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9

Let $\psi = -1/\phi$. Then the Fibonacci numbers $(F_n)$ and the Lucas numbers $(L_n)$ are given by $$F_n = \frac{\phi^n - \psi^n}{\phi - \psi}, \qquad L_n = \phi^n + \psi^n.$$ Using this, your number $\mathcal{E}_n$ can be written as $$\mathcal{E}_n = \frac{4^n + (-1)^n + 2^n L_n}{F_n} \in \Bbb{Q}.$$

7

Dividing your formula by $q=\pi$ gives $$n \cdot \frac{1}{\pi} = p + \frac{r}{\pi} ,$$ so $p$ is the integer part, and $\frac{r}{\pi}$ is the fractional part, of $n \cdot \frac{1}{\pi}$. As $n$ runs through the integers, this fractional part $\frac{r}{\pi}$ is uniformly distributed on $[0,1)$ by Weyl's equidistribution theorem, since $\frac{1}{\pi}$ is ...

6

Let $\infty_+$ and $\infty_-$ be the (real) points at infinity of the curve (corresponding to $y/x^2 = +1$ and $y/x^2 = -1$) In general (unless $p^2=1$), a parabola $y = px^2+qx+r$ intersects the elliptic curve at $4$ points. More precisely, $Div(y-px^2-qx-r) = [P_1]+[P_2]+[P_3]+[P_4]-2[\infty_+]-2[\infty_-]$. From a point $P$, you find the parabola that ...

5

That is pretty trivial. If $\mathbb{Q}$ is dense in $\mathbb{R}$, for any $(x_1,\ldots,x_n)\in\mathbb{R}^n$ we may find $(q_1,\ldots,q_n)$ such that $|x_i-q_i|\leq\varepsilon$, for any $\varepsilon>0$, so $\|x-q\|_1\leq n\varepsilon$ is arbitrarily small.

4

$$\frac29 = \frac29\cdot\big(1\big) = \frac29\cdot\left(\frac{11}{11}\right) =\frac{2\cdot 11}{9\cdot 11} =\frac{22}{99}$$ Placing rational numbers into decimal form you will find that they either terminate ($1/4$ for example), or repeat for ever, like the $2/9$ you found. I recommend trying some long division on integers divided by $11$, or $7$. You ...

3

$\sup S$ is shorthand for the least upper bound of $S$ in some set $T\supset S$ with respect to an order $\leq$ on $T$. However, we often omit $T$ and $\leq$ when they are obvious from the context. Consider the case when $T=\mathbb{Q}$ and $\leq$ is the usual order. Since $\sup\left\{ x\in\mathbb{Q}:x^{2}<2\right\}$ has an upper bound but no least upper ...

3

We have $r_m \leqslant 1 \iff m \equiv 1 \pmod{2}$. Suppose the set $E$ of positive rationals that don't occur in the sequence were nonempty. Choose an $r = \frac{a}{b}$ with minimal denominator from $E$. We know that $b > 1$ since $r_{2^{n-1}} = \frac{n}{1}$ for all $n$. If $r < 1$, then $\frac{1}{r} = \frac{b}{a}$ has smaller denominator than $r$, ...

2

As you wrote in a comment, the proposition you are trying to use doesn't apply directly to $\Bbb Q^k\times \Bbb Q$. However, if we know that $D$ is countable and $E$ is some other countable set, then there is a bijection $f: D \to E$, and the map from $D \times D$ to $D \times E$ given by $(x,y) \mapsto (x,f(y))$ is a bijection. Since by the proposition $D ... 2 More generally, the product of dense sets is dense in the product topology. For a proof, see this post. 2${{y^2 - y} \over 1 {}} \div {{y^2 - 1} \over 3}$Now whenever you divide by something, you are essentially multiplying the reciprocal of the something. So${{y^2 - y} \over 1 {}} * {{3} \over y^2-1}=\frac{y(y-1)}{1}*\frac{3}{(y-1)(y+1)}$Cancel$y-1$to get your required answer. 2 Let$P(x,y)$be the assertion: $$f(f(x)^2y)=x^3f(xy)\space\forall x,y\in\mathbb{Q^+}$$ We get (due to the injectivity): $$P(1,y): f(f(1)^2y)=f(y)\iff f(1)^2y=y\iff f(1)=1$$ $$P(x,f(y)^2): f(f(x)^2f(y)^2)=x^3f(xf(y)^2)=x^3y^3f(xy)$$ $$P(xy,1): f(f(xy)^2)=x^3y^3f(xy)$$ If we combine the last to results, we get: $$f(f(xy)^2)=f(f(x)^2f(y)^2)\iff ... 2 Ok, I'll complete with your solution...$$f\left ( f(x)^{2}y \right )=x^{3}f(xy) \tag{1}.f\left ( f(x)^{2} \right )=x^{3}f(x) \tag{2}.$$Now replace x by xy in (2) , now apply (2) twice, second time to \left (y, f(x)^{2} \right ) instead of (x,y)$$f\left ( f(xy)^{2} \right )=(xy)^{3}f(xy)=y^{3}f\left ( f(x)^{2}y \right )=f\left ( ... 1 As said in the answer of @HansLundmark this is Weyl's equidistribution theorem. It is an application of the ergodic theorem combined with the theorem that an irrational rotation of the circle is ergodic, and here's some details of how to reduce it to those two theorems. Consider the unit circle$S^1$in the complex plane. Given$q \in \mathbb{R}$, consider ... 1 Stolen from Michael Galuza's comment: When you "swap around" the second term you should also make it a multiplication, not a division. You then get a common factor$y-1$$$\frac{y^2-y}{1}\times\frac{3}{y^2-1}=\frac{3y(y-1)}{(y-1)(y+1)}=\frac{3y}{y+1}‌​$$ 1 This is an interesting way to write a continued fraction. Write the index in binary, then read it from right to left. Each segment of$00\dots01$represents $$+\cfrac1{n}$$ where$n$is the length of the segment. If the rightmost segment has no rightmost$1$, then that segment just represents $$n$$ where$n$is the length of the segment. For example ... 1 Your induction hypothesis should only consider$k \in \mathbb{N}$with$k \ge 2$, not all integer$k$. Aside from that, your proof is correct. 1 The set $$\left\{\sum_{k=1}^n\frac1{k!}:n\in\Bbb N\right\}$$ is contained in the rationals, and has upper bound. It is not difficult to show that$3$is an upper bound. But in the reals, the series converges to$e$, which is not rational. For a relatively easy proof, see here. 1 To answer in a more formal way than you are probably looking for, rational numbers are defined as an equivalence class. The form of the class looks like$[(a,b)]$where$a,b$are both integers and$b\ne 0$. In this sense,$[(a,b)]=\frac a b\$. We call it an equivalence class because there's a whole set of members that are equivalent to each other. For ...

1

In terms of repeating decimals $$\frac{1}{9} = .1111111111\cdots$$. Any multiple of this value leads to a repeating similar pattern. As examples defined by this problem: \begin{align} .2222\cdots &= \frac{2}{9} \\ &= \frac{22}{99} \\ &= \frac{222}{999}\\ &= \cdots \end{align}

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