Mathematics - 2016-12-12 (page 4 of 6)

Secret

13:01

How to describe the observation that: For a magma with 3 elements, letting S={a,b,c}. Multiplied a on the left on S gives aS={aa,ab,ac} which corresponds to the row a of the cayley table. Then if a is an identity, and K is some row in the cayley table, aK=K for all K. What is a doing here?

a when multiplied on K seemed to map entire rows of the cayley table to another row

Tobias Kildetoft

$a$ is not doing anything there (that is what being an identity means)

Secret

similarly if a is a left absorber, then aK={a,a,a} for all K, so a is mapping any rows of the cayley table to a row of a s

so it seems whatever I am doing here is similar to semigroup actions, but I am not sure if they are indeed semigroup actions

Tobias Kildetoft

I have no idea what you are trying to do.

Secret

That is, if I had a set X whose elements are the rows of the cayley table, then multiplication by any element in the magma seemed to map X to X if the magma is associative. Therefore multiplication on the left or right on any member of X seemed to be an endomorphisim of X (as endomorphism is defined to be a map from an object to itself, if I recall)?

Tobias Kildetoft

@Secret An endomorphism is a map from an object to itself which preserves whatever structure you have on the object

Secret

13:18

I wonder...?
Let X be the set of all rows of the cayley table of the magma $(M,\circ)$, with $\circ : M \times M \rightarrow M$ the binary operator. Define a binary operator $\cdot : M \times X \rightarrow X$. Therefore, multiplication on the left maps rows to rows hence $X$ is closed under $\cdot$.

Therefore $\cdot$ should preserve the structure of $X$?

Tobias Kildetoft

what structure?

Secret

I am not sure... Each element in $X$ has elements $y$ which are in $M$. Therefore the closure of the magma $(M,\circ)$ will be preserved by the action $\cdot$?

Secret

Actually using set notation, how do we write a set that consists of rows of a cayley table given a magma M?

Tobias Kildetoft

That depends on whether you consider the rows ordered or not

Secret

Even though the columns and rows of the cayley table is unordered by convention, I think in order for my above mapping formalism of rows to rows to work properly, the rows and columns has to be considered to be ordered. (this is specified by how the cayley table is drawn, and the ordering will not be changed by any operations after it is specified)

Otherwise for example, if S={a,b,c} is unordered and a is a left identity, then multiplied a on the left of S should give aS={a,b,c}, but if S is unordered, then one can equally say aS={b,c,a} or {c,b,a} or any permutation of the elements in S

Tobias Kildetoft

13:31

And why would that be an issue?

Secret

Thinking normally: if a is a left identity, then $ax=x$ for all x in $M$. That means we expect $a$ maps to $a$ by $a$, $b$ maps to $b$ by $a$ and so on. If the rows are unordered, than means $aS=\{b,c,a\}$ is a valid result (since unordered rows, which are unordered set of elements of $M$ all equivalent barring the positioning/permutation of the elements in the set/row.)

But if that's the case, then if $b$ permutes elements in a row (which is what happens in $\mathbb{Z}/n$ then $bS=aS$ since all rows are equivalent barring permutation.

Null

If I have to show well-definedness for a sequence, I only have to look if 0 lands in some denominator or? Or are there other things to consider?

Secret

thus for unordered rows, the result of a and b cannot be distinguished, and we could as well say that b is a left identity while actually it is not

DHMO

Exercise: prove that $\not\exists n \in \Bbb N: s(n)=n$ using just the Peano axioms

Exercise 2: Prove the same results using the von Neumann definition of $\Bbb N$ instead.

@Null You might be interested

Null

@DHMO what is $s$?

Secret

13:40

@TobiasKildetoft Therefore, let's say using $\mathbb{Z}/3$ as an illustration, if we want to distinguish the multiplicative action of $1$ or $2$ done on the set $S=\{0,1,2\}$, the rows must be ordered else $1S=2S$

Rakso

Is there jobs in mathematics ?

DHMO

@Null the successor function

Null

@DHMO so s(2)=3?

DHMO

@Null yes

Tobias Kildetoft

@Secret Sure

@Rakso Yes

Rakso

13:42

@TobiasKildetoft Reason for asking is that im thinking hard about switching my major from computer science to math

Antonio Vargas

There are jobs in computer science too.

Tobias Kildetoft

@Rakso Probably even more jobs in CS though if that is the main concern

Rakso

Sure but i've found a bigger interest for math than CS, thats why i want to switch

@AntonioVargas @TobiasKildetoft Would someone with a masters in pure Math have problems with jobs later on or how does it work?

