Can chatgpt accurately calculate expected lengths in Pascal's triangle?

[Office_Shredder]

TL;DR Summary

Prove you are smarter than the bot!

Chatgpt is actually pretty good at generating math problems. It's awful at solving them. I guarantee every question posted here cannot be solved by chatgpt, but maybe can be solved by a human? My plan is to spend a couple minutes getting a question I think it's cool and then posting it here - I don't know if I'll actually do it each day.

Since chatgpt is bad at knowing whether things are true or not, and I'm not going to try to solve all of these before posting, most will be of the form prove or find a counterexample
#5.) .I take a piece of string of length 1, and do the following k times: cut the string into a ratio of 2 to 1 (so a string of length 1 is cut into a string of length 2/3 and a string of length 1/3) I then throw out one of the pieces, leaving myself with a single piece that I can repeat this process on, until I have made k cuts.

What is the expected length of the final piece of the string in terms of k, if
a.) Each time I pick one of the two pieces to throw out randomly with equal probability
b.) Each time I "grab" a random spot on the string and throw that piece out - i.e. I have a 2/3 probability of throwing out the longer piece

6.) If you start an infinite random walk on the integers where each step is 50/50 to be one unit left or one unit right, show the expected number of times you return to the origin is infinite.

Bonus question: what about a walk in 2 or more dimensions? Slightly handywavy arguments are acceptable.

#11) given three points on a plane, not collinear, prove there exists a unique circle passing through them.

#12) Prove or disprove: Every graph of 30 vertices has two vertices with the same degree (same number of edges)

Is 30 important here? It actually asked me a question about friends that might have been a mix between a graph theory question and the birthday problem but this is what I reduced it to.

#13) Prove or disprove: Every infinite dimensional Banach space has a linearly independent sequence that is not a basis.

Solved problems (no solutions in the spoiler so feel free to take a crack at them)

Spoiler

#1.) Either prove every compact connected metric space is path connected, or find a counterexample.

fresh42 found a counterexample

#2.) Prove every finite group $G$ has a normal subgroup $H$ such that $G/H$ is simple, or find an example of $G$ for which this is not true. Bonus: if it's true, does it require that $G$ is finite?

solved by Infrared for the finite case

and the bonus infinite counterexample!

#3) if $f:\mathbb{C}\to \mathbb{C}$ is holomorphic, and $f(z)\in \mathbb{R}$ whenever $|z|=1$, either prove $f$ must be constant or find a counterexample

solved by Infrared

#4.) If $a,b,c$ are positive integers such that $a+b+c=72$, find the maximal possible value of $(a-b)^2+(b-c)^2+(c-a)^2$.

fresh_42 with a very elegant transformation

#7). Prove or give a counterexample: Every irreducible complex representation of a compact group is unitary

Infrared observes it doesn't have to be unitary under a random choice of inner product

but you can pick an inner product that makes this work

#8) Prove or give a counterexample:
Every ideal in a commutative ring is the intersection of the prime ideals containing it.

fresh_42 with a simple counterexample

#9) Prove every lebesgue measurable set of positive measures contains measurable subsets of arbitrarily small positive measure, or provide a counterexample.

Infrared shows there exists a subset of every smaller measure

#10) Prove or disprove: Every uniformly continuous function that is differentiable on a closed interval is necessarily Lipschitz continuous.

Counterexample by infrared

 

[fresh_42]

False. $$(X,\|\cdot\|_2)=\{(x,y)\in \mathbb{R}^2\,|\,(y=\sin(x^{-1}) \wedge 1\geq x> 0)\vee (x=0 \wedge -1\leq y\leq 1)\}.$$

 

[berkeman]

False.

QED. fresh is smarter than the bot.

 

[fresh_42]

QED. fresh is smarter than the bot.

I'm not sure it's closed, so it might fail to be compact. Topology is a b...
$\{(1/n,\sin(n))\}$ has no convergent subsequence that I see.

 

[Office_Shredder]

If you have an infinite cover of open balls, you only need finitely many to cover the y axis component, and I think those guarantee there exists $\delta>0$ such that $(x,\sin(1)x)$ is covered if $\delta<x$?

