Rédei László - Az euklideszi és nem euklideszi geometriák megalapozása F. Klein szerint -

 

Foundation of Euclidean and Non-Euclidean Geometries according to F. Klein, Volume 97

1st Edition

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Author: L. Redei

Editors: I. N. Sneddon M. Stark

eBook ISBN: 9781483282701

Imprint: Pergamon

Published Date: 1st January 1968

Page Count: 410 

 

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Interesanta prezentarea here
 

Anca PRECUPANU - ANALIZĂ MATEMATICĂ : Funcții Reale (1976)

 Editura Didactică și Pedagogică, BUCUREȘTI, 1976

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Equations in the Set of Positive Rational Numbers

Ecuații în Mulțimea Numerelor Raționale Pozitive

 

          Este vorba de Ecuații de/reductibile_la Tipul $a\cdot x +b=0,$ unde $a,b \in \mathbb{Q_{+}}, x \in \mathbb{Q_{+}}$. Pratic, la acest nivel analizăm ecuația de forma $a\cdot x =b$. Soluția generală este $x=\frac{b}{a}$, în unele cazuri apare cerința $x \in \mathbb{N}$.

GAZETA MATEMATICĂ Seria B N0 6-7-8/2019

 

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Lucrare de Control - MATEMATICĂ -cls a VII-a - până la ARII

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Rezolvare, 05.11.2002

CĂTĂNEȚ   Florina - clasa a VII-a A

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PROBLEM 4574 - CRUX MATHEMATICORUM vol. 46, no 8

En, p 415........................................................................................................................

 

 Fr, p 418.....................................................................................................................

 

ANSWER CiP

Equality $\Leftrightarrow x_{1}=\cdots=x_{n}=16$

 SOLUTION CiP

           LEMMA     If   $0\leqslant x <1$  then

(1)                                       $\frac{1}{1-x}\geqslant 1+4x^{2}$.

                               Equality if and only if  $x=0$ or $x=\frac{1}{2}$.

     Proof of Lemma

We perform simple calculation

$\frac{1}{1-x}-1-4x^{2}=\frac{x(1-2x)^{2}}{1-x} \geqslant 0$.

The conclusions result.

$\square$(LEMMA)

           Let $f(x)=\frac{1}{8-\sqrt{x}}$  where $0<x<64$. If we write

$f(x)=\frac{1}{8} \cdot \frac{1}{1-\frac{\sqrt{x}}{8}} \geqslant \frac{1}{8} \cdot(1+4(\frac{\sqrt{x}}{8})^{2})$

 where I applied the Lemma for $\frac {\sqrt{x}}{8}$ instead of $x$, then it turns out 

(2)                                $f(x) \geqslant \frac{1}{8}+\frac{x}{128}$.

      We put in the formula (2), one by one $x=x_{1}, x_{2}, \cdots x_{n}$ and add the obtained results.

So we have

$\sum_{i=1}^{n}f(x_{i})\geqslant \sum_{i=1}^{n}\left ( \frac{1}{8}+\frac{x_{i}}{128} \right )=\frac{n}{8}+ \frac{\sum_{i=1}^{n}x_{i}}{128}=\frac{n}{8}+\frac{16 \cdot n}{128}=\frac{n}{8}+\frac{n}{8}=\frac{n}{4}$

 
what needed to be shown.

$\blacksquare$

The equal sign occur when in (2) eqal signs occur, i.e. $\frac{\sqrt{x_{i}}}{8}=\frac{1}{2}$, so all $x_{i}=16.$

 

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Tracking Number: 10788
Received:        Mon Oct 12 16:50:40 2020

From:  Ciobanu Petre
       Scoala Gimnaziala "Samuil Micu" SADU
       Sibiu, Romania
Email: ptr.ciobanu@gmail.com

Type:  Solve a Crux (Numbered) Problem
       (problem 4574)

Files:
  yynwlsnl.pdf


Comments:
See my Blog
https://ogeometrie-cip.blogspot.com/2020/10/problem-4574-crux-mathematicorum-vol-46.html

Crux Mathematicorum <crux@cms.math.ca>	lun., 12 oct., 19:52
către eu

 =============

         Added April 7, 2021

 

           See their solution, V47n03, pages 159-160 (they give two solutions)

          The first solution starts from identity

$\frac{1}{1-t}=1+t^2+t+\frac{t^3}{1-t}$

 and for the interval $t\in ( 0,\;1 )$ we have 

$t+\frac{t^3}{1-t}\geq 3t^2\;(\Leftrightarrow t(2t-1)^2 \geq 0)$.

 N.CiP In fact, it is equivalent to inequality (1).

          

          The second solution uses that for $x \in (0, \; 64)$ we have

$\frac{1}{8-\sqrt{x}} \geq \frac {x+16}{128}\; \Leftrightarrow \sqrt{x}(\sqrt{x}-4)^2 \geq 0.$

N. CiP It is essentially the inequality (2).

= end added=

PROBLEM MA90 - Crux Mathematicorum V46N8

The statement of the problem

En (p 350)

 Fr(p 351)

ANSWER CiP

Solutions are the pairs $(x,y)=(k+1,k)\;\;and\;\; (2k+1,2k-1),\;\;\;k=1,2,\cdots$

SOLUTION CiP

          Let $z=x-y$. From

$\frac{y(x+y)}{x-y}=\frac{y(x-y)+2y^{2}}{x-y}=y+\frac{2y^{2}}{z}$

we see that $\frac{y(x+y)}{x-y}\in \mathbb{Z}$ $\Leftrightarrow$ $\frac{2y^{2}}{z}\in \mathbb{Z}$. Hence, if $z=1$ or $z=2$, we get respectively $x-y=1$, $x-y=2$.  In case $x-y=1$ we find $y=k,\;x=k+1$; in case $x-y=2$ we must $x,y$-odd numbers (otherwise $gcd(x,y)\neq1$) so $y=2k-1,\;x=2k+1$.

          Let now $z=x-y=2^{k}\cdot u$, $u$-odd number. 

$\frac{2y^{2}}{z}=\frac{y^{2}}{2^{k-1}\cdot u}$

If $k-1>0$, in order that the latter to be positive integer we must that $2 \mid y^{2}$ and $p \mid y^{2}$, where $p$ denote a prime divisor $p\geq 3$ of $u$. Hence $2 \mid y$ and $p \mid y$ and from $x=(x-y)+y=z+y$ it follow that $2 \mid x$ and $p \mid x$ which contradicts $gcd(x,y)=1$. So $k-1 \leq 0$ and $u=1$ and other solutions no longer exists.

$\blacksquare$

      Remark. The solutions are exactly those for which $\frac {x+y}{x-y}$ is a positive integer.

===========

Added April 7, 2021

          

          They notice more simply that $y$ and $x-y$ are coprime, so in identity $\frac{y(x+y)}{x-y}=y+\frac{2y^2}{x-y}$, $x-y$ must divide $2$, and so is equal $1$, or $2$.

        They express the answer 

$(x,y)=(u+1,u)\;,\;(v+2,v)$, $u-$positive integer, $v-$positive odd integer.

This coincides with our answer.

= end added=