Mathematics - 2017-03-24 (page 1 of 5)

Suppose you have three letters a,b,c. I'd like you to make some arrangement of these letters. For example:
abaaaaacbaabc
Now we make strings out of this by taking the n-th to the 2n-th term and lay these strings out (we don't take strings that go from n to k<2n, just leave those last characters ignored):
ab,baa,aaaa,aaaac,aaacba,...

Say I take the third string:
aaaa
And I remove the characters but don't rearrange them. I can get
aaaa,aaa,aa,a
These are the only possible combinations. If any of these strings are the same as previous strings, you're arrangement of letters is called "improper".

Find the longest arrangement of letters possible using two letters, three letters, and four letters that is not improper..

theDoctor

Okay, anyone know of any equations that begin with a real, non-rational number and end with an an integer? exponentiation to 0th power is already known....

Meow Mix

$e^{i\pi} = -1$

theDoctor

1:01 AM

end=equate

darn, that one is controversial to me....

Meow Mix

Why?

You have two irrationals in there, and even an imaginary.

theDoctor

because it is not well defined the interaction between C and R

for example what is 1/i?

Meow Mix

$i$

Simply Beautiful Art

A proper two letter string for example would be:
abbaaa
Not optimal of course, but this is a proper string with two letters

theDoctor

I should have anticipated Euler's....

Meow Mix

1:02 AM

$1/i = i$

theDoctor

i... that's an interesting answer. What does it mean philsophically to divide a unit, i times?

is 2/i = i?

Meow Mix

No, it's $2i$

theDoctor

interesting... how to arrive at these answers?

Meow Mix

$1/i = \sqrt{1} / \sqrt{-1} = \sqrt{1 / -1} = \sqrt{-1} = i$

theDoctor

Ha, nice one.

Meow Mix

1:04 AM

I recommend you watch 3b1b's video on complex numbers.

It brings up multiplication in the complex plane.

theDoctor

Thanks. will look for video

Meow Mix

No problem

Simply Beautiful Art

@MeowMix I think I figured it out

Meow Mix

@Simply Awesome, what is your solution?

Daminark

@Meow You dropped a negative

Meow Mix

1:08 AM

What? Where?

Simply Beautiful Art

@MeowMix 1/i = -i

You didn't check $\sqrt{-1}=\pm i$ is all

Meow Mix

good point

Daminark

$\frac{1}{i}=-i$, cross multiply and check

The way I prefer to think about things is that a complex number is a pair of real numbers with addition and multiplication defined as they are

The way to handle division by complex numbers is usually to make the denominator real by multiplying through with the conjugate

Simply Beautiful Art

After n visits, we have
the first person is at $(-1)^n$
the second person is at $(-1)^{2\lfloor n/2\rfloor}$
In general, the kth person is at $(-1)^{2^{k-1}\lfloor n/2^{k-1}\rfloor}$
where positive is dissatisfied and negative is satisfied.

Meow Mix

That's a very complicated answer.

But it might be right.

Would you like me to give you my answer?

Simply Beautiful Art

1:13 AM

Sure, and my answer isn't very complicated :P

Meow Mix

By that I mean,t more complicated than it has to be

It's the binary representation of $n$.

@Daminark Might think this is cool.

Simply Beautiful Art

0: 000000000000000000
1: 100000000000000000
2: 010000000000000000
3: 110000000000000000
4: 001000000000000000
5: 101000000000000000
6: 011000000000000000
7: 111000000000000000
8: 000100000000000000
9: 100100000000000000
10:010100000000000000
11:110100000000000000
12:001100000000000000
13:101100000000000000
14:011100000000000000
15:111100000000000000
16:000010000000000000

This is what I did

but that is cool

theDoctor

You know, that is so bizarre... i never saw anyone suggest that $\sqrt{-1}=\pm i$ even though it MUST be true, yes?

Simply Beautiful Art

@theDoctor Branches

Meow Mix

@theDoctor Yes, it's the same as suggesting $\sqrt{4} = \pm 2$

Would you like another problem @Simply?

Simply Beautiful Art

1:15 AM

Technically, you could replace $i$ with $-i$ in all of mathematics and it would stay consistent.