Tobias Shuxue Laoshi

@Rakso Switch! I switched from CS to maths. Best thing I've done. I wouldn't worry about jobs. If you're good student, you'll find 'em.

DHMO

@Null do we have the same axioms?

Tobias Kildetoft

13:45

@Rakso It is my understanding that those are fairly easy to come by, but I don't have much direct experience with it, as I have stayed in academia

Null

@DHMO wikipedia or proofwiki? otherwise just link :D

DHMO

@Null proofwiki.org/wiki/Axiom:Peano%27s_Axioms/Formulation_1

Rakso

@TobiasShuxueLaoshi Thats something i've wanted to hear for quite some time. Thanks. What kinds of jobs exist for peoples with math degrees?

Antonio Vargas

@Rakso, I've also stayed in academia but my impression is that it's harder to get a job with a pure math masters than with a more "applied" math masters like mathematical finance or statistics.

or actuarial science

DHMO

@Null the first two axioms are existences of successor function and zero, respectively

Rakso

13:47

@TobiasKildetoft @AntonioVargas What made you guys stay in academia? Also how hard would it to later with a degree in math get into something like finance?

Antonio Vargas

@Rakso, fear of real world responsibilities. (joke)

Tobias Kildetoft

@Rakso I stayed in academia because I like doing research, and nobody outside wants to pay me for it

Or at least they would probably want me to start doing some more useful research

Mike Miller

finance hires lots of people out of academia that are morally bankrupt enough to work in finance

Rakso

How is a life in academia then? Do you even get paid for research? At my Uni researchers doesn't get paid and have to educate students for cash

Tobias Kildetoft

@Rakso At most research universities, the researchers do get paid

Not that the job does not involve teaching, but it is mainly research

meow-mix

13:51

good morning

Rakso

programming is what brought me to CS but the main part i always liked within programming was problem solving and i feel i get to work more with problem solving in maths than in CS

Sophie

I'm failing really hard at this: let $\alpha\in\mathbb{C}$ prove that $f(z)=z^\alpha$ is holomorphic

meow-mix

i can change my name in just 6 hours! :)

Antonio Vargas

@Sophie That's because it's false unless $\alpha$ is a nonnegative integer.

Rakso

The choice of switching is hard because all my friends and family says its a bad idea mainly becayse of limited jobs

Mike Miller

13:53

It also doesn't even make sense unless $\alpha$ is an integer

meow-mix

@Sophie $\alpha = -1$...?

DHMO

@MikeMiller why?

meow-mix

you have an isolated singularity at $0$ for non-negative integers

DHMO

I thought $f: \Bbb C \to \Bbb C$

Antonio Vargas

@DHMO, it's multivalued unless you choose a branch.

DHMO

13:54

I see

Mike Miller

@DHMO Tell me how you would define it.

DHMO

@MikeMiller never mind

Secret

Attempt at formulate the mapping thingy again:
Given a magma $(M,\circ)$ with $n$ elements. Let $\mathcal{F}$ be an unordered set consists of ordered sets $K$ of size $n$ and an ordered set $S$ also of size $n$. Let $a,b \in M$. Define $\cdot : M\times \mathcal{F} \rightarrow \mathcal{F}$. Therefore $\cdot$ maps any $K\in \mathcal{F}$ to any $L\in \mathcal{F}$ as follows: $a\cdot S=K$ and $a\cdot K = L$ for all $a,b \in M$ and $S=\{a_1,a_2,\dots,a_n\}$, the set of all elements in $M$ before any operation $\circ$ is done.

Mike Miller

@AntonioVargas Let them think about it instead of telling them why it's wrong :)

Antonio Vargas

@MikeMiller you're right

Null

13:55

@DHMO what again stops me from defining s(0)=1,s(1)=3,s(2)=2?

Mike Miller

That's my preferred brand of pedagogy: only be so helpful as to get someone to answer their own question.

DHMO

@Null that's my question, and you're not tricking me into giving the answer directly ;)

Secret

Sorry there's a typo:

Null

@DHMO so your question is: why would the above definition of s hurt the axioms?

DHMO

@Null yes

Mike Miller

13:58

@Null You've been given the excellent hint that it's true. So it's time to just take a look at the axioms and try to work out how to apply them.

