@Srivatsan well, 2 being prime needs to be mentioned. Furthermore, the basic axiom is that if $p$ is prime and $p|ab$, then $p|a$ or $p|b$, nothing about $p^k$ shows up. That is where the induction is needed.
I answered the drama-inducing question too math.stackexchange.com/a/101553/1284 one thing i'm still stumped by is whether my argument implicitly uses induction and whether induction is necessary to prove this (i.e. is it a theorem of Q?)
With integers, there's always induction implicitly hiding somewhere. The only question is whether you can rope off the conclusion of the induction as a "well-known fact" or not.
@HenningMakholm I think that for a question so basic, I would only assume that if $p$ is a prime and $p|ab$, then $p|a$ or $p|b$. I think that induction is needed to get that if $p^k|ab$ and $p\not|b$ that $p^k|a$.
@DanBrumleve I really think so in a question this elementary. I also don't like the answers using the Fundamental Theorem of Arithmetic. It is too advanced for this.
I worry about circular reasoning when it comes to proving the FTA
i thought these sorts of problems were easy when i was learning number theory in high school, but it's a lot trickier now without the context of a textbook
Hi. I have just read this discussion and the discussion in question. I agree with both Bill and robjohn. It not pedagogical go give such answers to a person struggling at this level.
I think what the OP of that question needs most is some words about being careful and explicit about which assumptions (known facts, axioms, or whatever) he seeks to prove his property from. All of the answers so far just dive directly into proof details without being clear what stage they are playing on.
i agree with bill and yall that Paul's answer is missing some steps, but bill's answer although rigorous was tricky to follow. i was looking for a middle ground in my answer. my answer makes the most sense to myself. :)
Even if you haven't been able to check it out from a library or buy your own copy, I suggest you get the book from the `wire' and have a preliminary look!
@robjohn I don't think that would be accurate -- since the domains and codomains are different, the notion of "commute with" I know of does not apply om this situation.
@robjohn Yeah, so. My main fear was that I'd missed a conventional well-defined meaning of "carry over". But then again, it was a quote from Wikipedia. Not always a paragon of clear exposition.
@ymar Except for its name and origin, my institute and the courses it offers has nothing to do with Statistics, except possibly on a minor scale, like any other good Math dept!
Anyway, I should be going. Just came here to find out what was going on after I seemingly moved up a slot in the all-time rankings without passing anyone. 'Night.
Kannappan if K is a subgroup of G and also a strict subgroup of every subgroup of G then mustn't it strictly contain itself? i'm not very familiar with group theory so i'm probably missing something
@KannappanSampath On the question about the difference between groups and semigroups? I am. But I'm talking about another one now. I didn't get any answers. Just two comments -- helpful, but only comments.
@ymar Not really up my alley, but it looks like a good question. It may be too open-ended to get a good answer unless someone really expert happens upon it.
@KannappanSampath From the little that I’ve seen, I’d be very hesitant to ask anything on MO; it’s not a very welcoming place, though some folks there are perfectly reasonable.
MO has very high standards. i completely switched to math.SE when it launched because it is more welcoming to amateurs like myself. now i'm very hesitant to post anything there. many of the same people are here anyway. unless i'm certain my question is "research-level" i will stick to math.SE.
i'm comfortable enough with math.SE. i would blog too if i had the confidence but i think it will take me some time before i reach that point. if you are in university maybe you have fellow students who can review before you publish?
Point taken. (@MarianoSuárezAlvarez my concern was not so much as the trivial errors rather than coming out as crank or hack since my interest is phil. of maths. which can be receptive to varying degrees of taste...)
nobody will mistake you for a crank if you are honest about your own shortcomings, although self-indemnifying qualifiers don't help a curious reader, so keep it balanced.
Agreed about the typos...actually SE helps too. Because there is maturity in everything. I think I am asking more to the point questions now than vague ones like before...btw Silver Badge beckoning tomorrow ;) )
I remember when I asked in MO: Should there be concept of God in maths.. :facepalm: yet I am still alive.
