Geometric sequence from integer and decimal parts A positive number has a decimal part and an integer part.  The decimal part, the integer part, and the positive number itself form a geometric sequence.  Find this positive number.
My attempt: Let... 
*
*d=decimal part 
*i=integer part 
*e=the entire number 
Then $\frac{d}{i}=\frac{i}{e}$.  I also noticed $e=i+d$ so $\frac{d}{i}=\frac{i}{d+i}$.  But there are still too many variables and it seems I don't have enough information to solve this.
 A: HINT: Clearly $d<i<e$. Let $r=\frac{i}d=\frac{e}i$, so that $i=dr$ and $e=dr^2$. As you say, $e=d+i$, so 
$$dr^2=dr+d\;,$$
and we can divide through by $d$ to find that
$$r^2=r+1\;,$$
an equation that you can solve for $r$. Finally, you know that $i=dr$ must be an integer and that $0<d<1$. It’s possible to finish from here with a bit of persistence and just a little cleverness in working out what $d$ must look like if $dr$ is to be an integer. If you get completely stuck, check the spoiler-protected block below.

 Try $d=r-1$.

A: Hint: By dividing the $3$ numbers in the sequence by $i$ we get another sequence satisfying the conditions: $d/i, 1, 1+d/i$. So you can assume $i = 1$ to find a solution to your problem.
A: From $\frac{d}{i}=\frac{i}{d+i}$, we find that $i^2-di-d^2=0$ which implies that
$$i=d\cdot \frac{1\pm\sqrt{5}}{2}.$$
Since $e$ is positive, $i$ can not be zero (otherwise by the above equation also $d=0$). By the same reason, $d>0$. Therefore $i$ is a positive integer and $d\in(0,1)$. Thus  we have 
$$1\leq i=d\cdot \frac{1+\sqrt{5}}{2}<2,$$
that is $i=1$ and $d=\frac{2}{1+\sqrt{5}}$. Finally
$$e=i+d=1+\frac{2}{1+\sqrt{5}}=\frac{1+\sqrt{5}}{2}.$$
A famous number isn't it?
