The h Bar

Bernard Meurer

8:00 PM

I know that $$\forall_{x\in\mathbb{R},y\in\mathbb{P}}x^y\in\mathbb{P}$$ is false

0celo7

Jesus

How are you even defining $x^y, y\notin\Bbb Q$?

Bernard Meurer

@QuantumBrick Nice, will read, thanks!

Well isn't the exponential continuous?

Danu

Nathan For You - Gas Station Rebate - Daniel's Advice

►

QuantumBrick

@BernardMeurer I just noticed you're trying to prove that 4^sqrt{2} is irrational. Is that it?

0celo7

Dude, did you turn into a mathematician over night?

Bernard Meurer

8:02 PM

then there should be infinite possible values between $x^y$ and $x^{y+1}$

@QuantumBrick Yes, Precisely

0celo7

You're apparently doing basic logic, proving $x^y$ is continuous in $y$ is a long way away.

And that's not a definition.

@BernardMeurer Proof?

Have you proved that square roots exist?

user218912

@0celo7 @Danu the infinitesimal lorentz transformation matrix for boosts on the $x$ axis is symmetric but the infinitesimal lorentz transformation matrix must be antisymmetric.

user218912

I found the matrix in cahill.

ACuriousMind

@0celo7 One doesn't "prove" that, one defines exponentiation by irrational exponents as the limit of rational exponents, and typically these introductory classes do not do that. Rather, you're doing formal manipulations with the exponentiation rules.

user218912

but I showed that it has to be antisymmetric.

0celo7

8:04 PM

Why did you tell Danu that?

user218912

because he's good at physics

0celo7

I don't think he wants to help you.

Danu

@IceLord Sure

Bernard Meurer

@0celo7 I'm really liking my analysis course

0celo7

@ACuriousMind Of course one proves it.

user218912

8:05 PM

@Danu so then why is the one in the book symmetric?

Bernard Meurer

What do you mean "Showing square roots exist"?

0celo7

@ACuriousMind One defines it via logarithms and the natural exponential.

Bernard Meurer

They're axiomatically defined are they not?

0celo7

Showing that everything converges uniformly is then the key.

ACuriousMind

@0celo7 Not in my world :P

user218912

8:05 PM

@BernardMeurer I am really not liking it.

0celo7

@ACuriousMind One can define $e^x$ for any $x$, right?

using the power series

Bernard Meurer

@IceLord Mine or yours? :P

user218912

mine.

0celo7

Then $a^x=\ln(a)e^x$ or something.

ACuriousMind

@0celo7 I'm not saying your definition doesn't work, I'm saying I wouldn't use it.

Bernard Meurer

8:06 PM

$$\forall x_0 \in \mathbb{R}: \displaystyle \lim_{x \mathop \to x_0} \ \exp x = \exp x_0$$

user218912

@Danu what do you mean by sure?

0celo7

What is this limit now? How are you doing limits on day 1?

Bernard Meurer

@IceLord Why not?

Danu

@IceLord I agree

Bernard Meurer

@0celo7 Welcome to Europe

0celo7

8:08 PM

@BernardMeurer How did you define $\exp$?

user218912

@Danu @0celo7 read embed.gyazo.com/85b83c077799e3fdb176a713afb5066f.png

0celo7

@IceLord I'm not helping you

user218912

am I retarded or is the $B_1$ matrix not antisymmetric?

Bernard Meurer

@0celo7 $$exp(x) = e^x$$?

0celo7

What is $e^x$?

user218912

8:09 PM

@BernardMeurer it's boring.

Bernard Meurer

@0celo7 I haven't got there yet

@IceLord What are you seeing right now?

0celo7

whut

user218912

@BernardMeurer $inf$ and $sup$ and least upper bound and others.

Bernard Meurer

@0celo7 What?

0celo7

How are you doing this exponential stuff if you don't know what the exponential is?

user218912

8:13 PM

@BernardMeurer our problem set was easier than yours, we had to do dumb things like proving $\sqrt{3} + \sqrt{5}$ is irrational.

0celo7

How do you do that?

user218912

it's simple

Bernard Meurer

@0celo7 I don't have to prove that $4^\sqrt{2}$ is irrational, I want to because I'm just curious

@0celo7 LOL, Q.E.D

0celo7

@BernardMeurer How did you prove the other thing?

user218912

@0celo7 how :(

0celo7

8:15 PM

With the irrational exponent or something

Bernard Meurer

@0celo7 Which thing?

0celo7

That seems highly nontrivial

user218912

@ACuriousMind do you know what the contradiction is here?

