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9

It does repeat, but with a very long period: Factoring the denominator into primes gives $$998 = 2 \cdot \color{#3f3fff}{499};$$ since $2$ is a factor of $10$ and $10$ is a primitive root modulo $\color{#3f3fff}{499}$, the period of repetition is $\color{#3f3fff}{499} - 1 = 498$ digits long. Indeed, consulting WolframAlpha gives: $$\color{#bf0000}{ ...


7

Emphasis on "apparently". The same thing happens with $1/7$: $$\begin{array}{crrrrrrrrrl} & 0 & . & 14 \\ + & & & & 28 & & & & & & =2\times 14 \\ + & & & & & 56 & & & & & = 4\times 14 \\ + & & & & & 1 & 12 & & & & =8\times 14 ...


7

No, homeomorphisms of $\mathbb{Q}$ need not be monotone. For an irrational $c > 0$, let $$h_c(x) = \begin{cases}x &, \lvert x\rvert < c\\ -x &, \lvert x\rvert > c. \end{cases}$$ Then $h_c$ is a non-monotonic homeomorphism of $\mathbb{Q}$.


5

Let $r=s/t$ where $s\in\mathbb{Z}$ and $r\in\mathbb{N}$. Assume $e^r=p/q$ where $p,q\in\mathbb{N}$. Then $$ pqt^nJ_n(r)= p^2 t^n A_n\left(\frac{s}{t}\right)+ q^2 t^n B_n\left(\frac{s}{t}\right)\in\mathbb{Z} \tag{1} $$ Note that $$ 0<J_n(x)\leq\frac{x^{2n}}{n!}\int_{-x}^xe^tdt=\frac{2 x^{2n}\sinh x}{n!} $$ So $$ 0<pqt^nJ_n(r)\leq \frac{2 pq t^n ...


4

For all $q \in \Bbb Q$ there exists $n \in \Bbb Z$ such that $q + n = 271$. False. $\Bbb Q$ is the set of rationals, i.e. fractions: $\Bbb Q = \{\ldots, -\frac 1 3,- \frac 2 2,-\frac 1 2, 0, \frac 1 2, \frac 2 2, \frac 1 3, \ldots\}$. $\Bbb Z$ is the set of integers: $\{\ldots, -3, -2, -1, 0, 1,2,3,\ldots\}$. Take $q = \frac 1 2$ for an easy ...


3

As per TonyK's answer, it is only necessary to find the smallest denominator for which a fraction in $[L,U)$ exists. To avoid what he calls "messy" oscillations, we can proceed as follows: Start with $(a,b,c,d)=(0,1,1,1)$. *[Remark: We have $\frac ab<L<U\le \frac cd$, and $ad-bc=-1$, and any fraction between $\frac ab$ and $\frac cd$ has denominator ...


3

It is enough to find the smallest possible $b$, and then the smallest possible $a$ given $b$. This is not completely obvious, so I will try to prove it: Suppose we have positive integers $a,b$ such that (i) $b$ is the smallest denominator of all fractions in $[L,U)$; and (ii) $a/b$ is the smallest multiple of $1/b$ in $[L,U)$. This means ...


3

Hint $(\Bbb Q_{> 0}, \cdot)$ is isomorphic to $(\Bbb Z, +) \oplus (\Bbb Z, +) \oplus \cdots$.


3

We first consider the case $n$ odd, since the solution is simpler. Let $n = 2m + 1$. Choose $a_i: 1 \le i \le m$ nonzero distinct rational values such that the sum of squares is itself a square: $$a_1^2 + a_2^2 + \ldots + a_{m}^2 = A^2$$ (One way to generate such a list would be to iteratively use Pythagorean triples, ie something like $a_1 = 3, a_2 = 4, ...


2

It is possible for an irrational number to have a predictable pattern; consider $0.1101001000100001...$. It is also possible to have an irrational number that is another irrational number away from a rational; i.e. $x-y = r $, where $x,y$ irrational and $r $ rational; in fact the equivalence classes of such numbers are dense in the reals. So you can subtract ...


