## A math book on introducing proofs(?)

You're welcome. I can see where you're coming from. I think why a lot of students might be seeing the technique as the holy grail of proof-writing (which obviously leads people of higher stature to scoff at the book) is because it makes all the logic they've learned (which often comes far before real proof-writing) into something they can immediately use to prove theorems. Suddenly the students feel enlightened and elementary proofs go from extremely difficult to something that they actually have a feel for. This causes them to have unfounded confidence in this technique when it's in fact nothing more than a way to organize some thoughts.

Edit: To answer your last question, a lot of the stuff that comes before mathematical induction is perhaps a little unnecessary for the aim of the book. Every section from relations and after (up until mathematical induction) are mostly exercises in proof-writing mixed with a very basic introduction to some topics that undergrads will soon encounter.

 Quote by mathwonk "A real valued function f defined on the interval I, is said to be 'bounded' if and only if..........."
Sorry, if it's off topic, but what do you mean by this challenge? Are you saying that you don't want your students to say that "a real valued function $f(x), x \in X$ is bounded on $X_1 \subset X$ if for every $x \in X_1$ there exist $m$ and $M$ such that $m \leq f(x) \leq M$"? I think this theorem(?) doesn't holds only for monotone functions. So how can we define boundedness for all functions?

Would a better way to say this be that "A real valued function $f$, defined on the interval $I$, is said to bounded if and only if there exist $m$ and $M$ such that for every $x \in I_1 \subset I$ we have $m \leq x \leq M$"? EDIT: or maybe not, this sounds a bit circular (and wrong) to me...or is it okay to say that $f$ is bounded iff $I$ is bounded?

Or maybe the easier way would be to say that $f$ is bounded on $I$ iff there exist $f(m)$ and $f(M)$ ($m,M \in I$) such that $f$ is defined only for $x \in [m,M]$? This sounds messy too, but what I mean to say, is that $f$ is bounded on an interval if it is defined in that interval and not defined outside of it.

Don't really know... This may all be wrong, since I'm making this up in my head.

- Kamataat
 Recognitions: Homework Help Science Advisor For Bifodus. I didn't say Velleman uses truth tables to prove anything (other than results in predicate logic). Perhaps you only glanced at my post? I said it pointlessly starts with truth tables (and things that any student of mathematics at university ought to already know). What it intends the students to understand and what the students actually understand from it are not necessarily related, or are you accusing me of lying about the work my students hand in based upon reading this book? It is a complete waste of money, and is the recommended text for a course I teach, so yes, I have looked at it in order to see what my students are beign told to buy. And what I saw wasn't good mathematics. The best way to learn to write proofs is to read a lot of them and write a lot of them and there is no need to have a special book in order to do that.
 Recognitions: Homework Help Science Advisor kamataat: all of your statements are incorrect. did you mean for any of them to be correct? or are yopu just giving sample incorrect answers? the correct answer is the following: a real valued function f on a domain D is bounded if there exist numbers m,M such that for every x in D, m ≤ f(x) ≤ M. this is different from your statement because your quantifiers are placed incorrectly. a common mistake made by my students is indeed to confuse boundedness of the domain with the (correct) boundedness of the set of values taken over that domain, as some of your later statements do.

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 Quote by mathwonk I was never able to teach my proofs class to understand the logical structure of induction.
I had some success in this regard with a course called "Mathematical Structures". It is a terminal course in mathematics for liberal arts students, and it covers induction. Needless to say, I could have strangled my Department Chair for assigning me to it.

Anyway, I explained it by analogy, and the students were able to understand what they were doing, even if they didn't always do it correctly. Here goes.

You have an infinite line of dominoes. You have to prove that they'll all be knocked down after the first one is knocked down.

First, you have to establish that the first one is knocked down. This of course is the analog of proving P(1).

Then, you have to prove that if a domino at some arbitrary point along the line is knocked down, then it will knock down it's neighbor. This of course is the analog of proving P(k)-->P(k+1).

So when all is said and done, you've proven that all the dominoes go down in just two steps. When the first falls, it knocks down the second. When the second falls, it knocks down the third.....

