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

[Albert1]

(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 :

$[\dfrac {y}{2^1}+\dfrac {1}{2}]+[\dfrac {y}{2^2}+\dfrac {1}{2}]+[\dfrac {y}{2^3}+\dfrac {1}{2}]+-----+[\dfrac {y}{2^{10}}+\dfrac {1}{2}]=[y]$

 

[Albert1]

(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)

Spoiler

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

 

[Albert1]

hint of (1) and (2)

Spoiler

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:

Spoiler

(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)