In lecture I learned the following definition of valuation: Let $(K,+,\cdot)$ be a field and let $(G,+)$ be a totally ordered group. A map $v: K \longrightarrow G\cup\{\infty\}$ is a valuation if the following properties hold:  $v(ab) = v(a)+v(b)$ $v(a+b) \ge \min\{v(a),v(b)\}$ $v(a) = \infty \iff a = 0$  Then we proved that:  $v(1) = 0$ $v(a^{-1}) = -v(a)$ $v(-a) = v(a)$ $v(a - b) \ge \min\{v(a), v(b)\}$  Later in the lecture we used the &quot;easy fact&quot; that $$ \text{If }v(a) \ne v(b), \text{ then } v(a+b) = \min\{v(a), v(b)\}$$ I do not understand why this is true. I looked on  and there they define a valuation to so that it has this property. But is it possible to derive this from my definition?
8405
Let $F$ be a field and $v:F\rightarrow G\cup\{\infty\}$ be a valuation on $F$ so $G$ is a totally ordered abelian group with $\infty$ having the properties $\infty+\infty=g+\infty=\infty+g=\infty$ and $\infty\leq\infty$ for all $g\in G$, $\infty \notin G$.     Show if $v(a)\neq v(b)$ then $v(a+b)=\min(v(a),v(b)$).   I must be missing some trick here on how to show this is true.  My approach has been to use that $G$ is a totally ordered group so I start by assuming $v(a)\neq v(b)$ and letting my first case be $v(a)&lt;v(b)$ as trichotomy holds in $G$. By $v$ being a valuation on $F$, we then have that $v(a+b)\geq \min(v(a),v(b))=v(a)$. From here what I've tried doing doesn't seem to help much.  I've said lets assuming $v(a+b)&gt;\min(v(a),v(b))=v(a)$ and we also have $$v(a+b)&gt; v(a) \Rightarrow v(a+b)+v(a^{-1}) &gt; 0 \Rightarrow v(1+ba^{-1}) &gt; 0$$ and $$v(a)&lt;v(b) \Rightarrow 0&lt; v(b)+v(a^{-1}) \Rightarrow 0 &lt; v(ba^{-1})$$ also $$v(1+ba^{-1})\geq\min(v(1),v(ba^{-1}))=\min(0,v(ba^{-1}))=0.$$  How does one solve this?
39727
To make sure that we are talking about the same, I would like to post the relevant definitions I know first.  Definitions:  A pair $(G, +)$ where $G$ is a set and  $+: G \times G \rightarrow G$  is called a commutative group if it has all of the following three characteristics:   (associativity): $\forall x,y,z \in G: (x+y)+z=x+(y+z)$ (identity element): $\exists e \in G \forall x \in G: e + x = x = x +e$ (inverse elements): $\forall x \in G \exists x^{-1} \in G: x^{-1} + x = e = x + x^{-1}$ (commutativity): $\forall x,y \in G: x+y=y+x$.     A triple $(\mathbb{K}, +, \cdot)$, where $\mathbb{K}$ is a set and  $+: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$  $\cdot: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$  is called a field if it has all of the following three characteristics:   $(\mathbb{K}, +)$ is a commutative group $(\mathbb{K} \setminus \{0\}, \cdot)$ is a commutative group distributive properties: $\forall x,y,z \in \mathbb{K}: x \cdot (y+z) = (x \cdot y) + (x \cdot z)$ and $\forall x,y,z \in \mathbb{K}: (y+z) \cdot x = (y \cdot x) + (z \cdot x)$     $\mathbb{N^+} := 1, 2, 3, 4, 5 ...$ (all positive integers)  $\mathbb{N}_0 := \{0\} \cup \mathbb{N}$ (the natural numbers with zero)  $\mathbb{N}$ can be both: $\mathbb{N^+}$ or $\mathbb{N}_0$.    My question: Does a field exist that has $\mathbb{N}$ as its set?  I know that $(\mathbb{Q}, +, \cdot)$ is a field and I know that $(\mathbb{N}, + )$ is only a commutative semigroup.   But maybe it's possible to define two mappings $\circ, *$ for $\mathbb{N}$ in such a way that $(\mathbb{N}, \circ, *)$ is a field.    Related:     The smallest field containing the integers is the field of rational   numbers.   Source: en-Wikipedia:   I don't know if this is true. If somebody can prove (or at least scribble the proof of) the quote it would be a good answer, I guess.
