. Again maybe I'm looking at it wrong because of the late hour but it seems like it's falling apart. In general the proof seems a bit too chaotic in some places at best. Still thank you for content and something to think about.

[@nonameAccountable]

What if a_0 >= p ? I don't think that would ever be the case since you can always find a p great than both a_0 and c.

[@kokainum]

​@@nonameAccountableI guess you're right because p wasn't chosen to be the smallest so we can add this property. But again, it should be mentioned right away or at least later so we don't need to go back and check properties we have or think if we can add properties. That's what I meant when I said this proof is a bit lacking and chaotic. I suppose he doesn't make it up live and had it all prepared before making a video so the reasoning should be clear.

[@user-ne1nw6hw2q]

The proof is fine, the author of this channel just didn't tell it properly (which suggests he doesn't fully understand it neither) "Homework" part doesn't seem to follow from anything, but we are free to choose large enough p so that the condition is satisfied. > What if a_0 >= p You can always consider a_0 to be 0, because it doesn't change the fact if the sum is rational or not (you simply add a 1 to the whole sum). > So integer divided by integer is always integer? It's integer because it's the difference of two integers from the equality above

[@d-nize]

What I like on proof by contradiction is that you never know where the road ends. For me it was a surprise that p=2. -> contradiction I assumed something like 0 < integer < 1 or natural number < 0. Very nice.

[@andrewporter1868]

What species of rational number xy satisfies xy = a^n - 1 for natural x, y, a, and integer n? I'm trying to solve the issue of frac(a^n b) = 0 which is to say I'm trying to compute the multiplicative order of x or y modulo a; or alternatively, I'm looking for that function which enumerates the factors of z if z = a^n b. This is equivalent to determining what xy is congruent to -1 modulo a since some a^n - 1 in base a is a contiguous sequence of digits of value a - 1, but as I show below, this is digit-shifted by m (defined below) given our factor of a^m; and what this equation is saying is that either x or y added onto itself by the number of times given by the adjacent coefficient gives us some a^n - 1 since multiplication is repeated addition; therefore the first significant digit must be a number in the residue class x or y modulo a such that some kx or ky is congruent to -1 modulo a. Now I know that for a=2, there are indefinite solutions to this equation for arbitrary x or y, and in fact, the more general is a^m xy = a^n - 1, or alternatively xy = a^m (a^n - 1) given m is an integer, but high school math never taught anything useful beyond the basics that I can apply right now in my every day algorithms research, and I don't feel like broot feorcing Wolfram MathWorld to find the needle in the haystack theorems that can help me reason this out without reinventing the wheel on modular arithmetic. Any ideas that don't involve spending months self-teaching myself?

[@CTJ2619]

I went irrational trying to compute my Erdos number!

[@shruggzdastr8-facedclown]

...and how many peer-reviewed scientific/math proof papers have you published?

[@CTJ2619]

@@shruggzdastr8-facedclown 1 but not really math related

[@shruggzdastr8-facedclown]

@@CTJ2619: There's a video on one of Brady Haran's many YT channels (either Numberphile or Sixty Symbols -- can't remember which) on the actual Erdös Number, which is what I thought that you might've been referencing

[@nonameAccountable]

At 9:20 how is sum of 1/n! from n=0 to infinity greater than sum of a_N+1!/a_n! where n=N+1 to infinity?

[@user-uh9bo2im1h]

Bigger thing in the denominator

[@nonameAccountable]

​@user-uh9bo2im1h Take a_n = n and N =2 and a_N+1 = 3, and you will see that the inequality does not hold.

[@luggepytt]

I think you could prove this simpler by using a minimal tweak of the standard proof that e is irrational. Let’s denote the sum S, and assume that S is a rational number, b/c (b, c ∈ N) Then S×c = b, so S×c ∈ N Therefore, also S×c! ∈ N Now find aₙ such that aₙ ≤ c < aₙ₊₁ Split S into to parts S₁+S₂ where: S₁ = 1/a₀! + … + 1/aₙ! S₂ = 1/aₙ₊₁! + … From the above follows that (S₁+S₂)×c! ∈ N Prove that S₁×c! is an integer. (Easy!) Prove that 0 < S₂×c! < 1, thus not an integer (Homework! Hint: Prove that S₂×c! < ½ + ¼ + ⅛ + … = 1) Hence (S₁+S₂)×c! cannot be an integer, which contradicts the assumption.

[@nonameAccountable]

There may be an a_n that is not

[@riadsouissi]

This Note, S2 is equal to an integer plus something that can be shown to be between 0 and 1. Michael proof is not optimal. We don't need a prime p and we don't need c < a(n).

[@klausolekristiansen2960]

How do we know that an aN smaller than c exists?

[@giorgioleoni3471]

What if N=1, i. e. the sequence starts at something bigger than p?

[@andy-kg5fb]

Can sm1 tell the ans for the homework exercise

[@thanderhop1489]

This is easy if you construct things in a different order from how Michael did it and add an extra step. We suppose that the sum of reciprocal factorial a_n's is b/c with b,c natural numbers. Then find M so that c < a_M (possible since a_n is an increasing integer sequence). Then find an odd prime p such that a_M < p (possible since there are infinitely many primes). Then let N be the minimum integer with p

[@BlackIntegral]

@@thanderhop1489 Yeah. He didn’t say that p is the smallest prime larger than c. So it is kinda straightforward you can always find p and N satisfying that inequality.

[@assassin01620]

​@@thanderhop1489 What about C = 4 A_N = 3 P = 5 A_(N+1) = 6 Where N =1 We have: C < P A_N < P P < A_(N+1) An >= n for all n However C < A_N < P < A_(N+1) doesn't hold Why is this wrong? What am I not understanding??

[@ivanlazaro7444]

Just consider a whole sequence of a_n's with the properties given. Then, is it possible to find N, c and p satisfying this properties wrt the sequence. Of course for some numbers in the sequence this conditions will not follow. Need to seek for them but it is proven that those exist by construction of the sequence

[@Happy_Abe]

@@assassin01620you’re right with your values, but we can choose things in a different order, instead of choosing p first, choose a_N so that it’s greater than c and there’s a prime p between a_N and a_(N+1), possibly equaling a_N+1. Such an N will always exist since for any c and any a_n greater than c, there’s a prime p above it, take the minimal j such that a_(j+1) is greater than or equal to that prime p and that j will be your N.

[@natepolidoro4565]

Woo hoo

[@xaxuser5033]

i m so disappointed with your reasoning every time i watch your videos, your carelessness regarding rigor leads to fatal mistakes in your videos, for example here the way you wrote down things, it s not necessarily that we have c

[@eauna0029]

Technically the bound should be a_N>=c, but a big enough p achieves the strict inequality too, so it's no bother to the proof

[@xaxuser5033]

@@eauna0029 a_N >=c is not necessarily either the way he presented it in the video

[@xaxuser5033]

@@eauna0029 and yes if p is large enough, you will get what you want, but not the way he did it. I'm talking about rigor nothing else.

[@bobbobson6867]

But if a(n)=1 for all n then Sum(1/a(n)!) = -1/2 😮