# Tag Info

11

Assuming that you want the usual addition operation on $\mathbb{Z}$ to be its addition operation as a $\mathbb{Q}$-vector space, then no, it is not possible for it to be a $\mathbb{Q}$-vector space. For example, we would have a scalar $\frac{1}{2}\in\mathbb{Q}$ and an element $\mathbf{1}\in\mathbb{Z}$ of our supposed vector space (using bold-face to ...

7

You can proceed directly as follows: $2x = (x+y) + (x-y)$ which must be irrational as it is the sum of a rational and an irrational. So $x$ is irrational. Similarly $2y = (x+y) - (x-y)$ is irrational.

6

Continued fractions. $$0.385 = \frac1{2+\frac1{1+\frac1{1+\frac1{2+\frac1{15}}}}}$$ and $$\frac1{2+\frac1{1+\frac1{1+\frac1{2}}}}=\frac5{13}\approx 0.3846$$ so it could have been $13$ questions. It could not have bneen less because any fraction$\frac ab$ between $\frac38=0.375$ and $\frac25=0.4$ has $b\ge 13$: We have $\frac ab>\frac38$, which implies ...

6

At least in the topological setting, it's very hard to get any sort of manifold theory working over the rational numbers. The reason is that $\mathbb{Q}^n$ is homeomorphic to $\mathbb{Q}$ for all $n\geq 1$, as is any $\mathbb{Q}^n$-manifold that you might construct. In fact, every countable, regular, first-countable space without isolated points, and in ...

5

There's a very nice theorem due to Lang reproduced in Appendix 1 of his Algebra from which the Hermite-Lindemann theorem follows as a Corollary. Assuming Hermite-Lindemann which says that if $\alpha$ is algebraic over $\mathbb{Q}$ then $e^{\alpha}$ is transcendental, it follows pretty quickly that $ln(\alpha)$ is transcendental for rational $\alpha$, since ...

4

The answer is yes, since we have: $$x = \frac{1}{2} \left( (x+y)+(x-y) \right)$$ $$y = \frac{1}{2} \left( (x+y)-(x-y) \right)$$ So $x$ and $y$ can be expressed as the the sum and product of rational numbers

3

Maybe check out diffeology. Diffeological spaces are generalized smooth spaces that generalize smooth manifolds. The category of smooth manifolds embeds fully and faithfully into the category of diffeological spaces, which has the extremely nice property of being a quasi-topos. You could define the sort of spaces you seem to be interested in as diffeological ...

3

Hints: enumerate any countable subset of $\mathbb R$ as, say, $\{a_n\}_n \ge 1$. Now cover $a_n$ by an interval of length $\frac{\varepsilon}{2^n}$. What is the total measure of the cover? Notice that $\varepsilon>0$ may be chosen arbitrarily. What does that imply?

2

No, the values $f(x)$ are rational linear combinations of irrational numbers. Let's say you have a sequence $b_i=f(x_i)$ in rational points and let be $a_i$ the coefficients of the polynomial. Therefore you can solve for the coefficients if you have enough $b_i$ (1+polynomial order). This can be written in the form $$\vec{b}=A\vec{a}$$ where $A$ is an ...

2

Oh well, I will answer the question just because it seems to me that it is better to have as many as possible questions on the site that are answered. I would like to thank Erick Wong who pointed me in the right direction with his comment. In this paper Louis Joel Mordell proves much more general result in a theorem which he states in this way: A ...

2

Key points: Your set is closed under internal addition and subtraction (that is, if $x$ and $y$ are in your set then so is $x + y$ and $x - y$. Further, your set is closed under multiplication by arbitrary integers (clearly). Suppose that $0$ were a limit point of your set. That means that given an arbitrarily small $\epsilon > 0$ I can find integers ...

2

Your question is not well-defined because some fractions do not have a decimal expansion that consists only of repeating blocks after the decimal place, such as $\frac{1}{6}$. However, if the denominator $d$ is coprime to $10$, then what you are looking for is equivalent to the order of $10$ modulo $d$. In that case, we have reduced this to a previously ...

2

Explicit bijections are rather tedious to come up with. However, the Schroeder-Bernstein theorem implies that a subset of a countable set is countable (or finite, if your definition of countable excludes finite sets) and its proof does not require AC (see https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem). This provides an alternative to ...

2

You don't need the axiom of choice for the following statement: If $X$ is countable, and $f$ is a function whose domain is $X$, then the range of $f$ is countable. You also don't need the axiom of choice for the following statement: $\Bbb{N\times Z}$ is countable. Finally, define $f(n,m)=\frac nm$ or $0$ if $m=0$, and show that this is a ...

1

Every positive rational number can be written as a finite continued fraction$^*$, and every finite continued fraction can be associated with a string over the alphabet $\Sigma=\{0,1\}$. For instance: $$\frac{89}{13}=[6;1,5,2] \longrightarrow 11111101111100,$$ $$\frac{101}{47}=[2;6,1,2,2]\longrightarrow 1100000010011,$$ so, by reading that string as the ...

1

The solution is correct. I would do it only slightly differently, and more briefly: First, for any $r\in\mathbb Q$ the function $x\mapsto f(x+r)-f(x)$ only takes rational values and is continuous, so it must be constant; call it $c_r$. A simple iteration gives $f(x+kr)-f(x)=kc_r$ for any $k\in\mathbb Z$. For the different constants $c_r$ to be compatible ...

1

I don't really understand your problem... If $x$ is rational and $k\geq 2$ big enough, $f_{n,k}(x)=1$ because $\cos (k!\pi x)=\cos(2\pi k')=1$. If $x$ is irrational, $\cos(k!\pi x)<1$ and thus $(\cos(k!\pi x))^{2n}\to 0$ if $n\to \infty$, therefore you got your result.

1

1) A terminating decimal representation means a number can be represented by a finite string of digits in base $10$ notation, e.g. $0.5$, $0.25$, $0.8$, $2.4$ 2) A non-terminating decimal representation means that your number will have an infinite number of digits to the right of the decimal point. There are two sorts of non-terminating decimal numbers. ...

1

This comes from the fact that a measure is countably additive. Let $I$ be a countable set. You have then $$\mu(I) = \mu ( \bigcup_{x\in I} \{x\} ) = \sum_{x\in I} \mu( \{x\} )$$ Now if every singleton has measure zero, it follow that $$\mu(I) = 0$$ And this is the case for the Lebesgue's measure on the real line.

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