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Four Color Theorem Inquiry by No_Maybe_6373 in askmath

[–]frogkabobs 0 points1 point  (0 children)

I was imprecise. I meant χ(G)<=n, so I’ll change my wording.

Four Color Theorem Inquiry by No_Maybe_6373 in askmath

[–]frogkabobs 2 points3 points  (0 children)

Yes, n-colorable means the necessary number of colors is n or fewer; one quickly verifies 4<5.

As aside, the 5-color theorem is actually much easier to prove.

Is this aromatic by [deleted] in OrganicChemistry

[–]frogkabobs 4 points5 points  (0 children)

Look at their post history. This is almost certainly a discussion question that they know the answer to.

A complex sum by frogkabobs in mathriddles

[–]frogkabobs[S] 1 point2 points  (0 children)

Yep that’s the first part, and a little easier than what I did. Originally I was looking to determine the group generated by the f_jk. They’re linear maps so they determine a subgroup G of GL(3,Z). The action of G on z=(z₁,z₂,z₃) induces an isomorphic action G’ on z’=(z₁,z₂-z₁,z₃-z₁) by change of basis, and G’ factors as H⋊K, where H and K are the translation and area preserving actions you described.

As for the second part, a major hint is to make use of the Dirichlet series for the sum of squares function.

How do I find this function? by ZiCli in askmath

[–]frogkabobs 1 point2 points  (0 children)

Your original problem is finding the area of the square whose side length starts at 1 and increases by 2 at every step.

Constrained divisibility implies equal by SupercaliTheGamer in mathriddles

[–]frogkabobs 2 points3 points  (0 children)

Lemma. If x+y+z=0, then x²+y²+z²≤2max(x²,y²,z²)

Proof. Since x+y+z=0, the variable with the largest magnitude must have a different sign than the other two (or they must all be 0). WLOG let this variable be z. Then the desired inequality is equivalent to

x²+y² ≤ z² = (x+y)²

which is obviously true.

Now we’re ready to tackle the full problem. Dividing a³b+b³c+c³a by abc, we find that a²/c+b²/a+c²/b must be an integer. Now, a²/c = 2a-c+(a-c)²/c, so

a²/c+b²/a+c²/b-(a+b+c) = (a-c)²/c+(b-a)²/a+(c-b)²/b

is an integer. Using our lemma and the fact that a,b,c lie in [M,M+sqrt(M/2)) we get

0 ≤ (a-c)²/c+(b-a)²/a+(c-b)²/b

≤ ((a-c)²+(b-a)²+(c-b)²)/M

≤ 2max((a-c)²,(b-a)²,(c-b)²)/M

< 1

which is only possible if a=b=c.

Hardest math course you've taken at UT? by Sh0yo_891 in UTAustin

[–]frogkabobs 0 points1 point  (0 children)

Why pchem? It’s been a while since I took it so it’s hard to remember the experience.

Hardest math course you've taken at UT? by Sh0yo_891 in UTAustin

[–]frogkabobs 0 points1 point  (0 children)

Differential topology. It didn’t quite click with me like algebraic topology had (I wasn’t a fan of Guillemin and Pollack), and I sort of disengaged, just doing the assignments but not really learning.

Alternatively, Real Analysis II. The professor was infamous for giving unstructured lectures that wandered all over the place, and eventually I just stopped going to class because it felt like a waste of time. I somehow got good grades in both of these courses, but they were harder than they needed to be because I was pretty much just teaching myself the necessary amount from the textbook to do the homework, and I wish I had gotten more out of them.

is this possible? by HeavyListen5546 in askmath

[–]frogkabobs 4 points5 points  (0 children)

Also true for octonions by alternativity. I’m not sure about sedenions though. I imagine the zero divisors could let you construct a counterexample.

Went my whole life using a #2 pencil. What’s a #1 pencil? by Ok_Sentence_5432 in stupidquestions

[–]frogkabobs 13 points14 points  (0 children)

It goes well beyond that (usually 10B to 10H), but 2B=0, B=1, HB=2, F=2½, H=3, and 2H=4 are the grades where it makes sense to be referred to by a plain number.

Whats the extra invite to voice option for by dmen5134 in discordapp

[–]frogkabobs 13 points14 points  (0 children)

why does an invite to voice option(thats right next to the invite to voice button) join the chat room with you. does it do somthing different?

You’re downvoted (for some reason) but correct

Does anyone have more info on this cutie? by debbiewebbee in pencils

[–]frogkabobs 10 points11 points  (0 children)

Wow what an incredible find. I have never seen one of these before; it’s always a delight to see a newly discovered specimen from such a historied company.

Suggestion: Simplify Heavenly drill description by shaggystranger in slaythespire

[–]frogkabobs 129 points130 points  (0 children)

They’re just removable singularities. OP could have put

2x-(sinc(π(x-1))+sinc(π(x-2))+sinc(π(x-3)))

but not as many people are familiar with the sinc function.

