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  1. #1 Proofs 
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    I'm really new to the act of formulating proofs, and would like some pointers, even tutorials, in the basics of doing so. My present goals are to derive the proof for the fundamental theorem of Calculus, to prove that differentiation works, and to derive the rules for differentiation and integration. I think tackling differentiation would be a good starting point for me, and on that note I ask for a little help on where to go. I know what I need are limits of secant lines as the difference between the 2 points involved in the line go to 0, causing the line to be tangent.


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    Almost all proofs in calculus come down to two things: (1) The least upper bound property of the reals and (2) the triangle inequality.


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  4. #3 Re: Proofs 
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    Quote Originally Posted by Arcane_Mathamatition
    I'm really new to the act of formulating proofs, and would like some pointers, even tutorials, in the basics of doing so. My present goals are to derive the proof for the fundamental theorem of Calculus, to prove that differentiation works, and to derive the rules for differentiation and integration. I think tackling differentiation would be a good starting point for me, and on that note I ask for a little help on where to go. I know what I need are limits of secant lines as the difference between the 2 points involved in the line go to 0, causing the line to be tangent.
    You can do precisely what you have outlined. But don't be surprised if it is a little trickier than you think. The attempt to understand the essence of what makes calculus work was behind the development of point set topology in the early twentieth century.

    What salsaonline said is absolutely true. But those are fairly powerful ideas -- particularly the least upperbound property.

    The least upper bound property implies the completeness of the real line (it is equivalent) and the connectedness of intervals. Connectedness is what gives you the "intermediate value theorem", and that gives you the mean value theorem. With the mean value theorem you can prove the fundamental theorem of calculus.

    In many introductory calculus courses the intermediate value theorem is taken on faith with the aid of a few cartoons. If you can understand it from a more rigorous standpoint then you have won most of the battle.

    Try to prove the intermediate value theorem. Take all the time that you need. If you can prove that using the least upper bound property, then you are well on your way. If not you might want to take a look at a text on real analysis (see below), which is essentially calculus with proofs.

    I don't know of any good tutorials on how to do proofs. You can see some good examples in books with a title like "Real Analysis". Bartle's book Elements of Real Analysis is excellent, but it is genereally used for classes at the junior-senior level with students who have already gained at least a calculational facility with calculus. Walter Rudin's book Principles of Mathematical Analysis is also excellent, but Rudin's proofs are so slick that you will feel like someone picked your pocket. Apparently he was like that in graduate school -- I knew someone who was at Duke when Rudin was a student there and took his oral exams. He came up with at least one brand new proof of a standard theorem (I don't recall which one) while standing in front of the examining committee.
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    To add to what Rocket is saying, let me put it this way: Good look proving anything in calculus without the least upper bound property.
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    Quote Originally Posted by salsaonline
    To add to what Rocket is saying, let me put it this way: Good look proving anything in calculus without the least upper bound property.
    Absopositively. In fact I think I can prove that you can't do calculus without it. Calculus with just the rationals, for instance, doesn't work.

    For anyone reading this who doesn't know what we are talking about, here is the statement of the least upper bound property.

    Definitions. A subset of the real numbers is called bounded and is an upper bound if for every . x is a least upper bound for A if is an upper bound for A and for every upper bound of .

    Least Upper Bound Property: Every bounded subset of the real line has a least upper bound.

    Theorem. A set has at most one least upper bound.
    Proof. Suppose and are both least upper bounds for A. Then and hence
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    Quote Originally Posted by DrRocket
    Absopositively. In fact I think I can prove that you can't do calculus without it. Calculus with just the rationals, for instance, doesn't work.
    It depends on what you mean by "calculus". A lot of things still work. I mean, you still have the Archimedian property and the triangle inequality. You even have a notion of Cauchy sequences, except now they don't always converge.

    So you can still define limits (a lot of them will no longer exist), and you can therefore define the derivative of things like polynomial functions.

    Of course, you lose so much (like all the transcendental functions, connectivity, and sequential compactness of closed intervals), that staying inside the rationals is obviously of little merit.
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    I expect a lot of what I'd like to do to be rather much trickier than I expect, but I cherish the learning process and, like chemboy, really want to learn this stuff. Sadly, I don't have the resources available to me (full time student, rather broke) to get the books that are fantastic for this sort of thing, I'm kind of stuck with, my most advanced book, James Stewart's; early transcendentals, 5th ed. textbook. I kept all my textbooks from my classes. It's the understanding that I want, and I'd rather do it on this forum, where I can, at the very least, share my learning with those who are interested in it.
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    Okay, here's an exercise then:

    Using the least upper bound property of the reals, prove that for any positive real number , and any positive integer , there is a unique positive real such that . That is, we can always take the n-th root of a positive real number.

    Also, another good exercise: Prove the Archimedian property of the reals: For two positive numbers with , prove we can find a positive integer such that . (Hint, remember to use the least upper bound property of the reals.)

    As a reminder, the least upper bound property of the reals says that for any subset of real numbers which is bounded above, there exists a unique real (the supremum of S) which is the least upper bound. That is, sup S is (1) an upper bound for S, and (2) is less than or equal to all other upper bounds for S.

    Equivalently, the reals possess the "greatest lower bound property". I'll leave it to you to figure out what the greatest lower bound property means. The greatest lower bound of a set that is bounded below is called the "infinum" of the set, denoted inf S.
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    Okay, so I have a number a and I want to show that I can always define a unique x for every integer n as which is conversely for this, I'd want to show that any real number can be divided into n times, right? Also, that for every real number there is only ONE number that can be multiplied by itself n times to produce that original number a? Also given that all components will be positive.

    I'm not sure how to apply that theorem, as it says if there is a subset of the reals, that is bounded, then there is a 'least upper bound' and I honestly don't see how that makes sense, seeing as how if there is an upper bound, how can there be a least? Wouldn't all upper bounds be the same, since if there is a an upper bound greater than the least upper bound, wouldn't it nullify the position of said least upper bound?