Daminark

OH THAT TOTALLY WORKS

That's dank

Simply Beautiful Art

@MeowMix I have to head to bed :-(

Meow Mix

@SimplyBeautifulArt Uhhh not really I don't think..

Simply Beautiful Art

@MeowMix Posted it as a question and that's the answer I got

Meow Mix

Can I see it?

The question

PVAL-inactive

1:16 AM

@SimplyBeautifulArt but then we'd only know about one explicit automorphism of $\Bbb C$.

theDoctor

Here was my answer e^(ln 2)=2

Meow Mix

@theDoctor That's not very creative is it though, huh? :P

theDoctor

@simply: that is so true, yet who has ever taught it that way?

Simply Beautiful Art

6

Q: If $i^2=-1$, then what about $(-i)^2=-1$?

By definition, $i^2=-1$, right? But one can then clearly deduce that $(-i)^2=-1$. The only difference I see is that one is $-1$ times the second. So what allows us to differentiate between $i$ and $-i$? Can they be used synonymous? That is, does nothing happen if all of a sudden we were to s...

@theDoctor No-one, since no-one cares and it makes no difference

Daminark

Basically, $i$ was defined as one of the roots of $x^2 + 1 = 0$

theDoctor

1:17 AM

@MeowMix: why not? It is completely surprising that you can raise e to some, non-0 power and get a integer. (omitting eulder)

Meow Mix

@theDoctor Nope, it isn't because $\ln 2$ is defined as the exponent for $e$ which gives 2

Daminark

But you can't really distinguish between the roots of the polynomial since it's reducible over $\mathbb{R}$

Meow Mix

I could just as easily say $x^{log_x(y)} = y$ for any $x,y$ :P

Simply Beautiful Art

You guys are silly, everyone knows the correct answer is $e^{-\infty}=0$.

Real and non-trivial

Meow Mix

Totally! Just like how $1/0 = \infty$, right? Fuck limits :)

^ sarcasm

theDoctor

1:19 AM

@simply: you're starting to get to my question....

Simply Beautiful Art

@MeowMix gets a few laughs out of it

theDoctor

ln 2=some irrational, right?

Meow Mix

It is.

Daminark

Corollary: $-\infty$ is not an integer

Meow Mix

But it's defined as the power to $e$ which gives 2

I could just as easily give $e$ and $1/e$, then multiply them

theDoctor

1:20 AM

So you're taking e to an irrational power, and out drops an integer.

Meow Mix

Both of them are irrational

But yet, they give 1.

theDoctor

no, there's something different.

Simply Beautiful Art

@MeowMix Any luck on the puzzle btw? Before I head to sleep

Meow Mix

No, it isn't. What you're doing here is applying the inverse to the output of a function

which will just give you your input.

theDoctor

in my mind, there is an interesting philosophical issue that I'm calling "domain theory"

Meow Mix

1:21 AM

Similarly, I could say $e$ and $-e$, and add them

@theDoctor And, what does "domain theory" pertain to? :P

theDoctor

Domain theory deals with (to me) the intersection between different domains, say R and Q and C and geometry -- very different with different sets of axioms in some cases.

So there has to be some rules about how they interact

Meow Mix

What do you mean by "interact"?

theDoctor

well, for example, does 1.0 = 1? (R -> Z)

Daminark

I mean essentially, a complex number with imaginary part $0$ isn't strictly speaking a real number

theDoctor

@daminark: why do think so?

In what ways does it fail in the reals?

Daminark

1:23 AM

I mean because a complex number is an ordered pair of real numbers

That's the domain thing you're speaking of

Meow Mix

@Daminark That's like saying the group of symmetries of a tetrahedron isn't $S_4$

theDoctor

a complex number can be reduced to a real, however.

Meow Mix

It isn't technically, but they're the same thing.

@theDoctor 1.0 is the same as 1

theDoctor

(by making the imaginary part 0...)

Daminark

So $(1,0)$ is not technically $1$. This distinction is of absolutely no relevance, though

Meow Mix

1:24 AM

$\Bbb Z \subseteq \Bbb R$, so yes.

Daminark

Like, an embedded copy of something is just as good as the real thing

Simply Beautiful Art

Well, good night, and good lucks on the problem

Meow Mix

@Daminark Well "rotate around z-axis" is not technically some $S_4$ permutation

theDoctor

My argument is that they are not, and here's why...