You get the idea.
 Recognitions: Homework Help Science Advisor well that is certainly one of the many analogies i tried using. There is not a great deal of difficulty conveying the general idea, such as "to be able to climb a ladder, first you must get on the bottom rung of the ladder, and then you must be able to climb from any rung to the next rung." and so on. the problem i had was getting the students to actually be able to write up an induction proof logically correctly. e.g. here is a statement of the principle of induction: Principle of (ordinary) induction: Let P1, P2,....,Pn,.... be an infinite sequence of statements, one for each natural number n. If we can prove: 1) P1 is true, and 2) for each k>=1, whenever the statement Pk is true, then also Pk+1 is true; then all the statements P1, P2,....,Pn,... are true for every n. My class would restate this for instance by saying in step 2, that one begins by assuming Pk for all k. of course then it is moot to bother proving P(k+1), since it has already been assumed. i.e. they would start by assuming all the dominoes had been knocked down. this kind of confusion over the precise use of induction. essentially never went away. the contrast between assuming P, and assuming the statement "P implies Q" is very subtle.
 Recognitions: Homework Help Science Advisor well my research reveals the fact that my view of a book differs from my students' view as follows: Comparing the book of Bond and Keane say, with that of Robert S. Wolf, I greatly prefer Wolf, but my students may prefer Bond and Keane. I.e. Wolf writes intelligently, and authoritatively, as if he actually understands the material, whereas Bond and Keane make substantive oversights, and present canned proofs apparently copied without reflection from other sources. But, although the actual material presented by Wolf is superior, the WAY it is presented is more difficult for stduents to access. E.g. Bond and Keane present a definition, and then immediately follow if by several simple examples, (not always correctly justified, but so what, the student does not know that). Wolf on the other hand, prattles on verbosely like a college professor with tenure. Hence I have to choose whether I want my students to have higher level material that they find ahrder to read, or inferior material that is laid out more simply. Unfortunately the intelligent and knowledgeable author finds it hard to identify with a reader who does not read well. And I admot that even though i think I know how to read, I also have trouble organizing mathematics presented in a discursive style, rather than in simple outline form. For instance my own graduate algebra book, carefully written though I may think it is, has no index. Hence students find it hardert to use and trefer to quickly, because it is writtten for someone who plans to sit down with it and go through it. Now writing an index requires no mathematical knowledge at all, but someone willing to make one adds value to his book.

 Quote by mathwonk kamataat: all of your statements are incorrect. did you mean for any of them to be correct? or are yopu just giving sample incorrect answers? the correct answer is the following: a real valued function f on a domain D is bounded if there exist numbers m,M such that for every x in D, m ≤ f(x) ≤ M. this is different from your statement because your quantifiers are placed incorrectly. a common mistake made by my students is indeed to confuse boundedness of the domain with the (correct) boundedness of the set of values taken over that domain, as some of your later statements do.
You shouldn't really take note of all those statements... I didn't know what I was doing...

So in my first statement it's wrong to say that "for every $x \in X_1$ there exist $m$ and $M$". Thank you for pointing that out to me.

But what does one do in the case when one has $y=\sin x, X_1 = [\pi; 3\pi]$? The function is bounded on $X_1$, but $m=M=0$. Obviously it's not bounded according to your definition, but nevertheless $y$ is bounded from two sides, in the sense that the function ceases to have any values beyond two points on either end of the line. Is this then not a bounded function, although it looks like that on a picture? Should I stop taking pictures/graphs to literally and concentrate more on the words and symbols presented to me?

- Kamataat
 What makes you think m = M = 0? Because those are the values that sin(x) takes at x = pi and x = 3pi? A function defined on a closed and bounded interval doesn't need to assume its maximum or minumum values at the endpoints of the interval. You seem to be confused on what boundedness refers to. It doesn't refer to the domain of the function, it refers to its range (i.e. the image of its domain). To get back to your example, try m = -1, M = 1. Then, given some x in [pi, 3pi], is m <= sin(x)? Yes. Is sin(x) <= M ? Yes. Hence the function is bounded (using mathwonk's definition). (Of course, lower/upper bounds are not unique, and there is no need to restrict ourselves to [pi, 3pi] in the case of the sine function).

