Algebraic/Transcendence Theory

DHMO

02:08

Six exponentials theorem (6E):

Let $x_1,\ldots,x_m$ be complex numbers that are linearly independent.

Let $y_1,\ldots,y_n$ be complex numbers that are linearly independent.

Let $mn > m + n$.

Then, the set $\{e^{x_i y_j}\vert 1\le i\le m, 1\le j\le n\}$ contains at least one transcendental number.

DHMO

02:26

5. If $2^t$, $3^t$, and $5^t$ are all integers, then $t$ must be an integer.

Lemma: $\ln 2$, $\ln 3$, and $\ln 5$ are linearly independent over the rational numbers.

Equivalently, they are linearly independent over the integers.

Consider the equation $a\ln2+b\ln3+c\ln5=0$, with $a$, $b$, and $c$ being integers.

Applying the exponential function to both sides, we obtain $2^a 3^b 5^c = 1$.

The fundamental theorem of arithmetic tells us that $a$, $b$, and $c$ must be zero.

Therefore, $\ln 2$, $\ln 3$, and $\ln 5$ are linearly independent over the rational numbers.

Now, if $t$ is a rational number but not an integer, then $2^t$ cannot be an integer as well.

If $t$ is irrational, then $1$ and $t$ are linearly independent over the rational numbers.

Let $x_1=1$, $x_2=t$, $y_1=\ln 2$, $y_2=\ln 3$, $y_3=\ln 5$.

Confirm that $2 \times 3 > 2 + 3$.

Applying 6E, there is at least one transcendental number among $e^{\ln 2}$, $e^{\ln 3}$, $e^{\ln 5}$, $e^{t\ln 2}$, $e^{t\ln 3}$, $e^{t\ln 5}$.

That is, there is at least one transcendental number among $2$, $3$, $5$, $2^t$, $3^t$, $5^t$.

Which contradicts with the fact that they are all integers.

Therefore, $t$ must be an integer.

Secret

07:56

So, algebraic independence is a special case of linear independence where the coefficients are algebraic numbers from an algebraic field?

That is, the argument of the polynomial $x_i$ is the number that we are interested in checking for transcendence, while the $z_i$ are like the coefficients in a linear combination?

DHMO

08:21

@Secret I don't understand your question. I think I can clarify by an example.

$\pi$ and $i\pi$ are not algebraically independent, because they satisfy the polynomial $z_1^2 + z_2^2 = 0$

here $x_1 = \pi$ and $x_2 = i\pi$.

Also, algebraic independence implies linear independence

Secret

Ah I see, so two numbers are algebraically dependent if they satisfy the same nonzero polynomial, whereas they are algebraically independent if the only polynomial where they can satisfy together is the zero polynomial.

DHMO

yes, that's it

Secret

cool thanks

Anything else I am still digesting btw

Secret

08:45

For ln(-1) are we using the principal value given how ln(z) is multivalued?

DHMO

@Secret For algebraic $a$, any value of $\ln a$ is transcendental, as long as $a$ is neither zero nor one.

But sure, $i \pi$ is the principle value of $\ln(-1)$.

Secret

I see

@DHMO I am not really getting what g is doing, is it taking some kind of polynomial complex conjugate of the form $w\overline{w}$ thus it must become zero for imaginary arguments?

DHMO

@Secret yes, and in fact I was shocked when I learnt the proof that this is valid

the proof is that g(x)=f(ix)f(-ix) must be even since it is invariant under negation

therefore the imaginary terms all have coefficient 0

Secret

09:55

ok so that means, in the hierarchy of the exponential theorems, we have proved the ground level of the theorems, we then noticed that the unproven conjectures actually implies the proved theorems thus we are convinced that the conjectures have to be true?

(report the pics here so people know what we are talking about)

in Mathematics, 4 mins ago, by DHMO

DHMO

@Secret I think the story is that people tried to generalize theorems and made conjectures from the theorems

Secret

Hmm I see

I wonder if they will ever get to two exponential theorem and have that proved...

DHMO

Consider $x_1 = 1$, $y_1 = 2i\pi$, $y_2 = \ln 2$.
It is clear that $x_1$ is linearly independent.
Also, since $y_1$ is purely imaginary and $y_2$ is wholly real, they are also linearly independent.
However, $e^{x_1 y_1} = 1$ and $e^{x_1 y_2} = 2$.
Therefore, the two-exponential conjecture is false.

Secret

I see, always need to be mindful of counterexamples, lol

DHMO

@Secret Exercise: must $a$ and $e^a$ be linearly independent, whenever $a$ is non-zero?

