joi, 29 octombrie 2020
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}.
miercuri, 28 octombrie 2020
luni, 26 octombrie 2020
luni, 12 octombrie 2020
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.
Raspunsul de confirmare

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Received: Mon Oct 12 16:50:40 2020
From: Ciobanu Petre
Scoala Gimnaziala "Samuil Micu" SADU
Sibiu, Romania
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Type: Solve a Crux (Numbered) Problem
(problem 4574)
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See my Blog
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| lun., 12 oct., 19:52 | ||
|
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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=
marți, 6 octombrie 2020
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=