# Tag Info

4

Suppose that $x + y$ is an integer, but that $y$ is not an integer, while $x$ is. Then $y = \left( {x + y} \right) + \left( { - x} \right)$, and since the sum of integers is an integer (you've shown that a sum of integers is an integer, and $x$ is an integer implies $- x$ is an integer), and ${x + y}$ and $x$ are both integers, we conclude that $y$ is also ...

2

Questions like these are fundamental and intuitive enough that we need a set of axioms which we can assume. Suppose, for contradiction, that there exists an integer $x \in \mathbb{Z}$ and a $y \not\in \mathbb{Z}$ such that $x+y \in \mathbb{Z}$. Then, since integers have additive inverses, $\exists -x \in \mathbb{Z}$ such that $x+(-x)=(-x)+x=0$. Since ...

2

It's not true as stated. You could take the function $$f(x)=\begin{cases}1&x\le 0\\2&x>0\end{cases}$$ which has derivative $0$ on the open set $(-2,-1)\cup (1,2)$. Let's assume path-connectedness as suggested in the comments. Then between two points $a,b\in U\subset \mathbb{R}^n$ we have a smooth path $$x(t),\;0\le t\le 1$$ By the mean value ...

2

The linked-to notes were maybe a little unclear on this point, perhaps in the interest of not being too pedantic. Anyway, for a given face $F$ of a connected plane graph, a boundary walk of $F$ is a closed walk that contains every edge on the boundary of $F$. This means that a boundary walk must start and stop at the same vertex. The degree of a face is ...

2

In combinatorics, you focus on the structure of the problem and then work out the formula. For example, when you say there are $k \choose 2$ ways to choose 2 flavors, that's the true number by definition even if you don't know the formula for $k \choose 2$. Some combinatorial quantities like Stirling numbers of the second kind don't even have any simple ...

2

Very simply it can be done like this: $gcd(a,b)=d$. Now we ask can: $gcd(\frac{a}{d},\frac{b}{d})=e$ for $e>1$? Well, this implies $e|\frac{a}{d},e|\frac{b}{d} \Rightarrow em=\frac{a}{d},en=\frac{b}{d} \Rightarrow dem=a,den=b \Rightarrow de$ is a common divisor of $a,b$ which is greater than $d$, thus a contradiction as $d$ by definition was supposed as ...

1

This is a special case of the GCD distributive law ($3$ proofs of it are below). Namely $$\color{#c00}c = (a,b) = ((a/c)c,(b/c)c) = (a/c,b/c)\color{#c00}c\overset{\rm\large cancel\ \color{#c00}c}\Rightarrow 1 = (a/c,b/c)\qquad\qquad$$ Below are sketches of three proofs of the gcd distributive law $\rm\:(ax,bx) = (a,b)x\:$ using various approaches: ...

1

You'ra on the right track. Every edge of the dual is corssed by an edge of $G$, hence a cycle of length $k$ in the dual graph gives us $k$ edges of $G$ with one endpoint in the interior and one endpoint in the exterior of the cycle. Thus removing theses $k$ edges from $G$, we obtain a nonempty interior and a non empty exterior component of $G$ (or maybe even ...

1

Let $a,b$ have the following prime factorisations: $$a= \prod_{n=1}^\infty p_n^{\alpha_n} ,b=\prod_{n=1}^\infty p_n^{\beta _n}.$$ (Here $(p_n)$ is the ascending sequence of prime numbers). We then have $$\gcd(a,b)=\prod_{n=1}^\infty p_n^{\min(\alpha_n,\beta_n)},$$ and consequently $$\frac{a}{\gcd(a,b)}=\prod_{n=1}^\infty ... 1 The simplest way to prove a statement of the form \;(\exists z) P(z)\; is by transforming it to something of the form \;(\exists z)(z = Q)\; where \;Q\; does not contain \;z\;: such a statement is always true. (This is a special case of the "one-point rule"). In this case, we can transform \;x = y \circ z\; by the following calculation: ... 1 The goal of your \epsilon_f is to make sure |f(x)| is bounded below for all x near a. So it doesn't matter what this is, as long as it's positive. Then write something like: "Let \epsilon_f>0 be arbitrary" or "Fix \epsilon_f:=x". Here x can be any real number such that x<|L|, since we want |L|-\epsilon_f>0. Apart from that, it ... 1 If f has an inverse, than this automatically implies that f is injective and surjective: f is injective because f(x)=f(y) implies x=y, applying f^{-1} to both sides. f is surjective, because given any y\in Y we have f(f^{-1}(y))=y. That is f(X)=Y. But: Notice how we needed both equations, f(f^{-1}(x))=x and f^{-1}(f(x)) in that ... 1 Let x, y \in \Bbb R \setminus \{0\} be arbitrary. Then x^{-1} and y^{-1} both exist. Then use the commutative law which states that xy = yx. Choose which one you want to multiply either side with. Then see that each side equals 1, the multiplicative identity. More explicitly you need to prove that the multiplicative inverse of (xy) is ... 1 I think you answered your own question. The intervals [a_j, c_j] and [c_j, b_j] for 1 \le j \le k determine 2^k k-cells, that is, there are 2^k k-cells$$J_1 \times J_2 \times \dots\times J_k$$where J_i is either [a_i, c_i] or [c_i, b_i], such that their union is I. 1 Let me answer the following (tangential) question. Is there even such a thing as a degree of belief in a proof? There is such a thing as degree of belief in a theorem. For example, suppose T is a sentence in the language of set theory expressing a principle of arithmetic. Then if a computer verifies that there is a proof of T from the axioms of ... 1 It'd do this by induction. Let d_n\ldots d_0, e_n\ldots e_0 be the two numbers in base b. For the right-most digit, as you say, the worst case is d_0=e_0=b-1, and you get$$ a_0 + b_0 = (b-1) + (b-1) = 2b -2 = 1\cdot \mathbf{b^1} + (b-2)\cdot \mathbf{b^0} \text{,} $$i.e. the carry c_0=1. For the n-th digit, the worst case is d_n=e_n=b-1 ... 1 Well, to some extent one could argue that there is nothing to show since this is obvious. Still, for a formal proof: Multiply both sides of x<y by y^{-1} (NB: this is a positive number) to obtain y^{-1}x<1, then multiply both sides by x^{-1} to obtain y^{-1}<x^{-1}. 1 I got a little confused reading some of the later stuff in your proof... but I did get that you say for functions from A to B the cardinality is$$|B|^{|A|}$$So in this case the cardinality is$$|A|^2$$since there are two elements in (x,y). You just have to show that the product of two countable sets is countable. EDIT: Ok let's start a new area ... 1 If you expand this integral you should get:$$\int_0^\pi \cos(nt)\sin(t)dt = \frac{1}{4i}\int_0^\pi \left(e^{int}+e^{-int}\right)\left(e^{it}-e^{-it}\right)dt= \frac{1}{4i}\int_0^\pi e^{i(n+1)t} - e^{i(n-1)t} + e^{-i(n-1)t} - e^{-i(n+1)t}dt$$Note that this is exactly the sum of sines:$$= i\int_0^\pi \sin((n+1)t)-\sin((n-1)t)dt Now perform the ...

Only top voted, non community-wiki answers of a minimum length are eligible