Null

@DHMO well, P5 is a candidate imo

DHMO

@Null yes it is

Secret

Attempt at formulate the mapping thingy again:
Given a magma $(M,\circ)$ with $n$ elements. Let $\mathcal{F}$ be an unordered set consists of ordered sets $K$ of size $n$ and an ordered set $S$ also of size $n$. In addition $K,L \subset M$. Let $a,b \in M$. Define $\cdot : M\times \mathcal{F} \rightarrow \mathcal{F}$. Therefore $\cdot$ maps any $K\in \mathcal{F}$ to any $L\in \mathcal{F}$ as follows: $a\cdot S=K$ and $a\cdot K = L$ for all $a,b \in M$ and $S=\{a_1,a_2,\dots,a_n\}$, the set of all elements in $M$ before any operation $\circ$ is done.

Sophie

@AntonioVargas everywhere? That must be why I can't prove it

DHMO

Is there any proof that the solution to $x^{x^x} = 2$ is irrational?

Antonio Vargas

14:00

@Sophie, well, see @MikeMiller 's comment

Null

@DHMO do you mean $x^{x^{x}}$ or the powertower?

Mike Miller

@Sophie You need to start by asking "What does $z^\alpha$ mean?"

DHMO

@Null I do mean $x^{\left(x^x\right)}$

Sophie

@MikeMiller for $x,y\in\mathbb{C}$, $x^y=e^{y\ln(x)}$ where the exponential function is defined by the power series and the logarithm is defined as its inverse(which is multivalued)

Tobias Shuxue Laoshi

@Rakso Same here. But if you really love maths, it wont be an issue. I'm still in grad school so I don't know squat really, but if I'd be worrying about jobs, I probably wouldn't be able to focus hard enough on the maths.

Mike Miller

14:02

You've got the terms in the exponential backwards, $y \ln(x)$.

@Sophie OK. So we have the problem that $\ln$ is not well-defined, right? How do you get around that?

Sophie

so you have to chose one of the branches of the log, but it doesn't matter which so you can just choose the main one

Mike Miller

It does matter which! You'll get a different answer depending on which one you choose.

Secret

More illustrative example:
Suppose I have the group $(\mathbb{Z}/3,\circ)$ Let $\mathcal{F}$ be an unordered set consists of the following ordered sets: $\{0,1,2\},{1,2,0},\{2,0,1\}$. Therefore $S=\{0,1,2\}$. Define $\cdot : \mathbb{Z/3}\times \mathcal{F} \rightarrow \mathcal{F}$ as follows: Given $K\in \mathcal{F}$ and $a\in \mathbb{Z/3}$, $a\cdot K \in \mathcal{F}$. For example $2\cdot S=\{2,0,1\}$.

Mike Miller

But sure, feel free to choose the main one. Then the point is that $z^\alpha$ is only well-defined where the branch of $\ln$ is well-defined (for the usual branch, whenever $x$ is not a nonpositive real).

Sophie

@MikeMiller well yes, $x^y$ will evaluate to something else but then you can think of $x^y$ as a set instead of a number

Rakso

14:04

@TobiasShuxueLaoshi Whats grad school? Like for bachelor degree?

Mike Miller

@Sophie What could it possibly mean for a set-valued function to be holomorphic? :P

Tobias Shuxue Laoshi

@Rakso Master and PhD. I'm still at my master, so there's still many years of fun left :) And beyond PhD it's hopefully even more fun.

Sophie

@MikeMiller functions are just sets of ordered pairs, ordered pairs are just sets of sets, it's basically sets all the way down. I might as well add another layer

Secret

Therefore what exactly is $\cdot$ doing, is it some kind of group action?

Mike Miller

@Sophie err, you go ahead and do that and then let me know when you can say what a holomorphic set-valued function is.

meow-mix

14:07

@MikeMiller would there exist such a thing as analytic set-valued functions?

Mike Miller

I can imagine a definition. It's not interesting and reduces to thinking about branch cuts.

DHMO

@Secret are you interested?

meow-mix

@TobiasShuxueLaoshi that is a truly badass name

Sophie

I thought the idea of complex exponentials had a canonical definition

Tobias Shuxue Laoshi

@meow-mix 有一点点。Hehe.

