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11

Let $a_n = \frac{4^n}{n+1}$ and $b_n=\frac{(2n)!}{n!^2}=\binom{2n}{n}$. Then $a_1=b_1$ and: $$ \frac{a_{n+1}}{a_n} = 4 \frac{n+1}{n+2},\qquad \frac{b_{n+1}}{b_n} = 2\,\frac{2n+1}{n+1}\tag{1} $$ hence we just need to check that: $$ \forall n\geq 1,\qquad \frac{2n+2}{n+2}< \frac{2n+1}{n+1} \tag{2} $$ holds to prove our claim by induction. Also notice ...


7

It is certainly not the case that 1+1=0 in every field; e.g., the rationals, the reals, the complex numbers, the $p$-adics, the integers modulo 17, . . . There are fields in which 1+1=0 - this property is called characteristic 2. More generally, for any prime $p$, there are fields in which 1+1+ . . . +1 ($p$ times) equals 0; see ...


6

Use the Euclidean Algorithm ( I mean the Division Algorithm ): Given $n$-an interger. Divide $n$ by $5$: $n = 5q+r$, then $r$ as the remainder, it must be non-negative, and less than $5$. Thus: $r = 0,1,2,3,4$. So $n = 5q, 5q+1,5q+2,5q+3$ or $5q+4$, proving the statement.


6

The main result is Euclid's lemma: If $p$ is prime and $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$. Now $6$ divides $m$ iff $2$ and $3$ divide $m$. Apply these two facts to $6$ divides $n^2$ to conclude that $2$ and $3$ divide $n^2$ and so that $2$ and $3$ divide $n$.


5

Note you're using a slightly incorrect definition: it's $$\lim_{x\to 3}\frac{x^2-9}{x-3}=6$$ iff for all $\epsilon>0$ you can find a $\delta>0$ such that $$\displaystyle{0<|x-3|<\delta\implies \left|\frac{x^2-9}{x-3}-6\right|<\epsilon}$$ and not $$\displaystyle{|x-3|<\delta\implies \left|\frac{x^2-9}{x-3}-6\right|<\epsilon}$$ ...


5

You will know that these are mathematical statements when you can assign a truth value to them. That is, if you can look at it and say "that is true!" or "that is false!" then it is a mathematical statement. It doesn't mean anything else, it doesn't require numbers or symbols are anything commonly designated as "mathematical."


5

Guten Morgen, fellow German here. I was in the lucky position, to have professors in the undergrad courses who started their lectures with a quick review of the last lecture (5 minutes) which was given in english. That way, you started to learn the basic math terminology already in the first semesters. But I also had to look up names of theorems which I ...


4

Your narrative in English is good. Two things I would suggest are: Van Nostrand Mathematical Encyclopedia is available both in English and German. The key to English version is your German original.. and vice-versa. A request for time at Chat room with German mathematicians here may help. As you keep on working at writing you learn slowly, friendly ...


4

Usually, a proof by contradiction of the statement $p \implies q$ is when you assume that the opposite of the desired conclusion is true (i.e., assume the negation of $q$ is true), and follow a few logical implications until you reach a statement that somehow explicitly or implicitly contradicts an initial assumption from the statement $p$. Meanwhile, a ...


4

Just show a pattern of tautologies that clearly has infinitely many wff's. For example, $$A\implies A,\ B\implies B,\ C\implies C,\ldots$$ or $$A\implies A,\ (A\land A)\implies (A\land A),\ (A\land A\land A)\implies (A\land A\land A),\ldots$$ or many other patterns. Make one that you like.


4

I would write it as: Let $P(n)$ stand for the expression: $$\forall x\leq n(x\not\in A)$$ Then use the assumption that $A$ has no least element to prove that $P(1)$ and $P(n)\implies P(n+1)$. Thus, we've shown that $\forall x:x\not\in A$, which means $A$ is empty. That's essentially the same as your proof, but uses less set notation.


4

We have $|f| < P$, $|g| < Q$, and $|1/g| < R$ for some $P$, $Q$, and $R$, so $$\left|\frac{f}{g} \right|= \left| f\frac{1}{g} \right| < PR.$$


4

We can prove this for $n\ge 0$ by induction on $n$. We first check that $n=0=5k$, for $k=0$, as our base case. Suppose now, our inductive hypothesis, that $n-1=5k+j$, for some $0\le j\le 4$. If $0\le j\le 3$, then $n=5k+(j+1)$, is of the desired form. If instead $j=4$, then $n=5(k+1)+0=(5k+4)+1=(n-1)+1$. I leave those $n<0$ for you to do; they can be ...


4

A simple proof is based on the observation that $\dfrac{(2n)!}{(n!)^2}$ is the central binomial coefficient $\displaystyle{ {2n} \choose n}$. Look at row $2n$ in the Pascal triangle. The sum of all $n+1$ terms is $2^{2n}= 4^n$. Now, the central binomial coefficient is the largest number in that row and so $4^n \le (n+1){{2n} \choose n}$. [I've used $a_1 ...


4

Let's start by the definition of convergence of a sequence. Definition A a sequence $x_n$ converges to some number $a$, if for any $\epsilon >0$, there exits a number $N$, such that $n > N$ implies $|x_n - a| < \epsilon$. Now, here's my attempt to prove "easily" that the sequence $s_n = \frac{1}{n}$ really converges to $0$. Discussion ...


