I state a theorem first, and want to write the proof in the later section. And I want to restate the theorem when I am going to prove it. So in the whole paper, I write the theorem for two times, but I want them have the same number. How can i do this?
31592
Often in my papers I want to mention theorems in the introduction that are introduced in more detail and proved in a later section.  I'd like to include the statement of these theorems twice over, each time numbered as the theorem would naturally appear in its later section.  My current solution (using amsthm) is, e.g.  %(in the introduction) \newtheorem*{thm:associativity}{Theorem \ref{thm:associativity}} \begin{thm:associativity} Lorem ipsum ... \end{thm:associativity}  %(in a later section) \begin{thm} \label{thm:associativity} Lorem ipsum ... \end{thm}      Can anyone think of a cleaner solution?
54859
When writing formal proofs in abstract and linear algebra, what kind of jargon is useful for conveying solutions effectively to others? In general, how should one go about structuring a formal proof so that it is both clear and succinct? What are some strategies for approaching problems that you will need to write formal proofs for?
65008
Learning about proof by induction, I take it that the first step is always something like "test if the proposition holds for $n = \textrm{[the minimum value]}$"  Like this:       Prove that $1+2+3+...+n = \frac{n(n + 1)}{2}$ for all $n \ge 1$.   Test it for $n = 1$:  $$n = \frac{n(n + 1)}{2} \implies 1 = \frac{2}{2}\textrm{, which holds.}$$  * The rest of the proof goes here *    So, I do it all the time (like a standard). But I never really thought about why. Why do I do such test? I can see that if the test does not hold, it cannot be proven by induction, I guess. But is there another reason we do this?
101131
I am trying to work on the numbering of the theorems/definitions/lemmas etc., and I have some problems with the numbering.   I would like the theorems, propositions, corollarys, definitions, conjectures, examples to follow the same numbering, and to reduce the numbering. For example, in my code below, the stuff goes as follows:     Chapter 1      Addition      1.1 Basics (SECTION)      Definition 1.1.1.      Theorem 1.1.1.      Theorem 1.1.2.      Example 1.1.1.      1.1.1. Some tips (SUBSECTION)      Definition 1.1.2.      1.2 Advanced stuff (SECTION)      Definition 1.2.1.   But how would I get it to be like this below?:     Chapter 1      Addition      1.1 Basics (SECTION)      Definition 1.1      Theorem 1.2      Theorem 1.3      Example 1.4      1.1.1. Some tips (SUBSECTION)      Definition 1.5      1.2 Advanced stuff (SECTION)      Definition 1.6   And for each new chapter, how would I get the numbering to be resetted? Like it would be Definition 2.1, Theorem 2.2, Theorem 2.3 for chapter 2.  I like the layout of my table of contents, i.e. Part I is addition and subtraction, Part II is Multiplication and division, and I am happy with the way the sections and subsections are working, but I just want to definitions, theorems etc. to be corrected.  \documentclass[11pt,a4paper]{report} \usepackage{amsmath,amsfonts,amssymb,amsthm,epsfig,epstopdf,titling,url,array}   \theoremstyle{plain} \newtheorem{thm}{Theorem}[section] \newtheorem{lem}[thm]{Lemma} \newtheorem{prop}[thm]{Proposition} \newtheorem*{cor}{Corollary}  \theoremstyle{definition} \newtheorem{defn}{Definition}[section] \newtheorem{conj}{Conjecture}[section] \newtheorem{exmp}{Example}[section]  \theoremstyle{remark} \newtheorem*{rem}{Remark} \newtheorem*{note}{Note}      \begin{document} \tableofcontents  \part{Addition and Subtraction}  \chapter{Addition} \section{Basics} \begin{defn} Here is a new definition \end{defn} \begin{thm} Here is a new theorem \end{thm} \begin{thm} Here is a new theorem \end{thm} \begin{exmp} Here is a good example. \end{exmp} \subsection{Some tips} \begin{defn} Here is a new definition \end{defn} \section{Advanced stuff} \begin{defn} Here is a new definition \end{defn} \subsection{Warnings} \begin{defn} Here is a new definition \end{defn}  \chapter{Subtraction} \section{Basics} \subsection{Some tips} \section{Advanced stuff} \subsection{Warnings}  \part{Multiplication and Division}  \chapter{Multiplication} \section{Basics} \subsection{Some tips} \section{Advanced stuff} \subsection{Warnings}  \chapter{Division} \section{Basics} \subsection{Some tips} \section{Advanced stuff} \subsection{Warnings}    \end{document}
101588
I am self studying number theory from Introduction to Analytic number theory by Apostol.   I have a doubt in one argument of proof of Theorem 5.14 and I am posting its image.      My doubt is in 2nd line of 2 nd paragraph how Apostol writes - If y is a solution of (5) then ay $\equiv $ at ( mod m) . How does y being a solution of (5) implies that?       Can someone please help
99260
So I see that the process for proof by induction is the following (using the following statement for sake of example:  $P(n)$ is the formula for the sum of natural numbers $\leq n$:  $0 + 1 + \cdots +n = (n(n+1))/2$ )   Show that the base case is trivially true: $P(0)$: let $n = 0$. Then $0 = (0(0+1))/2$, which is true. Show that if $P(k)$ holds then $P(k+1)$ also holds. Assume $P(k)$ holds for some unspecified value of $k$. It must be then shown that $P(k+1)$ holds.   This is the part I don't get: 'Assume $P(k)$ holds'  We are just assuming something is true and then 'proving' that something else is true based on that assumption. But we never proved the assumption itself is true so it doesnt seem to hold up to me. Oviously proof by induction works so am I viewing the process incorrectly?