2

Suppose you run a numerical analysis program, and it outputs $$0.5323$$ What does that even mean? Does it mean: $$x = 0.5323$$ $$0.5323 \le x < 0.5324$$ $$0.53225 \le x < 0.53235$$ $$\text{Probability}\left(\frac{|x - 0.5323|}{0.5323} < 0.01\right) > 95 \%$$ The meaning of a numerical program still must be expressed and proven ...


1

I'm not sure what you are allowed to assume. Do you know that $e$ is transcendental and can you use that? If so consider $e^{\frac{p}{q}}=l$ where $l\in \mathbb{Q}$. What happens when you look at the polynomial $x^p-l^q$ ? Looking at the first part of your question I'm assuming it's unlikely you are allowed to use the fact that $e $ is transcendental ...


1

In the comments I proposed following formula for $K\gg 1$ (about $15$ digits precision for $K>30$) $$K-\gamma-\log(K+1/2)-\frac 1{24K^2}+\frac 1{24K^3}-\frac{23}{960K^4}+\frac 1{160K^5}+\frac {11}{8064K^6}+\frac 1{896K^7}$$ and to compute directly $\,K-H_K\;$ for small values of $K$ ($H_K=\displaystyle\sum_{i=1}^K\frac 1i\;$ is an harmonic number). This ...


1

$$\sum_{\alpha=2}^{K} \frac{\alpha-1}{\alpha}=K-\digamma^{(0)}(K+1)-\gamma$$ With $\digamma^{(0)}$ is the derivative of the digamma function and $\gamma$ is the eulergamma


1

No there isn't, because finding a compact form for the sum is equivalent to find a compact form for the harmonic sum $$ \sum_{k=1}^n \frac1k $$ which we know that doesn't exists. But you can approximate your sum by using $$ \sum_{k=1}^n \frac1k=\log(n)+\gamma+O(n^{-1}) $$


1

Yes, such a field exists. It's sometimes called $\mathbb{Q}(i)$. It's a degree two extension of $\mathbb{Q}$. It has the property that some polynomials with coefficients in $\mathbb{Q}$ have roots in $\mathbb{Q}(i)$. For instance, $X^2+1$. However, the complex numbers are algebraically closed—any polynomial with coefficients in $\mathbb{C}$ has roots in ...


1

People do indeed define the "complex rationals", i.e. the extension of the rationals by $i$, often cqlled the Gaussian rationals. The question is what you want to do with this.


1

In order for you to be able to convert the denominator to a number of the form $2^n 5^m$, it needs to a number that is divisible by $2$ or $5$ and no other primes, and that is not the case for $264600$. But, as it turns out, you can make the denominator that way by "simplifying" the fraction: $$\frac{1323}{264600} = \frac{1}{200}.$$ Now observe that $200 = ...


1

Why is your algorithm exponential? Here's an option: Don't think about the numbers as fractions, but as pairs. Your goal is to find the pair $(a,b)$ such that $L\leq a/b\leq U$ such that $a\times b$ is minimal. In this case, we do NOT reduce fractions. For a fixed $b$, we can easily find the smallest possible $a$ that satisfies the conditions by ...


1

The standard proof that the real numbers are uncountable does not need to be phrased as a proof by contradiction. To say that $\mathbb{R}$ is uncountable is to say that if $f : \mathbb{N} \to \mathbb{R}$ is a map, then it is not a bijection. The standard proof in fact proves that $f$ is not a surjection by explicitly exhibiting a real number which is not in ...


1

It's easier to first talk about endomorphisms because these have a ring structure. In fact, as has been mentioned a few times already, this group is a countable direct sum $\bigoplus_p \mathbb{Z}$ of a copy of $\mathbb{Z}$ for every prime, and hence in some respects it behaves like a vector space. In particular, its endomorphism ring $\text{End}(\bigoplus_p ...


1

My intuition says that since distinct integers differ by at least one and rationals are arbitrarily close, the density of the integers should be zero and the density of the rationals should be positive. If you throw in the reals, I'll throw in the towel.



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