Secret

14:10

@DHMO Well I used that to elucidate the associative laws and to check them as I build the structure. Tobias clarified that mapping is not an endomorphism. In trying to describe the idea hoping to find out what mathematical term for such mapping (if any), we concluded that the rows and columns of the cayley table must be ordered

Mike Miller

@Sophie Sure. But logarithms don't.

meow-mix

well now we have 2 tobias'

DHMO

@Secret I mean, my two exercises

Sophie

$\ln(x)=\{y\in\mathbb{C}:e^y=x\}$ move fast and define things

3

Secret

Suppose the solution to $x^{x^x}=2$ is rational, that is $x-\frac{a}{b}$ for some $a,b\in \mathbb{Z}$. Then ${\frac{a}{b}}^{\frac{a}{b}^{\frac{a}{b}}}=2$. Taking ln both sides gives

Antonio Vargas

14:13

"most fast and define things" is my new motto

Mike Miller

@Sophie Again, you're going to have trouble defining the notion of a set-valued analytic function.

No matter how fast you move.

What you can successfully do is prove that after choosing a branch of $\ln$, $z^{\alpha} = e^{\alpha \ln(z)}$ is analytic.

Secret

$\frac{a}{b}\ln(\frac{a}{b}^{\frac{a}{b}})=\ln 2$. Now

Sophie

@MikeMiller I'm taking these as challenges

Mike Miller

That's fine, I just won't bother engaging you with it.

DHMO

@Secret (a^b)^c $\ne$ a^(b^c)

Secret

14:16

wait so $(a^a)^a\neq a^{(a^a)}$?

DHMO

ya

41 mins ago, by DHMO

Exercise: prove that $\not\exists n \in \Bbb N: s(n)=n$ using just the Peano axioms

40 mins ago, by DHMO

Exercise 2: Prove the same results using the von Neumann definition of $\Bbb N$ instead.

@Secret ^ are you interested?

Mike Miller

@DHMO It's almost certainly true that your $x$ is transcendental. It's almost certainly false that any professional mathematician knows how to prove that.

DHMO

@MikeMiller but if you see here:

30

A: Not especially famous, long-open problems which anyone can understand

Let ${^n a}$ denote tetration: ${^0 a}=1, {^{n+1} a}=a^{({^n a})}$. It is unknown if ${^5 e}$ is an integer. It is unknown if there is a non-integer rational $q$ and a positive integer $n$ such that ${^n q}$ is an integer. It is unknown if the positive root of the equation ${^4 x}=2$ is ratio...

It starts with $n > 3$

implying that the case for $n = 3$ is known

Mike Miller

Ah, good call. It follows from Gelfond-Schneider, the only real 'big tool' in transcendental number theory. The point is that if $a$ is rational, then $a^a$ is always irrational but still algebraic. That it's irrational follows because if $a = p/q$ in lowest terms, $q$ cannot be a $q$th power unless $q=1$.

DHMO

and then?

Mike Miller

14:25

Immediate application of GS. If $a$ is rational, $a^a$ is irrational but algebraic, so $a^{a^a}$ is transcendental.

(I'm assuming of course that $a$ is not an integer.)

DHMO

thanks

Secret

Let $n=0$. Then $S(0)=0$, which is false as there is no succesor of 0 in the natural numbers. By the axiom of induction (or by repeatly applying $S()$), this means there exists no S(n) for all $n\in \mathbb{N}$ which contradict to the existence of S(n) for all $n\neq 0$.

actually... what is $S(\emptyset)$...?

DHMO

@Secret $S(\emptyset)$ = $\emptyset$

Secret

in that case. the proof is ok

DHMO

we are being very dangerous here

Mike Miller

14:29

The empty set is not a natural number!

DHMO

since in von Neumann, the empty set is 0

Mike Miller

@Secret Repatedly applying $S$ is not something you can do in Peano arithmetic. For one, there are nonstandard models where not every number is a successor of 0.

@DHMO Then in that model $S(\varnothing) = 1$.

DHMO

@MikeMiller but that would violate P5

proofwiki.org/wiki/Axiom:Peano%27s_Axioms/Formulation_1

Mike Miller

No, it doesn't. How do you intend to define "the set of all numbers which are successors of 0" in the language of Peano arithmetic alone?

DHMO

@MikeMiller $\left({0 \in A \land \left({\forall z \in A: s \left({z}\right) \in A}\right)}\right)$

just as in P5

@MikeMiller could you construct such a model?

Mike Miller

14:32

That says nothing about whether or not $z$ is of the form $S(S(S(\dots(S(0)\dots)$.

Secret

@MikeMiller In that case, it is unclear how can we reach n since S is only injective, with 0 not an image and obey induction

Mike Miller

Me? Nah. But it's been done many times. Any such model is necessarily pretty gross.

DHMO

@Secret just use induction without applying S many times

and if you make your proof more formal, you would find a hole somewhere

Mike Miller

IIRC in any such model addition is uncomputable.