30673
I am asked to prove the second mean value theorem:  Let $f$ and $g$ be defined on $[a,b]$ with $g$ continuous, $f\ge 0$, and $f$ integrable. Then there is a point $x_0 \in (a,b)$ such that $$ \int_{a}^{b}f(x)g(x)dx = g(x_0) \int_{a}^{b}f(x)dx. $$  My (incomplete) proof goes as follows:  Since $ [a,b] $ is compact and $ g $ is continuous, we know there exist $ x_1, x_2 \in [a,b] $ such that $ g(x_1) \le g(x) \le g(x_2) $ for any $ x \in [a,b] $. Thus, we have the following string of inequalities $$ f(x)g(x_1) \le f(x)g(x) \le f(x)g(x_2), \, \, \forall x \in [a,b], $$ $$ \Rightarrow g(x_1)\int_{a}^{b} f(x)dx \,\le \, \int_{a}^{b} f(x)g(x) dx \, \le  \, g(x_2)\int_{a}^{b}f(x)dx. $$  My problem now is to show the final equality. It seems to me the intermediate value theorem is somehow involved, but I cannot see how to use it. Furthermore, how do I make use of the $f\ge 0$ condition in the hypothesis?
119914
The theorem to prove is: $X$ is a positive continuous random variable with the memoryless property, then $X \sim Expo(\lambda)$ for some $\lambda$. The proof is explained in this , but I will type it out here as well. I would like to get some clarification on certain parts of this proof.  Proof Let $F$ be the CDF of $X$, and let $G(x)=P(X&gt;x)=1-F(x)$. The memoryless property says $G(s+t)=G(s)G(t)$, we want to show that only the exponential will satisfy this. Try $s=t$, this gives us $G(2t)=G(t)^2,G(3t)=G(t)^3,...,G(kt)=G(t)^k$. Similarly, from the above we see that $G(\frac{t}{2})=G(t)^\frac{t}{2},...,G(\frac{t}{k})=G(t)^{\frac{1}{k}}$. Combining the two, we get $G(\frac{m}{n}t)=G(t)^\frac{m}{n}$ where $\frac{m}{n}$ is a rational number. Now, if we take the limit of rational numbers, we get real numbers. Thus, $G(xt)=G(t)^x$ for all real $x&gt;0$. If we let $t=1$, we see that $G(x)=G(1)^x$ and this looks like the exponential. Thus, $G(1)^x=e^{xlnG(1)}$, and since $0 &lt;G(1) \leq  1$, we can let $lnG(1)=-\lambda$. Therefore $e^{xlnG(1)}=e^{-\lambda x}$ and only exponential can be memoryless.  So there are several parts that I am confused about:  Why do we use $G(x)=1-F(x)$ instead of just $F(x)$? What does the professor mean when he says that you can get real numbers by taking the limit of rational numbers. That is, how did he get from the rational numbers $\frac{m}{n}$ to the real numbers $x$? In the video, he just says that $G(x)=G(1)^x$ looks like an exponential and thus, $G(x)=G(1)^x=e^{xlnG(1)}$. How did he know that this is an exponential?
8620
I'm having trouble proving this. This is what I have so far:  Let $F$ be a field. Let $v(x) \rightarrow 1$ for all $x$ not equal to $0$.  So if we let $x$ be in $F$ where $x$ not zero then we can write $x$ as: $x=qy+r$ for some $y$ in $F$. If $r$ not zero then $V(r)=1$.  Not sure what to do from here. I know eventually I'm supposed to get no remainder $(r=0)$ but I'm stuck at applying the definition and valuation map.