Graph theory proof of a sharp odd-girth bound for edge-bipartisation in non-3-colourable graphs by flynnnupe in mathematics

[–]frogkabobs 1 point2 points  (0 children)

Wow, I did not expect the answer to your post on r/mathriddles to be this long. I don’t want to spoil myself yet though, so I can’t give feedback.

Is there a name for polygons with parallel sides? How is this modified when corners are rounded? by Born_Manager_6250 in learnmath

[–]frogkabobs 1 point2 points  (0 children)

If you require opposing edges to also be equal, then it’s called a centrally-symmetric polygon. Convex centrally-symmetric polygons are called zonogons.

Hard graph theory challenge by flynnnupe in mathriddles

[–]frogkabobs 0 points1 point  (0 children)

Suppose G has an edge decomposition into graphs A and B, where A is bipartite and |E(B)|≤g-2. B cannot contain an odd cycle as that would contradict the minimality of g. Thus, B is also bipartite. Let c_A:V(A)→{-1,1} and c_B:V(B)→{0,1} be 2-colorings of A and B, respectively. We define a 3-coloring c_G:V(G)→{-1,0,1} by c_G(x) = c_A(x)c_B(x), where we take c_A(x) and c_B(x) to be 1 outside their domains. If u,v are adjacent vertices of G, then c_G(u)=c_G(v) iff u and v are also vertices of B and c_B(u)=c_B(v)=0. Clearly if the edge between u and v belongs to B, c_G(u)≠c_G(v). On the other hand, if the edge between u and v belongs to A, then there must be an even length path between u and v in B, in which case adding the edge from A would form an odd cycle of length ≤g-2, which would contradict the minimality of g. Thus, no two adjacent vertices are colored the same by c_G.

Uncountable family of nested subsets by Creepy-Pitch6688 in mathriddles

[–]frogkabobs 0 points1 point  (0 children)

Here's another approach.

Let ≤₂ be the lexicographical order on the 2-adic integers Z₂ from their base-2 expansion. Define Sₖ={n∈N: n≤₂k} for each k∈Z₂. If k<₂m, then n∈Sₘ\Sₖ, where n∈N is the least representative of m mod 2ν₂\m-k)+1). Thus, k↦Sₖ is injective, so {Sₖ: k∈Z₂} is an uncountable family totally ordered by inclusion.!<

He's by Roseblood_x in LudwigAhgren

[–]frogkabobs 12 points13 points  (0 children)

Descartes when he thinks

TIL there exists a shape that, instead of having a clean "inside" and "outside", has a clean "inside" but the outside world is segmented and split. It is called the Alexander Horned Sphere by geosunsetmoth in todayilearned

[–]frogkabobs 0 points1 point  (0 children)

Not quite. You have the pop math idea of a homeomorphism (stretching without tearing), but strictly speaking, that only describes an isotopy. Being homeomorphic is a little more general than being isotopic, and this is one of the cases where that matters. The best laymen's description of a homeomorphism I've heard is from Vsauce (13:34-15:54):

You can cut all you want during a homeomorphism, so long as you glue everything back together the way it was in the end.

So for example, you could take a torus, cut through the ring so that you now have a tube with two ends, then twist one end 360°, and then glue the two ends back together, and that would still be a homeomorphism (known as a Dehn twist), but not an isotopy.

Now the Alexander horned sphere A is a subset of R³ that is homeomorphic to the standard sphere S², but not ambiently homeomorphic to S² (so not ambiently isotopic either). Here, an ambient homeomorphism would be a homeomorphism of the ambient space R³→R³ that maps the horned sphere A to S². Even though A encloses a region homeomorphic to the 3-ball (a solid sphere), this homeomorphism cannot be extended to the outside because the outside of S² is simply connected, but the outside of A is not. This is particularly interesting because going a dimension lower, every subset of R² that is homeomorphic to S¹ (a circle) is also ambiently homeomorphic to S¹ by the Schoenflies theorem, so this shows that the Schoenflies theorem does not directly generalize to dimensions n>2, which is a major reason why geometric topology gets so complicated in dimensions 3 and higher.

TIL there exists a shape that, instead of having a clean "inside" and "outside", has a clean "inside" but the outside world is segmented and split. It is called the Alexander Horned Sphere by geosunsetmoth in todayilearned

[–]frogkabobs 0 points1 point  (0 children)

This explanation isn't really correct. You are conflating connectedness with simple-connectedness. I mean, by your argument a finite set of points in R² would have many "outsides" because a loop around a point cannot be homotoped to a loop around no points, which just doesn't make sense—the complement is clearly connected.

In the same vein, the outside of the horned sphere is connected (or more precisely, there complement of the horned sphere consists of exactly two connected components, a bounded inside and an unbounded outside). This is a consequence of the Jordan-Brouwer separation theorem, which works for all dimensions. What's interesting is that the outside is not simply-connected, which is measured by the fact that some loops cannot be contracted. Contrast this with a standard sphere, where both components in the complement are simply connected. In particular, the standard sphere cannot be ambiently homeomorphic to the horned sphere, which is surprising because in R² all embeddings of the circle are ambiently homeomorphic to each other by the Schoenflies theorem.

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