 

[fresh_42]

If you have an infinite cover of open balls, you only need finitely many to cover the y axis component, and I think those guarantee there exists $\delta>0$ such that $(x,\sin(1)x)$ is covered if $\delta<x$?

Yes, I forgot that the y-axis balls have to have a positive diameter.

 

[Office_Shredder]

Number 2 is posted!

#2.) Prove every finite group $G$ has a normal subgroup $H$ such that $G/H$ is simple, or find an example of $G$ for which this is not true. Bonus: if it's true, does it require that $G$ is finite?

 

[Swamp Thing]

TL;DR Summary: Prove you are smarter than the bot!

Chatgpt is actually pretty good at generating math problems. It's awful at solving them.

An experiment combining the power of GPT with that of Wolfram Alpha:-

 

[Office_Shredder]

#3 if $f:\mathbb{C}\to \mathbb{C}$ is holomorphic, and $f(z)\in \mathbb{R}$ whenever $|z|=1$, either prove $f$ must be constant or find a counterexample.

 

[topsquark]

#3 if $f:\mathbb{C}\to \mathbb{C}$ is holomorphic, and $f(z)\in \mathbb{R}$ whenever $|z|=1$, either prove $f$ must be constant or find a counterexample.

Does it always include the counterexample comment? If so it seems a bit lukewarm to me.

-Dan

 

[Office_Shredder]

Does it always include the counterexample comment? If so it seems a bit lukewarm to me.

-Dan

I ask chatgpt to generate questions. Often the question is wrong (for problem 2 it asked something that was flat out incorrect and i adjusted it). The point of this exercise is I can easily find questions that I think are interesting, and hopefully a lot of them actually are. I'm trying to filter out dumb things (the original version of problem 3 was about the real axis instead of a circle, which is uh, obvious) and things that are just dumbly hard, but I am generically not solving every problem before posting it here. The only way we will know if these claims are true is if someone here solves the problem.

 

[Infrared]

3. Looks true. The imaginary part of such a holomorphic function is harmonic and vanishes on the unit circle and so vanishes everywhere in the interior (for example, by Green's identities). By Cauchy-Riemann, the real part is constant there. So $f$ is constant inside the unit circle and hence everywhere

2. You need to assume $G\neq\{1\}.$ Take $H$ to be a maximal proper normal subgroup, which obviously exists if $G$ is finite. By a correspondence theorem, the normal subgroups of $G/H$ are of the form $A/H$ where $H\subseteq A\subseteq G$ and $A$ is normal in $G,$ so $A/H$ is either trivial or the whole quotient $G/H$, meaning $G/H$ is simple. Finiteness is not required: $\mathbb{Z}/2\mathbb{Z}$ is simple.

Note: By Zorn's lemma, I think every group should have a maximal normal proper subgroup. I don't think finiteness is required. (Edit: this last part is wrong as I later realize)

 

[Office_Shredder]

I think everything @Infrared wrote is right! I was thinking about 3 a bit more. If you draw a curve that splits the complex plane into two, and ask is there a non constant holomorphic function on all of $\mathbb{C}$ that is real valued on this curve, the answer is no if the curve creates a bounded section, yes if the curve is a line (you can create a linear function that satisfies this). What about other curves? Is there a holomorphic function that is real on all points of the form $x+x^2i$ for example?

#4
If $a,b,c$ are positive integers such that $a+b+c=72$, find the maximal possible value of $(a-b)^2+(b-c)^2+(c-a)^2$. Don't worry, I didn't forget about the high school level problems! Chatgpt claims this is a classic high school question, I'm curious if anyone has ever explicitly seen it before.

 

[fresh_42]

How about $(A_5\times A_5) \ltimes_\varphi (V_4 \times \mathbb{Z}_5)$ where the automorphism $\varphi$ acts in a way that you cannot get $A_5$ factored out? The idea is to give it a semisimple part that cannot be factored into simple ones. Or going directly to where the idea stems from:
$$
\left(\mathfrak{sl}(3,\mathbb{F}_7) \times \mathfrak{sl}(11,\mathbb{F}_7)\right)\ltimes \mathfrak{R}
$$
with a radical $\mathfrak{R},$ say $\mathfrak{R}\cong \mathbb{F}_{7}^{33},$ that acts on both direct factors in a way that doesn't allow to factor out a single simple subalgebra. This is a matrix algebra and therefore the Lie algebra of a matrix group with the required properties.