DHMO

and you would need to invoke another axiom

so please make your proof more formal

Mike Miller

14:33

If you know a little logic, the existence of such a thing follows from the compactness theorem.

DHMO

@MikeMiller in such a model, does there exist another non-successor element?

Null

beware, atomic bomb incoming: proofwiki.org/wiki/…

DHMO

@Null hey, I asked you to prove it as an exercise, not find answers online

Mike Miller

@DHMO No. That 0 is the only element without a successor follows from the induction axiom.

Null

<- doesnt understand this

@DHMO i don't understand how one can say anything about $s$ by the axioms

Mike Miller

14:34

Consider the set $\{z \mid z = 0 \vee \exists n : s(n) = z\}$.

DHMO

@Null well every axiom is about $s$

Null

only that $s(n)$ is certainly >0

DHMO

@Null order is an atomic bomb

Mike Miller

It's patently obvious that this satisfies the assumption in the induction axiom.

DHMO

in fact I'm writing a proof about why order is a trichotomy on proofwiki

Null

14:35

@DHMO ah, see, then my approach is in general false

Tobias Shuxue Laoshi

@Rakso So you're swedish? Byt till matte nu!

DHMO

@Null just use induction

Null

@DHMO how can we if ordering is nonexistant?

Mike Miller

Induction is more or less all you can use to prove anything. So feel free to use it.

DHMO

@Null the axiom of induction is P5

proofwiki.org/wiki/Axiom:Peano%27s_Axioms/Formulation_1

Null

14:39

@DHMO do you understand it?

Akiva Weinberger

Business idea: a scarf you can tie like a tie.

But, like, not made out of tie, made out of scarf.

Mike Miller

would not purchase

DHMO

@Null I would write a better proof

Null

@DHMO I find the formulation of p5 confusing, but that doesn't say much :P

DHMO

@Null just treat P5 as induction

just use induction

Akiva Weinberger

14:41

I wonder if there's any sort of space where points divide the space into three sections instead of two

Mike Miller

all points?

Akiva Weinberger

like a sort of quadrachoomy rather than a trichotomy

Null

@DHMO so, basecase would be n=0, s(n)=0 by axiom p4

DHMO

@AkivaWeinberger could you give an example of a "trichotomy"?

Alessandro

A space where removing any point leaves you with 3 connected components?

Akiva Weinberger

14:42

like a sort of quadrachotomy rather than a trichotomy

DHMO

@Null yes

Mike Miller

@Null You can always define sets by whether or not they satisfy certain properties. So it's equivalent to say "if $0$ satisfies property, and every successor satisfies property, then everything satisfies property"

Secret

@Alessandro A topological space where every point is a 3-vertex?

Mike Miller

That's an interesting question. We have a resident point set topologist but I forget what name he uses right now.

Alessandro

That's an interesting question

Mike Miller

14:46

My mild suspicion is the answer is no if the space is Hausdorff... but I'm really not the person to ask about something like that.

Secret

I am confirmed by many to knew basically nothing about topology in 2016

Mike Miller

I do not understand that sentence.

Secret

many=sufficient topology maths chat users, as most discussion end up me almost completely confused and decided to go back to book again later

Null

@Secret i am confirmed to know basicly nothing about math.

^and about english language :D

Alessandro

Such a space would necessarily be $T_1$ though if I'm not mistaken

Mike Miller

14:51

Topology is big and hard.

How?

Null

@DHMO so my quest is showing $s(n+1)\not=n+1$ basically?

Alessandro

If you have a finite number of connected components they're clopen so you can write the complement of every point as an union of open sets so the singletons are closed

Null

@DHMO and the set $\mathbb{N}$ is only defined by peano, so we have no knowledge?

Alessandro

Hm, wait, nevermind, that doesn't work

Einer

@Null What do you mean when you say that $\mathbb{N}$ is defined by peano? Do you mean by Peano Arithmetics? Because that's definately not the case - there are many non-standard models of PA

Astyx

14:58

Hi

DHMO

@Null your quest is to show $s(n)\ne n \implies s(s(n)) \ne s(n)$

Alessandro

Ah, well, a $4$ elements set with the discrete topology works though

Mike Miller

@Alessandro I demand connected

>:(

Alessandro

Fair enough

DHMO

@Null and it is a direct result of one of the axioms

Mike Miller

15:00

@Einer He means to say that all we know about it is the axioms

DHMO

@MikeMiller every point in a discrete topology is connected

Null

@DHMO enough tips, i come back later :)

Alessandro

Gotta run for a physics lecture :/ I'll think about that question later though, it seems interesting

Mike Miller

@DHMO ...