    BUT, going on the idea that there is a subset A of the reals, bounded by a such that for all numbers , there exists one x that would be the nth root of a. Would it be sound to state that, in the subset of the reals, if and only if , if a=1, x=1=a, because , no matter how many ones you have. ummmmm..... Stuck, sorry, not sure how to prove it from here. this one seems almost like it should be proved by definition.

    the second exercise, prove that, if , there exists an integer n such that is equivalent to or, if y is the least upper bound of S, then x, by definition, is less than y and, subsequently contained in S.

    for every real number n, there exists a number n+1 that is larger than n. So, I can define n outside of my subset, that there exists a real number such that, as long as since , since, by definition both x and y are positive. Is that right?
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  11. #10  
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    Quote Originally Posted by Arcane_Mathamatition
    Okay, so I have a number a and I want to show that I can always define a unique x for every integer n as which is conversely for this, I'd want to show that any real number can be divided into n times, right? Also, that for every real number there is only ONE number that can be multiplied by itself n times to produce that original number a? Also given that all components will be positive.

    I'm not sure how to apply that theorem, as it says if there is a subset of the reals, that is bounded, then there is a 'least upper bound' and I honestly don't see how that makes sense, seeing as how if there is an upper bound, how can there be a least? Wouldn't all upper bounds be the same, since if there is a an upper bound greater than the least upper bound, wouldn't it nullify the position of said least upper bound?

    BUT, going on the idea that there is a subset A of the reals, bounded by a such that for all numbers , there exists one x that would be the nth root of a. Would it be sound to state that, in the subset of the reals, if and only if , if a=1, x=1=a, because , no matter how many ones you have. ummmmm..... Stuck, sorry, not sure how to prove it from here. this one seems almost like it should be proved by definition.

    the second exercise, prove that, if , there exists an integer n such that is equivalent to or, if y is the least upper bound of S, then x, by definition, is less than y and, subsequently contained in S.

    for every real number n, there exists a number n+1 that is larger than n. So, I can define n outside of my subset, that there exists a real number such that, as long as since , since, by definition both x and y are positive. Is that right?
    Perhaps the least upper bound property is a bit mysterious to you. It is, however, very important and you need to start to understand what it says.

    Consider the set {}. That set is bounded above. For instance 4 is an upper bound. In the least upper bound is . However, if you limit your attention to only the rationals, , it is still bounded above but there is no least upper bound in while there is a least upper bound in . That is because there are "holes" in but there are no "holes" in

    We say that is complete because it has no holes, but is not complete because there are holes in it (the irrational numbers plug the holes and give you ).

    It is because is complete that you can find in but not in .

    What salsaonline is asking you to do is to use the least upper bound property of to prove the existence of nth roots. That is a good exercise and will help to understand the least upper bound property and why it differentiates the real numbers from the rational numbers. That is also why you can do more calculus with the reals that with the rationals.
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    Sorry, but not correct. Keep at it. A rule of thumb with proofs: when you get it, you won't even have to ask if it's correct. It will just feel right.

    Reply directed at Arcane.
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    what I should really be trying to prove, is that always exists for every positive real number 'a' and every integer 'n'. I still don't quite get how to prove this one, it almost seems like a definition in and of itself.
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    Quote Originally Posted by Arcane_Mathamatition
    what I should really be trying to prove, is that always exists for every positive real number 'a' and every integer 'n'. I still don't quite get how to prove this one, it almost seems like a definition in and of itself.
    No. Fix n. Then show that given any positive real number A there is some positive real number x such that .

    I suggest that just to make this more concrete for you that you do the case n=2 first and then do the proof for an arbitrary positive integer n. The proof is really the same anyway.

    Use the least upper bound property. It is essential.

    Remember that the least upper bound property does not hold for the rational numbers. Remember also that you can't take square roots and stay in the rational numbers either. for instance is a problem for the rationals.

    Here is a hint. Fix A and consider the case n=2. Consider the sets {} or {}. See what you can do with the least upper bound property or the greatest lower bound property.

    Don't give up too soon. Think about this a while. It is only difficult until you see how to think about the problem. You need to work it out for yourself.
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    okay, lets say I have a subset of . I let x be the least upper bound of A such that for all . I will define . if then and so now, for all other upper bounds of s, I will denote them to be y, such that if then and so , therefore . since there is only one least upper bound of S, only that value will, when raised to the 2nd, give A.

    replacing 2 for n would be the proof of the general case, true? All assuming every number involved is positive.
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    I'll write the full proof for n=2 so you can see how these go.

    We wish to show that for any positive real number , I can find a real number such that .

    Let . Clearly is bounded above, since , so any number bigger than cannot be in this set.

    Since is bounded above, it must have a least upper bound. Let be the least upper bound of . I claim that .

    Suppose this is not the case. Then either or . Suppose then that . Then by definition, is in the set . Now for any , . Choose so that . Then must lie in , contradicting the fact that is an upper bound for .

    Next, suppose that . Then, analogous to the above argument, one can pick an sufficiently small so that . But this would contradict the fact that was the LEAST upper bound.

    It follows that we must have .
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    Quote Originally Posted by Arcane_Mathamatition
    okay, lets say I have a subset of . I let x be the least upper bound of A such that for all . I will define . if then and so now, for all other upper bounds of s, I will denote them to be y, such that if then and so , therefore . since there is only one least upper bound of S, only that value will, when raised to the 2nd, give A.

    replacing 2 for n would be the proof of the general case, true? All assuming every number involved is positive.
    First you need to define some specific set that you think has something to do with the square root of A. Then you need to show that it is bounded. Then you need to show that the least upper bound or the greatest lower bound actually is a square root of A.