 Quote by Muzza What makes you think m = M = 0? Because those are the values that sin(x) takes at x = pi and x = 3pi? A function defined on a closed and bounded interval doesn't need to assume its maximum or minumum values at the endpoints of the interval. You seem to be confused on what boundedness refers to. It doesn't refer to the domain of the function, it refers to its range (i.e. the image of its domain). To get back to your example, try m = -1, M = 1. Then, given some x in [pi, 3pi], is m <= sin(x)? Yes. Is sin(x) <= M ? Yes. Hence the function is bounded (using mathwonk's definition). (Of course, lower/upper bounds are not unique, and there is no need to restrict ourselves to [pi, 3pi] in the case of the sine function).
Thank you. I realize I was wrong. I should really learn to read mathematics carefully. For example I wasn't careful enough to notice that the "less than or equal" sign is used and instead I assumed that it's equal only and thus f(x) must equal m and M when all it has to do is to be between m and M.

- Kamataat
 Sorry matt, I must be having trouble parsing the sentences "I think my issue with these attempts to teach proofs by truth tables are that the students get good at showing some logical rules but..." and "There is a book with the same name as Polya's that appears to think explaining what truth tables are is the best way to teach people how to prove things, which is just utter nonsense." And I'm not accusing you of lying about something you haven't said. If it's true from your experience that this book causes people to misunderstand mathematics, then I'll give you the authority on this topic. You undoubtedly have more experience than I have, matt, for I am just an upper division math major. But I am an upper division math major who, while doing elementary calculus in the early years, studied proof-writing with this book in preparation for more rigorous mathematics. These upper division classes have been significantly easier for me than for the rest of the class. Whether or not I can chalk this up to having read this book and having worked out all the problems I don't know. What I do know is that the book doesn't include anything that one should find controversial, and I have found it unique in that it actually includes commentary in the many many proofs it contains.
 Recognitions: Homework Help Science Advisor well it is entirely possible that Matt does not see anything in the book that he thinks should be useful, and that his students have not performed well after using it, and yet it can still be true that it was useful for you. everyone has their own epiphany experience, where they "get" something for some reason, at a certain time. that is why it is hard for one person to select a book that will help another. What some of us older heads do try to do, is recommend a book, which may be hard to read, but which, if eventually mastered, will at least promise to contain the insights we think important. It may still be the case that a student will require some other source to get ready for these books. you notice a lot of people keep recommending polya, and courant and robbins. there is a reason for this.
 Recognitions: Homework Help Science Advisor Velleman does attempt to get people to start proving things with truth tables. Namely results about truth tables, or simple connective logic. Ie they learn how to prove logical rules, tautologies, whatever, but the way they do this is useless for proving anything else. Ie anything in almost all of mathematics. I will only recommend books that I think a student will be reading in three years time whether it be for work or pleasure, and Velleman does not fit that criterion. Poya does. It is more important to figure out what the argument is to prove it first, rather than know how to negate a string of quantifiers. Writing out the proof is then simply writing down the solution to the problem. I don't find the idea of spending a huge amount of time learning these "methods" in artificial ways helpful for any student. Certainly, I may be thinking that the fact that in order to show an assertion is false for all x, it is sufficient to find an x for which it fails, is obvious, and that is a fault in my view of what students should know: common sense is unique amongst virtues in that everyone possesses it. Perhaps some students (most?) need to have simple logical rules explained, however it took me three years of students to find a satisfactory way of explaining why the slogan "false implies false" is true that everyone grasps. And afterwards, they come to write "show the sequence 2 - 1/n^2" tends to 2 and they can't because their problem solving skills (Ie ways of finding a proof) are lacking. I'd rather have students who've never seen connectives and quantifiers but understand that if x has property, and having property P implies having property Q, and having property Q implies having property R, then x has property R, than someone who can write out the truth table of (P=>Q)/\(Q=R) => (P=>Q) So perhaps it would behove me to differentiate between the skills of solving a problem, and the skill of writing it on paper. I tihnk the former is far more important than the latter, since the latter is a mechanical process, like doing things with truth tables. I would also point out that students in learning, artificially, the methods of proof then struggle to figure out which one to apply in any given situation. ie for a question that doesn't say "use induction to solve".