# Tag Info

0

It may be easier to see if you use the sequence approach to continuity: Pick an $p\in \mathbb Q$ and from each neighborhood $B_{1/n}(p)$ choose an irrational number $x_{n}$. Then you get a sequence with the property that $x_{n}\to p$ and $f(x_{n})=d\quad \forall n\in \mathbb N$. Since $f(p)=c\neq d$, $f$ is not continuous at $p$. Simlarly, if $x\in ... 0$\mathbb{Q}$is dense in$\mathbb{R}$, which means in this case that: $$\forall x,y \in \mathbb{R} \hspace{2mm} \text{with} \hspace{2mm} x < y \hspace{2mm} \exists r \in \mathbb{Q}: x < r < y$$ So, for two real (particularly irrational) numbers, there is always a rational number in between, no matter how close the two reals are to each other. So ... 0 Say we have a continuous function$f :\Bbb{Q}\to \Bbb{Q}$for which we know$f(a)<0<f(b)$for fixed rational numbers$a<b$. Now set $$M:=\{x\mid a\leq x\leq b \text{ and } f(x)<0\}.$$ Clearly,$M$is bounded above. Suppose that$M$has a least upper bound$\alpha$. It is not hard to see that this entails$f(\alpha)=0$(if you don't believe ... 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. 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 ... 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 ... 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$r$, ... 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 $$\mathcal{E}_n = \frac{4^n + (-1)^n + 2^n L_n}{F_n} \in \Bbb{Q}.$$ 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 ...

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.

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.

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 $y-1$ $$\frac{y^2-y}{1}\times\frac{3}{y^2-1}=\frac{3y(y-1)}{(y-1)(y+1)}=\frac{3y}{y+1}‌​$$

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 ( ...

0

$f(f(x)^2f(z)^2)=x^3f(xf(z)^2)=x^3z^3f(xz)$ so $f(x)f(z)=f(xz)f(1)=f(xz)$ because $f$ is injective. So $f(x)$ is defined by all the values of $f(p)$ for $p$ prime. $f(f(x))^2=x^3f(x)$ so $f(p)=g(p)^2/p$ for a rational number $g(p)$. $$f(f(p))=f(g(p)^2/p)^2=p^3g(p)^2/p=p^2g(p)^2\\ f(g(p)^2/p)=pg(p)\\ f(g(p))^2=(g(p))^3$$ So $g(p)$ is a square number, ...

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 ...

2

More generally, the product of dense sets is dense in the product topology. For a proof, see this post.

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.

0

If the fractions have the same value (whether expressed in decimal or not), then they are equal, because 'equal' means 'having the same value'. So there is no question.

-3

These fractions are different representatives of the same class of equivalence of the field of fractions $\mathbb{Q} .$

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}

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 ...

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