e*pi is irrational ? Proof via complex number by a 14 year old subscriber

For a better reason, see the pinned comment by Fematika.

e*pi is irrational ? Proof via complex number by a 14 year old subscriber, Ron.

blackpenredpen,
math for fun,

[Fematika]

Sorry to burst his bubble, but this proves that it can't be represented as any fraction of any real or complex numbers, not just integers. So not only would this prove that e * pi is irrational, but also that it isn't a real or complex number. The problem here is that in the complex numbers, (a^b)^c =/= a^(bc) necessarily, so some steps may be invalid without further explanation. This is also true with their free use of the natural logarithm as if it were the real-valued logarithm. It is also invalid to say that 2ai = 0 in that step, because the eponential function is not bijective. Anyways, I don't mean to discourage him, it's just that the proof is wrong.

[dhanarsantika]

e = π = 3
e * π = 9
It's rational according to engineers

[DaanSnqn]

Not what I was doing at that age. Despite of the mistakes, he is making a great start for becoming a good mathematician.

[gchtrivs]

Congratz to the 14-yo for being curious! There are several things to note.
1. In the proof, the assumption that both a and b are integers was never used. a and b might as well been complex numbers, and the proof would not change. Obviously, pi*e is expressible as pi*e/1, so we automatically get nonintegers a and b, which automatically shows that the proof presented is wrong, without us having to find a reason why it is wrong and flawed.
2. Complex exponentiation and logarithms are very intricate and manipulating them like that doesn't work as well as with positive real numbers. For example ln((-1)^2) is not the same as 2 ln(-1)=? 2 i pi. You have to be EXTREMELY careful with the details when dealing with complex exponentiation and logarithms, one of the reasons being that e^x is not even injective for complex numbers, which is also a mistake that was made in the proof( e^x = e^y does not imply x=y, though i think it does imply that x=y + 2pi)
3. Your explanation that the proof doesn't work because the implication does not go back( the two equations are not equivalent ), in my opinion, is just plain wrong. Take, for example simple proof by contradiction that (-3)^2 is positive. Assume that it's negative.(1) x=-3, (2) x^2 = (-1)(-1)*9 = 9 > 0, which contradicts our assumption that it is negative, thus it's positive( i know that proof by contradiction is not the best way to prove this fact, but i'm only using to illustrate a flaw in your thinking) . Notice how while (1) implies (2), (2) does not imply (1), but the proof works anyway, because that's just how proof by contradiction works. In a proof by contradiction, you only need implications to hold one way and lead to a falsehood. Assumption => consequence 1 => .... => consequence n => something obviously false, thus assumption = false.
4. Now knowing that your explanation is wrong, let me suggest my explanation why the proof is wrong. Its mistake most likely lies in, as I have explained before, complex exponentiation and logarithms. ((-1)^(eb))^2 does not equal to 1. Generally, ((-1)^(x))^2 does not equal to 1. To prove this, take x = 1/4, so (-1)^(x) = sqrt(i) ( maybe some plusminus signs, again, complex exponentiation is difficult :D ), and so in this case ((-1)^(x))^2 = (sqrt(i))^2 = i.

There are probably more mistakes in the proof, my bet all of them related to complex exponentials, but yeah, don't feel bad about this, I just wanted to clear some things up :))

P.S. Blackpenredpen, i'm very much enjoying your videos, keep bringing the GOOD content ;) Cheers

[paulgrimshaw]

A big clue that the proof is invalid is that at no point does it use the fact that a and b are integers.

[vlix]

One small thing is that the natural log function for complex values is multivalued

[anonymoushumanoid]

The proof may not be correct, but DANG this guy was creative! Sparks are going, and a bright future lies ahead of him! Congratulations on consuming enough knowledge to get where you are, Ron! And don't stop!

[我是-hu]

e^a =e^b does not mean a=b
e^(2(pi)i)=e^0
2(pi)i=0

[Zarunias]

He never used the fact that a and b have to be natural numbers. So he accidently proved that e*pi is not only irrational but also cannot be written as any fraction, not even as (e*pi)/1

[guitarheroprince]

When I was 14, all I did was fail in every single subject possible.