BAYMAX

goodnight@SimplyBeautifulArt

Simply Beautiful Art

1:24 AM

@BAYMAX :D

Meow Mix

@theDoctor Do you understand how to construct $\Bbb R$?

theDoctor

1.0 is generally a referenced differently than integer 1. It's like absolute reference frame vs. relative.

1.0 relates implicitly to 0.0, but 1 does not.

Daminark

@Meow My point was more on your side, like sure it's not technically the case that they're the same set, the objects are set-theoretically different, but also no one cares

theDoctor

no, i don't think so...

Meow Mix

@Daminark Yeah I was just being a douche for the sake of it.

Daminark

1:26 AM

@theDoctor The thing is, in each construction, you're creating an embedding of an old set

Meow Mix

@theDoctor 1.0 isn't any different from 1

Because this is not programming.

@Daminark Soooooorrrryyyyyy :P

skull petrol

1.000... = 1

Meow Mix

Oh hey @Akiva

Akiva Weinberger

@MeowMix I did see the problem before

Meow Mix

@skullpetrol Thanks for the wonderful insight.

theDoctor

1:27 AM

@skullpetrol: that was my point

when you say 1.0 are you saying 1.000....?

Akiva Weinberger

They actually sent out the problems to the alumni for "playtesting" before they officially posted them

Meow Mix

@Akiva Oh, alright.

theDoctor

In physics, you only have 2 signifiant digits in the first number.

skull petrol

0.999... = 1.000... = 1

Akiva Weinberger

(That's the best word I can think of)

Meow Mix

1:27 AM

@theDoctor That's most definitely not true.

theDoctor

It's not true that physics only counts 2 significant digits?

Daminark

That is false

theDoctor

see, there something strange there...

Meow Mix

@Akiva I hope I can go next year.

theDoctor

you're wrong, 1.0001 would be 5 significant digits, for example, perhaps a more precise number than 1.0

Meow Mix

1:29 AM

It looks like the world is trying it's hardest to fuck me over with math as of late.

theDoctor

@MeowMix: what's not true?

Meow Mix

@theDoctor That physics goes to 2 significant digits

Daminark

In actual experimental science, you look at your error resolution, and then you handle significant digits in your calculations so that you get maximal precision (most number of digits) without being inaccurate.

theDoctor

no, your professors probably simplyi have not thought much about the nature of number.

and domain theory

Daminark

Anything in your error margin won't affect how many digits you went down

theDoctor

1:30 AM

as far as they go is defining and input domain and an output domain for a given function.

Daminark

@theDoctor The thing is this, each of these number systems is different, but you get an isomorphic copy

So for example

We build the real numbers out of equivalence classes of Cauchy sequences (or Dedekind cuts if you hate yourself)

To be precise, Cauchy sequences of rational numbers

theDoctor

No, you only get an isomorphic copy out of convenience, not out of rigor, or axiom.

Meow Mix

@Daminark I hate myself and don't use Dedekind cuts /s

Daminark

Oh no it's very rigorous

theDoctor

lol

Meow Mix

1:31 AM

No, @theDoctor, you can obtain it with rigor.

BAYMAX

Dekind cuts are rigorous and interesting!

Daminark

See the thing is, you can create an algebraic isomorphism between $\mathbb{Q}$ and the set of equivalence classes containing $(q,q,\ldots)$ where $q\in\mathbb{Q}$

theDoctor

really, and what if I change the quanity of the distance of 0.0 to 1.0? so that it is twice the quantity between zero and 1?

quantity

BAYMAX

constructing irrational numbers using rationals

Daminark

@theDoctor Well now you've redefined the absolute value function

Or the distance function

Akiva Weinberger

1:32 AM

I've always preferred Dedekind cuts to equivalence classes of Cauchy sequences, but it seems like I've been outvoted

Daminark

So that you no longer have an isometry as you did before

theDoctor

@perhaps, but nothing mathematically prevents me from defining my range of quantities differently.

Physics just happens to try to make them the same, by (I think) tacitly referring to Imperial Units.

Mike Miller

:36249408

Akiva Weinberger

The addition function on $\Bbb R$ is that induced by the addition function on $\Bbb Q$

Same for the ordering

So, the only arbitrary thing is the notion of "inducedness", I guess

theDoctor

You'll have to educate my on Chaucy sequences, because I'm still trying to rigorously demarcate Q and R.