ACuriousMind

@BernardMeurer In general, establishing the (ir)rationality of a certain number is hard. In this case, Gelfond-Schneider tells you it's irrational, but I don't see an "easy" proof off the top of my head

user218912

@ACuriousMind chat.stackexchange.com/transcript/message/32450668#32450668

Bernard Meurer

8:16 PM

@ACuriousMind No. Gelfond-Scheneider tells me it's Transcedental, which implies irrationality

ACuriousMind

@IceLord Dude, I cannot hold two conversations and read that picture and think about how to prove things at once

Bernard Meurer

@0celo7 Oh, that's trivial

user218912

@ACuriousMind but you're ACM

Danu

@IceLord I think your issue is the following

ACuriousMind

@BernardMeurer Correct, but I don't see a direct way to prove it irrationality "by hand"

Bernard Meurer

8:17 PM

$$q = \sqrt{2}^\sqrt{2}\\q^\sqrt{2}$$

Danu

It depends on whether indices are raised or not.

Bernard Meurer

@0celo7 There, QED

Danu

Look at 10.228 of Cahill. He has the condition you're probably referring to

ACuriousMind

@BernardMeurer Don't expect him to reply for the next half hour :P

@BernardMeurer Is that supposed to be the proof idea for "there are two irrational numbers such that the exponent is rational"?

Bernard Meurer

@0celo7 $$\frac{\log 2}{\log 3}\in\mathbb{P} \\ 2^{\frac{\log 3}{\log 2}}\in\mathbb{Q}$$

@ACuriousMind He told me to tell you to pleasure yourself

user218912

8:21 PM

@Danu the matrix representation of ${\omega^\mu}_\nu $ depends on the indices in what way?

user218912

what how do you do spaces in latex i forgot

ACuriousMind

Do {\omega^\mu}_\nu for better (not perfect) spacing.

Also, spaces are \, and \;

user218912

@ACuriousMind yep i remember now

user218912

thanks

Bernard Meurer

@ACuriousMind Indeed, proving irrationality through transcendence becomes easy with Gelfond-Schneider, but the direct proof of irrationality is the real bitch

Danu

8:23 PM

@BernardMeurer Proving transcendental is REALLY hard IIRC

As in an unsolved problem for numbers related to $\pi$ and $e$

ACuriousMind

@BernardMeurer I'm supposing he did not phrase that so carefully compliant with the Be Nice policy. :P

Danu

Not for $\pi$ and $e$ themselves, sorry

Bernard Meurer

@Danu note the > with Gelfond-Schneider

how to quote

ACuriousMind

Multiline still breaks markup

Bernard Meurer

halp

ACuriousMind

8:25 PM

> was correct to quote, but shift+enter broke it

Bernard Meurer

Broke everything

@Danu $a^b$ is transcendental if $a,b$ are algebraic, and $b$ is irrational

Danu

What, no?

$2^2$ is not transcendental

Bernard Meurer

HA

ACuriousMind

@BernardMeurer and $b$ is irrational.

Bernard Meurer

Ooops

Danu

8:27 PM

lmao

Bernard Meurer

I GOT A PROOF

ACuriousMind

@Danu Did you know we don't even know whether $\mathrm{e}$ and $\pi$ are independent?

Bernard Meurer

For $4^{\sqrt{2}}\in\mathbb{P}$

We know that $$\forall_{x\in\mathbb{R};y\in\mathbb{P}} xy\in\mathbb{P}$$

Danu

@ACuriousMind Right, that was the thing. Wiki link?

I remembered it had somethign to do with $e$ and $\pi$... :P

Bernard Meurer

Wait

no

ACuriousMind

8:29 PM

@Danu Here

Bernard Meurer

My proof is through transcendence

Sigh

ACuriousMind

Why on earth do you call the irrationals $\mathbb{P}$ by the way?

Danu

^

THIS IS WRONG :D

The P is reserved for projective spaces

Bernard Meurer

19

Q: Is there an accepted symbol for irrational numbers?

$\mathbb Q$ is used to represent rational numbers. $\mathbb R$ is used to represent reals. Is there a symbol or convention that represents irrationals. Possibly $\mathbb R - \mathbb Q$?

notation

Because of this

Danu

@ACuriousMind @ACuriousMind Right... The most impressive one is that we don't know if $\pi+e$ is irrational!

> should be preceded by a clear statement as to the fact that it is being used to denote the set of irrational numbers

So not a good idea to use for standard

Bernard Meurer

8:31 PM

Don't we know that they are both transcendental?

user218912

so nobody knows what my problem is?

Bernard Meurer

at least $\pi$ I know is transcendental

QuantumBrick

@BernardMeurer You can use that for a^x there's an inverse function log_a? This can be proved by expressing a^x = e^(x ln a ) and using the inverse function theorem.