Secret

10:26

Suppose $a$ is algebraic. By (L), $e^a$ is transcendental. Now consider the linear combination:

$$c_1a+c_2e^a=0$$
$$\frac{c_1}{-c_2}a=e^a$$

Since $a$ is algebraic and algebraic numbers are closed under $*$. Therefore, $\frac{c_1}{-c_2}$ is transcendental. This means at least one of $c_1$ or $c_2$ is transcendental. Therefore $a$ and $e^a$ are linearly independent over the algebraic numbers.

Now suppose $a$ is transcendental, then $e^a$ is unknown since there exists at least one $e^a$ whose transcendence is unknown ($e^e$). Suppose $e^a$ is algebraic, then using closusre of algebraic numb…

"can they be linearly dependent", sorry typo

DHMO

@Secret You have narrowed the question, but you still haven't answered the question.

Secret

what are we aimming the linearly independence over in this question, just the algebraic numbers or all of $\mathbb{C}$?

DHMO

@Secret Right. I forgot to mention that. Actually, I'm just interested in linear independence over the rational numbers.

From now on, whenever I say linear independence, I'm referring to the rational numbers.

Secret

12:02

For all nonzero $a$ algebraic, $e^a$ is transcendental by (L). Therefore $\{a,e^a\}$ are algebraically independent over $\mathbb{Q}$ and hence must be linearly independent over $\mathbb{Q}$.

For $a$ transcendental, for a simple approach to a counterexample, consider finding an $a$ such that $a=e^a$. This can be done by doing as follows:
$$ae^{-a}=1$$
$$-ae^{-a}=-1$$
Now the equation is in the form that can be solved by the Lambert W function:
$$-a=W(-1)$$
$$a=-W(-1)$$
which is a complex number. Since a solution is found, there exists at least one nonzero $a$ where $a$ and $e^a$ are linearl…

DHMO

@Secret brilliant. In addition, $x+e^x=0$ does have a real (transcendental) solution.

Also, you found a fixed point of the function $f(z) = e^z$.

Secret

More generally, $\{a,e^a\}$ are linearly independent over $\mathbb{Q}$ if given $\frac{c_2}{c_1}$ rational, $a=-W(\frac{c_2}{c_1})$ exists for the given $c_2,c_1$.

DHMO

Which is a result you proved in the main room as well as the third result in this room.

Secret

actually is the lambert W function defined over all $\mathbb{C}$?

DHMO

yes

Secret

12:15

my typo here: are linearly dependent over $\mathbb{Q}$

DHMO

that's a new result then

Secret

wait so that means $a$ and $e^a$ are always linearly dependent when transcendental? since the lambert W function exists for all rationals $\frac{c2}{c1}$??

DHMO

what do you think?

Secret

12:32

I guess that will mean given a transcendental number $s$. If $s=-W(K)$ has rational solutions $K$, then $e^s$ and $s$ will be linearly dependent over $\mathbb{Q}$. Since from a proof in the main room that algebraic numbers times transcendental numbers of the form $exp()$ must be transcendental, that means both $a$ and $e^a$ must be transcendental.

I am not sure how that will be helpful given W() is not an elementary function (perhaps there are theorems to find the rational solutions $x$ of W(x) given $y=W(x)$

But the implication for this is the following:

$$e=-W(R)$$

If there exists a rational $R$ such that the above equation is true, then $e^e$ is transcendental

DHMO

@Secret Consider the set $\{x \in \Bbb C \vert x\text{ and }e^x\text{are linearly dependent}\}$.

One can easily show that it is countable.

However, the set of transcendentals is uncountable.

Secret

so there's a contradiction, now to find out what went wrong with my proofs...

DHMO

Therefore, $a$ and $e^a$ are not always linearly dependent when they are both transcendental.

23 mins ago, by Secret

wait so that means $a$ and $e^a$ are always linearly dependent when transcendental? since the lambert W function exists for all rationals $\frac{c2}{c1}$??

The latter sentence is correct.

Secret

ok I see

DHMO

However, $a$ and $e^a$ being both transcendental does not mean that $a=-W(q)$ where $q \in \Bbb Q$.

Secret

12:47

Linear dependence: Nonzero $c_1,c_2 \in \mathbb{Q}$ (there cannot be just one of the $c_i$ be zero else the other must go to zero as the numbers are nonzero)
$$c_1a+c_2e^a=0$$
$$c_1a=-c_2e^a$$
$$-\frac{c_1}{c_2}ae^{-a}=1$$
$$-ae^{-a}=\frac{c_2}{c_1}$$
$$a=-W(q)$$

The only way I can think of that foils this is given two transcendentals $c_1$ and $c_2$, their ratio is a rational $q$, thus blocking the converse of the claim to be true always.