109064
I'm kicking myself over this one, but I just can't seem to make the argument rigorous. From Axler's Linear Algebra Done Right:  for a vector space $V$ with an underlying field $F$:  Take an element $a$ from $F$ and $\vec{v}$ from $V$. $a\vec{v}=\vec{0}\implies a=0 $ or $ \vec{v}=\vec{0}$  After only being able to come up with half of a direct proof, I tried doing this by proving the contrapositive $a\neq 0 \wedge \vec{v} \neq \vec{0} \implies a\vec{v}\neq \vec{0}$  Say $a\vec{v}=\vec{u}$. Since $a$ is non-zero, we can divide both sides by $a$.  $$\vec{v}=\frac 1 a \vec{u}$$  If $\vec{u}$ were $\vec{0}$ then by  $$\frac 1 a \vec{0}=\frac 1 a (\vec{0}+\vec{0})\implies\frac 1 a \vec{0}=\vec{0}$$ $v$ would be $0$ as well. Since it isn't by assumption, $\frac 1 a \vec{u}$ cannot be zero and so $\vec{u}$ cannot be as well.   Is this fully rigorous? It seems like a very simple question, but I'm not sure about it. Namely, the last step of $\frac 1 a \vec{u}\neq 0 \implies \vec{u}\neq 0$ doesn't seem obvious. I think I need to use the $1\vec{v}=\vec{v}$ axiom, but I'm not sure how.  Is there a more direct proof? This whole contrapositive business seems a bit clunky for something so simple.
24760
Let $K$ be a finite extension of $\mathbf{Q}_p$, i.e., a $p$-adic field. (Is this standard terminology?)  Why is (or why isn't) an algebraic closure $\overline{K}$ complete?  Maybe this holds more generally:  Let $K$ be a complete Hausdorff discrete valuation field. Then, why is $\overline{K}$ complete?  I think I can show that finite extensions of complete discrete valuation fields are complete.
64888
If $f,g$ are continuous at $a$, show that $h(x)=\max\{f(x),g(x)\}$ and $k(x)=\min\{f(x),g(x)\}$ are also continuous at $a$. Here is my attempt at a proof. It feels very elaborate and I am not sure if it is correct. Can someone please point out any mistakes or places where I may improve. Thanks!  By the definition of continuity at $a$ of $f,g$ we have that $\lim\limits_{x\to a}f(x)=f(a)$ and $\lim\limits_{x\to a}g(x)=g(a)$. Suppose that $f(a)&gt;g(a)$. Then there exists a $\delta&gt;0$ such that $f&gt;g$ for all $x$ satisfying $|x-a|&lt;\delta$. Then for $x$ satisfying $|x-a|&lt;\delta$ we have $h(x)=f(x)$ and $k(x)=g(x)$ and thus $h$ and $k$ are continuous at $a$ because $f$ and $g$ are continuous at $a$. Now if $f(a)&lt;g(a)$ we simply relabel $f=\tilde{g}$ and $g=\tilde{f}$, so that $\tilde{f}(a)&gt;\tilde{g}(a)$ and we apply the previous result to show that again $h$ and $k$ are continuous at $a$.\  Now if $f(a)=g(a)$ then we can distinguish three cases. $f\geq g$ in a small neighborhood about $a$, $f\leq g$  in a small neighborhood about $a$ or $f\geq g$ on one side of $a$ and $f\leq g$ on the other side. In the first case we assume that $f\geq g$ in some small neighborhood about $a$. Then, since $f(a)=g(a)$ we have $h(x) = f(x)$ in this neighborhood and $k(x) = g(x)$ and again we see that $h$ and $k$ are continuous at $a$ due to the continuity of $f$ and $g$ at $a$. Similarly if $f\leq g$ in a neighborhood around $a$ then $h(x)=g(x)$ and $k(x)=f(x)$ in this neighborhood and $h$ and $k$ are continuous at $a$.   Suppose that just to the left of $a$ we have $f\geq g$ and just to the right of $a$ we have $f\leq g$. Then $\lim\limits_{x\to a^-}h(x)=\lim\limits_{x\to a^-}f(x)=f(a)=h(a)=g(a)=\lim\limits_{x\to a^+}g(x)=\lim\limits_{x\to a^+}h(x)$. Similarly $\lim\limits_{x\to a^-}k(x)=\lim\limits_{x\to a^-}g(x)=g(a)=k(a)=f(a)=\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a^+}k(x)$. So $\lim\limits_{x\to a}h(x)=h(a)$ and $\lim\limits_{x\to a}k(x)=k(a)$ which means that $h$ and $k$ are continuous at $a$. Lastly if $f\leq g$ just to the left of $a$ and $f\geq g$ just to the right of $a$ we can relabel $f=\tilde{g}$ and $g=\tilde{f}$ and apply the last result to show $h$ and $k$ are continuous at $a$.   Again, please point out any mistakes I may have made. Thanks!!
