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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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