Examples for $ \text{If } \; x-\lfloor x \rfloor + \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor = 1 \text{, then } x \text{ is irrational.}$ I saw the answers here about how to prove:
$$ \text{If } \; x-\lfloor x \rfloor + \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor = 1 \text{, then } x \text{ is irrational.}$$
I understood the proof. But, can someone give me an example for an irrational $x$ that makes this equality true?
 A: If $x>1$ then $0< \frac 1x < 1$ so $\lfloor \frac 1x \rfloor = 0$ and the equation reduces to
$$x + \frac{1}{x} = 1 + \lfloor x \rfloor$$
Any number $x>1$ can be written on the form $x = n + r$ where $n$ is an integer and $0<r<1$. Take this form for $x$ in the equation above. Since $\lfloor n + r\rfloor = n$ the equation then reduces to
$$ r^2 + (n-1)r - (n-1) = 0$$
Solve the quadratic equation for $r$ and pick the root that is in $(0,1)$. $x = n + r$ will then be an irrational number that satisfy your equation for any value of the integer $n > 1$.
A: Clearly,$$x+\frac1x$$ is an integer, let $n$, and solving the equation for $x$,
$$x=\frac{n+\sqrt{n^2-4}}2,\\
\frac1x=n-x.$$
For large $n$, we have by the Taylor development
$$x=\frac{n+\sqrt{n^2-4}}2=\frac n2\left(1+\sqrt{1-\frac4{n^2}}\right)\approx n-\frac1n,\\\frac1x=n-x\approx\frac1n.$$
Thus the integer parts are respectively $n-1$ and $0$, and the initial equation is 
verified:
$$x-(n-1)+(n-x)-0=1$$
(actually it works for any $n>2$).
A: Hint:  If $k\gt2$ is an integer, then
$$\left\lfloor k+\sqrt{k^2-4}\over2\right\rfloor=k-1$$