114169
Why does $\bigcup \emptyset = \emptyset$ but $\bigcap \emptyset$ is not defined?    If I had to guess, I'd say it's also equal to $\emptyset$.
69296
Let $K$ be a number field and let $R=\mathcal{O}_K$ be its ring of integers. Let $a_1, \ldots, a_h$ be ideals of $R$ generating the class group $Cl(K)$. Let $S$ denote the set of valuations corresponding to prime ideals dividing $a_1\cdots a_h$. Let  $$R_S = \{a \in K \mid v(a)\geq 0,\ v \text{ non-Archimedean valuation on } K, \  v \notin S \}$$ Silverman (p.213 Arithmetic of Elliptic Curves) claims that $R_S$ has trivial class group. I’m sure this is a standard and simple fact, but I’m having trouble proving it.   We have an inclusion $R \to R_S$ and a map $Cl(R) \to Cl(R_S)$. Let $I$ be an ideal of $R_S$, we want to show it is principal. We have $I \cap R \equiv \prod a_i^{e_i}$ in the class group so there exists principal ideal $P,P’$ of $R$ such that  $$P(I\cap R) = P’ \prod a_i^{e_i}$$ $$P(I\cap R)R_S = P’ \prod a_i^{e_i}R_S$$  Got stuck. Somehow the $a_iR_S$ become principal in $R_S$. I know  $$R=\bigcap\limits_\mathfrak{p} R_\mathfrak{p}$$ $$R_S=\bigcap\limits_{\mathfrak{p} \notin S} R_\mathfrak{p}$$ Where $\mathfrak{p}$ ranges over all maximal ideals of $R$
112058
Let $r\in \mathbb{R}$ and let $0\neq s \in \mathbb{R}$. Define operations $\#$ and $\&amp;$ on $\mathbb{R}$ x $\mathbb{R}$ by $(a,b) \mathbin\#(c,d)=(a+c,b+d)$ and $(a,b) \mathbin\&amp;(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)$ as addition and multiplication, respectively. Do these operations form a field?  I have shown that every axiom of the field is satisfied except for the "existence of multiplicative inverses" axiom. This is where I need help. We either need to show that this axiom fails and so we don't have a field or that the axiom $aa^{-1}=1$ is satisfied for every $a\neq 0 \in \mathbb{R}$  The multiplicative identity in this problem is $(1,0)$.  Any solutions or hints are greatly appreciated.
42406
My book says  \begin{equation} (a,b)=\{\{a\},\{a,b\}\} \end{equation}  I have been staring at this for a bit and it doesn't make sense to me. I have read several others posts on this, but none made any sense to me.  For example,   Based on how my ignorant brain is viewing this, I don't see why the definition could not be.  \begin{equation} (a,b)=\{\{a\},\{b\},\{a,b\}\} \end{equation}  aka the power set. What is the significance of the {a} in that definition? Please keep things simple if possible. Normally definitions have a valid and clear reason for being defined that way.    Clarification  First, I understand what an ordered pair is. I just don't see how the set notation says that.   Second,  \begin{equation} (a,b) = \{\{a\},\{a,b\}\}=\{\{a,b\},\{a\}\} \end{equation} Sets don't preserve order, but ordered pairs do. How does the third part of the equality apply to the definition?  Third, another issue with the notation that I have starts with the Product Property of Sets  \begin{equation} \text{Let $X$ and $Y$ be sets} :\ X=\{a,b,c\}\text{ and }Y=\{a,d,e\}. \end{equation} \begin{equation} \text{Then }X \times Y = \{(a,a),(a,d),(a,e),(b,a),(b,d),(b,e),\dots,(c,e)\} \end{equation} If we look at the first ordered pair and our given definition we have  \begin{equation} (a,a)=\{\{a\},\{a,a\}\} \end{equation}   How can this be so, you can't have duplicates in sets? I guess what I am looking for in an answer, is not a proof or a definition of ordered pairs, but rather something like, "This notation says what it says because...". Except for the second to last point I get the terminology, I just don't get the connection between the two different uses of notation.