    Take it slow and easy, One step at a time. Make certain that your explanation is crystal clear to you before you bring your proof out.

    One more hint, the last one. Consider the cases A>1, A<1 and A=1 separately. That might make things easier. A=1 ought to be very easy, and if you prove the case A>1 or A<1 then the other case should follow immediately by looking at 1/A.

    Edit: I missed the fact that salsaonline had posted a proof. His proof is very efficient and eliminates the need to consider the cases that I suggested. Perhaps if you try to construct a proof while considering those cases, and using salsaonlines proof to help you along the way you might gain some insight. Or a better way might be to take salsaonline's proof and extend it to the case of nth roots. Either exercise would probably help.
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    I have a feeling it's a good thing that I see similarities between the stuff I regurgitated and the proof that Salsa wrote out. And it would be fair to say I have little exposure to proofs, and the notations used, but I think I understand the concepts behind what needs to be done here. I definitely need to have more definitions in what I'm doing and need to have a better grasp of sets of numbers. I think my big problem right now is the whole concepts in the idea of subsets of a set of numbers, the idea was only glazed over in my current and previous classes, so I think I haven't been properly introduced to what now seems to be a very central part of mathematics. I kind of wonder if I've been jipped out of a proper education on the topics I was taught.
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    Quote Originally Posted by Arcane_Mathamatition
    I have a feeling it's a good thing that I see similarities between the stuff I regurgitated and the proof that Salsa wrote out. And it would be fair to say I have little exposure to proofs, and the notations used, but I think I understand the concepts behind what needs to be done here. I definitely need to have more definitions in what I'm doing and need to have a better grasp of sets of numbers. I think my big problem right now is the whole concepts in the idea of subsets of a set of numbers, the idea was only glazed over in my current and previous classes, so I think I haven't been properly introduced to what now seems to be a very central part of mathematics. I kind of wonder if I've been jipped out of a proper education on the topics I was taught.
    My guess is that you have seen the rather standard elementary classes in mathematics. In those classes the emphasis is on "solving equations" and "finding the answer".

    You need to be able to do those things, but that is not really what mathematics is about. It is about studying whatever kinds of order the human mind can recognize and then developing a deep understanding of the implications of that order. That requires a profound change in mindset from what you typically find in the elementary classes. It is something that is fairly mysterious until you "get it". It don't know how to tell you to "get it" except to keep trying to understand the subject through an understanding and creation of proofs.

    The subject is really all about proofs, but that is often hidden until you are in some relatively advanced classes.

    My best advice is this. Keep trying to develop a crystal clear proof of the theorem that salsaonline stated for you. He was giving you good advice. It is probably not going to be easy, but if you persist and succeed it will be worth the effort. I will hazard a guess that if you work hard and get this proof correctly the next one will be much easier. But you will have to "hurt your head" to get the first one.
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    Quote Originally Posted by salsaonline
    Also, another good exercise: Prove the Archimedian property of the reals: For two positive numbers with , prove we can find a positive integer such that . (Hint, remember to use the least upper bound property of the reals.)
    I will consider the set and since I will use the greatest lower bound of S since it is obvious that it is bounded below in that n-1 could be less than which would not be a valid n for S.

    we will say that k is the greatest lower bound of S such that

    assume there is some such that
    if then it must be true that which would mean that k is not the greatest lower bound of S, which contradicts the definition of k, and as such must not be true. It is then proper to say that there must be some

    I'm pretty sure this looks right, but again, never really encountered proofs before this last class I've had, with the exception of the 2-column proofs of geometry from HS. Please point out any mistakes I may have made, thanks.
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    A priori, you don't even know if S has any elements at all.

    Btw, as a rule of thumb, one does not need the least upper bound prop (or greatest lower bound prop) for a subset of the integers.
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    Quote Originally Posted by salsaonline
    A priori, you don't even know if S has any elements at all.

    Btw, as a rule of thumb, one does not need the least upper bound prop (or greatest lower bound prop) for a subset of the integers.
    But often it is useful to use the well-ordering property --- that every subset of the natural numbers has a least element. The extension to bounded (above or below) subsets of integers is clear.

    It might not be obvious to all readers (salsaonline certainly knows this), but the well-ordering principle is equivalent to the validity of mathematical induction.
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    My point was that if someone says "hint, use the least upper bound property", it's a good sign that you'll be using something stronger than the well-ordering principle of the naturals. It stands to reason therefore that the set you'll be constructing for the proof will probably not be a subset of the integers.
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    I do know S has some elements though, and that all elements of S will satisfy the required problem. I know this because I can approximate any Irrational number to be between 2 rationals, and I can always find an integer larger than a rational number, since a rational number is the ratio of two integers. I know I can prove it for the integer being larger than a rational, but I don't know how to show that there is always a rational larger and lesser than an irrational, so I'm sure my argument is weak as it is. But I know it's true, just not how to show that it is true.
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    Quote Originally Posted by salsaonline
    My point was that if someone says "hint, use the least upper bound property", it's a good sign that you'll be using something stronger than the well-ordering principle of the naturals. It stands to reason therefore that the set you'll be constructing for the proof will probably not be a subset of the integers.
    Can't argue with the hint. The reals are the reals for a reason, and the lub property is a big part of that reason.

    If you are trying to prove something that is true for the reals but not for the rationals, there is a good chance that the proof will use the least upper bound property in some essential way. I can't think of any counter-examples off the top of my head.

    On the other hand, from the perspective of what is fundamental, the well-ordering principle is at the root of a lot of things. If you start with the Peano axioms (or Zermelo Frankel for the purists) then the well-ordering principle, or something equivalent, is an axiom. From the Peano Axioms you can construct the reals and prove the least upper bound property. The well-ordering principle is the centerpiece of the Peano Axioms. Hence there is a valid perspective from which one can make a case that the least upper bound property is the result of the well-ordering of the naturals plus some definitions and constructions. Give me that naturals and the well-ordering property and I will build for you the reals with the least upper bound property.