Sorry I mean in that case $a$ and $e^a$ can be linearly independent

DHMO

6. Let $a$ be a complex number. $a$ and $e^a$ are linearly dependent over the rational numbers iff there exists a rational number $r$ such that $a = -W(r)$

Proof: see above.

Secret

Well, I mean, having two transcendentals whose product or sum send them back into the algebraic numbers is strange enough (but true), would it be even more mind blending if they can go as far for their product or sum to send back to the rationals and that the two numbers in question are not some simple multiples of each other?

DHMO

@Secret If $t_1 t_2 = \dfrac p q$ for integers $p$ and $q$, then $t_1 = \dfrac p {q t_2}$.

Secret

Well in that case, both sides can be transcendental no problem then (although I am thinking about something analogous to 1=0.99999... or $e^{i\pi}$, that $t_1$ and $t_2$ may have a very unrelated notation, expression or something so that people don't immediately see they are rational multiples of each other).

DHMO

13:02

I don't have an answer for that.

Secret

I guess that's one reason why it is so hard to check $e^e$, $\pi e$ etc.

DHMO

@Secret It is noteworthy that S implies the transcendence of them both.

Secret

Ah I see

DHMO

13:18

7. Let $c$ and $d$ be transcendental numbers. Then, at least one of $c+d$ and $c \times d$ is transcendental

@Secret can you prove the above?

Sophie

14:14

@DHMO this is a good one

DHMO

@Sophie can you prove it?

Sophie

yes

DHMO

please prove it here then

Sophie

if both of $c+d$ and $cd$ were algebraic then the polynomial $(x-c)(x-d)$ has algebraic coefficients, but polynomials with algebraic coefficients have algebraic roots

Secret

Consider
$f(x)=(x-c)(x-d)=x^2-(c+d)x+cd=0$
Suppose the sums and products are both algebraic, then $f(x)$ is an algebraic polynomial with $c$ and $d$ as solutions, which by the definition of algebraic independence, $c$ and $d$ are algebraic numbers, a contradiction. Therefore at least one of the sum or products will be transcendental

DHMO

14:18

nice @Secret @Sophie

Balarka Sen

have we decided on proving anything nontrivial yet?

Secret

According to DHMO, result 6 is new but is shot down

DHMO

@Secret what does "shot down" mean?

@BalarkaSen well, do we have anything nontrivial yet?

Sophie

proving $e$ is transcendental with Lindeman-Weiestrass is kind of cheating

Balarka Sen

these results you quote seem like bookkeeping with the actually nontrivial theorems.

DHMO

14:22

I don't think I found 3 and 6 from a book

Secret

Well, I want to use result 6 to prove $e^e$ and its cousins, but the possibility of having two transcendental being rational multiples of each other seemed to prevent me from doing that

Balarka Sen

@DHMO it's still just bookkeeping

why not understand those fundamental theorems these are the consequences of?

DHMO

@BalarkaSen do you have any non-bookkeeping results?

well, I didn't pin the theorems lol

Balarka Sen

yes. Lindemann-Weierstrass theorem is not bookkeeping, neither is Gelfand-Schneider. the real number theory lies in proving those: the relatively easy consequences are bookkeeping.

DHMO

should I pin the theorems as well?

Balarka Sen

14:23

pin and prove

Secret

I always having trouble working with polynomials especially when they start to multiply with each other

Balarka Sen

by prove, I also mean understand the proofs. I'll engage if you start a discussion on understanding them (I don't know the proofs)

Secret

and the proof of the Lindermann weierstrass theorem is full of that on the first page, plus some lemmas that give some upper bound of something involving integers

(at least for wikipedia's version)

Sophie

@DHMO are you reading any interesting book on the subject?

DHMO

@Sophie not really

Lindemann–Weierstrass theorem

In transcendental number theory, the Lindemann–Weierstrass theorem is a result that is very useful in establishing the transcendence of numbers. It states that if α1, ..., αn are algebraic numbers which are linearly independent over the rational numbers ℚ, then eα1, ..., eαn are algebraically independent over ℚ; in other words the extension field ℚ(eα1, ..., eαn) has transcendence degree n over ℚ. An equivalent formulation (Baker 1975, Chapter 1, Theorem 1.4), is the following: If α1, ..., αn are distinct algebraic numbers, then the exponentials eα1, ..., eαn are linearly independent over t...