31902
Question : If $p$ is prime, and $0&lt; k&lt; p$ show that $ p \mid {p \choose k}$  ${p \choose k}$ can be rewritten as:  $${p(p-1)(p-2)\dots(p-(k-1))(p-k)! \over (p-k)!\cdot k(k-1)(k-2)\dots3\cdot2\cdot1}$$  Now the $(p-k)!$ cancels out. Since in the numerator we have k consecutive  integers, k divides one of them (not p as p is prime and $k &lt;p$). Then k-1 divides another one or two terms in the k consecutive terms (again not p). This reasoning continues until  3 and 2.   This means that ${p \choose k}$ can be rewritten as pq implying $p | {p \choose k}$  What if any of this proof is wrong. I am worried about any 'overlaps', i.e. say k-6 divides terms that k-3 divides invalidating the argument.  P.S. I hope the above is legible as I am new to this site. Thank you!
110109
I find myself in an awkward situation. Nearly two years ago I submitted a paper to a reputable mathematics journal. After 15 months it was rejected with the rationale that there had been papers published (most since the submission date) that contained results that I should have referenced. I found that a little spurious, but moved on and put my work together with some colleagues and got our new paper accepted (pending some language issues) by a different journal.  However, one of my co-authors has found that my exact result (barring some small notational changes, but with the exact same language; I can recognise it as mine, I slaved over finding just the right words!) from my original paper has had the proof published in a book by one author of the references that caused the paper to get originally rejected!  My assumption is thus that the paper was sent to this person to referee and they liked the result so much that they kept it for themselves, which also explains the delay in refereeing... I believe that this author has also presented this result at a conference as their own work. They have a number of papers published on the topic in question from their doctoral thesis.  My question is what steps can I take now to re-assert my authorship?  I believe that it would have been simpler if I had also submitted it to the arXiv, but I wasn't sure (at that time) how that would be seen by the journal. I did give a presentation of my results at a local conference, so I would be able to produce credible witnesses to back up my claim, if necessary. The book published containing my work is largely the doctoral thesis of the author, so I am attempting to find a copy of the thesis since it was deposited before I submitted my paper to the journal. Can anyone suggest any other recourse?  My co-authors would prefer to err on the side of caution and include a reference to the result in our revised paper, but I would really like to reclaim my work...
60269
In many mathematics papers, I read sentences like:     With an appeal to Theorem 4.5 we get …   Is the capitalization of theorem in this case correct? If it is correct why do we capitalize the word?
40563
This was the last part of an exam question. I have spent a long time looking for such an example but have failed (sequences when one of the three conditions is dropped are easily found but I couldn't find one when all three are required). Could you also give the way you thought of the example in your answer please?   Thanks!
112244
I saw the following "theorem" and its "proof".  I can't explain well why the argument is wrong. Could you give me clear explanation so that kids can understand.  Theorem: All natural numbers are equal. Let $a, b \in \mathbb{N}$, then a=b.  Proof by induction. Let $m=\max\{a, b\}$. We will prove that the theorem holds for all $m\in \mathbb{N}$. If $m=1$, then $\max\{a,b\}=1$, so $a=b=1$. Now assume that it holds for $m=k$ for some number $k$. Now let $\max\{a, b\}=k+1$. Then $\max\{a-1, b-1\}=k$ and thus by assumption $a-1=b-1$, so $a=b$. Therefore, the proof is complete.
115339
I want the theorems to be numerated considering the depth, namely:  Chapter 1.  Theorem 1.1.  Section 1.1.  Theorem 1.1.1.  Subsection 1.1.1.  Theorem 1.1.1.1.  Theorem 1.1.1.2.  Theorem 1.1.1.3.  I have used different theorem types like  \newtheorem{th}{Theorem}[chapter] \newtheorem{thh}{Theorem}[section] \newtheorem{thhh}{Theorem}[subsection]   but if I change part of a section into a subsection I have to change all the theorem types within.  Is there any way to unify this?
