[yequalsx]

I teach math at a community college. Your question is not so simple to answer. Much of mathematical teaching involves lying and not justifying statements. The details are often way more complicated than the idea.

It is "obvious" in the real numbers that if you multiply two numbers and get 0 then one of them must be zero. I doubt you could prove this. It's obvious simply because you are used to it being true. But it is not true for all algebraic systems. The algebraic structure of all 2x2 matrices can be viewed as an extension of the real number system and in the set of 2x2 matrices you can multiply two matrices to get 0 in which neither matrix is 0.

One of the goals of abstract algebra is to understand under what conditions certain properties hold in an algebraic system. To truly understand these things requires the oft mentioned mathematical maturity. But to get to the point of gaining this maturity requires just accepting what you've been told is true is indeed true.

We tell students in Calculus I that the function 1/x is discontinuous at 0. There's a break in the graph there. But, in reality, it is meaningless to talk about a function being continuous (or not being continuous) at a number not in the domain of the function. Indeed, in the standard subspace topology the function 1/x from R-{0} to R is continuous. But this nuance is way too complicated to get across to students in Calculus I so we fudge things a bit. This happens a lot at lower levels of math.

EDIT: So my point is that if your goal is to truly understand things then yes, Abstract Algebra is necessary. If your goal is to be operationally functional in working with polynomials over the real numbers then it isn't.

[bumbledraven]

> It is "obvious" in the real numbers that if you multiply two numbers and get 0 then one of them must be zero. I doubt you could prove this.

Theorem: If x & y are reals and xy=0, then at least one of x and y is zero.

To see this, assume x and y are both nonzero. Divide both sides of xy=0 by x (this is valid because x is nonzero). Then y=0; contradiction. Therefore, at least one of x and y is zero.

[yequalsx]

First you'd need that the integers are an integral domain and then build up to the fact that the reals are a field. At least that's how I'd go about it. Maybe one can start with the reals themselves. Here's the proof that the integers have no zero divisors:

https://proofwiki.org/wiki/Ring_of_Integers_has_no_Zero_Divi...

[eru]

Are you sure that the assumptions that your proof relies on are more fundamental than than what you are trying to prove?

(Eg if memory serves right, you might use the property you are trying to prove to justify why you can even divide by non-zero x on both sides.)

[JadeNB]

> Are you sure that the assumptions that your proof relies on are more fundamental than than what you are trying to prove?

Yes, at least to the extent that "fundamental-ness" can be defined. It is a well, and probably universally, established convention that the field axioms allowing division by non-0 elements (that is to say, allowing multiplication by arbitrary elements, and multiplicative inversion of non-0 elements) is taken as part of the definition of a field, and the fact that the set of non-0 elements in a field is closed under multiplication as a theorem about fields.

[yequalsx]

I believe the point the person you were responding to was getting at is that the person's proof assumes that the reals are a field. It's not a convention that cancellation holds for fields. It's a theorem one proves about integral domains. That a field is closed under multiplication is not a theorem; it's part of the definition of being a ring.

[JadeNB]

> It's not a convention that cancellation holds for fields. It's a theorem one proves about integral domains. That a field is closed under multiplication is not a theorem; it's part of the definition of being a ring.

Yes, that's what I said. My point was that, between

(A) making an axiom the right to divide in a field by non-0 numbers, and proving the closure of non-0 numbers in a field under multiplication,

and

(B) just making an axiom the closure of non-0 numbers in a field under multiplication,

it is only convention (rooted in the deeper convention of not making an additional axiom out of something we could prove from existing ones) that we choose (A) instead of (B). (Also notice that I was talking about the closure under multiplication of the set of non-0 numbers, which is (usually) not part of the definition, rather than of the entire set, which is indeed always part of the definition.)

[qubex]

I studied applied mathematics, but it took me ages to shake away (false) intuition engendered in me about “Calculus” and “infinitesimals” at high school. Sure, it worked, but learning “differentiation from first principles” with the limit taken when δ︎x→︎0 just by cancelling out did an unmeasurable amount of damage to my ability to absorb the formal Weierstrss formulation in terms of limits.

[romwell]

Congratulations. Now that you're a grown-up, you can re-do the damage by learning https://en.wikipedia.org/wiki/Non-standard_analysis

On a more serious note, you can understand most, if not all, of Calculus by saying that dx=0.0001, and that A ~= B if they don't differ by more than, say, 0.01 (say, that's the instrument error).

Then you get your limits, FTC, and so on, and verify the results with a four-function calculator.

Example: f=x^2, f' = ?

(f(x+dx) - f(x))/dx = (x^2 + 2x*dx + dx^2 - x^2)/dx = 2x + dx = 2x + 0.0001 ~= 2x

The mental effort you have to make here is that things on the LHS of ~= are "actual" values, and on the RHS are "measured" values, and that ~= is not an equivalence relation.

On a yet more serious note, learning about differential forms will help justify some of that high-school notation.

On a philosophical note, Weierstrass is not the end-all of Calculus. Neither Newton nor Leibniz did it that way. By adding rigor, some argue that the essence has been obscured (hence the non-standard analysis above).

[qubex]

Yeah, I’m aware of the numerics... but I’m also aware that if you do the same process backwards (naive numerical integration), for example for simulating planetary motions, you get into ridiculous situations where the collective momentum of a closed system rises exponentially after a ”close flyby”. This is precisely the kind of situation where ”false intuition” created by these shallow teachings cause the most harm.

[romwell]

I don't think it's harmful - without stumbling onto an example like that, it's hard to justify why we need solid foundations and exact solutions.

On the other hand, very brute numerics work for an awful lot of scenarios - that's why epsilon-delta came centuries after Calculus was invented.

For instance, "first-order optics" and "third-order optics" rise from chopping off the Taylor series after the 1st and 3rd term, resp. And it works! In many places, 1st order approximations are just good enough. A lot of scenarios are inherently stable.

So I don't think the intuition you build up is wrong - it just has a scope. There's nearly always a place for counter-examples where "things work the way you think they should" wouldn't apply, however you think about things :)

On a philosophical note, the continuity is a human construct - down there, things seem to be discrete, just with a very small step size. Continuity models these pretty well, until it doesn't - but that doesn't mean the intuition you build up is wrong. Just limited in scope.

[qubex]

I see what you are saying, but to each his own: I’m one of those kids (there's one in every mechanics class) that grow irate at the sin(ϴ︎)≈︎ϴ︎ approximation in the pendulum solution and then waste weeks using the full series expansion, only to find out the results differ only in the third or fourth decimal place. The thing is, I emerged from the experience thinking to myself “that was a useful learning moment”.