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7

We can prove this using the well known identity $$\cos (a-b) = \cos a\cos b + \sin a\sin b$$ Here $a = \dfrac{\pi}{2}, b = x$ so now we have \begin{align*} \cos\left(\frac{\pi}{2}-x\right) &= \cos\frac{\pi}{2}\cos x + \sin\frac{\pi}{2}\sin x \\ &=0\cdot\cos x + 1\cdot \sin x \\ &= \sin x \end{align*}


6

The Principle of Mathematical Induction says that for all "properties" $P$, $$\left(P(0)\land\forall k\in \mathbb N\left(P(k) \implies P(k+1)\right)\right)\implies \forall n\in \mathbb N(P(n)).$$ So you're basically asking how to write the $\forall k\in \mathbb N\left(P(k)\implies P(k+1)\right)$ bit. It's a universal statement. It's common to start those ...


5

Given an injection $f : S \to \mathbb{N}$, where $S$ is infinite, you can define a function $h : \mathbb{N} \to S$ by, for each $n \in \mathbb{N}$, defining $h(n)$ to be the element $s \in S$ for which $f(s)$ is the $n^{\text{th}}$ least element of $f(S) \subseteq \mathbb{N}$. This is possible because $S$ is infinite, so such an element $s \in S$ always ...


4

1. You just bound in by making the denominator smaller and numerator bigger, this is a common trick in these proofs. $$\frac{11}{2(2n+7)}<\frac{11}{2(2n)}=\frac{11}{4n}<\frac{12}{4n}=\frac3n$$ These proofs have TWO STEPS. Basically, you want to do the "prep work", then the proof. Let me write out this proof in full. The above is the PREP WORK and not ...


4

Geometrically: given a right triangle with angles $A$, $B$, and $C=\pi/2$ and sides $a$, $b$, and $c$ opposite the respective angles, you have $$\sin(A)=\frac{a}{c}$$ and $$\cos(B)=\frac{a}{c}$$ And we know that $$B=\pi-A-\pi/2$$ so the result is immediate. Using Euler's formula: We know that $e^{i\theta}=\cos(\theta)+i\sin(\theta)$, so that ...


4

Consider the problem systematically: Starting with $1$, the sequence cannot contain the number $2$ until there are $2$ adjacent $1$s. Similarly, the sequence cannot contain $3$ until there are either $3$ adjacent $1$s or $3$ adjacent $2$s. Since the former will occur first, this occurs when there are $3$ adjacent $1$s. Continuing this trend, it is not ...


3

Your wording is completely confusing. Your statement: Suppose that the sequence of prime numbers is finite, so the sequence $R_n$ is finite is wrong. Also, you then say that $R_{n+1}$ is not a prime number, but you did not determine what the value of $n$ is. Furthermore, you then proved that $R_{n+1}$ is divided by $1$ and itself, and concluded that ...


3

HINT: Note that $$\left|a_n-\frac{1}{3}\right|=\left|\frac{2n+5}{6n-3}-\frac{1}{3}\cdot\frac{2n-1}{2n-1}\right|=\left|\frac{2n+5-(2n-1)}{6n-3}\right|=\left|\frac{6}{6n-3}\right|=\left|\frac{2}{2n-1}\right|$$ Taking $n_0\ge\max\left(\frac{1}{\epsilon}+1,2\right)$ we have the inequality, since $$n\ge ...


3

Part (1) Essentially whoever wrote the proof wants to get an $n$ in the denominator (to be explained shortly). Note that $\frac{11}{2(2n+7)}=\frac{11}{4n+14}<\frac{11}{4n}<\frac{3}{n}$, so that we got a nice bound with $n$ in the denominator. Part (2) Note that we have $|a_{n} - \frac{3}{2}| < \frac{3}{n}$ but really what we want is to show that ...


2

I'll run through your questions. Where did $\frac{3}{n}$ come from? It is an upper bound on $\left|a_n-\frac{3}{2}\right|=\frac{11}{2(2n+7)}$, and one way it can pop up is as follows: $$\frac{11}{2(2n+7)}=\frac{11}{4n+14}<\frac{11}{4n}<\frac{12}{4n}=\frac{3}{n}.$$ It is meant as some simple, vanishing upper bound of the difference ...


2

Consider the range of the injection $f:S\to\Bbb N.$ It will necessarily be an infinite (why?) subset of $\Bbb N,$ which can readily (and order-isomorphically!) be mapped onto $\Bbb N$ by sending its least element to the least element of $\Bbb N,$ its second-least element to the second-least element of $\Bbb N,$ and so on. Edit: Basically, all you have to do ...


2

Since $f$ is an injection, we have that $f:S\rightarrow f(S)$ is a bijection. Since $f(S)\subset N$. It suffices to find a bijection between some infinite subset $A\subset N$ and $N$. Now, we can construct the following bijection $g:N \rightarrow A$: $g(1):=\min A$, $g(2):=\min A-\{g(1)\}$, $g(3):=\min A-\{g(1),g(2)\}$ and so on. It remains to show that ...


1

For #1, that doesn't work, because you don't know that $d_B$ is induced by a norm. In fact it is not, because all norms on positive dimensional spaces are unbounded, but that metric is bounded. (I might edit if you post an attempt and show where you get stuck.) For #2, one open cover that you always have of a set $A$ in a metric space is $$\{ B_\epsilon(x) ...


1

$\textbf{Hint}$: Let $p_n$ be a prime divisor of $R_n=n!+1$, and show that $p_n$ must be larger than n. Then conclude from this that there are infinitely many primes.


1

I believe you meant to say that if the sequence of prime numbers is finite, then the values of $n$ for which $R_n$ is prime must also be finite. From here, you can simply take $n$ to be the product of all the primes (only finitely many by assumption), and then following the same argument as in Euclid's proof, there must be some prime that divides $R_n$ but ...


1

This is a typical proof by induction. The base case is just that the first term ($1$) does not contain any digit greater than $3$ nor more than $3$ consecutive, equal digits. If some term had a digit greater than $3$, it would have had to come from more than $3$ (that is, $4$ or more) consecutive, equal digits in the previous term. But the first four of ...



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