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What is high level mathematics really like?

  1. Jul 21, 2011 #1
    I've been self-studying calculus and I got to say I really love it (part of the reason why I'm leaning toward theoretical physics). I fell in love with the way they state extremely powerful mathematical concepts through rationale and logic and just because I find it pretty darn cool.

    But do I have an overly romanticized view of mathematics? I've been seeing a lot of whining and complaining about higher level mathematics and that its proof based (true?).

    Anyhow, what is higher level mathematics really like?
  2. jcsd
  3. Jul 21, 2011 #2
    Higher level mathematics is extremely proof based!! Everything that one does in mathematics relies on rigorous proofs. Proofs are sometimes difficult for people, but proving can be quite fun :smile:

    Higher mathematics basically has the following steps in its development:
    - Finding the right tools to describe something
    - Proving that the tools work under certain circumstances.
    - Extending the tools so that they work under more general circumstances
    - Proving that they work there also.
    - Extending
    - etc.

    This is a very broad description, but let me give you an example of what pure mathematics does:

    Problem: finding the area under a curve.
    Solution: the integral

    So, in the first step, we prove that the integral actually does the right job and behaves exactly how we want to.
    However, we cannot integrate every function! So, one tries to extend the notion of an integral and this gives us the Lebesgue integral. We prove that the Lebesgue integral corresponds to the original integral in the usual cases and we prove that the Lebesgue integral works how we expect it to work.

    All that pure math is about it to rigorize intuition and to abstract the familiar.
  4. Jul 21, 2011 #3
    To add to micromass' answers, I would say that mathematics simply relies on the laws of logic. And in order to use any mathematics, you certainly need to prove that all your statements are logical!

    It would be like using Taylor's theorem on some polynomial, and when some one asks why you can do that, you'd just say, "well the book says so." Well, why does Taylor's theorem work in the first place? "Higher-level mathematics" proves it.
  5. Jul 21, 2011 #4
    Math is proof-based because they need to know that each step of reasoning is absolutely correct. For example in calculus they tell you what a limit is. But given some function or sequence, how do you know that the limit even exists? So in a course called Real Analysis, which you take after the first two years of calculus, they go back to square one. They give a logically rigorous construction of the real numbers, and they prove the least upper bound property: any nonempty set of real numbers that is bounded above, has a least upper bound.

    With that principle in hand, you can be certain that functions and sequences that "should" converge to a limit, actually do. Then they can make rigorous definitions of continuity and differentiability, give a rigorous definition of the integral, etc. So it becomes totally proof-based.

    But you don't need to worry about that right now. However it's true that once you get past the first two years of calculus, the nature of math classes changes substantially. It's all definition/theorem/proof. But the concepts you study are very interesting in themselves, so don't let the idea of proof put you off.
  6. Jul 21, 2011 #5
    Thanks for the replies.

    I guess there is no concrete way of telling how my interest will play out in higher level mathematics, I'll have to wait and see. :)

    Though I have a question: Is proof based mathematics anything like the proofs I see in my calculus books? Because I find them pretty cool in themselves, though I feel that they don't correspond to the rigor needed in a proof based math class.
  7. Jul 21, 2011 #6
    I think elementary calculus is where I began to get used to the level of abstraction and rigor employed in pure mathematics. Proofs of stuff like "product rule" and "chain rule" aren't that fun because you just end up rolling out the definition of a derivative and doing some basic algebra. Still, it gets you started with out-of-the-box thinking in some ways.

    A lot of proofs involve that so-phrased "rolling out definitions" I mentioned before. I think the rigor shows when you have to prove existence and uniqueness -- properties of some mathematical objects that are really taken for granted.
  8. Jul 21, 2011 #7
    Sounds cool.
    You said proof of stuff like product rule and chain rule aren't that fun (which I thought were interesting to an extent). So how is the proof later more fun?
  9. Jul 21, 2011 #8
    Well, to prove the product rule you pretty much just assume functions f and g are differentiable at x and let h(x) = f(x)g(x), then find h'(x) from first principles. By "more interesting", I meant proofs you can't carry out in such a direct manner. You have to be a bit clever and creative to get the result you want. http://en.wikipedia.org/wiki/Euclid's_theorem#Euclid.27s_proof" is a good example of such an interesting "fun" proof -- he shows that the opposite of his hypothesis leads to a contradictory statement. If you haven't seen something like that before, check it out!

    edit: You have a PhD? >_> Not in mathematics then, I suppose?
    Last edited by a moderator: Apr 26, 2017
  10. Jul 21, 2011 #9
    Higher math not only encompasses the theoretical side but also other areas like numerical analysis & applied math. Many or most differential equations, nobody knows how to solve. So we use computers to approximate solutions. Or you've got a matrix equation with 50 variables. Again, the methods people program into computers are interesting. The theory helps guide us- is there a solution? are there infinitely many solutions? Can knowing one solution help us narrow in on the other solutions more quickly?