    You have probably suffered through the construction of the reals from the Peano Axioms, but if not Edmund Landau's little book Foundations of Analysis lays it out very efficiently (but dry barely describes it). You can read it in one rainy morning.
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    I've long ago made peace with the construction of the reals by working it out on my own. (By defining the reals to be a equivalence classes of Cauchy sequences of rationals, where two Cauchy sequences are equivalent if their difference converges to zero).

    To be precise, we let {q_n} be Cauchy if for any N > 0 |q_n-q_m|< 1/N for n,m sufficiently large. We say that q_n ---> 0 if for any N > 0, q_n < 1/N for n sufficiently large.

    Define the reals to be the set of Cauchy sequences {q_n} modulo ~, where {q_n} ~ {r_n} if q_n-r_n ---> 0.

    The advantage to this approach is that all algebraic properties of the reals are almost automatic from this definition. The disadvantage is that the least upper bound property requires a proof, whereas it's more or less automatic in the Dedekind cut approach.

    Actually, it's for this reason that I prefer this Cauchy sequence construction. Philosophically, I think that the normal algebraic and ordering properties of the reals should be transparent in the definition. That is, given a definition for the reals, I don't like the idea of having to "work" to prove them. I'm happy to work to prove the least upper bound property if it means that all the more basic properties of the reals come for free.
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    Quote Originally Posted by Arcane_Mathamatition
    I do know S has some elements though, and that all elements of S will satisfy the required problem. I know this because I can approximate any Irrational number to be between 2 rationals, and I can always find an integer larger than a rational number, since a rational number is the ratio of two integers. I know I can prove it for the integer being larger than a rational, but I don't know how to show that there is always a rational larger and lesser than an irrational, so I'm sure my argument is weak as it is. But I know it's true, just not how to show that it is true.
    You don't know anything a priori (though I'll grant you as given all well-known properties of the rationals that make no reference to irrationals).
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    Isn't this a proof of differentiation?

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    What is ?
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    actually, it would be I interpret the d in the equations to roughly mean , or change. i.e. which is, in an instantaneous sense, what the derivative is.
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    Quote Originally Posted by salsaonline
    What is ?
    It is a small increase in x, more specifically the limit of that inrease as it approaches zero.
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    Quote Originally Posted by esbo
    Isn't this a proof of differentiation?

    isn't it defined as:

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    Quote Originally Posted by esbo
    Quote Originally Posted by salsaonline
    What is ?
    It is a small increase in x, more specifically the limit of that inrease as it approaches zero.
    I know the intended meaning. But what is it precisely?
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    Quote Originally Posted by esbo
    Quote Originally Posted by salsaonline
    What is ?
    It is a small increase in x, more specifically the limit of that inrease as it approaches zero.
    The question is what definition do you use for dx that permits a rigorous definition for the derivative in terms of real numbers.

    "A small increase in x" is not a real number. "The limit of that increase as it approaches zero" IS zero.

    So you have defined dx to be zero, and that does not provide a defintiion of the derivative that is meaningful.

    Neither does it help to define dx as "an infinitesimal change in x". You cannot do that and work within the real numbers. There is a way to do this in the context of what are called the "non-standard real numbers" but that requires a LOT more sophistication to handle even the basic definitions -- you basically need to know what an ultrafilter is just for starts.

    The point here is that to define concepts rigorously one needs to be fairly careful and then work strictly withing the concepts that have been defined to construct definitions and proofs that advance the subjectd.
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    Quote Originally Posted by salsaonline
    I've long ago made peace with the construction of the reals by working it out on my own. (By defining the reals to be a equivalence classes of Cauchy sequences of rationals, where two Cauchy sequences are equivalent if their difference converges to zero).

    To be precise, we let {q_n} be Cauchy if for any N > 0 |q_n-q_m|< 1/N for n,m sufficiently large. We say that q_n ---> 0 if for any N > 0, q_n < 1/N for n sufficiently large.

    Define the reals to be the set of Cauchy sequences {q_n} modulo ~, where {q_n} ~ {r_n} if q_n-r_n ---> 0.

    The advantage to this approach is that all algebraic properties of the reals are almost automatic from this definition. The disadvantage is that the least upper bound property requires a proof, whereas it's more or less automatic in the Dedekind cut approach.

    Actually, it's for this reason that I prefer this Cauchy sequence construction. Philosophically, I think that the normal algebraic and ordering properties of the reals should be transparent in the definition. That is, given a definition for the reals, I don't like the idea of having to "work" to prove them. I'm happy to work to prove the least upper bound property if it means that all the more basic properties of the reals come for free.
    That would be the modern approach, and it has the advantage of extending unchanged to the completion of an arbitrary metric space. It is clean and efficient.

    But I guess I think everyone ought to at least once feel the pain of fighting through the exercise of the Dedekind cut approach since it is the way it was done originally and has survived in many text books. But I also think they should only have to do it once. If you have done Dedekind cuts once I think you will appreciate the use of equivalence classes of Cauchy sequences a bit more. (Kind of like banging away on your thumb with a hammer --- it feels so good when you stop.)
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by esbo
    Quote Originally Posted by salsaonline
    What is ?
    It is a small increase in x, more specifically the limit of that inrease as it approaches zero.
    The question is what definition do you use for dx that permits a rigorous definition for the derivative in terms of real numbers.

    "A small increase in x" is not a real number. "The limit of that increase as it approaches zero" IS zero.

    So you have defined dx to be zero, and that does not provide a defintiion of the derivative that is meaningful.

