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## Hot answers tagged rational-numbers

5

$1000y = 273.273273...$. Therefore, $1000y-y = 273$. Solving for $y$, $y = \frac{273}{999}$.

5

Let us work with the closed interval $[0,2]$, let $\{q_n\}$ be a enumeration of $\mathbb{Q}\cap [0,2]$, and define $$I_n := \bigg(q_n -\frac{1}{2^{n+1}}, q_n+ \frac{1}{2^{n+1}}\bigg)\bigcap [0,2]$$ then $\{I_n\}$ is an open cover of the rationals in the space $[0,2]$. (I used $\frac{1}{2^{n+1}}$ so that the sum of the length is $1$, there is a small typo ...

5

There are many decimals between 0.5 and 0.625 - in fact, there are more than you could ever count! So it's a little hard to know exactly what you're asking for. Possibly, you're looking for the number exactly between 0.5 and 0.625, which you find by adding them together and dividing by two to get $(0.5 + 0.625) \div 2 = 1.125 \div 2 = 0.5625$. On the other ...

4

Since $y$ has a repeating cycle of $3$ digits, just write: $\;1000y=273+y$, solve for $y$, and simplify. This procedure, though intuitively clear, is not rigourously justified, but it can easily be justified with the interpretation of the positional notation for numbers via a geometric series.

4

One consequence would be that $e \pi$ is transcendental because, for any $z$, if $A=e+z$ and $B=ez$ are both algebraic then $e$ is a solution of $x^2-A x+B=0$, which makes $e$ algebraic. But it isn't.

4

Leibniz formula tells us: $$\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1}=\frac{\pi}4=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\dots$$ Group the terms by two, the difference is positive, so there is an obvious increasing sequence of rationals that converges to $\pi$: $$4\left(1-\frac{1}{3}\right)+4\left(\frac{1}{5}-\frac{1}{7}\right)+\dots$$ Or more formally ...