    Most (90%?) of Calculus students struggle to simply understand Calculus, much less the proofs, which some professors don't emphasize.
    If A. you have the ability to see the lack of rigor in some Calculus books, and
    B. you desire to see more rigorous proofs
    then C. you are probably well suited for higher level math. I'm assuming you're in an engineering based calculus course & not a business calculus or simplified calculus course.
    The students who complain about higher math, may have been just as likely to complain about Calculus when learning Calculus.
    Last edited: Jul 22, 2011
  11. Jul 22, 2011 #10


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    You generally deal with more abstraction and in the pure part, you have to prove things.

    Personally I find the stuff I'm doing now really cool. The whole idea of wavelets being applications of hilbert spaces, which are a newly developed framework of analysis is great.

    The higher you go, the more abstract it gets and by abstract I mean that you are dealing with things that have the capacity to represent more "things" (or if you are familiar with set theory, "larger sets").

    As far as math goes, my opinion is that math is more or less created out of necessity. It is usually the case due to reasons like necessity to understand something (like in physics, or any other applied science), or to help solve existing problems in other developed areas of math (think of all the areas of mathematics and the new fields that crop up to help solve past problems).

    The thing I find great about math is that when you see the reason why it exists and what questions it can answer, you just look back in awe and appreciate the labor of people gone by to contribute to our understanding of a tiny fraction of the world.

    But also remember that a lot of things in math are tedious, but everything has its tedious parts, so don't get discouraged by that (but also be aware that not everything is "romantic").
  12. Jul 22, 2011 #11
    Not really. "Proofs" in HS/AP textbooks are more derivations than anything. It gets from point A, the use of several specific formulas and definitions, and we miraculously get to point B, the "proved" method. However, we're often left out on why we carried out each step to get from point A to point B. That is, how did we know that if we randomly thew in such steps, that we could discover such a formula?

    I guess I'm saying that proofs in higher-level mathematics are easier to follow in terms of reasoning out "why this", "why that", etc. You're more at peace with how the process came to be and how the concept was discovered.
  13. Jul 22, 2011 #12
    The proofs in higher math are more akin to the proofs I saw in Honors Geometry. When it's done they are less trick ("why we carried out each step") based. One definitely sees some tricks, +b -b, *b/b, and more, but it's clearer earlier why we use them. However, finding an ironclad reasoning to get from start to finish without assuming anything was a challenge.
  14. Jul 22, 2011 #13
    I think, if you love "extremely powerful mathematical concepts through rationale and logic", that you will love proof-based math. (And I agree, calculus is pretty darn cool. :-) Proofs can be very beautiful.

    It's possible that the whining and complaining you heard was not really about it's being proof-based, but about what I would call excessive rigor. I think good math teaching should balance rigor with intuition. But some mathematicians (Rudin's Real and Complex Analysis comes to mind) seem to feel that it's positively evil to give their students even the slightest understanding of the subject. In that regard, I think there's some substance to the complaints.
  15. Jul 22, 2011 #14
    Proofs are more than just a way of confirming something, in my opinion learning a proof is the fastest and most direct (and sometimes, the only) way of learning a subject. You can differentiate a million different functions without ever appreciating the meaning of what you're doing, then one day learn it from first principles and understand it instantly.

    Also they're a far more intuitive and natural way of learning than simple memorisation of abstract formulas.
  16. Jul 23, 2011 #15
    I don't have a PhD, but under degree in the "edit your details" section it says "Degree (or working towards) in what field?"...

    So I'm working towards a degree of a PhD so I put that down. Maybe I should change it?

    Fyi, I'm going to be a sophomore in college next semester and leaning toward theoretical physics because it fulfills my math and physics need =].

    Thanks! And your right. I see people complain about calculus and claiming differential equations to not make any sense.. but that simply means they don't understand it! A lot of people go through math at a superficial level and fail to dig deep to understand the underlying mathematics.

    I am aware that anything can become tedious at times. Beauty doesn't come cheap. :devil:

    And I don't clearly understand your definition of abstract. To me I've thought abstract mathematics was something you can't visualize and doesn't have an immediate connection to the real world.

