Tag Info

Hot answers tagged

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