MHB Solving $y>0$ given $[y]^2=y\times(y)$ and Proving Inequalities

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Inequalities
AI Thread Summary
The discussion focuses on solving the inequality $y>0$ under the conditions that $[y]^2 = y \times (y)$, where $[y]$ is the integer part and $(y)$ is the decimal part of $y$. It establishes that $y$ can be expressed as $y = [y] + (y)$, leading to the exploration of proving the equation $[y] + [y + \frac{1}{2}] = [2y]$. Additionally, it examines the case where $0 < y < 2^{10}$ to prove a summation involving the integer parts of fractions of $y$. The thread emphasizes the relationships between integer and decimal parts of real numbers in the context of inequalities and their proofs.
Albert1
Messages
1,221
Reaction score
0
(1) Given :$y>0$,let $y=[y]+(y)$

where we define $[y]$ the integer part of $y$

and $(y)$ the decimal part of $y$

here $0≤(y)<1$

if $[y]^2=y\times(y)$

find $y=?$

(2) $y\in R,y=[y]+(y)$

the definition is the same as (1)

prove :$[y]+[y+\dfrac {1}{2}]=[2y]$

(3) if $0<y<2^{10}$

using (2) prove :

$[\dfrac {y}{2^1}+\dfrac {1}{2}]+[\dfrac {y}{2^2}+\dfrac {1}{2}]+[\dfrac {y}{2^3}+\dfrac {1}{2}]+-----+[\dfrac {y}{2^{10}}+\dfrac {1}{2}]=[y]$
 
Last edited:
Mathematics news on Phys.org
Albert said:
(1) Given :$y>0$,let $y=[y]+(y)$

where we define $[y]$ the integer part of $y$

and $(y)$ the decimal part of $y$

here $0≤(y)<1$

if $[y]^2=y\times(y)$

find $y=?$

(2) $y\in R,y=[y]+(y)$

the definition is the same as (1)

prove :$[y]+[y+\dfrac {1}{2}]=[2y]$

(3) if $0<y<2^{10}$

using (2) prove :

$[\dfrac {y}{2^1}+\dfrac {1}{2}]+[\dfrac {y}{2^2}+\dfrac {1}{2}]+[\dfrac {y}{2^3}+\dfrac {1}{2}]+-----+[\dfrac {y}{2^{10}}+\dfrac {1}{2}]=[y]$
hint of (1) and (2)
let $y=[y]+(y)=[y]+t=n+t ,0\leq t<1,\,\,and \,\, n\in N$
from (1)$n^2=(n+t)t$ find $n=?$ and then we get the values of $t\,\, and \,\,y$
(2)$[y]+[y+\dfrac {1}{2}]=[2y]$
we get $2n+[t+\dfrac{1}{2}]=2n+[2t]$
you have to prove $[t+\dfrac{1}{2}]=[2t], 0\leq t<1$
(3) can be easily proved from (2)
y replaced by $\dfrac{y}{2},\dfrac {y}{2^2},-------,\dfrac {y}{2^{10}}$ respectively
 
Last edited:
Albert said:
hint of (1) and (2)
let $y=[y]+(y)=[y]+t=n+t ,0\leq t<1,\,\,and \,\, n\in N$
from (1)$n^2=(n+t)t$ find $n=?$ and then we get the values of $t\,\, and \,\,y$
(2)$[y]+[y+\dfrac {1}{2}]=[2y]$
we get $2n+[t+\dfrac{1}{2}]=2n+[2t]$
you have to prove $[t+\dfrac{1}{2}]=[2t], 0\leq t<1$
(3) can be easily proved from (2)
y replaced by $\dfrac{y}{2},\dfrac {y}{2^2},-------,\dfrac {y}{2^{10}}$ respectively
solution:
(1)$n^2=(n+t)t, n\in N ,and\,\, 0\leq t<1$
we have :$t^2+nt-n^2=0,t=\dfrac {-n\pm \sqrt {n^2+4n^2}}{2}$
$\therefore n=1,t=\dfrac {\sqrt 5 -1}{2},y=1+t=\dfrac {1+\sqrt 5}{2}$
(2) prove $[t+\dfrac {1}{2}]=[2t],0\leq t<1$
(i) $0\leq t<0.5, [t+0.5]=[2t]=0$
(ii)$0.5\leq t<1, [t+0.5]=[2t]=1$
(3) from (2) $[y+\dfrac {1}{2}]=[2y]-[y]$
$y$ replaced by $\dfrac {y}{2},\dfrac {y}{2^2},-----,\dfrac {y}{2^{10}}$
$[\dfrac{y}{2}+\dfrac {1}{2}]=[y]-[\dfrac {y}{2}]----(a)$
$[\dfrac{y}{2^2}+\dfrac {1}{2}]=[\dfrac {y}{2}]-[\dfrac {y}{2^2}]----(b)$
-------
$[\dfrac{y}{2^{10}}+\dfrac {1}{2}]=[\dfrac {y}{2^9}]-[\dfrac {y}{2^{10}}]----(j)$
(a)+(b)+----+(j) we get the proof of (3)
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top