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## Hot answers tagged rational-numbers

10

The rational number? Rational numbers are a dense subset of $\mathbb{R}$, there are infinitely many of them between $\sqrt{2}$ and $\sqrt{3}$. For instance, since the difference between $\sqrt{2}$ and $\sqrt{3}$ is between $\frac{3}{10}$ and $\frac{1}{3}$, there is for sure an integer number between $4\sqrt{2}$ and $4\sqrt{3}$: $6$, for instance. By ...

8

A quick and inelegant approach is to use the (beginnings of the) decimal expansions $\sqrt2=1.41\dots$ and $\sqrt3=1.73\dots$. Any terminating decimal between these two will solve your problem, for example $1.5$ or $1.6$ or $1.7$. You could also use any periodically repeating decimal between the two, like $1.6666666\dots$. Once you've obtained an answer ...

4

I just learnt this, it is not about the decimal expansion, but I think it is clearly related to your question : A simple proof that $e$ is irrational is using its factorial expansion $2+\sum_{n=2}^\infty \frac{1}{n!}$. The theorem is that $$\sum_{n=1}^\infty \frac{a_n}{n!}, \qquad a_1 \in \mathbb{Z},\quad a_n \in 0 \ldots n-1$$ is irrational if and only if ...

3

As $\sqrt2 = \sqrt{200/100}$ and $\sqrt3= \sqrt{300/100}$, we need to find a rational number $x$ such that $$\frac1{10}\sqrt{200}< x<\frac1{10} \sqrt{300}$$ Choose any perfect square such as $225$ or $256$ in between 200 and 300. Then $x=\sqrt{225}/10 = 15/10=5/3$ and similarly $16/10=8/5$ would be the numbers with the desired property.

3

As @m_t_ says, just write down a formula for any number that has a non-repeating decimal expansion and you've shown that number is irrational. For example $0.1010010001000010000010000001\ldots$ Such a number is obviously immediately irrational from inspection of the decimal expansion.

3

Apologies for the length of the argument. I tried to format it to be more comprehensible. Multiplying out $(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$, so we have the equations: $1)$ $a=-a-b-c$ $2)$ $b=ab+ac+bc$ $3)$ $c=-abc$ Notice that if $a = 0$ or $b = 0$, then by the $(3)$, $c=0$ contradicting the fact that they must be distinct. Therefore ...

2

There is a generalization of vector spaces called "modules" which allow any ring to serve as scalars. When you use the integers as the ring of scalars, a "module" is the same thing as a "abelian group". The group of 'factorizations' is indeed a free abelian group, which is the kind of abelian group that behaves most similarly to a vector space. ...

2

It is a simple consequence of Szemeredi's theorem. The set of numbers of the form $[n\pi]$ has a positive ($\geq\frac{1}{4}$) upper density in $\mathbb{N}$, hence such a set contains arbitrarily long APs. On the other hand, if we take a convergent $\frac{p_m}{q_m}$ of the continued fraction of $\pi$, we have: $$\left|\pi-\frac{p_m}{q_m}\right|\leq\frac{1}{... 2 Counterexample: If one takes p = 2 and q=3 then we get x \in \left[\frac{1}{3}, \frac{2}{3}\right]. But x = \frac{1}{2} \in \left[\frac{1}{3}, \frac{2}{3}\right] and 2 \not\geq 3. 2 Note that I do not need the conditions \gcd(a,b)=1 and \gcd(c,d)=1 here. Therefore, these conditions are omitted in the proof below. Let \textbf{A}:=\begin{bmatrix}a&b\\c&d\end{bmatrix}. Suppose that \textbf{x}:=\begin{bmatrix}x\\y\end{bmatrix}\in\mathbb{Q}^2 satisfies \mathbf{A}\,\mathbf{x}=\mathbf{b}, where \mathbf{b}=\begin{bmatrix}u\... 2 For n=3, taking x_3 = 1 - x_1 - x_2 we get -1 + x_1 x_2 - x_1 x_2^2 - x_1^2 x_2 = 0. This is an elliptic curve with Weierstrass form s^3+(23/48)s-181/864+t^2, according to Maple. I think it ought to be a finite computation (not one that I know how to do) to find generators for the rational points. EDIT: Well, apparently this is an unsolved ... 1 HINT: Try \sqrt2 and -\sqrt2. More generally, try any two irrationals whose sum is rational. 1 An alternative approach: \sqrt2 and \sqrt 3 are non but solutions to the equations x^2-2 and x^2-3. Consider their graphs and notice that any positive rational solution to the polynomial x^2-a, for a\in (2,3), will satisfy your requirement. We know the solutions to x^2-a are given by \pm\sqrt a. Then consider only the positive and write ... 1 Let's take a shot at this. Suppose \sqrt2 had repeating digits. By the standard approach, we can find integers m, k such that$$\sqrt2 = \frac{m}{10^k-1}.$$Squaring both sides and expanding, we have$$m^2 = 2\color{red}{(10^k-1)^2}.$$We see that the term in red cannot have a factor of 2; therefore, the integer m^2 appears to have a single ... 1 You are basically asking for the number of (standard) lattice points inside the fundamental parallelepiped F of the lattice generated by the matrix A. Since the vertices of F are on the standard lattice, the number of lattice points in F is the same as the number of points in F+x for any translate of F by an integer vector x. Therefore, if you ... 1 Here's an explanation via an application to a conrete example.$$ x = 0.15\ \overbrace{504}\ 540\ 540\ 540\ \ldots \qquad (\text{“540'' repeats.}) $$Since the repetend has three digits, we multiply by 1000 by moving the decimal point over three places:$$ \begin{array}{rcc|c|cc|c|c|c|c|c|c|c|c|c} 1000x & = & 1 & 5 & 5 & . & 4 & ...

1

I asked myself the same a while ago, and I have an interesting answer: that representation is, for instance, the key argument of Rusza's proof for the existence of an infinite (almost optimal in terms of density) Sidon set.

1

Every couple of fractions $c/a$ and $d/b$ in the Stern-Brocot tree can be represented (in the inverse notation according to Concrete Mathematics) by the matrix ${\bf M} = \left\| {\,\begin{array}{*{20}c} a & b \\ c & d \\ \end{array}\,} \right\|$, where, iff the fractions are the generators of another fraction in the tree, then the ...

1

...But since $\delta < \epsilon$ , $\forall \epsilon > 0, \epsilon \in \mathbb Q$... This part is not right. Make sure to use the right argument with the right inequality!

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