    Neither does it help to define dx as "an infinitesimal change in x". You cannot do that and work within the real numbers. There is a way to do this in the context of what are called the "non-standard real numbers" but that requires a LOT more sophistication to handle even the basic definitions -- you basically need to know what an ultrafilter is just for starts.

    The point here is that to define concepts rigorously one needs to be fairly careful and then work strictly withing the concepts that have been defined to construct definitions and proofs that advance the subjectd.
    But then when you write it becomes undefined or infinite or whatever. So you have the same problem.

    So maybe that is the answer, it is impossible to find the slope of a graph at a point because a point does not have a slope, I am sure we can all agree on that one

    So basically 90% of the maths you think you know is wrong

    Probably explains a few exam results!
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    Quote Originally Posted by esbo
    Quote Originally Posted by DrRocket
    Quote Originally Posted by esbo
    Quote Originally Posted by salsaonline
    What is ?
    It is a small increase in x, more specifically the limit of that inrease as it approaches zero.
    The question is what definition do you use for dx that permits a rigorous definition for the derivative in terms of real numbers.

    "A small increase in x" is not a real number. "The limit of that increase as it approaches zero" IS zero.

    So you have defined dx to be zero, and that does not provide a defintiion of the derivative that is meaningful.

    Neither does it help to define dx as "an infinitesimal change in x". You cannot do that and work within the real numbers. There is a way to do this in the context of what are called the "non-standard real numbers" but that requires a LOT more sophistication to handle even the basic definitions -- you basically need to know what an ultrafilter is just for starts.

    The point here is that to define concepts rigorously one needs to be fairly careful and then work strictly withing the concepts that have been defined to construct definitions and proofs that advance the subjectd.
    But then when you write it becomes undefined or infinite or whatever. So you have the same problem.

    So maybe that is the answer, it is inpossible to find the slope of a graph at a point because a point does not have a slope, I am sure we can all agree on that one

    So basically 90% of they maths you think you know is wrong

    Probably explains a few exam results!
    dy/dx is not a quotient of some quantity dy by some quantity dx. It's just a symbol. It's meant to suggest the meaning that you intuit, but you could replace every instance of dy/dx with a symbol like :P (y,x), and it wouldn't change anything.

    Indeed, for y = f(x), people often write f'(x) for the derivative of y with respect to x. Again, it's just notation. The actual meaning of the derivative comes from its definition as the limit



    provided such a limit exists.
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    Quote Originally Posted by esbo

    But then when you write it becomes undefined or infinite or whatever. So you have the same problem.
    No it becomes well defined if you make your definitions carefully and meaningfully, which is what is done when one develops the theory of the derivative properly. See any calculus text or, better yet, a text on real analysis. As an alternative you might cogitate until you understand the definition provided by salsaonline.

    Quote Originally Posted by esbo
    So maybe that is the answer, it is impossible to find the slope of a graph at a point because a point does not have a slope, I am sure we can all agree on that one ;
    The notion of the slope of a the graph of a function at a point is DEFINED to be the derivative of the function at that point. It is apparently possible to find that quantity. It is done by calculus students on a daily basis.

    Quote Originally Posted by esbo
    So basically 90% of the maths you think you know is wrong
    You have chosen the wrong pronoun. 100% of the math that I know is correct, and I know why it is correct and can prove it. You,however, may have your own problems.

    Quote Originally Posted by esbo
    Probably explains a few exam results!
    Maybe it explains why I passed the qualifying exam. the general exam and the dissertation defense. Is that what you meant ? Or were you refering to some specific exams that gave you trouble ?

    "It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so." Mark Twain

    "Under certain circumstances, urgent circumstances, desperate circumstances, profanity provides a relief denied even to prayer." Mark Twain
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    the personal attacks are definitely getting old quick, please refrain from anymore argumentum ad hominem fallacies, if you will, esbo, please. all you are doing is goading Dr. Rocket to sink to the same level and debase you as you have debased him. Please stop, at least in this thread.
    Wise men speak because they have something to say; Fools, because they have to say something.
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    yeah, I have to admit that the hostility of the some of the posts here is a bit baffling.
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    Quote Originally Posted by Arcane_Mathamatition
    the personal attacks are definitely getting old quick, please refrain from anymore argumentum ad hominem fallacies, if you will, esbo, please. all you are doing is goading Dr. Rocket to sink to the same level and debase you as you have debased him. Please stop, at least in this thread.
    I am not really sure what you are talking about I speak english not latin .
    My post did nont goad anyone I just expresed my views which I am entitled to do.
    The only thing I said which could pospsible be seen as an attack was
    "So basically 90% of the maths you think you know is wrong"
    Bbut that was not directed at Dr Rocket it was a general statement derived from the
    fact that a point does not have a slope, and intended as a joke.

    If a point has a slope what is the slope of a point at x=2, y=7?

    I would like an answer!
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    Quote Originally Posted by salsaonline
    Quote Originally Posted by Arcane_Mathamatition
    I do know S has some elements though, and that all elements of S will satisfy the required problem. I know this because I can approximate any Irrational number to be between 2 rationals, and I can always find an integer larger than a rational number, since a rational number is the ratio of two integers. I know I can prove it for the integer being larger than a rational, but I don't know how to show that there is always a rational larger and lesser than an irrational, so I'm sure my argument is weak as it is. But I know it's true, just not how to show that it is true.
    You don't know anything a priori (though I'll grant you as given all well-known properties of the rationals that make no reference to irrationals).
    the definition of a rational number is that it can be written as the ratio of two integers, where as an irrational number can NOT be written in that way. Thats the bulk of what I know to be true of the rationals and the irrationals. I'm sure there is some way to prove that some specific irrational number can be bounded, by that I mean there is one number not equal to the rational both above and one below, by rational numbers. I guess another way of saying this, is that some irrational number can always be approximated to be between 2 rational numbers.
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    Quote Originally Posted by esbo
    If a point has a slope what is the slope of a point at x=2, y=7?