If you can use this, than maybe I can hint for a proof.

ACuriousMind

@BernardMeurer Yes, but we don't know whether there's a rational such that $\pi = r-\mathrm{e}$ or not.

Danu

@BernardMeurer We do

@IceLord You don't think my suggestion is right?

Bernard Meurer

8:34 PM

@QuantumBrick Use $foo$ to enable MathJax please :)

Danu

Look, Cahill has the equation that you probably used (an analog of it) to show that $\omega$ is antisymmetric.

There cannot be a contradiction there.

ACuriousMind

@BernardMeurer Then you also should advise @QuantumBrick that there's a link to activating MathJax in the upper right corner of the room ;)

user218912

@Danu yes I agree with that.

user218912

but why is the matrix symmetric then?

Bernard Meurer

@Danu Isn't $$\forall_{x,y\in\mathbb{P}} xy,x+y\in\mathbb{P}$$ true?

QuantumBrick

8:36 PM

@ACuriousMind I'm looking for it haha

user218912

@Danu sorry if i'm being dumb, stuff just isn't obvious to me.

Danu

@BernardMeurer Why would it?

ACuriousMind

@BernardMeurer Take $x=\sqrt{2}$, $y=-x$.

Danu

What you're saying would imply that, if I subtract an irrational number from a rational one, I never get an irrational one.

That makes absolutely no sense.

Also what ACM says.

Bernard Meurer

@Danu I don't know

Danu

8:40 PM

@BernardMeurer Do you agree that what I said after that also makes sense? Also ACM's counterexample of course.

Obliv

@bernard what is $\mathbb{P}$

Bernard Meurer

@Obliv $$\mathbb{P}=\mathbb{R}\setminus\mathbb{Q}$$

user218912

@BernardMeurer everyone is ignoring me because of you :(

user218912

fine i'll come back in a few hours

Bernard Meurer

Jeez I need better notation for Irrationals

Danu

8:42 PM

@IceLord ?!?!

I'm not ignoring you

ACuriousMind

@BernardMeurer Just say "is irrational". But I get that you're currently in that first-sight-of-math phase where you think math is all about fancy notation ;)

user218912

@Danu

user218912

6 mins ago, by IceLord

but why is the matrix symmetric then?

Bernard Meurer

@ACuriousMind It's easier to put it in the LaTeX already though :/

Danu

@IceLord Did you try to follow Cahill?

user218912

8:43 PM

from 10.228?

user218912

onwards?

Danu

Yeah

Obliv

@0celo7 my last one was like ~1500 words

Danu

That's how you proved antisymmetry right?

Bernard Meurer

Okay

I see why I was wrong

$$\forall_{x,y\in\mathbb{P}} xy,x+y\in\mathbb{P}\cup\{0\}$$

Obliv

8:44 PM

but i'd say 10% of the words were meaningful. It's these stupid minimum requirements for each section that beef up the word count. @0celo7 I'd consider going deeper into the physics of the experiment but I'm not able to do the derivations alone/comprehend them so it'd be stupid if i used them in a lab report.

Even with citations.

Bernard Meurer

@ACuriousMind Take that!

:p

Obliv

why would you put the x,y in P as a subscript @bernard

ACuriousMind

@BernardMeurer Nope, consider $x=\sqrt{2}+1,y=-\sqrt{2}$.

Take that.

:P

Obliv

wouldn't it just be easier with $\forall x,y \in \mathbb{P}$

user218912

@Danu I proved it by plugging in $\Lambda^\mu_\nu = \delta^\mu_\nu + \omega^\mu_\nu$ into $\eta_{\mu\nu} \Lambda^\sigma_\mu \Lambda^\tau_\nu$ and showing it equals to $\eta_{\mu\nu}$.

Bernard Meurer

8:46 PM

@Obliv To make it harder to read

QuantumBrick

@BernardMeurer Let $y= a^\sqrt{n}$, where $\sqrt{n} \notin \mathbb{Q}$. By using the inverse function theorem one can show that $a^x$ has an inverse given by $\log_a$. Then $\sqrt{n} = \log_2 y$. Can I go on? Am I talking nonsense?

Danu

Bernard, if you read my previous messages better you'd stop trying to dodge ACM's counterexamples.

Bernard Meurer

@ACuriousMind Dang it

Danu

7 mins ago, by Danu

What you're saying would imply that, if I subtract an irrational number from a rational one, I never get an irrational one.

That's just absurd ^

Obliv

i'm going to have a mental breakdown with all of this tex

Bernard Meurer

8:46 PM

@Danu I was just trying to beat the 0 one

Danu

@IceLord Right.