DHMO

@MikeMiller i'm just kidding

Non-standard model of arithmetic

In mathematical logic, a non-standard model of arithmetic is a model of (first-order) Peano arithmetic that contains non-standard numbers. The term standard model of arithmetic refers to the standard natural numbers 0, 1, 2, …. The elements of any model of Peano arithmetic are linearly ordered and possess an initial segment isomorphic to the standard natural numbers. A non-standard model is one that has additional elements outside this initial segment. The construction of such models is due to Thoralf Skolem (1934). == Existence == There are several methods that can be used to prove the existence...

@MikeMiller Please see the section "From the compactness theorem"

is $x$ a successor of any number?

Mike Miller

15:02

Yes.

DHMO

which?

Mike Miller

the one before it!

DHMO

fair enough

Mike Miller

Like I said, it's just a straightforward application of induction. I claim that every number is either 0 or a successor. As proof: consider the set of numbers that are either zero or a successor. Then zero is in this set, and every successor is in this set. So it's everything.

Rakso

@TobiasShuxueLaoshi Yea im swedish! Vet inte om det är rätt val, måste tänka igenom det mer antar jag.

Tobias Shuxue Laoshi

15:05

@Rakso Understood.

DHMO

@MikeMiller Refer to $0 \in A \and n \in A \implies s(n) \in A$

if $x$ is not a successor of $0$

how do you know it is in $A$?

Rakso

@TobiasShuxueLaoshi Were you the guy emailing me? :)

Mike Miller

What? The only set in sight is $A$.

Tobias Shuxue Laoshi

@Rakso Indeed so. So the other people wont be bothered by all the Swedish.

Rakso

@TobiasShuxueLaoshi Ah, i've replied to your e-mail. We can talk there no problems. Were cunfused by your last name there and on your email account.

Tobias Shuxue Laoshi

15:07

@Rakso Good.

DHMO

@MikeMiller I mean, we construct a set $A \subseteq P*$

where $0 \in A \and n \in A \implies s(n) \in A$

can you prove that $x \in A$?

Null

@DHMO \wedge $\wedge$

DHMO

oh thanks

Rakso

@TobiasShuxueLaoshi Nice to make some Swedish mathematician contacts :)

Tobias Shuxue Laoshi

@Rakso Doing my best to fill the gap :)

Rakso

15:12

@TobiasShuxueLaoshi Do you study here in Sweden?

Sophie

Gud kommer det

Null

@DHMO assume: s(s(n))=s(n). Then by p3, s(n)=n. right direction?

DHMO

@Null yes

Tobias Shuxue Laoshi

@Rakso I do indeed. KTH/SU combo.

DHMO

@TobiasShuxueLaoshi do you speak Mandarin?

Rakso

15:14

@TobiasShuxueLaoshi KTH must be much better than were im currently at, Linnéuniversitetet

Null

@DHMO are we allowed to make a second assumtion in the step to lead it to a contradiction?

DHMO

@Null which assumption?

Null

@DHMO my first assumtion, for my induction is: $s(n)\not=n$

Tobias Shuxue Laoshi

@DMHO 会说，但是很久没有说。

Null

since that is the case for n=0

DHMO

15:16

@TobiasShuxueLaoshi 为什么?

Null

I don't know if inequalities are even helpful here

DHMO

@Null what is your second assumption?

Null

@DHMO In the step, i assume (to contradict it): s(s(n))=s(n)

DHMO

and then?

Null

then i get by p3, s(n)=n

DHMO

15:17

therefore?

Null

which obv contradicts my first assumtion

Tobias Shuxue Laoshi

@DHMO 因为我住在瑞典。我住在中国的时候很有跟中国人说话的机会。

DHMO

@Null Yes, it's called proof by contradiction

Null

but to me this looks not right :/

Tobias Shuxue Laoshi

@DHMO 这里没有那么多。

DHMO

15:19

@TobiasShuxueLaoshi 你学了中文多久？

Rakso

@TobiasShuxueLaoshi What was the reason you first choose CS if you liked math much more?

Mike Miller

@DHMO Sorry, let me rephrase. Consider the set of 0 and all successors. Then 0 is in this set, and if n is a successor, so is S(n). (Indeed, we don't even need to assume that n is a successor!)

