duality-theoremsterminology
I'm not a math expert but I know a little bit of calculus and theorems. I've heard things like "this result is "dual"", or this "theorem is "dual"". Often people say "dual comes for free". Like you swap variables or something and you get another result.
I've never understood this. Can anyone explain what "dual" means precisely? Examples would be nice as well.
Thanks!
I've been in your situation and will address several points from your question.
You write that you've been studying Russian for a year and are proficient. What does that actually mean? For example, how well do you understand Russian verbs: aspect, verbs of motion, transitive vs. intransitive, conditional using бы? Reading a lot will definitely improve your understanding of these grammatical points over time, but I would not have considered myself proficient after just 1 year of study and you do, so I'm curious how you judge that you are proficient now. Can you have conversations with native speakers about random topics already?
The link you give from the Univ. of Bonn has all the accent marks stripped off the Russian words. The person who created that webpage explains at the top -- in Russian -- why that was done, but for English speakers it's a bad idea not to have the accent marks because you should learn how to correctly pronounce words. (There will still be important subtleties, e.g., конечно in math.) You should get the offline version of the Lohwater dictionary, which has the accent marks. Also look at Gould's Russian for the Mathematician.
You say that your understanding is enhanced by connecting different disciplines in meaningful ways. I don't understand what you really have in mind here. Do you expect to understand the snake lemma better by knowing Russian? Frankly I don't think you're going to improve your understanding of math in a meaningful way by learning Russian unless your goal is to read something in Russian that has not been translated into English, and what might that be? The textbooks you mentioned by author names are already translated into English and you're not going to have a series of amazing mathematical insights by reading the original book instead of the translation. (I'm talking about math, not literature.) [Edit: I would agree that you can improve your Russian by reading math which interests you in Russian, so in the direction math ==> Russian I see that one discipline can assist the other. Reading math in Russian improved my grasp of several grammatical concepts, which I then used in conversations about everyday non-math topics.]
You say that you want to be able to speak, write, and read math in Russian. You left out listening, which goes together with speaking. These are four very different skills. Reading is the easiest: start reading Russian books with an English translation nearby and that way you can pick up new terms -- paying attention to the grammar along the way. Dictionaries are not as helpful for writing (correctly) as they are for reading because to write (correctly) you need to generate the Russian yourself, and in particular you must know mathematical phrases instead of individual words. One resource for this is a book by Sosinskii, Как написать математическую статью по-английски. You can find it online at http://www.ega-math.narod.ru/Quant/ABS.htm, although there are some glitches in the text-to-html conversion, so I think a physical copy of the book is better. Buy it at the Independent Univ. of Moscow bookstore. :) At the end of the book Sosinskii lists many mathematical phrases in English followed by their translation into Russian. That was meant to help Russian readers (the intended target audience of the book) learn what standard math phrases in English mean, but you can use it in reverse to learn how to write English phrases in Russian. To speak math in Russian you have to overcome an additional hurdle: learn how to pronounce mathematical notation in Russian (Latin and Greek variables, integrals, infinite series, rational functions, etc.), and that is essentially impossible without having native speakers around to help.
You write that one of your goals is to speak with Russian colleagues without having to pester them about Russian terminology all the time. The most efficient method to avoid bugging them about their terminology is to speak to them in English. For the most part they don't need you to learn Russian in order to communicate, since most (not all, but most) of them are going to speak English already. They'll certainly all be able to read English.
Some parts of what I've written above are kind of harsh, but that's because I think you need a reality check: reading Russian math after you know the grammar is a pretty straightforward skill to pick up by reading a few math books or articles where you are already familiar with the subject matter (so you can focus on the grammar and words and not the depth of the math). However, you seem fixated on the idea of speaking math in Russian and I think learning that skill is a complete waste of time for a non-fluent speaker unless you are going to visit Russia for mathematical reasons soon. Any Russian mathematician you meet in the US -- you are a student now in Vermont -- will speak English and trying to talk about math with them in Russian is not necessarily going to work out the way you expect: I know one prominent Russian mathematician in the US who forgot how to speak about math in Russian because he had done it only in English for so many years. Even in Russia you can get by discussing math with most mathematicians in English (at least in Moscow and St. Petersburg). Until you have a good reason to be speaking math in Russian you're probably also not going to find much practical use in writing math in Russian either (e.g., preparing lecture notes). The only exception I can think of is if you find a Russian pen pal for a language exchange (you write in Russian, the pen pal writes in English, and you correct each other) and the pen pal is interested in discussing math. That would give you a good reason to write math in Russian.
On a personal note, which explains where I am coming from in my assessments above, I could read math in Russian and have non-math conversations in Russian for about 20 years without having much of an idea about how to speak math in Russian (esp. pronouncing the notation) until I taught a math course in Russia last year to undergraduates, some of whom were not comfortable hearing English (but all of them could read English). While there I had a pretty strong incentive to learn how to talk about math in Russian because I had to do that on a regular basis with students. All the professors I met there spoke English.
It's the product and coproduct in an arbitrary category $C$ that are dual to each other, and this is because their definitions are categorically dual: in the dual category $C^{op}$, the product is the coproduct in $C$ and vice versa.
In a particular category $C$, though, the product and coproduct may not have dual properties, and this should merely be taken as evidence that $C$ is very different from $C^{op}$. For example, in the category of sets, the product is the Cartesian product and the coproduct is the disjoint union. Decategorifying Cartesian product and disjoint union on finite sets, we then obtain multiplication and addition on non-negative integers (and this, as far as I know, is the main reason why we call products products and occasionally call coproducts sums). As you correctly observe, product distributes over coproduct but coproduct does not distribute over product in this situation, and all this means is that $\text{Set}$ is not equivalent to $\text{Set}^{op}$.
Three more examples:
In the category $\text{CRing}$ of commutative rings, coproduct distributes over product but not the other way around. This is one indication that $\text{CRing}^{op}$ behaves similarly to $\text{Set}$ (or at least more similarly than $\text{CRing}$ does), and this is one abstract way to justify scheme theory.
In the category $\text{Ab}$ of abelian groups, finite products and finite coproducts agree, and neither operation distributes over the other. ($\text{Ab}^{op}$ turns out to be equivalent to the category of compact abelian groups through Pontrjagin duality.)
In a poset regarded as a category in the usual way ($a \le b$ means that there is a single arrow $a \to b$), the product is the infimum (if it exists) and the coproduct is the supremum (if it exists); in particular, the two-element product is the minimum and the two-element coproduct is the maximum. A poset is a lattice if finite products and coproducts exist and a distributive lattice if either distributes over the other (the two conditions turn out to be equivalent for lattices). In lattice theory, therefore, it is very natural to regard "addition" (supremum) and "multiplication" (infimum) as dual.
Best Answer
A duality is a pair of related concepts that display a one-to-one translation symmetry, usually (not always) as the result of some form of involution operator.
In classical logic, the operators, $\vee$ and $\wedge$ form a dual, and negation is their involution operator. This is expressed through deMorgan's Laws:$$\neg(A\vee B) = \neg A\wedge \neg B\\\neg(A\wedge B)=\neg A\vee \neg B$$
Similarly in set algebra, the $\cup$ and $\cap$ operators form a dual, with complementation being the involution. $$(A\cup B)^\complement = A^\complement\cap B^\complement\\ (A\cap B)^\complement=A^\complement\cup B^\complement$$
In predicate logic, the quantifiers, $\forall$ and $\exists$ are a dual, again with negation being the transformation. $$\neg \forall x\;P(x) \iff \exists x\;\neg P(x)\\\neg \exists x\;P(x) \iff \forall x\;\neg P(x)$$
And such.