Suppose a problem "Find a factor for the term 1+y in a problem such as 'x+xy'". Here for example x(1+y) so the asked factor is x. What is the name for this "asked factor"?
I heard that 2nd order PA has only one model, the "true natural numbers".. yet there is an axiomatic extension which is not inconsistent yet as no model
PA with a new symbol $\infty$ and the axioms $\infty \not = Z$, $\infty \not = SZ$, $\infty \not = SSZ$, ...
its consistent because any proof of false has to use finitely many axioms, but any finite portion of the axioms are modelled by the standard model
@hhh that question is (a) of a different type than the other one and (b) doesn't make sense. in a sum a1+a2+..., the collection of summands {a1,a2,...} has a common factor which is the gcd of the terms. in the case of 1*2+3, you don't say "common factor of 2," nor do you say "common factor of 1 is 5": you say the common factor of 1*2 and 3 is 1 (since gcd(1*2,3)=1). note that the common factor of a sum depends on how it's presented, e.g. the c.f. in 1*2+3*2 is 2 but the c.f. in 2*2+2*2 is 4.
3*4*5*5 with respect to 5 should return 3*4 3*4*5*5 with respect to 4 should return 3*5*5 3*4*5*5 with respect to 3 should return 4*5*5 (B) This factor called?
3*4*5*5 with respect to 5 should return 3*4 3*4*5*5 with respect to 4 should return 3*5 3*4*5*5 with respect to 3 should return 4*5 (C) This factor called?
Just 4 fun with PSTricks.
\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-eucl,multido,fp}
\FPset\Width{4.00}% paper width
\FPset\Height{6.00}% paper height
\FPset\Step{0.50}% interline skip
\FPeval\Lines{round(Height/Step-1:0)}% number of lines
\def\X{2}% abscissa of the t...
What is interesting is that I thought I was ugly at the time.
I know how to solve the problem: Until now I have evidence that I'll find myself beautiful in the future, by knowing that I know that I'm beautiful now.
The older term for number theory is arithmetic. By the early twentieth century, it had been superseded by "number theory".[note 2] (The word "arithmetic" is used by the general public to mean "elementary calculations"; it has also acquired other meanings in mathematical logic, as in Peano arithmetic, and computer science, as in floating point arithmetic.)
The use of the term arithmetic for number theory regained some ground in the second half of the 20th century, arguably in part due to French influence.[note 3] In particular, arithmetical is preferred as an adjective to number-theoretic.
I'm reading Shawn Hedman's A First Course in Logic: An Introduction to Model Theory, Proof Theory, Computability, and Complexity:
Definition 1.18 Formula $G$ is a consequence of formula $F$ if for every assignment $A$, if $A\models F$ then $A\models G$. We denote
this by $F\models G$.
...
@GustavoBandeira what are they used for? are the validity of logical statements ever that complicated that in order to phrase them rigorously you need to introduce all this notation
mathematicians also want $\forall$ and $\exists$ in their work. yes, being familiar with propositional logic is extremely useful for anyone who wants to do serious math. I'd recommend it as mandatory even.
@GustavoBandeira Would you be able to explain what first-order logic is in a nutshell? I head of a weaker notion of isomorphism between algebraic structure is elementary equivalence, where two objects (say, groups) share the same set of true first-order statements that can be said about them - in a sense, this says they are approximately the same, at the resolution of "first-order" descriptions, and I'd like to know what this means.
I wish I'd learned logic much, much earlier. Obviously young students couldn't handle much depth, but at least a basic introduction to a few concepts would be nice. Just understanding the concept of axioms and deductive rules would put all of math into some perspective. When I finally understood ...
@GustavoBandeira I'm sorry, but I perhaps don't understand the question fully, but I don't see an area where any tool (computers included) might not be useful
@robjohn: is this limit familiar to you? $\lim_{x\to\infty} \int_0^x \frac{\log(1+t)}{t} \mathrm{d}t - \frac{\log^2(1+x)}{2} = \frac{\pi^2}{6}$ I evaluated it in 2 different ways, one involves dilogarithm identities and the other one(rather hard) involves mutivariable calculus. I was wondering if we can avoid the dilogarithm identities and the multiple integrals.
Since $$\zeta(x) \Gamma(x)=\int_0^{\infty} \frac{y^{x-1}}{e^y-1} \mathrm{d}y$$ $$\int_0^{\infty} \frac{y}{e^y-1} \mathrm{d}y=\zeta(2) \Gamma(2)=\frac{\pi^2}{6}$$ and the last integral is obtained by letting $1+t=e^y$
@PeterTamaroff: If a $n \times n$ matrix a satisfies A^k = 0, then its minimal polynomial is of the form x^t. Here t <= n since the minimial polynomial has degree at most n by Cayley-Hamilton. Does this work?
@PeterTamaroff Prove this: A linear map $T$ is nilpotent iff there is a basis for $V$ such that the matrix for $T$ is upper triangular with all zeros on the diagonal.
How many ways can I map {1,2,...,n} to {1,2,3} so that the range has two elements? Would this be 3C(2n,n) since there are 3 pairs of length 2 in {1,2,3}?
I'm starting to doubt that it is a combination, and I doubt my parameters (2n, n)
[About this question](http://math.stackexchange.com/questions/315073/how-can-commuting-with-frobenius-imply-the-order-of-an-element-in-the-inertia-gr) Maybe commuting with the Frobenius implies that the representation is one-dimensional, hence lying in the ground field, thus the order statement?