[Sam_on_YouTube]

1 reason I disagree with you on the flaw and 1 way to improve the proof (I'm assuming there is a flaw somewhere, but I'm not sold on what you said bein the flaw):

First, I don't have a ton of experience with mathematical proofs, but I have plenty of experience with analytoc philosophy including formal first order and modal logic. If A implies B and not B, then not A. You don't need A if and only if B to get from not B to not A. Going in one direction is sufficient.

Second, you can't conclude from 1=e^2ai that 0=2ai. Complex powers rotate vectors around the complex plain. They have multiple solutions. However, you can conclude from that that a is an integer multiple of pi, since e^2npii =1. So a=npi.

So epi=npi/b, so e=n/b, so e is rational.

This is still a contradiction.

I'm guessing the actual problem is of a similar nature to what I pointed out. What I pointed out doesn't hurt the proof, but I suspect something is in there that stems from the same issue with complex powers that will make this proof fall apart. I don't see what it is though.

Very nice work regardless, especially by a 14 year old.

[wiiwillRule]

This was such a joy to watch! A 14 year old attempting proofs... woah.

[thomashoffmann]

Proof by contradiction doesn't need to go back. You start from something and start concluding. If the result is wrong the starting point must be wrong. You can deduct a true sentence from a wrong one but not the other way around. Squaring a term might make it true but it will never make it wrong. So the statement that each line must be equivalent is wrong. Concluding in one direction is sufficient for a proof by contradiction.
A => B => ... =>C
If c is wrong, a must be wrong and ^a is then true

[rhversity]

Tbf I’m also 14 and i can also understand this stuff so i understand how he knows this stuff. He probably also has a huge interest in math and watched a lot of maths videos like me

[SlingerDomb]

comment section in this vid just help me much more understand complex numbers and expo log. thanks nerd.

[BrunoVisnadi]

There are a bunch of mistakes in the proof, but none of them is related to 'not being able to go back'.

[ericlizalde]

You cannot deal so happily with complex logs. You cannot pass the complex log from one side to another because it's not a function itself, it has multiple values

[brunoandrades]

proof that 2πi = 0

e^2πi = cos(2π) + isin(2π)
= 1 + i(0) /ln on both sides
2πi = 0

I know the mistake is obvious, but this bothered me a lot when I was 12

[staffehn]

NO, NO, NO, NO, NO (sry, no offense, but) the thing you point out is NOT the mistake!

First: You do NOT need to be able to "go back". The implication "e*pi is rational" => "e*pi=0" is totally enough to find out, that in fact e*pi cannnot have been irrational in the first place.

Second: To point out some ACTUAL mistake: The first one I’m seeing is that the proof uses "2 e b ln(i) = ln(i^(2 e b))". That kind of rule doesn't apply to the complex logarithm’s "principal value" of log(i), because the actual inverse of the complex exp function is not uniquely defined, but just defined modulo 2*n*pi*i. (Remember, the complex exp function is periodic in the imaginary direction, with period 2*pi*i !!).

Let me rephrase: If you (artificially) make the complex log function uniquely defined by only taking "principal values", that is: complex number results z with -pi < Im(z) <= pi, then the logarithm rules DO break, that is you DON’T have for all x, y, a:
a*ln(x) = ln(x^a)
or
ln(x)+ln(y)=ln(xy)


Third: There are a few more mistakes, but they all boil down to the same problem. For example, because complex exp is periodic, you cannot go from e^0=e^(2ai) to 0 = 2ai. Other than that I stopped checking this “proof” because I felt like I found enough flaws already.

Lastly: I would be happy, if you did a followup video, actually explaining these problems about complex exp and log.

For example featuring the counterexamples for the log rules on complex principal value logarithm that I found on the German Wikipedia:

(i) ln(-1) + ln(-1) = pi*i + pi*i = 2*pi*i =/= 0 = ln 1 = ln ((-1)*(-1))

(ii) 2*pi*i ln(e) = 2*pi*i =/= 0 = ln 1 = ln(e^(2*pi*i))

[rishabhdhiman]

I proved, 2pi=0 when I initially learned complex exponentiation. Complex logarithm is multivalued.