@Dan yeah i was actually looking at interesting math blogs questionn... but it's chosen now, i will put it on my page to boost profile views and pique curiosity :+]
i've heard many times something like "math is interesting, but i can't handle all the numbers". imo it is a problem with math education not people (no not that imo)
I hope that there will be not fourth assignment, then I'll only have to prepare a lecture on the Whitehead problem, and I'm done with the course.
Today is the last lecture in algebraic topology too. I have to solve 7 questions and take an exam on 3 of those (in which I have to solve only one) and then I am done with this course as well.
Yeah, the exam is only 20% of the grade. He gave us the choice of three questions from the homework assignments and he'll give them priority 100%/66%/33% and we can only solve one question of those. However remembering the proof for three questions is not a big deal.
Oh, don't be. I've had plenty of hard exams... this exam is only a formality since they are trying to eliminate all the "grade purely on a final assignment" method in elementary grad-level courses.
I'm not sure, I've not had to deal with $[ \cdot]$ before. Is $\lim_{x \to 0^+} [x] = 0$ the first question in the book? I'm asking because I'm not sure what one would have to do to show that the equality indeed holds.
Also, the second limit looks as if it evaluated to the same value as the expression without the $[ \cdot ]$ but again I'm not sure how to rigorously show it.
What should I do when I'm not good enough at solving limit questions @MattN? like the ones I said or limits of trigonometric functions. My book has bunch of limit questions but without solution
But then I think we have it. What do you think of this: Use $x^3 - 1 \leq [x^3] \leq x^3$ to get $-1 \leq [x^3] - x^3 \leq 0$ then $$ \lim_{x \to 10^-} \frac{1}{x - [x]} \leq \lim_{x \to 10^-} \frac{[x^3] - x^3}{[x] - x} \leq 0$$
No. That doesn't lead anywhere.
@AndreyVihrov It's really a simple matter of copy pasting the js code into a bookmark of your browser. Takes 30 seconds. : )
In mathematics, Descartes' rule of signs, first described by René Descartes in his work La Géométrie, is a technique for determining the or negative real roots of a polynomial.
The rule gives us an upper bound number of positive or negative roots of a polynomial. It is not a complete criterion, i.e. it does not tell the exact number of positive or negative roots.
Descartes' rule of signs
Positive roots
The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial ...
@AsafKaragila Well, I have you to thank. I was going to write an answer, which in hindsight may be ill-advised, but you closed the question before I finished composing.
(I can't learn lattice theory at the same time as I write a MathRev, and lattice theory requires me being awake, and most math papers go down much better with a bit of alcohol...)
Russian I could not read, and German only a little bit. So I decide to try my hand first at this seemingly harmless statement.
I spent the better part of the afternoon getting nowhere. So I hit the library and looked for the German paper.
The title said something about transfinite induction.
The section in which the theorem I wanted was proven began with "Let us recall Zorn's Lemma".
So I figure, ah! I didn't use transfinite induction nor Zorn's Lemma, that must be the missing ingredients.
So I went back and spent the better part of the evening and still got nowhere.
The next day I checked the Handbook of Analysis and found a proof of a slightly weaker statement, but if you do the right substitutions of definitions could be copied word for word to prove the statement I want.
And it emphatically did not use either transfinite induction or Zorn's Lemma.
So I've been trying to figure out using my limited German, why the heck the author of that article mentioned either of those two things when the main result of his paper required neither.
I should probably add that this German paper was published before the independence of AC was shown, in the era when people were fairly allergic to using AC when they don't have to...
Okay, the theorem goes something like this: Let $(E,\leq)$ be a complete lattice. Let $\mathcal{C}$ be a subset of $E$ closed under infimum. Let $h:E\to E$ be the hull operator sending elements $x\in E$ to $h(x) = \inf \{ c\in \mathcal{C} | x \leq c\}$.
Then if $(E,\leq)$ is an algebraic lattice (every element is the sup of all preceding compact elements) and if $h\circ sup = \sup\circ h$ applied to sets of compact elements, then $\mathcal{C}$ is inductive (that it is closed under chain supremum).
The interesting thing is that the partial converse, that $\mathcal{C}$ being inductive implies that $h$ commutes with $sup$ on compact elements, holds without the algebraicity assumption on the lattice.