+170 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, \(\{x\} = x - \lfloor x \rfloor.\)
+9468 { 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