DanielSank

Any reason this discussion isn't on Math SE?

;)

Bernard Meurer

I know that wasn't a full solution

DanielSank

Folks there know stuff.

Danu

@DanielSank Because there they do real math ;)

Bernard Meurer

8:47 PM

@DanielSank It'd end too quickly

ACuriousMind

@Danu Bwahahahahahahahahahahahahahahahahaha

DanielSank

@Danu No way, duder. They're super-duper helpful.

ACuriousMind

This chat is so funny today :P

0celo7

Who decided to suspend me? Not nice.

Bernard Meurer

@QuantumBrick let me think

ACuriousMind

8:47 PM

@0celo7 Auto-suspension for a validated flag, you should be used to those by now :P

user218912

@0celo7 why, you didn't say anything bad.

Bernard Meurer

@QuantumBrick Can you show me the inverse function theorem?

Wait

Obliv

@0celo7 rip that's like 2 flags in the past week i think

Bernard Meurer

$\log_a$ of what?

Oh

Wait

Hmm

Danu

@DanielSank I know.

I ask them about complex geometry all the time.

My favorite users actually returned (ocassionally)! :D

user218912

8:49 PM

@0celo7 oh that.

QuantumBrick

@BernardMeurer If I can go on, then, since $\sqrt{n} \notin \mathbb{Q} \Rightarrow \log_a y \notin \mathbb{Q}$. Suppose $ y \in \mathbb{Q}$ even though we know $ \log_a y \notin \mathbb{Q}$. This means that, for some $p,q \in \mathbb{Q}$, we have $\log_a y = \log_a p/q \Rightarrow y = p/q$, which is absurd. This proves $y \notin \mathbb{Q}$.

Danu

→ 4 messages moved to Trash

ACuriousMind

@Danu Interesting, I'm able to reply to this. I wonder what purpose that serves, since one can't reply to deleted messages.

Danu

@ACuriousMind To comment on my actions? :P

ACuriousMind

lol, that's usually not the SE way for moderator actions

QuantumBrick

8:54 PM

@BernardMeurer Wait, that's not a valid proof.

user218912

@Danu so ideas?

user218912

if you don't know I can ask my TA

user218912

i'm going there right now

Obliv

Oh man.. @johnR I've been trying to understand derivations of Einstein's and Debye's model of a solid which involves $\frac{\partial S}{\partial U}$ and I had no idea why they were determining the different arrangements that a system could be in to solve for temperature.

Danu

@IceLord So what exactly was your conclusion?

Obliv

8:57 PM

Until now, when I decided to read the next chapter..

Danu

$\omega^\mu_{\ \nu}=-\omega^\nu_{\ \mu}$?

Obliv

@bernard learn to teX brah

user218912

yes

Danu

So how does that translate to 10.228?

Bernard Meurer

@0celo7 $$\frac{\log 3}{\log 2} \notin \mathbb{Q} \\ \text{Let's say} \frac{a}{b}=\frac{\log 3}{\log 2}; a,b\in\mathbb{Z}; (a,b)=1 \\ 2(\frac{a}{b})=3 \\ 2^a = 3^b \\ \text{It is trivial to see that }2^a\text{ will always be even, while }3^b\text{ will always be odd}$$

This work?

user218912

8:59 PM

@Danu I think.

user218912

that's in matrix form.

ACuriousMind

@IceLord "I think" is not an answer to "How does...?".

Danu

Right, so why does Cahill still have some $\eta$'s around? Can you reconstruct that?

Bernard Meurer

@ACuriousMind Is that a proof by absurd?

John Duffield

Anybody, is there a proper name for the angle at which a precessing gyroscope is leaning? It's theta in the picture here.

user218912

9:00 PM

@ACuriousMind i didn't see the "how does" part

Bernard Meurer

What would I call that?

user218912

@ACuriousMind i'm blind sometimes

John Duffield

Is it inclination angle?

ACuriousMind

@BernardMeurer Something has gone wrong in your third line.

Bernard Meurer

yes, yes just noticed

$$\frac{\log 3}{\log 2} \notin \mathbb{Q} \\ \text{Let's say} \frac{a}{b}=\frac{\log 3}{\log 2}; a,b\in\mathbb{Z}; (a,b)=1 \\ 2^{\frac{a}{b}}=3 \\ 2^a = 3^b \\ \text{It is trivial to see that }2^a\text{ will always be even, while }3^b\text{ will always be odd}$$

user218912

9:02 PM

oh @Danu

Bernard Meurer

Ta-daa

user218912

I figured it out

Danu

@IceLord Was it what I said?

user218912

yes

user218912