I wouldn't bet that such a construction is impossible.

However, with $H=G$ we always have the simple factor $\{1\}.$ So it is either trivially true, or if we rule out the simple solution, probably false.

 

[fresh_42]

Just a possibly stupid question: Is $f(re^{i\varphi })=\varphi\cdot r$ holomorphic?

 

[fresh_42]

4) $2\cdot 69^2=9522$

 

[Infrared]

Just a possibly stupid question: Is $f(re^{i\varphi })=\varphi\cdot r$ holomorphic?

Aside from the fact that $\varphi$ is not uniquely defined given $e^{i\varphi}$, still no. Your function is always real with nonconstant real part, so it can't possibly satisfy the Cauchy-Riemann equations.

I think everything @Infrared wrote is right! I was thinking about 3 a bit more. If you draw a curve that splits the complex plane into two, and ask is there a non constant holomorphic function on all of $\mathbb{C}$ that is real valued on this curve, the answer is no if the curve creates a bounded section, yes if the curve is a line (you can create a linear function that satisfies this). What about other curves? Is there a holomorphic function that is real on all points of the form $x+x^2i$ for example?

Since every harmonic function is the real (or in this case imaginary) part of a holomorphic function, your question should be equivalent to asking whether there is a nonconstant harmonic function vanishing on your curve. Google finds me a paper considering exactly this question: https://www.ams.org/journals/tran/1...-1966-0197755-5/S0002-9947-1966-0197755-5.pdf
In particular, if you go to section III.2, they show that it is possible for a parabola.

 

[Infrared]

However, with $H=G$ we always have the simple factor $\{1\}.$ So it is either trivially true, or if we rule out the simple solution, probably false.

The definition I'm familiar with (and I think is standard) excludes the trivial group from being simple. Anyway I don't see anything wrong with my argument in post 12?

 

[fresh_42]

The definition I'm familiar with (and I think is standard) excludes the trivial group from being simple. Anyway I don't see anything wrong with my argument in post 12?

#2.) Prove every finite group $G$ has a normal subgroup $H$ such that $G/H$ is simple, or find an example of $G$ for which this is not true. Bonus: if it's true, does it require that $G$ is finite?

I want to construct a finite, non-simple group $G$ such that subgroups $H$ aren't normal, or if they are normal, then $G/H$ isn't simple. My idea was to take a semisimple, non-simple subgroup, and a solvable normal subgroup with an action that doesn't allow splitting the semisimple parts into single normal simple ones. Something like $\left(SL(3,\mathbb{F}_7)\times SL(3,\mathbb{F}_7)\right) \ltimes (\mathbb{F}_7^*)^9$ or similar with a sufficiently complicated automorphism. If you exclude $G=H$ then Zorn's lemma runs into an empty set!

 

[Infrared]

If you exclude $G=H$ then Zorn's lemma runs into an empty set!

Okay you're probably right here but anyway for finite groups (which it looks like yours is) there are obviously maximal normal subgroups. No Zorn's lemma necessary! I really don't see where my argument can fail.

 

[fresh_42]

Okay you're probably right here but anyway for finite groups (which it looks like yours is) there are obviously maximal normal subgroups. No Zorn's lemma necessary! I really don't see where my argument can fail.

I see where the problem is. The groups you get by the correspondence theorem do not need to be isomorphic to normal subgroups of $G$.

Say $G=A_5^2\ltimes_\varphi S$ where $S$ is a solvable normal subgroup. $G/S\cong A_5^2$ is not simple, and I do not see that $A_5$ can be normally embedded in $G$ if only $\varphi$ runs between the two simple factors, i.e. is complicated enough, or maximal away from the identity. So we have simple subgroups that are not normal, and normal subgroups that are not simple and have no simple factor groups. you simply cannot lift the $A/H.$

 

[Infrared]

I see where the problem is. The groups you get by the correspondence theorem do not need to be isomorphic to normal subgroups of $G$.

I only used the last bullet point here: https://en.wikipedia.org/wiki/Correspondence_theorem
(that normal subgroups of $G/H$ are exactly $A/H$ for $A$ a normal subgroup of $G$ containing $H.$)

Say $G=A_5^2\ltimes_\varphi S$ where $S$ is a solvable normal subgroup.