    I would like an answer!
    give me a line that goes through that point and I'll tell you.
    Wise men speak because they have something to say; Fools, because they have to say something.
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    Quote Originally Posted by salsaonline
    yeah, I have to admit that the hostility of the some of the posts here is a bit baffling.
    I think you misunderstand them.
    There was no hostility in my last post.
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    Quote Originally Posted by Arcane_Mathamatition
    Quote Originally Posted by esbo
    If a point has a slope what is the slope of a point at x=2, y=7?

    I would like an answer!
    give me a line that goes through that point and I'll tell you.
    I wil give you two!!

    x=2 and y=7!!

    You have a 50% chance of getting it right, perhaps

    But that is the whole point of it, a line is not a point.
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    Quote Originally Posted by esbo
    Quote Originally Posted by salsaonline
    yeah, I have to admit that the hostility of the some of the posts here is a bit baffling.
    I think you misunderstand them.
    There was no hostility in my last post.
    I was referring to Dr R, although, to be fair, you have been bating him.
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    Quote Originally Posted by esbo
    Quote Originally Posted by Arcane_Mathamatition
    Quote Originally Posted by esbo
    If a point has a slope what is the slope of a point at x=2, y=7?

    I would like an answer!
    give me a line that goes through that point and I'll tell you.
    I wil give you two!!

    x=2 and y=7!!

    You have a 50% chance of getting it right, perhaps

    But that is the whole point of it, a line is not a point.
    you are failing to give me a line. The point of a slope is that it exists on lines, curves, surfaces, etc. etc. not just a point. BUT, if the point is on a line, then I can tell you the slope of that line AT that point. which is what differentiation is.
    Wise men speak because they have something to say; Fools, because they have to say something.
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    Quote Originally Posted by salsaonline
    Quote Originally Posted by esbo
    Quote Originally Posted by salsaonline
    yeah, I have to admit that the hostility of the some of the posts here is a bit baffling.
    I think you misunderstand them.
    There was no hostility in my last post.
    I was referring to Dr R, although, to be fair, you have been bating him.
    Both Dr. R and esbo have been going at it in around 3 threads, just each one insesently bashing the other for bashing them. From what I've seen, esbo started the whole mess, and Dr. Rocket is responding in kind. the 90% of math crap that esbo stated, promptly followed by that crack about exams, seemed to me to be a blatent attack on Dr. Rocket. the whole thing is totally ridiculous and is helping no one at all, which is why I'm repectfully requesting that BOTH of you two just stop! esbo with the bating and attacks and Dr. Rocket with the retorts.
    Wise men speak because they have something to say; Fools, because they have to say something.
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    Quote Originally Posted by Arcane_Mathamatition
    Quote Originally Posted by esbo
    Quote Originally Posted by Arcane_Mathamatition
    Quote Originally Posted by esbo
    If a point has a slope what is the slope of a point at x=2, y=7?

    I would like an answer!
    give me a line that goes through that point and I'll tell you.
    I wil give you two!!

    x=2 and y=7!!

    You have a 50% chance of getting it right, perhaps

    But that is the whole point of it, a line is not a point.
    you are failing to give me a line. The point of a slope is that it exists on lines, curves, surfaces, etc. etc. not just a point. BUT, if the point is on a line, then I can tell you the slope of that line AT that point. which is what differentiation is.
    Yes you can tell the slope of a straight line I agree, but I don't think you can find the slope of a point.
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    Quote Originally Posted by esbo
    Quote Originally Posted by Arcane_Mathamatition
    Quote Originally Posted by esbo
    Quote Originally Posted by Arcane_Mathamatition
    Quote Originally Posted by esbo
    If a point has a slope what is the slope of a point at x=2, y=7?

    I would like an answer!
    give me a line that goes through that point and I'll tell you.
    I wil give you two!!

    x=2 and y=7!!

    You have a 50% chance of getting it right, perhaps

    But that is the whole point of it, a line is not a point.
    you are failing to give me a line. The point of a slope is that it exists on lines, curves, surfaces, etc. etc. not just a point. BUT, if the point is on a line, then I can tell you the slope of that line AT that point. which is what differentiation is.
    Yes you can tell the slope of a straight line I agree, but I don't think you can find the slope of a point.
    Any line, any curve, doesn't have to be straight. Consider the orbit of the earth around the sun. If, at some point in time, you wanted to know the direction and magnitude of Earths travel, you could determine that i.e. the slope of the orvit at that point. And, in this example, it just so happens that if gravity instantaneously disappeared, then BOOM, the slope of the curve at that point becomes the slope the direction that the earth would then be traveling on, in a straight line. So YES, you can take the slope of a point ON A LINE.
    Wise men speak because they have something to say; Fools, because they have to say something.
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    Quote Originally Posted by esbo
    [
    Yes you can tell the slope of a straight line I agree, but I don't think you can find the slope of a point.
    So what?
    No one has ever tried to make sense of the notion of "the slope of a point", that notion is simply not foundin differential calculus.

    What is discussed, what has always been discussed is" the slope of a line tangent to a curve at a point". No more, no less.
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    Quote Originally Posted by Arcane_Mathematician
    From what I've seen, esbo started the whole mess, and Dr. Rocket is responding in kind. the 90% of math crap that esbo stated, promptly followed by that crack about exams, seemed to me to be a blatent attack on Dr. Rocket. the whole thing is totally ridiculous and is helping no one at all, which is why I'm repectfully requesting that BOTH of you two just stop! esbo with the bating and attacks and Dr. Rocket with the retorts.
    Moderator agrees. I think it would be a crime to lock this thread, as Arcane, salsa and Rocket are doing this subforum a great service with their exchanges, but if the ad homs continue, I shall have no choice.