Sophie

14:26

I tried reading Ireland/Rossen

DHMO

why on earth is the proof so long

this isn't even the theorem

Secret

because it is the foundation of everything

To truly understand what it means for a number to be transcendental, one has to dig into that proof and understood it

DHMO

planetmath.org/sites/default/files/texpdf/39438.pdf

@BalarkaSen is what you want to discuss the above?

Balarka Sen

yup. I want to understand how the proof goes and what's the intuition behind the proofs

DHMO

I see

Sophie

14:33

you should star Roth's and Liouiville's theorems too

DHMO

i haven't stated it

you can state them and I'll pin them

Balarka Sen

Theorem: $e$ and $\pi$ are transcendental.

Let's start by something simple like that.

DHMO

I agree

but I need some time to understand the paper lol

Balarka Sen

Yeah, sure.

DHMO

@BalarkaSen so my job is basically to copy the proof here in my own words and discuss the proof?

Balarka Sen

14:50

@DHMO nah. I want to understand what the intuition behind the proof is

Jack Don

Are we doing continued fractions and dedekind domains and such in here, how are we doing our proofs?

Liouville's theorem makes me think ye

But I don't see any

DHMO

@JackDon anything related to transcendence.

you can talk about that of course

Jack Don

Do you know continued fractions and such?

DHMO

not much

Jack Don

If you are interested in the title of this chat, I think they are invaluable

DHMO

14:56

@JackDon could you teach me?

Jack Don

It's a bit of work, but finite continued fraction is rational and infinite is irrational, so that's some motivation

DHMO

I know that

Sophie

$e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10\dots]$

Jack Don

Wait was it Liouville theorem or Liouville numbers you referred above

Sophie

both maybe

DHMO

14:59

I think you want Liouville numbers

kayak

Hi guys.

Jack Don

The latter is interestingly approached via continued fractions

DHMO

welcome

Jack Don

The former, no known to me relation

Hello Kayak

kayak

I have a agony.

Jack Don

15:00

What is your agony?

kayak

I got a worst credit in one of my class

Jack Don

Was the credit irrational?

kayak

It's not about mathematics.

I'm a grad student in grad school

And I'm attending some classes until now.

I tried hard to get a good point and be better on my mathematics ability. I really tried.

But the professor gave me the worst credit.

I'm going to meet the prof. tomorrow, and I'm going to ask him about how I can do better on math, and What is the method of studying mathematics.

TT

Is there any suggestion on my studying math?

What is the way to study math.

Is there any link for me?

Sophie

@kayak this is a very broad question, I don't think you're going to get a good answer

kayak

Because this is very broad, I can't get a good answer?

You mean I have to specify..

What is the general way to study math?

Jack Don

15:07

Sit down, back straight on a chair in an office away from anyone else, doing mathematics for 8 hours per day.

kayak

Seriously...

I studied for over 8 hours per day week day weekend and it's been over a year since I studied like that.

Jack Don

That should be sufficient

I would say in that time, you don't distract yourself with music, and that you ensure you are really studying

I like to stand at a whiteboard and talk out loud to an empty class room.

kayak

I never listen to music, I do not do SNS or something. I really tried hard.

I think I need to change my way to study math.

There is something wrong on my studying method.

Sophie

most people who have these kinds of complaints are lacking a key per-requisite for the class they're in

kayak

15:23

What is key per-requisite? @Sophie

Sophie

I don't know what class you're in so I don't know

kayak

per-requisite? or Pre-requisite?

Sophie

@kayak something you need to know before you take a class, say you need to know high school math before taking calculus

kayak

Ah.. I see. Your right in some sense.

I attended 'Ergodic thoery' class without attending 'functional analysis'

And I think 'combinatorics' is also needed for Ergodic thoery.

DHMO

22:51

@BalarkaSen I think I got the proof, but what the intuition behind the proof is is a much harder problem...

Balarka Sen

@DHMO I'm heading off to bed for today but maybe you'd want to tell me a sketch of the proof tomorrow? We can fight together to understand the intuition then.

DHMO

Ok, goodnight.

Balarka Sen

I mean there's gotta be a way the person (was it Euler? Shrug) who came up with it actually did come up with it.

Yeah, night.

DHMO

@BalarkaSen It is quite hard if you want me to dig out the original proof by Hermite though...

Moreover, even if you have the proof, you still don't have the intuition behind.

Transcript for

$Mathematics$ Algebraic/Transcendence Theory