Can you make a concrete example out of this? Specify $S$, the conjugation action of $A_5^2$ on $S$, and give me a maximal normal subgroup of this semi-direct product.

 

[fresh_42]

I only used the last bullet point here: https://en.wikipedia.org/wiki/Correspondence_theorem
(that normal subgroups of $G/H$ are exactly $A/H$ for $A$ a normal subgroup of $G$ containing $H.$)Can you make a concrete example out of this? Specify $S$, the conjugation action of $A_5^2$ on $S$, and give me a maximal normal subgroup of this semi-direct product.

I do not have an example, but you claim there is none. That is a strong statement. My idea comes from the decomposition of Lie algebras into a solvable radical and a semisimple subalgebra. Take a matrix algebra, that guarantees the existence of the corresponding groups, and - if required - make it finite. I simply cannot see why this should fail regarding the myriad of examples.

 

[Infrared]

I do not have an example, but you claim there is none. That is a strong statement.

Then can you please tell me exactly where in my proof you think there is a flaw? I can't tell which claim of mine you say you are refuting in your post 21.

 

[fresh_42]

2. You need to assume $G\neq\{1\}.$

$G=A_5^2\ltimes H$ with $H$ solvable.

Take $H$ to be a maximal proper normal subgroup, which obviously exists if $G$ is finite.

Given.

By a correspondence theorem, the normal subgroups of $G/H$ are of the form $A/H$ where $H\subseteq A\subseteq G$ and $A$ is normal in $G,$ so $A/H$ is either trivial or the whole quotient $G/H$, meaning $G/H$ is simple. Finiteness is not required: $\mathbb{Z}/2\mathbb{Z}$ is simple.

This means for our example:

The normal subgroups of $G/H\cong A_5^2$ are of the form $A/H$ where $H\subseteq A\subseteq G$ and $A$ is normal in $G,$ so $A/H$ is either trivial or the whole quotient $G/H$, meaning $G/H$ is simple.

There are no normal subgroups that contain $H$ in my example. So $A\in \{G,H\}.$ Now $G/H\cong A_5^2$ is not simple and $H/H=\{1\}$ isn't simple either.

Note: By Zorn's lemma, I think every group should have a maximal normal subgroup. I don't think finiteness is required.

I don't think finiteness has anything to do with it. I started with infinite matrix groups and made them artificially finite to fulfill the requirements. Drop the requirement and I still have an example.

 

[Infrared]

There are no normal subgroups that contain $H$ in my example.

Why not? (and I assume you mean proper)

 

[fresh_42]

Why not? (and I assume you mean proper)

Sure, $H$ contains $H$. Because I have a proper semidirect product and
$$
G=A_5^2\ltimes H \ncong A_5 \times \left(A_5\ltimes H\right)
$$

I admit that I have to prove existence, but so do you have to prove non-existence. I gave you a construction proposal via matrix groups. At least a bit. Say we have ($31$ in order to have some scalars)
$$
G=\left\{\begin{pmatrix}
&&&&&&0\\
&\operatorname{SL}(3,\mathbb{F}_{31})&&&0&&0\\
&&&&&&0\\
&&&&&&0\\
&0&&&\operatorname{SL}(3,\mathbb{F}_{31})&&0\\
&&&&&&0\\
\mathbb{F}_{31}&\mathbb{F}_{31}&\mathbb{F}_{31}&\mathbb{F}_{31}&\mathbb{F}_{31}&\mathbb{F}_{31}&\mathbb{F}_{31}^*
\end{pmatrix}\right\}
$$

 

[Infrared]

Sure, $H$ contains $H$. Because I have a proper semidirect product and
$$
G=A_5^2\ltimes H \ncong A_5 \times \left(A_5\ltimes H\right)
$$

I can't tell how this is an argument that there are no proper normal subgroups of your group that strictly contain $H.$

I admit that I have to prove existence, but so do you have to prove non-existence.

I already gave a proof. You haven't told me where you think it's wrong, only that you don't think it's true for some class of examples. Please point out specifically which step in my logic you believe is wrong and why.