    Let's just get back to mathematics, as it is commonly understood.

    Deal?
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    if people are going to talk in latin is it OK if I talk in Gaelic?
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    Anyway I don't think the 'proot' or definition of differentiation speaks of a tangent,
    what it actually speaks of is 'delta' x or whatever, a small increase in x.
    It then speaks of the limit of this value as it approaches zero, a somewhat vague term, because if it is more than zero it is inaccurate and if it is zero, and as anyone knows, you run into problems.
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    esbo: You are clearly confused about the theoretical basis of the differential calculus.

    Say what, if you ask a genuine question on this subject, phrased like, oh, I dunno, "please help me on this, I am so confused....", we have members here who would be more that willing to help you. Listen to them, learn from them, they are experienced teachers.

    And you are confused.
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    The limit operation is a perfectly well defined concept. Suppose that is a function. Then the statement



    simply means the following. For any error , we can find a distance so that whenever , we have that .

    In the situation when we are defining the derivative of a function f(x) at a point x, we let



    The derivative of f(x) at x is then .

    Of course, for some functions, this limit doesn't exist. For those functions where the limit does exist, it should be clear that there is no ambiguity in the given definition. I won't claim that this epsilon/delta approach is the most intuitive way of understanding the derivative at first glance. That's the price you pay for rigor.
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    Quote Originally Posted by Guitarist
    esbo: You are clearly confused about the theoretical basis of the differential calculus.

    Say what, if you ask a genuine question on this subject, phrased like, oh, I dunno, "please help me on this, I am so confused....", we have members here who would be more that willing to help you. Listen to them, learn from them, they are experienced teachers.

    And you are confused.
    I am not confused at all, I am just pointing out that a point is not a line.
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    after looking at these two exercises, I think I need something more rudimentary. I think the best way for me to do this, is if I started from scratch. If you guys could help me out from the beginning, allotting to me only the definitions of arithmetic, and making me develop all of the principles about the reals, rationals, integers and so forth
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    Do as you wish, but my advice is to try to move forward rather than back. Don't refuse to move on just because you don't understand everything completely. As long as you keep at it, the gaps in understanding fill in eventually.

    I didn't know how to do those problems either when they were first presented to me. I moved on because I had no choice--I was enrolled in the class. I took a few months for the material to begin to make sense.
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    I honestly don't know where to go. I figure backing up would give me a better understanding of each of those principles and theorems, and having to construct them would cause me to be able to use them better and actually know how to apply them. I really feel like my problem with those two exercises is the fact that I don't understand the step before hand, so to speak. It may be that I just need to expose myself to a lot of material, and just keep looking at it all, but I truly don't know where to go in this topic, as it is one I haven't had any experience in, and I apparently jumped in mid-stream
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    Quote Originally Posted by Arcane_Mathematician
    I honestly don't know where to go. I figure backing up would give me a better understanding of each of those principles and theorems, and having to construct them would cause me to be able to use them better and actually know how to apply them. I really feel like my problem with those two exercises is the fact that I don't understand the step before hand, so to speak. It may be that I just need to expose myself to a lot of material, and just keep looking at it all, but I truly don't know where to go in this topic, as it is one I haven't had any experience in, and I apparently jumped in mid-stream
    Salsaonline is giving you good advice. If you get humg up on this rather dry stuff, that is, after all, material that you have known since childhood, you will become discouraged and probably forget about mathematics.

    Press on. The development of the number systems is something that you can do at your leisure later, after you have learned a bit more mathematics.
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    Will do. Any other exercises that may be a good place to go?

    As I'm not familiar with it, something involving the Triangle inequality?
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    Quote Originally Posted by DrRocket
    If you get humg up on this rather dry stuff, that is, after all, material that you have known since childhood, you will become discouraged and probably forget about mathematics.

    Press on. The development of the number systems is something that you can do at your leisure later, after you have learned a bit more mathematics.
    Totally agree.

    As for exercises...

    Here's a wild thought. Feel free to veto it. But bear with me.

    Why not put the math down for a little while, and learn some physics? This won't help you learn how to do proofs, but it will force you to get really comfortable with the techniques you've learned in math. And this comfort may help you with proofs down the road.

    By learning physics, you'll also get introduced to, and acquire intuition for, more advanced mathematics. That intuition will serve you very well when you revisit those topics in a rigorous math class.
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    I actually am. I've gotten through all the introductory classes. In terms of the college I'm going to, I'm just waiting for the next course to open up, not exactly a popular field. My major is actually Mechanical engineering, simply because I was told it would be more useful than a mathematics degree. I'm really looking to get far in both fields, as they are rather intricately related (physics to math at least)
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    Quote Originally Posted by Arcane_Mathematician
    I actually am. I've gotten through all the introductory classes. In terms of the college I'm going to, I'm just waiting for the next course to open up, not exactly a popular field. My major is actually Mechanical engineering, simply because I was told it would be more useful than a mathematics degree. I'm really looking to get far in both fields, as they are rather intricately related (physics to math at least)
    Mechanical engineering is not necessarily more useful than a mathematics degree. But it is certainly perceived as more useful by typical management in manufacturing industries, unless and until the specific mathematician demonstrates his usefulness. I spent nearly 25 years doing that in the manufacture of rockets and munitions. The ability to understand what the mathematics really means, what is says, and more importantly what it does not say is very valuable. If you understand just a little probability you will have a huge leg up on the engineers and even on the statisticians.

    The chief engineer at the rocket facility where I started was a mathematician. I had technical responsibility for several facilities. But my engineering degrees didn't hurt either.

    The same prejudice regarding engineering vs mathematics also holds for engineering vs physics, but once the physicist does his stuff, he gets respect.

