Amazing - thanks so much! Very clever argument!

You can also use integration by parts.

What are these second 2's in there? Does that mean 2Sec^2(x)? or 2*Sec(2x) ? idk whats going on with these twos in this picture so im kinda at an impass.

What second 2's? They are just plugging in 2 for y and 0 for x. On the last column, 2sec2x means 2*sec(2x).

Trying to prove that the set (ℝ-{1},∘) with operation a∘b=a+b-ab is an abelian group.

Is proving commutativity as simple as the following:

Let x,y be elements of ℝ-{1}.
Then x∘y=x+y-xy=y+x-yx=y∘x.
Therefore ∘ is commutative.

Edit: the specific question would be: I can simply appeal to the fact that addition and multiplication are commutative on ℝ-{1}, correct? So in stating x+y-xy=y+x-yx, I'm not begging the question in any sort of way. Or do I have to demonstrate that addition and multiplication are commutative on ℝ-{1}?

You're fine. If you wanted to be extra pedantic, you could say:

"Since ordinary addition and multiplication commute on R, we see that x∘y=x+y-xy=y+x-yx=y∘x. Hence R-1 commutes under ∘."

Thanks.

WRT the bolded: If certain properties are true of R with respect to an operation (EG commutativity on R, associativity on R ), are those properties always going to be true of subsets of R?

If a statement is true for all elements of R, then it is true for all elements of any subset of R.

Sneaky proof:

Seems obvious when you put it that way .

I just peaked forward a few pages and saw "isomorphism" and "homomorphism" mentioned, so I'll have to look at the def closer to make sense of the proof here.

Yes, but not the closure property. Multiplying and adding real numbers always produce another real number, but that's not true of R-{1}.

Ace --

Don't get too far ahead of yourself, but an "isomorphism" is a structure preserving map. Basically a function f : (G, *) --> (H, **) [G, H groups] is an homomorphism of groups if f preserves the group structure of G.

f(g*h) = f(g) ** f(h)

If f is also a bijection (so we can map G one-to-one onto H and vice versa, and f creates basically an exact copy of G -- even with the same group structure and all -- in H), then we call f an isomorphism.

For now, G and H are "isomorphic" if they are the same group, maybe with different color hair and makeup.

I.e., the additive reals (R, +) are isomorphic to the positive reals under multiplication (R+, *) via the map f(x) = e^x. [This map is a bijection, and f(x+y) = e^(x+y) = e^x * e^y = f(x) * f(y).]

For a homomorphism, don't you also need that it maps inverses to inverses? For finite groups it follows, but otherwise I'm not sure.

It follows always. If f: G -> H is a h-ism and e is the unit of G, then f(e) is a unit of H (in fact the unit as the unit and inverses are always unique in a group) because f(g) = f(g*e) = f(g) ** f(e) for any g from G, and f(g)^{-1}=f(g^{-1}) for any g from G because f(g) ** f(g^{-1}) = f(g * g^{-1}) = f(e).

I have what I think is a measure theory related question.

Scenario 1
Bob has an infinite BR. Alice has 1 dollar.
They will flip a fair coin and bet $1 on each flip.
Bob will keep flipping until he has Alice' dollar.

According to measure theory, Alice's EV is -1, aka her edge is -100% (I would say Bob's EV is +1 but you can't add to infinity). I didn't believe this before (I tried to argue that it was 0), but then people mentioned measure theory and I trust that it's rigorously founded though I myself haven't learned it.

Bob can just flat-bet and there's a 100% chance he'll eventually cross the 0 mark to the +1 mark.
Or he can martingale. His betting pattern is irrelevant, either way Alice loses her dollar.

Scenario 2
Instead of fair coin-flipping, he's betting red/black in Roulette, or any game in which he wouldn't have a 100% chance of eventually pulling ahead by flat-betting.

Now, Alice has +EV if Bob flat-bets.

But what, according to measure theory, is Alice's EV if Bob martingales?

I have a feeling you guys will say the answer is -1, but then that would mean Alice's edge is affected by the betting pattern

There's no reason for the It's -1 in a useless technical sense based on the definition of EV given by the Lebesgue integral. Probabilist jason1990 once sent me a formal analysis of this, and he used it to explain the difference between these scenarios, but unfortunately it seems to have gotten lost when I backed up all my PMs along with other important things. I'm going to see if he has a copy, and if he does, I'll post it here. What's really important in both scenarios is not the EV but the limit of the EV as the number of plays goes to infinity. In both cases of scenario 2, the EV for Alice goes to infinity, and in scenario 1 it is always 0. The worthless technical results are due to a discontinuity at infinity. That commentary about what is useless and what is important is mine, not Jason's.

And I happen to agree with you. But I'm also interested in the technical answer for the philosophical and pure mathematical aspect.

Although, you seem to be saying it's the answer by definition rather than by application of axioms. In that case it's not interesting, because humans can define something however they want, doesn't necessarily provide insight into the nature of things.

I don't think I'll have a chance to take a measure theory course, since I plan to study Combinatorics. So maybe I'll read about it on my own but I imagine it's a difficult subject.

Incidentally, grad school is an excellent time to learn things unrelated* to what you're studying. The point is that you'll always learn what you need to in the area you're currently working. Take courses in other stuff while you have the time, energy, motivation, etc. Go to tons of seminars. Really learn math. Not just combinatorics. Also, I'd be really surprised if you don't have to take a measure theory course as a general requirement.

* asterisk due to the fact that you may somehow, someday want to relate the two. Or ask a question involving the "unrelated" guy, or ...

PM if you don't want to discuss here, but what field of combinatorics (/where are you going)? What kind of problems interest you, etc? Why combinatorics?

Ah, I wasn't sure how grad school worked. I kinda thought my whole schedule would have to be devoted to my area of specialization. If that's not the case then cool, I get to learn more math.

Don't really know yet. Most likely enumerative or probabilistic. My graph theory professor says University of Delaware would be a great choice for combinatorics. Though, I didn't think to ask about whether my choice of school should depend on the branch of combinatorics I'm seeking.

I'm just weirdly drawn to it, don't really know why. Learning any new little thing related to combinatorics excites me. For instance, just realizing that there's a different way of solving or understanding a familiar easy problem makes my day. When I get into a problem, it's like an obsession and all other things get put on hold. Combinatorics is the only subject that has that effect on me or that I have that kind of passion for. Though I do find pretty much all math interesting.

I can confirm that

I sent a PM already, but I think this really depends on the type of combinatorics you want to study (and where you want to end up in your career, how competitive you are in the application process, etc).

For me, Delaware would have been a pretty decent choice. I really like some of the problems that folks there work on. But the problems for the most part aren't objectively sexy, so it's unlikely one would end up at a top tier research university doing that kind of work. But it all depends on your goals, etc.

When I left 10 years ago they had world class experts on finite planes, extremal graph theory, and difference sets. I'm not sure if all of them are still there.

Yes afaik, and these folks really are excellent mathematicians.

My point was just that the Harvards, Berkeleys, and MITs of the world aren't really hiring guys doing difference sets. I think they are awesome and that they give rise to some excellent problems (I did an undergrad thesis 10 years ago (ok, 8 yrs ago) in the area). Career-planning is a giant optimization, and this is just a piece of data to factor in.

I think it was my idea. I told him I might be a combinatorialist if I could do my life over, but I'm too old now. So he's going to do that. Machine learning is cool too.