 

[fresh_42]

I can't tell how this is an argument that there are no proper normal subgroups of your group that strictly contain $H.$I already gave a proof. You haven't told me where you think it's wrong, only that you don't think it's true for some class of examples.

You did not. There are examples where there is no normal $A$. Hence the correspondence theorem doesn't provide normal subgroups. What's wrong with my example? $A_5^2$ is not simple, and $A_5$ not normal. And even if $A_5$ does not work, what about my matrix group? And I can extend it with additional rows at the bottom to make sure that none of the $\operatorname{SL}(3)$ becomes normal. Same as with the $A_5$ example, which I only took since it's easier to type. The principle stands: no normal groups between $H$ and $G$ and $G/H$ not simple.

 

[Infrared]

You did not. There are examples where there is no normal $A$. Hence the correspondence theorem doesn't provide normal subgroups.

I don't understand this at all. The correspondence theorem goes both ways. Every (normal) subgroup of $G/H$ is of the form $A/H$ where $A$ is a (normal) subgroup of $G$ containing $H.$ Since there are no normal subgroups of $G$ containing $H$ besides $G$ and $H$ (by maximality), a normal subgroup $A/H\subset G/H$ must be either trivial or the full group, i.e. $G/H$ is simple.

What's wrong with my example?

You haven't justified why there are no normal subgroups strictly between $H$ and $G.$

 

[fresh_42]

I don't understand this at all. The correspondence theorem goes both ways. Every (normal) subgroup of $G/H$ is of the form $A/H$ where $A$ is a (normal) subgroup of $G$ containing $H.$ Since there are no such normal subgroups besides of $G$ containing $H$ besides $G$ and $H$ (by maximality), a normal subgroup $A/H\subset G/H$ must be either trivial or the full group, i.e. $G/H$ is simple.

Not if $A=H$ is the only possibility. Then we get $H/H=1$ and the correspondence theorem is useless.
Why can't all subgroups with $H\subseteq A$ fail to be normal?

You haven't justified why there are no normal subgroups strictly between $H$ and $G.$

Because the semidirect product isn't a direct product. Look at my matrix example, possibly extended by another row at the bottom. I haven't checked, but I think you cannot factor a single copy of $\operatorname{SL}(3).$

 

[Infrared]

Not if $A=H$ is the only possibility.

$A$ can always be either $H$ or $G.$

Then we get $H/H=1$ and the correspondence theorem is useless.

I don't see the relevance of this statement.

Why can't all subgroups with $H\subseteq A$ fail to be normal?

It can certainly happen that $H$ and $G$ are the only possibilities for $A.$ This is exactly when my argument says $G/H$ is simple.

Because the semidirect product isn't a direct product.

I know what a semidirect product is. Why does this mean there aren't normal subgroups between $H$ and $G?$

 

[fresh_42]

I know what a semidirect product is. Why does this mean there aren't normal subgroups between $H$ and $G.$

And I knew that $H\subseteq H.$

Why not? (and I assume you mean proper)

I'm only saying that there are examples where it is the case that the semisimple part is only a subgroup and does not provide normal factors. It is an opinion driven by the analog situation with Lie algebras. This is also a kind of correspondence.

The correspondence principle simply does not apply in the case that there are no normal subgroups in between. So your proof fails in the situation I gave. $G/H$ is not simple and all subgroups $H\subseteq A\subseteq G$ are not normal. Why should this be impossible?

 

[Infrared]

And I knew that $H\subseteq H.$

I think you misunderstood my earlier post. $G$ is always a normal subgroup of $G$ that contains $H.$ I was suggesting that you meant there are no proper normal subgroups (of $G$) that strictly contain $H.$

The correspondence principle simply does not apply in the case that there are no normal subgroups in between.

This is untrue. The correspondence theorem holds for any group quotient.

 

[Office_Shredder]

.
The correspondence principle simply does not apply in the case that there are no normal subgroups in between. So your proof fails in the situation I gave. $G/H$ is not simple and all subgroups $H\subseteq A\subseteq G$ are not normal. Why should this be impossible?

Is your assertion here that there is a subgroup $K\subset G/N$ that cannot be written as $K'/N$ for some $K'\subset G$? I think this is false - $K'=KN$ should work (also the statement of the correspondence theorem is that it is a bijection)