A positive number, it's integral part and decimal part form a geometric progression. Then the greatest integer less than or equal to the number is?

Question

Let the number be $x$.
Integral part of the number $=[x]$
Decimal part of the number $=x-[x]$

Now as per the question $x,[x] $ and $x-[x]$ form a geometric progression. So :-
$[x]^2=x(x-[x])$

Now I am not able to proceed from here. How can we solve this equation? Please help !!!

Thanks in advance !!!

Answer 1

Let $[x]=rx$ and $\{x\}=r^2x$, adding them we get

$$[x]+\{x\}=x(r^2+1)$$$$\implies x=x(r^2+r)$$$$\implies r^2+r=1$$$$\therefore r=\frac{\sqrt 5 -1}{2}$$

This means that $x$ has to be in the form of $\frac{2n}{\sqrt 5 -1}$ where $n$ is a positive integer, since $[x]=rx$. This means that $[x]=n$. But again, $\{x\}=r[x]$. $$\therefore r[x] < 1$$ Substituting their values, we get

$$n < 1.618$$ or $n=1$ since $n$ is an integer.

So $$x=\frac{2n}{\sqrt 5 -1}=1.618$$

Answer 2

Let the ratio be $\alpha$ then $\alpha x=\lfloor x\rfloor$ and $\alpha \lfloor x\rfloor = x-\lfloor x\rfloor$ gives us $\alpha^2 x=x-\alpha x$. Using the quadratic equation, the solution for $\alpha$ is $x$ times the golden ratio, $\phi=1.61803...$, or its inverse $1/\phi=0.61803...$, and the integer requirement for $\alpha x$, and the requirement for $\alpha^2 x<1$ because of the decimal, gives us $x=\phi$, thus the sequence must be $\phi, 1, {1\over \phi}$. The greatest integer part of the question is left as an exercise for the reader.