Floor and Ceiling

{ a^-1 }  =  { a² }

 

Using the definition of the  { }  function, we can say

 

a^-1 - floor( a^-1 )  =  a² - floor( a² )

 

Now since  2 < a² < 3 ,  a  can't be  1 .  And so  floor( a^-1 )  =  0

Also since  2 < a² < 3 ,  we can say for sure that  floor( a² )  =  2

 

a^-1 - 0  =  a² - 2

                              Subtract  a^-1  from both sides of the equation.

0  =  a² - 2 - a^-1

                              Multiply through by  a  and note  a ≠ 0

0  =  a³ - 2a - 1

 

0  =  a³ - a  -  a - 1

 

0  =  a³ - a² - a  +  a² - a - 1

 

0  =  a(a² - a - 1) + 1(a² - a - 1)

 

0  =  (a + 1)(a² - a - 1)

 

a + 1  =  0	___ or ___	a2 - a - 1  =  0
a  =  -1		a  =  ( 1 ± √[ 1 - 4(-1) ] ) / 2
		a  = ( 1 ± √5 ) / 2
		a  =  (1 + √5 ) / 2	___ or ___	a  =  (1 - √5 ) / 2
		a  ≈  1.618		a  ≈  -0.618

 

We only want the middle solution because the other two violate the inequality  2 < a² < 3

 

So     a  =  (1 + √5 ) / 2