3

$(\Rightarrow)$ Assume $A\cap B\subseteq C$. Let $x\in A$. Then if $x\in B$, $x\in C$ by hypothesis. Otherwise, $x\in \bar B$. Thus, $x\in \bar B\cup C$, and $A\subseteq \bar B\cup C$. $(\Leftarrow)$ Assume that $A\subset\bar B\cup C$. Let $x\in A\cap B$. Then $x\in A$, so by hypothesis, $x\in \bar B$ or $x\in C$. But we know that $x\in B$ by assumption, so ...


3

As has been pointed out in comments, we do not need to add a constant symbol for every element of $\mathbb{R}$. Let $L$ be the usual first-order language for ordered fields. Extend $L$ to $L'$ by adding a constant symbol $c$. Let $T$ be the set of all sentences of $L$ true in the reals, under the usual interpretation. Make a new theory $T'$ by adding to $T$ ...


3

Instead of compactness, you can try an ultrapower argument. Let $\cal U$ be a nonprincipal ultrafilter on $\Bbb N$. Then $\Bbb{R^N}/\cal U$ is elementarily equivalent to the reals. And it is not hard to show it has the wanted cardinality, and that it is non-standard.


3

i) Given $$y=\frac{3x^2+2y}{x^2+2}$$ cross multiplying we have, $$y(x^2+2)=3x^2+2y$$ $$\implies yx^2+2y=3x^2+2y$$ canceling $2y$ in both side we have, $$yx^2=3x^2$$ Since $x\neq 0 \implies x^2\neq0$ So dividing both side by x^2, we have $$y=3$$ ii) On contrary suppose that $x=0$ then we have $x^2=0$ so $$0\cdot y=2\cdot 0 +y$$ $$\implies y=0$$ Which is a ...


3

You won't be surprised to learn that in the last seventy-plus years since Tarski's book was first published in English, many other books have been appeared which will perhaps serve better as introductions to modern logic. And if you have downloaded my Teach Yourself Logic, you will have seen my "entry-level" suggestions on formal logic at the beginning of ...


3

The direction is correct in general. But I recommend skipping the step in 2nd equation, $B=\frac A{\tan\theta}$, as when $\theta=0,\tan\theta=0$ and $B$ is not defined. It'd be better if you directly write $B=C\cos\theta$, in which $B$ is well-defined for all $\theta$. Btw, have you also considered the case beyond $0\le \theta \le \frac{\pi}2$?


3

I think it's no different from learning English in general (or indeed any other language), you learn both by seeing and by doing. The more English-language material you read, the quicker you will "get used to it" and naturally mimic what you have read in your own writing. As for "chances to write in English", it's really up to you. There's nothing stopping ...


3

The most important thing is practice. As you calculate more and more limits, you'll start to develop some form of intuition regarding what method to try first, and the correct approach will probably come faster to you. You might see a problem and be reminded of some other limit you did before, so you'll try a similar method and maybe find it also works. As ...


2

No, for example: let $A = \{\frac 1 k \mid k \in \mathbb N\}$, $N = \mathbb N$. Then $$ \exists n \in N \forall x \in A \colon xn \ge 1 $$ is false, since for a fixed $n \in \mathbb N$ we have $\frac 1 {n+1} \in A$ and $n \cdot \frac 1 {n+1} < 1$. On the other hand $$ \forall x \in A \exists n \in N \colon nx \ge 1 $$ is true. (Given $x = \frac 1 k \in A$ ...


2

Technically, there is an implicit issue of existence of limits which is being swept under the rug in the presentation you have given. The assumption of differentiability at $x_0$ says that the limit $$\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$ exists as a finite number. The limit $\lim_{x \to x_0} x-x_0$ exists and is zero regardless of our ...


2

There's some relevant content at mathineurope.eu mathineurope.edu - Maths in foreign-languages mathineurope.edu - German to English math terms mathineurope.edu - How to write mathematical English


2

If you must use a proof by contradiction, then you could use something along these lines. Suppose there were only a finite number of tautologies, let them be $A_i$ for $1 \leq i \leq n$. Since the $A_i$ are finite in number we can form their conjunction, another wff, $A_1 \land A_2 \land A_3 \land ... \land A_n$ (suitably bracketed!). But then if each ...


2

As far as I see you already got the holomorphic function $v(z)$ such that $$h(z) = (\frac{v(z)}{(z-z_0)})^m $$ so $$ h(z) = \frac{1}{(\frac{(z-z_0)}{v(z)} + z_0 - z_0)^m} \, .$$ Set $\psi(z) := \frac{(z-z_0)}{v(z)} + z_0 $ and notice that $\psi\, '(z_0) \neq 0$. Then $\psi$ is invertible near $z_0$ i.e. there is $g(z)$ such that near $z_0$ you have ...


2

In a field $(F, +, \cdot)$, the nonzero elements must form a group with multiplication. Hence, $(F-\{0\}, \cdot)$ must be an Abelian group. As such, this group must have a neutral element, $1$. Since $1 \in F-\{0\}$, $1$ cannot be $0$.


2

One of the field axioms is that $0 \neq 1$. Q.E.D.



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