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Find the positive real number \(a\) such that \(\left\{a^{-1} \right\} = \{a^2\}\) and \(2 < a^2 < 3\)

Here \(\{x\}\) denotes the fractional part of \(x\) that is, \(\{x\} = x - \lfloor x \rfloor.\)

 Jul 20, 2019
 #1
avatar+8756 
+5

{ 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

 Jul 20, 2019

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