Why do professors go through proof after proof with no rhyme or reason?
One theory is that this is an "easy" way to give a lecture (to be negative about it, a "lazy" way.) This may be true in some cases. But on the other hand, much of the instructor's education might have been this way, and maybe they even think the experience is valuable. So, they might actually be giving the students the best path they know of. Some students might even feel like that is the way they are most comfortable learning. So to be fair, such instruction may be given in good faith, and may have good points.
The fact is that really good exposition requires a really skillful teacher, and it's not easy to do. Incidentally, I found Artin a very good expositor, but I did observe that by doing this, some less dedicated readers might get bored or distracted during his exposition.
One of my books learning abstract algebra was Martin Isaacs' Algebra. At the time I did not like it very much, but looking back on it now I think I do like its exposition. This just goes to show that reasonable exposition is not always easy to evaluate.
My question is How do I prevent this from happening to me in future?
Oh, well that's easy! Go skim through a lot of alternative books on the same topic and soak up whatever you can! Don't pretend like it's your teacher's responsibility to put text on your plate. You already applied this when you picked up Artin's book and learned something from it.
why are textbooks like Gallian's popular in math instruction?
The "like" part here makes this a loaded question, but I could just say that this book is probably considered basic, safe and affordable. It probably also depends upon the teacher's experience with texts too.
How are you supposed to read them?
This varies a lot from person to person. Personally I discovered that I learn best by having three or four texts on the same topic that I can use to cross-reference topics. Usually at least one of the authors is going to say something that makes things click.
And most of all, this sets me up with a big supply of problems. Doing problems does a lot more than plain reading, for me. Of course you have to spend some time reading or you won't know what tools you have at hand, and you won't see the themes in the proofs.
Is math supposed be learnt like you are learning a game where you are given all the rules ( definitions) and you have to play the game according to those rules( theorems, exercises) ?
I guess ideally "no", but for some people, that's how mathematics first begins! Those who persist eventually find their own appreciation for the subject matter, and develop their ability with it. This "game" analogy certainly doesn't paint a pretty picture of pedagogy, but it's very rare to find teachers with enough ability to get the beauty of mathematics across from the very beginning.
Luckily, it sounds like you at least know mathematics is more than a string of memorized definitions and theorems and proofs, so you, my friend, are already well ahead of many other students. The rest are in the even sadder situation of thinking "Yes, that's all mathematics is. Isn't it awful?!"
In every ring (with unit) $R$ you have a $1$. And you have thus $1+1$, and $1+1+1$, &c. You can consider the collection of all these elements and their additive inverses, $-(1+1), -(1+1+1)$, &c. For $n>0$ call these elements $n1$ and set $(-n)1=-(n1)$, if $n=0$; $n1=0$. You should convince yourself that $m1+n1=(m+n)1$ for any pair of integers $n,m$, and $n1\cdot m1=(nm)1$. Of course $1\cdot 1=1$. This means the collection of such elements is a subring of $R$. It looks like $\Bbb Z$, since every element is just a sum of $1$ (or difference), but note that these elements needn't be distinct. For example, in $\Bbb Z/3\Bbb Z$, $2\cdot 1=-1\cdot 1$.
At any rate, now we can define a function $f:\Bbb Z\to R$ for any ring $R$ with unit that sends the integer $n$ to $n1$, as defined above. What you checked above is precisely the claim that $f(n)f(m)=f(mn)$ and $f(n+m)=f(n)+f(m)$, $f(1)=1$. This means $f$ is a morphism of rings. Since $R$ was arbitrary, we have shown that every ring $R$ admits a morphism of rings $f:\Bbb Z \to R$.
It remains to see that $f$ is unique. Now if $g:\Bbb Z\to R$ is another morphism of rings, we know that $g(1)=1$ (i.e. $g$ sends the unit of $\Bbb Z$ to the unit of $R$), hence $g(-1)=-1$ (why?). But any nonzero integer is $n$ is $1+\dots+1$ ($n$ times) or $-1-\cdots-1$ ($n$ times), so using $g$ is $\Bbb Z$-linear ($g(a+b)=g(a)+g(b)$) gives $g$ sends $n$ to $f(n)$ as defined above, i.e. $f=g$.
Best Answer
Your interpretation is correct.
Here is another way to interpret what $na$ means. First, solve this exercise: show that there is a unique ring homomorphism $\iota$ from $\mathbb{Z}$ to any ring $R$.
The homomorphism $\iota$ allows us to canonically identify any integer $n \in \mathbb{Z}$ with the element $\iota(n) \in R$. Thus, we can interpret $na$ as the product $\iota(n)a$, which is just the usual multiplication operation of $R$. If you have solved the above exercise, it should be easy to show that this agrees with the interpretation of $na$ as repeated addition.
If $\iota$ is injective, it identifies $\mathbb{Z}$ as a subring of $R$, which you alluded to in your post. In fact, the uniqueness of $\iota$ implies that the image of $\iota$ is the unique subring of $R$ that is isomorphic to $\mathbb{Z}$. However, note that $\iota$ is not necessarily injective. To get some intuition for $\iota$, I'd recommend computing it for a few examples (e.g., $\mathbb{Z} / n\mathbb{Z}$, $M_2(\mathbb{Z})$, $\mathbb{Z}[x]$).