51190
This is a question I have been thinking about for a while, which seems especially important as I hope to transition from undergraduate to graduate study in the next year, where studying for an exam becomes less important and studying for actual learning becomes more important.   I have always been required as part of my degree to learn proofs. Seemingly, basically, just memorising them line-by-line - there might be other methods for answering "prove [random proposition in the lecture notes]" questions, but memorisation is most effective and effectively forced by a tight time limit.   I always thought there had to be a good reason these kind of questions were put - in fact, apparently required by guidelines to be put - in the examinations. Therefore, even when teaching myself maths, I put in the effort to memorise as many proofs of theorems and propositions as I could. This especially seemed useful for maintaining understanding over long periods of time.   Now, I wonder if this was all just a bit of a waste of time. I have read many people commenting here that doing problems is far more useful - "you learn maths by doing it".   Am I wrong? Is there value in memorising proofs (after having read and understood them, not as opposed to skipping them entirely)? Or is this just a waste of time? What about when reading research papers?
101412
I have a section that is devoted to the proof of a theorem, and hence I used  \section{Proof of Theorem~\ref{thm}}   Of course, because of the headers that are capitalized, it asks for \ref{THM} when producing the header, and I get an obvious error. What is the proper way to handle such a case? Below is a MWE:  \documentclass{amsbook} \usepackage{lipsum} \newtheorem{thm}{Theorem} \begin{document} \chapter{First Chapter} \section{The Theorem} \begin{thm}[Lipsum Theorem] \label{thm} \lipsum[1] \end{thm} \section{Proof of Theorem~\ref{thm}} \lipsum[1-23] \end{document}   When compiling I get:  LaTeX Warning: Reference `THM' on page 3 undefined on input line 14.   (each time it produces a header), and the header shows     `2. PROOF OF THEOREM ??'
69970
The main question is :  Prove that $n^n&gt;1\times3\times5\times7\times\dots\times(2n-1)$  My approach :  We can write the R.H.S as,  $$\frac{(2n-1)!}{2\times4\times6\times\dots\times2(n-1)}$$ We can write $(2n-1)!$ as $(2n-3)!(2n-1)2(n-1)$  Thus, in this manner we can cancel out even terms till $n$ appears. I'm having trouble doing this. What I intend to do is say that any positive number's factorial is less than the number raised to itself. I welcome any alternate method. Please help.  P.S. I am still in high school and do not understand binomial theorem and induction.
108689
I have several of these types of problems, and it would be great if I can get some help on one so I have a guide on how I can solve these.  The question is:      Prove $A \setminus (B\setminus C) \neq  (A\setminus B) \setminus C$   I know I must prove both sides are not equivalent to each other to complete this proof. Here's my shot: We start with left side.   if $x$ is in $A$, then $x$ is not in $B$, not not in $C$ so $x$ is in $A$ and $C$ if $x$ is in $A$, then it's not in $B$ $\rightarrow$ $(A\setminus B)$ but if $x$ is in $A\setminus B$, then it must not be in $C$ however, earlier we stated $x$ is in both $A$ and $C$ we see that the two sides are not equal   Is this the right idea?  Should I then reverse the proof to prove it the other way around, or is that unnecessary?  Should it be more formal?  Thanks!
53756
Can you check if my proof is right?  Theorem.  $\forall x\geq8, x$ can be represented by $5a + 3b$ where  $a,b \in \mathbb{N}$.  Base case(s): $x=8 = 3\cdot1 + 5\cdot1 \quad \checkmark\\ x=9 = 3\cdot3 + 5\cdot0 \quad \checkmark\\ x=10 = 3\cdot0 + 5\cdot2 \quad \checkmark$  Inductive step:  $n \in \mathbb{N}\\a_1 = 8, a_n = a_1 + (x-1)\cdot3\\ b_1 = 9, b_n = b_1 + (x-1)\cdot3 = a_1 +1 + (x-1) \cdot 3\\ c_1 = 10, c_n = c_1 + (x-1)\cdot3 = b_1 + 1 + (x-1) \cdot 3 = a_1 + 2 + (x-1) \cdot 3\\ \\ S = \{x\in\mathbb{N}: x \in a_{x} \lor x \in b_{x} \lor x \in c_{x}\}$  Basis stays true, because $8,9,10 \in S$  Lets assume that $x \in S$. That means $x \in a_{n} \lor x \in b_{n} \lor x \in c_{n}$.  If $x \in a_n$ then $x+1 \in b_x$,  If $x \in b_x$ then $x+1 \in c_x$,  If $x \in c_x$ then $x+1 \in a_x$.  I can't prove that but it's obvious. What do you think about this?