Meow Mix

1:35 AM

I'm just a bill, yes I'm only a bill

And I'm sitting here on capitol hill

Daminark

Sure, that just becomes a question of inconvenience. The point is that the standard way of defining operations and orderings and all is well-defined, and gives an isometric isomorphism between one number system and some subset of the other

theDoctor

In truth the R^1 number line is orthogonal to the Z number line.

just as R and C.

Akiva Weinberger

It's important to note that the notions of real numbers and rational numbers are way older than either Dedekind cuts or Cauchy sequences

theDoctor

And this is why e^ln 2=2 is interesting to me.

Daminark

That's pretty false if we're dealing "orthogonality" in the strict sense of the word

Meow Mix

1:36 AM

But $\ln 2$ is defined as the power of $e$ which gives 2...

theDoctor

Physics (or whatever) may overlay R on Z, but that is just a unspoken agreement.

Meow Mix

Like how $1/e$ is defined as the number which multiplies with $e$ to give $1$

theDoctor

Right @AkivaWeinberger

Akiva Weinberger

(The existence of which requires proof)

Meow Mix

And $-e$ is defined as the number which adds with $e$ to give $0$.

Daminark

1:37 AM

The point here is this, there's the set theoretic notion of everything, and then there's the picture

Now, the set theoretic construction of $\mathbb{Z}$ gives you objects that are different when you consider representatives of integers in $\mathbb{R}$

theDoctor

@MeowMix: no, that is just the way you speak to yourself about it in your head. You could also say that ln x is the natrual logarithm and define the number e from it.

(or derive)

Akiva Weinberger

I think that it's better to think of the line as a collection of overlapping open intervals, rather than a collection of points.

Essentially, its topology.

theDoctor

@Daminark: Well, define your notion of orthogonality.

In my "strict" definition, except possibly at 0.

Daminark

But you can come up with a well-defined bijection, and with the way addition, orderings, etc are defined on $\mathbb{R}$, you can get an isometric embedding. Now, we can call that new set something else, and it behaved exactly like $\mathbb{Z}$ did

So in our picture and by abuse of notation, we say that $\mathbb{Z} \subset \mathbb{R}$

More like abuse of terminology

Akiva Weinberger

@theDoctor In any case — the definitions are never unique. Yours are almost certainly different than mine. But, if they're provably equivalent, why care? The question is, what new results can we find using the things we've defined.

Daminark

1:40 AM

But the point is, there's something there which provably corresponds directly to $\mathbb{Z}$ in all the ways that anyone cares

So this becomes an entirely moot point

Now, if you have an inner product space, orthogonality means the inner product is $0$

theDoctor

However, if you accept Rienmann surfaces as different from Euclidean planes, then you have to accept that R's number could be completely independent of Z except by symbolic similarities of the Arabic numbers.

Akiva Weinberger

Unless someone says, "Hey, look at this cool alternate definition of $\Bbb R$ I found"

That could be interesting

(Speaking of: Hey, look at this cool alternate definition of $e$ I found)

(if you care)

theDoctor

@AkivaWeinberger: Ha, I think that was mine ...;^>

Daminark

If you're looking at $\mathbb{Z}$ as a set theoretically different structure than $\mathbb{R}$, then unless you manufacture an entirely new and well-defined inner product structure so that everything in each set is orthogonal, then congrats, sure

Akiva Weinberger

@theDoctor You think what was yours?

theDoctor

1:43 AM

That cool definition of e.

Akiva Weinberger

What definition are you thinking about

Daminark

But if you're referring to the standard pictures and all, then well, pictures are representations, and what we're representing by $\mathbb{Z}$ in this picture (the copy of it embedded in $\mathbb{R}$) is not orthogonal to what we're representing by $\mathbb{R}$

@MikeMiller ?

Haha, you like it.

@theDoctor ?

1:46 AM

I'm formulating a better way to say it...

If e^(ln 2)=2, then ln(e^(ln2))=ln(2), and....

ln 2 != 2, so what went wrong here....?

sorry, that was sloppy I started over at "Let's see", dismiss the previous line of thought...