3

To build on my comment, there are probably a few ways to define that sequence. One "lazy" way to do it is to recognize that $\pi$ as an irrational has the property that for any $n \in \mathbb{N}$, we can find the $n^{th}$ decimal in $\pi$'s decimal expansion. So for $b_i \in \{0,1,\ldots,8,9\}$ we can write $$\pi = b_0.b_1b_2b_3b_4\ldots ... 3 Hint: Try finding a sequence in \mathbb{Q}\cap[0,1] that does not have a convergent subsequence in \mathbb{Q}\cap[0,1]. 3 Hint. Think about the quadratic formula for a quadratic equation with integer coefficients but two irrational roots. 3 This answer is kind of cheating, since it's really just extracting the essential pieces of measure theory needed to prove this. But it's self contained and doesn't need to use the word "measure". Fix a closed interval such as [0,2] whose length is at least as long as the sum of the lengths of your intervals I_n (as currently formulated, that sum is 2 ... 2 \mathbb{Q}\cap[0,1] is dense in [0,1] 2 Let n_0:=\lfloor\pi\rfloor and define inductively$$n_{k+1}:=\left\lfloor10^{k+1}\left(\pi-\sum_{i=0}^k\dfrac{n_i}{10^i}\right)\right\rfloor.$$Then$$\lim_{m\to\infty}\sum_{k=0}^m\dfrac{n_k}{10^k}=\sum_{k=0}^\infty\dfrac{n_k}{10^k}=\pi.$$2 The sequence in graydad's answer can be written in terms of the floor function as$$n \mapsto 10^{-n} \lfloor 10^n \pi \rfloor.$$This suggests some cute generalizations. Instead of the decimal expansion for \pi, we can use the expansion in any base q:$$n \mapsto q^{-n} \lfloor q^n \pi \rfloor$$converges to \pi for any whole number q \ge 2. In ... 2 The kicker is that, among real numbers, there is no such thing as two numbers that are "infinitesimally close" to each other. For example, consider the set S of numbers greater than 0. Take any s\in S. Now, note that \frac12s\in S, and is strictly smaller than s. Likewise, \frac14s is an even smaller element of S, and so on. In fact, given any ... 2 It suffices to prove the result for a partition into two parts, A and B, say. Also, it suffices to show that one of A,B contains a countable dense subset without endpoints (cf. ''Brian M. Scott's'' comment). Of course, the endpoint condition is harmless as we can simply remove them (we want to show only that one of the parts contains an isomorphic copy ... 2 The point of this exercise is to make you realize that there is no such thing as "the next rational number". You did well to realize \frac{a+1}b does not work. But the problem is more general: suppose you do decide that some number \frac c d is the "next" one after \frac a b. But ask yourself: what is the average of \frac a b and \frac c d? And ... 2 You can apply an "argument from authority" here :). As we all know, Dedekind constructed a one-to-one correspondence between the real numbers and certain sets of rational numbers, which he called "cuts". We all accept that Dedekind's construction of the real numbers is mathematically flawless. By the way these cuts are constructed (see his 1872 paper), if ... 1 The key is in the dots. It is tough to get a handle on an infinite thing, so usually, it would be a process: Start with x_1=5 Calculate x_2=5+6/x_1 Calculate x_3=5+6/x_2 Carry on; what is the limit of x_n? Suppose x_n=6+y_n. Then$$y_{n+1}=x_{n+1}-6\\ =6/x_n-1\\ =6/(6+y_n)-1\\ =-y_n/(6+y_n)$$Since -1\leq y_1\leq1, we ... 1 Notice, we can simplify the given expression as follows$$\frac{5}{x+3}-\frac{7x}{x-1}=\frac{5}{x+3}-\frac{7x-7+7}{x-1}=\frac{5}{x+3}-\frac{7(x-1)+7}{x-1}=\frac{5}{x+3}-\frac{7(x-1)}{x-1}-\frac{7}{x-1}=\frac{5}{x+3}-7-\frac{7}{x-1}=-7+\frac{5}{x+3}-\frac{7}{x-1}$$or Using partial fraction$$\frac{5}{x+3}-\frac{7x}{x-1}$$... 1 Just divide and see:$$\frac{367}{900}=0.407777777\ldots=\frac{4037}{9900}$$1 Suppose that we have (\frac{a}{b})^n = 2. WLOG a and b are in lowest terms (we can do this because we can effectively "cancel" out any common factors and obtain c and d in lowest terms with \frac{a}{b} = \frac{c}{d}). Then we have a^n = 2b^n. So a^n is even and hence a is even. Write a = 2k. Then 2^n k^n = 2b^n and hence b^n = ... 1 Let me elaborate a bit on my comments: Suppose n is coprime to 10. Then we have 10^{\varphi(n)}\equiv 1\pmod n, and thus it follows that 10^k\equiv 1\pmod n for some k dividing \varphi(n). Choose the smallest such k. This corresponds to 3. in your suggested algorithm. Then n must divide 10^k-1 and so all prime factors of n can be found ... 1 Hint: Find a sequence of closed, non-empty sets C_n\subseteq [0, 1]\cap \Bbb Q for n \in \Bbb N such that C_n \subseteq C_{n-1} and \bigcap_{n = 1}^\infty C_n = \emptyset. 1 You can cover it with infinite number of open sets such that no open set overlaps (open intervals with irrational endpoints). If \bigcup U_j = X, but U_j\cap U_k = \emptyset if k\ne j you can't reduce the sets to a finite covering. 1 You said, "A closed subspace of a compact space is compact." and then you tried using [0,1] as the compact space. But [0,1] is compact as a subset of the reals, as a subset of \mathbb{Q}, it's what you are trying to understand. The countable covering of [0,1] in \mathbb{Q} by point sets in [0,1] \cap \mathbb{Q} has no finite subcover in ... 1 Let c be any real number. What you want to show is that there is some subsequence that converges to c. Let \epsilon_{n}=\frac{1}{n} (really, anything that is positive and converges to 0). Now, pick n_1 so that x_{n_1}\in (c-\epsilon_1, c+\epsilon_1). Then, pick n_2 so that x_{n_2}\in (c-\epsilon_2, c+\epsilon_2) AND so that n_2>n_1. ... 1 If \sqrt[n]{q} is rational, then it can be written uniquely as:$$\sqrt[n]{q}=\prod p_i^{r_i}$$where r_i are integers. But then$$q=(\sqrt[n]{q})^n = \prod p_i^{nr_i}$$By uniqueness, then nr_i=q_i, so you can't have q_i not a multiple of n. 1 One can identify \{ 0,1 \}^{\mathbb{N}} with P(\mathbb{N}), where i \in f(s) if s_i=1. You can then identify P(\mathbb{N}) with [0,1] through the map:$$A \mapsto \sum_{i \in A} 2^{-i}.$$This amounts to binary expansion of a number in [0,1]. This is not quite a bijection, because for instance 1/2 has two binary expansions, namely ... 1 Hint. You may just observe and directly prove that$$ 0.9999999999999999\cdots =1  Have a look at this.

1

You should exclude the case $b=0$. It's enough to let $a$ belong to $\mathbb {Z}$ and $b$ belong to $\mathbb{N}$. You can define the equivalence relation you need by $(a, b) \sim (c, d)$ if and only if $ad = bc$. Then yes, $\mathbb {Q}$ is the quotient of the big set by the equivalence relation.

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