    Oh great!! I was following a couple of proofs in my calculus books such as the product rule and it turned me off. It said that the proof involved subtracting and adding the same quantity. >----> -f(x+@x)g(x)+f(x+@x)g(x) <------<

    I was left with a big WHY??

    I would like to understand something to the point that I could self-construct the proof through intuition. Not through memorizing that I should subtract and add the same quantity!

    I hope higher level mathematics is in a way that if you try hard enough you can master the art of proofing and not through guesswork!


    I haven't been exposed to rigor let alone excessive rigor but time will tell how I will feel about it.

    Wow that sounds awesome =D.
    Last edited by a moderator: Apr 26, 2017
  17. Jul 23, 2011 #16


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    When I try and measure abstractness, I am talking about classes of definitions that generalize a particular set of sub-definitions. Let me explain:

    Think of numbers. You start off with integers. Then you learn about rationals, reals, and finally complex numbers. Each definition is going up the chain in terms of abstractness.

    Then think about functions. You start with one variable and then you go into multiple variables. With calculus you start with one variable and then you get into frameworks that deal with any number of variables. Again in this example the class of things that you are defining and attempting to analyze is getting larger and larger.

    In a lingual kind of sense it's like going from defining and analyzing a house to defining and analyzing "buildings" in a general sense. All analysis of "buildings" should apply to those of "houses".

    It is true that some abstraction involves going from visual to algebraic descriptions (geometry is a good example when going past 3 dimensions), but it isn't always like this.

    In a nutshell, if there is room for some generalization of some sort, then that potential will involve more mathematical abstraction in regard to that particular area of mathematics.
  18. Jul 23, 2011 #17
    ^ I first learned the concept of "abstraction" in my first year CS class. The lecturer said that abstraction is filtering out all the unimportant information from separate ideas and making note of the similarities.

    One of my favorite examples is the idea of a metric space. It turns out that some results involving subsets of the reals obtained by using the properties of the usual Euclidian metric, d(x, y) = |x - y|, can be proven just by using the properties of this metric (triangle inequality, symmetry, etc). We forget the specific definition of the function d(x, y) in this case and just consider any function that is symmetric, positive definite, and subadditive. Then we consider sets of items of any quality on which such a metric can be defined.

    Abstraction is simply cutting away the irrelevant information from an idea and maintaining the useful stuff, in order to be more general and encompass more objects in the scope of your idea.
  19. Jul 24, 2011 #18
    I don't understand how this part is valid:

    "If q is not prime then some prime factor p divides q. This factor p is not on our list: if it were, then it would divide P (since P is the product of every number on the list); but as we know, p divides P + 1 = q. Then p would have to divide the difference of the two numbers, which is (P + 1) − P or just 1. But no prime number divides 1 so there would be a contradiction, and therefore p cannot be on the list. This means at least one more prime number exists beyond those in the list."

    1*2*3*4 = 24 = P
    24+1=25 = q

    25 is not prime... why would "p have to divide the difference of the two numbers, which is (P + 1) − P or just 1."

    And how does "no prime number divides 1 so there would be a contradiction, and therefore p cannot be on the list." prove that there must be at least one more prime number on that list?
    Last edited by a moderator: Apr 26, 2017
  20. Jul 24, 2011 #19
    First, if you want to follow Euclid's logic, you should be using the product of primes, not of all integers. That doesn't make a big difference. It takes a little more work, but one can still find an example of your point. The first is 2*3*5*7*11*13+1 = 30031 = 59*509.

    More important, remember that you're assuming that there is no prime greater than pn (which is 13 in the example). Euclid's reasoning shows that no prime <= pn divides q. Under the assumption that there IS no prime greater than pn, there must therefore be no prime that divides q. The example doesn't contradict this, because the smallest prime that divides 30031 is in fact bigger than 13.

    You started by assuming that every prime was on the list. You showed that this leads to a contradiction. That means that the assumption was WRONG. If the assumption "all primes are on the list" is wrong, then there must be at least one prime not on the list. (It doesn't have to be q, though. That's what your example shows.)
  21. Jul 25, 2011 #20
    Thanks for the reply, it's cleared up some things and I think I get it now.

    Euclid's proof shows that if P + 1 is not prime, then P + 1 can not be divided by a prime number in the finite set of prime numbers since when you divide P + 1 by a factor of P, that factor will end up dividing 1.

    [itex]\frac{(P + 1)}{factor of P}= factor of P\times(something) + \frac{1}{factor of P}[/itex]
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