Tobias Shuxue Laoshi

@DHMO 差不多三年。两年很努力地。

@Rakso Like so many other people in this oblong country -- maths anxiety.

DHMO

@TobiasShuxueLaoshi 三年就能看那么多字 :o

Null

@DHMO I don't understand why I can do this. I never did that in an induction proof before. So it could be just that I'm unfamiliar with it. Proof by contradiction I use normally somewhere else.

Rakso

15:22

@TobiasShuxueLaoshi So you started out with it because u weren't that good at maths? Because thats basically my reason for thinking about it

DHMO

@Null ok

Tobias Shuxue Laoshi

@Rakso I thought I wasn't good at maths. Turns out I was.

DHMO

@MikeMiller well, could you consider my set?

$0 \in A \land (n \in A \implies s(n) \in A)$

Tobias Shuxue Laoshi

@DHMO 不是很多。

Akiva Weinberger

@Null @DHMO Also \land ("logic and")

Null

15:23

$\land$

(it works)

Akiva Weinberger

Compare \lor $\lor$ and \lnot $\lnot$

Tobias Shuxue Laoshi

Gotta go back to my commutative algebra.

Null

@AkivaWeinberger that actually makes sense :)

Akiva Weinberger

The alternative for \lor is \vee $\vee$

So wedge=land and vee=lor

Null

$\neg$, \neg

Akiva Weinberger

15:25

Cool, didn't know that one

Null

didn't knew yours, yours are better for beginners I assume!

Akiva Weinberger

A cool thing also is that \in $\in$ backwards is \ni $\ni$

4

Null

$\not\ni$

DHMO

The Speed of Light is NOT About Light | Space Time | PBS Digital Studios

►

Null

@DHMO so a proof that proves that $s(n)\not=n$ for all $n\in\mathbb{N}$ really proves that this holds for any peano-set or?

DHMO

15:29

@Null yes

Akiva Weinberger

Also \ne and \neq are shortcuts for \not=

$\ne$ $\neq$ $\not=$

DHMO

Mike Miller

@DHMO I'm about to fly and I don't really much see the point. I gave a proof.

DHMO

@AkivaWeinberger wow they're really the same

Null

$\not\geq$

DHMO

15:31

10 mins ago, by Mike Miller

@DHMO Sorry, let me rephrase. Consider the set of 0 and all successors. Then 0 is in this set, and if n is a successor, so is S(n). (Indeed, we don't even need to assume that n is a successor!)

@MikeMiller I mean, we have $P^*$ right

does $P$ satisfy that?

Sophie

@DHMO I wish I had been taught physics this way. More method and less computation. At school we basically learned to replicate the examples we were shown

DHMO

@Sophie I agree

MathematicsStudent1122

Hi, I was wondering if anyone here has read "The Art of Computer Programming" by Knuth. How long will the entire work take to read?

Sophie

Depends. You can blitzkrieg the whole thing, but I recommend going slowly and reading and re-reading it at your own pace and doing all the exercises

Mithlesh Upadhyay

I answered this question, is all ball necessary to pick from the bag? The similar post does not say.

http://math.stackexchange.com/questions/2055508/a-bag-contains-9-balls-3-of-which-are-blue

Mike Miller

15:37

@DHMO I don't understand your question. I encourage you to figure it out.

DHMO

@MikeMiller are we assuming that apart from 0,1,2,... we have x?

Mike Miller

I cannot parse that sentence.

In every model of Peano arithmetic, the only non-successor is 0.

DHMO

Non-standard model of arithmetic

In mathematical logic, a non-standard model of arithmetic is a model of (first-order) Peano arithmetic that contains non-standard numbers. The term standard model of arithmetic refers to the standard natural numbers 0, 1, 2, …. The elements of any model of Peano arithmetic are linearly ordered and possess an initial segment isomorphic to the standard natural numbers. A non-standard model is one that has additional elements outside this initial segment. The construction of such models is due to Thoralf Skolem (1934). == Existence == There are several methods that can be used to prove the existence...

Please look at this

Mike Miller

Sigh. I know what's in the article.

DHMO

sorry

to me, it is saying that apart from 0,1,2,... we define x

Mike Miller

15:42

@DHMO In a nonstandard model of arithmetic, there is always a number greater than any standard natural, yes.

DHMO

and does the set of the "standard naturals" satisfy the five Peano's axioms?

Mike Miller

Yes.

DHMO

then by P5 x is out of concern

proofwiki.org/wiki/Axiom:Peano%27s_Axioms/Formulation_1

Mike Miller

I have no idea what that sentence means.