106246
Problem: Let $g:[0,1]\to\mathbb{R}$ be a continuos function such that $g(1)=0$. Show that $\sup\limits_{x\in[0,1]}|x^ng(x)|\to0$ as $n\to\infty$.  My approach: Let $f(x)=x^ng(x)$. Then, $f(x)$ is continuos on $[0,1]$. Using the extreme value theorem we can say, that $|f(x)|$ is bounded on $[0,1]$ and attains its bounds, i.e., $\exists x_0\in[0,1]$ such that $|f(x_0)|=\sup\limits_{x\in[0,1]}|f(x)|$.  Now, $|f(x_0)|=|x_0^ng(x_0)|$. If $x_0=0$, then we are done. So, let us assume $x_0\neq0$. By the Archimedean property of $\mathbb{R}$, $\exists$ a $y\in\mathbb{R}$ such that $y=x_0^n\leq1$, since $x_0\in[0,1]$.   Hence, $\lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}|x^ng(x)|=\lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}|f(x)|=\lim\limits_{n\to\infty}|f(x_0)|=\lim\limits_{n\to\infty}|yg(y^{\frac{1}{n}})|\leq\lim\limits_{n\to\infty}|g(y^{\frac{1}{n}})|=g(1)=0.$  I will be glad if anyone can tell me if I have done any mistake and if so please help me to correct it.  Thank you.
78410
I am reading a book on probability theory and I have troubles understanding why the following holds  $$    \sup_{k \ge n} A_k = \bigcup_{k\ge n} A_k $$  $$ \inf_{k \ge n} A_k = \bigcap_{k\ge n} A_k $$  I know how to define infimum and supremum, but I have troubles proving the above expressions. Unfortunately, the book states those expressions as a definition without proving them. A graphical explanation in terms of a Venn diagram will be very helpful.  Thank you in advance
89508
A question from my homework I'm having trouble understanding. We are given:  $A(1) = \{\varnothing\}$, $A(n+1) = A(n)\cup (A(n)\times A(n))$  $A=A(1)\cup A(2)\cup A(3)\cup \cdots \cup A(n)\cup A(n+1) \cup \cdots$ to infinity  The questions are:  1) show that $A\times A \subseteq A$  2) Is $A \times A = A$?  Thank you for your help.  I've tried writing $A(2)$ but it gets really complicated and I'm having trouble understanding what the sets are. Let alone solve the question.
84290
Let $$f(x)=\begin{cases} \sin \tfrac 1 x &amp;\text{if $x\ne 0$}\\ 0 &amp; \text{if $x = 0$.}\end{cases}$$ I have to show that $f$ has the intermediate value property. That is, for any $a &lt; b$, if $y$ is any real number such that $f(a) &lt; y&lt; f(b)$ or $f(a)&gt;y&gt; f(b)$, then there exists a $c \in (a,b)$ such that $f(c)=y$.  I feel like I kind of know how to go about completing this. I just am curious as to if I have to create a bound such as letting $a = -1$ and $b = 1$, or keep $a$ and $b$ both arbitrary.
56500
Motivation for the question: in Neukirch - Algebraic Number Theory (Chapter II - Section 6 - Henselian Fields) the author discusses the context of a nonarchimedean valued field $K$ and its completion $\widehat{K}$, and considers the separable closure $K_v$ of $K$ in $\widehat{K}$, so that $K \subseteq K_v \subseteq \widehat{K}$.  Then he writes the following:     When $K_v$ is algebraically closed in $\widehat K$ - hence in particular $\text{char}(K) = 0$ - this is immediately obvious [...]   (emphasis mine).  My question is about the implicit claim that if $K_v$ is algebraically closed,  then $\text{char}(K) = 0$.  Now we know that $K_v$ is algebraically closed if and only if $K$ is perfect (since every $K_v$ will be the algebraic closure of $K$ and so every algebraic extension of $K$ is separable).  So we're asking whether it is true that if $K$ is perfect, then it must have characteristic $0$.  I know that all fields of characteristic $0$ are perfect. I also know that we have the example of finite fields which are also perfect fields but have prime characteristic. This does not contradict the claim above because the finite fields have only the trivial valuation.  So is there something about having a nonarchimedean valuation on a perfect field that forces its characteristic to be zero?