    Salsaonline's suggestion to go learn physics is a good one. It is not the only way to go, but it is a good way to go. Quite a few mathematicians received their undergraduate training in physics. In my opinion, a physics undergraduate degree is probably a better degree than a typical undergraduate mathematics degree. You can take about the same mathematics classes as a math major, and you get to see the mathematics used on a daily basis. You can even switch from engineering to mathematics in graduate school -- I did.

    A word of caution. Don't believe everything that you hear regarding mathematics in physics classes. Physics books and engineering books quite commonly have false statements regarding some parts of mathematics -- particularly with regard to convergence of Fourier series and with statements regarding things like the Dirac delta function. Physicist can be a bit sloppy with the mathematics and get away with it because they deal with functions that are "nice" and therefore invoke unseen hypotheses.

    An alternate approach might be to take a class or read a book on introductory "Real Analysis". A very good text, usually available at a decent price on the used book market (Alibris.com for instance) is Elements of Real Analysisby Robert Bartle. He simply postulates the real numbers as a complete Archimedean field and then develops calculus correctly. A class from Bartle's book is typically a one-year class at the junior-senior level.

    Real analysis is fun stuff, and I think you would enjoy it. Once you have done that then you can take a rainy weekend and read Edmund Landau's Foundations of Analysis which goes through the development of the number systems starting from the Peano axioms. It is an incredibly dry book, but it is very short and very efficient in getting through that stuff. Then spend another rainy weekend and read Paul Halmos's Naive Set Theory (you can skip the parts where he develops the number systems) and you will learn what you need to know about cardinal and ordinal numbers.
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  66. #65  
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    I'd rather someone had a physicists' understanding of the Dirac delta function than no acquaintance with it whatsoever.

    But I get your point. Physicists, on occasion, do appear to not understand the mathematics very well. You really need to see both sides.

    Even to do mathematics though, I think it's important to know physics. A lot of work in math these days is motivated from problems in physics.

    Let me clarify what I really mean by "physics": I mean Lagrangian and Hamiltonian mechanics, and most importantly, classical quantum mechanics. It's possible to study mathematics without knowing those subjects, but I don't recommend it.
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  67. #66  
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    Quote Originally Posted by salsaonline
    I'd rather someone had a physicists' understanding of the Dirac delta function than no acquaintance with it whatsoever.

    But I get your point. Physicists, on occasion, do appear to not understand the mathematics very well. You really need to see both sides.

    Even to do mathematics though, I think it's important to know physics. A lot of work in math these days is motivated from problems in physics.

    Let me clarify what I really mean by "physics": I mean Lagrangian and Hamiltonian mechanics, and most importantly, classical quantum mechanics. It's possible to study mathematics without knowing those subjects, but I don't recommend it.
    I basically agree.

    Physicists not only sometimes appear to not understand the mathematics very well, they flaunt it. I have seen presentations of the Dirac delta that basically begin with "mathematicians don't like this very much but we don't care". That is simply ridiculous since the theory of distributions not only explains the monkey motions that follow that introduction it provides quite a bit more including a nice foundation for the study of partial differential equations.

    I also agree that physics provides good questions for mathematics research. But there are other sources as well.

    I'm not so sure that I would want someone's introduction to Hilbert space to come from a physics class in quantum mechanics. But exposure to QM is a good thing if you don't let the physicists looseness ruin functional analysis.

    Most importantly though, I cannot imagine trying to learn partial differential equations without some background in physics. Trying to solve PDEs withouth knowing why you want to do such a thing is futile. -- but I have seen just that attempted by trying to teach that subject to sophomores who have not had a good physics class.
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  68. #67 Alternate approach if you are having trouble with proofs 
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    For those who are being exposed to theorem proving for the first time and are having some problems getting the hang of it, there is perhaps a book and an alternate approach that might help. This is not the only way to learn to do proofs, it is not the way that I got into them, and I suspect that it is not the way that many other professional mathematicians did either. But it is not a bad way.

    There is a text written for a class designed to introduce students to rigorous mathematics and to doing proofs. It has lots of exercises, and starts with basic logic. It is A Transition to Advanced Mathematics by Douglas Smith, Maurice Eggen and Richard St. Andre. I have only skimmed the book but at that level it looks pretty good.

    The point of the book is not the material itself but rather learning what a proof is and how to do them. This is a gentle introduction to theoretical mathematics, as contrasted with the traditional "sink or swim" approach. It may appeal to those who think that they may be sinking.
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  69. #68 Re: Alternate approach if you are having trouble with proofs 
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    Quote Originally Posted by DrRocket
    For those who are being exposed to theorem proving for the first time and are having some problems getting the hang of it, there is perhaps a book and an alternate approach that might help. This is not the only way to learn to do proofs, it is not the way that I got into them, and I suspect that it is not the way that many other professional mathematicians did either. But it is not a bad way.

    There is a text written for a class designed to introduce students to rigorous mathematics and to doing proofs. It has lots of exercises, and starts with basic logic. It is A Transition to Advanced Mathematics by Douglas Smith, Maurice Eggen and Richard St. Andre. I have only skimmed the book but at that level it looks pretty good.

    The point of the book is not the material itself but rather learning what a proof is and how to do them. This is a gentle introduction to theoretical mathematics, as contrasted with the traditional "sink or swim" approach. It may appeal to those who think that they may be sinking.
    I went the sink or swim way, but that's because I didn't know any better. Real analysis out of Rudin's little blue book was my first proof class. I enrolled because the course was called "Intro to Analysis." And the course number was the smallest of the upper divs. So I thought, ok, I'm supposed to take that first. (I thought Intro to Analysis was like Intro to Math. I didn't know "analysis" was a particular kind of math.) Had I known better, I probably would have done things differently, but in retrospect, it was probably good getting one of my hardest classes out of the way so early in the game.

    There's also a book called "Chapter Zero"; I don't know if it's good or bad, but it's another book that introduces the student to the technique of proofs.
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