DHMO

P5 means the fifth Peano axiom

Mike Miller

15:44

Please stop linking me to that page. I know the axioms.

"x is out of concern" is the problem phrase. What does that mean?

Alessandro

You don't just add one x, when you do you add a lot more stuff to respect the axioms

DHMO

P5 states that P* is equal to the set of the standard naturals

Mike Miller

No it does not state that.

DHMO

Is the standard naturals a set $A$ such that $0 \in A$ and $n \in A \implies s(n) \in A$?

Mike Miller

Yes. There are other such sets equipped with a successor operation and element 0 satisfying the axioms.

Kushal Bhuyan

15:46

hey guys

DHMO

@MikeMiller I'm talking about the "standard naturals" inside our extended peano structure

Mike Miller

@DHMO How do you write down the standard naturals as a subset using the language of Peano arithmetic?

DHMO

@MikeMiller no idea

Alessandro

You can't

Mike Miller

Yes, that's the problem.

Kushal Bhuyan

15:50

Can anybody suggest me a good book on Commutative algebra ?

Mike Miller

Matsumura. Eisenbud-Harris. Atiyah-MacDonald.

Kushal Bhuyan

ok thanks @MikeMiller

abenthy

Are there suggestions on how to denote the maps $H_*(X;R) \to H_*(X;S)$ and $H^*(X;R) \to H^*(X;S)$ induced by a coefficient homomorphism $\varphi \colon R \to S$ for a space $X$? I would go by $\varphi_\#$ and $\varphi^{\#}$, but they are both covariant...

Null

16:09

so welldefinedness is simply another word for properly defined?

DHMO

Paraffin Paradox

►

Akiva Weinberger

In any model of arithmetic, if $x$ is a (nonzero) number, so are $x+1$, $x-1$, $2x$, $\lfloor x/2\rfloor$, $\lfloor x\sqrt2\rfloor$ (exercise), etc.

This is true even for nonstandard models of arithmetic

DHMO

@AkivaWeinberger the problem is you've invoked the reals by using $\sqrt{2}$

which model of arithmetic are we talking about?

Akiva Weinberger

Thus, the amount of nonstandard numbers in the model must be infinite or zero

@DHMO That's why it's not immediately obvious. There is a way to do it without invoking the reals. Peano Arithmetic

DHMO

@AkivaWeinberger $\sqrt{2}$ itself is in the reals

it isn't defined in Peano Arithmetic per se

Akiva Weinberger

16:21

I know this

Essentially, the idea is that one can prove that $\forall x\exists y:y^2\le2x^2\land (y+1)^2>2x^2$ (once you've defined $\le$ and $>$), and that that $y$ is unique

DHMO

@AkivaWeinberger this formulation is better

Akiva Weinberger

That $y$ is defined to be $\lfloor x\sqrt2\rfloor$.

And it exists even if $x$ is nonstandard, in a nonstandard model.

DHMO

@AkivaWeinberger than this formulation

I see

Semiclassical

hi chat

DHMO

hi

@AkivaWeinberger Well I would like to supplement that our addition and multiplication and ordering are all based on the Peano arithmetic so there's really nothing special about non-standard Peano arithmetics

$a \le b \iff \exists n: a + n = b$

abenthy

16:24

Maybe I should just mentally attack my readers by writing $C_*(X;\varphi)$ for the map $C_*(X;R) \to C_*(X;S)$ induced by $\varphi \colon R \to S$.

Balarka Sen

@MikeMiller IIRC Forever Mozart found a metric example.

But I don't know either.

Alessandro

I have a contruction that should work for that problem but haven't had time to think about it properly

Balarka Sen

I think I had one when Forever first told me but I can't remember anymore :(

Alessandro

But I'm thinking about something with a dense partial order that "forks" at every point when going up

Antonio Vargas

16:42

hi @Semiclassical

trilolil

Hello quick question here

Is a linear combination a special type of convolution?

arctic tern

huh?

trilolil

@arctictern huh!

arctic tern

16:58

I don't see any connection between the two. (Besides the fact that with finite settings, convolutions involve linear combinations).

You might as well ask if a banana is a special type of vehicle.

trilolil

@arctictern "convolutions involve linear combinations" absolutely.

so I was wondering whether one can consider calculating a linear combination the same as calculating a (special case) comvolution.

arctic tern

how so?

trilolil

Actually it seems you already answered my question. Convolution uses linear equations.

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$Mathematics$ Mathematics