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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.

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  • $\begingroup$ Note 0 < d < i < e. Forming a geometric sequence simply means there is a $\frac di = \frac ie \implies de = i^2$ combine that with i) $e = i + d$ and ii) $i \in \mathbb Z$ and iii) 0 \le d < 1$ and you should be able to solve. $\endgroup$ – fleablood Aug 14 '16 at 19:05
  • $\begingroup$ $de = i^2$. $d^2 + id - i^2 = 0$.$d = -i/2 + \frac{\sqrt{i^2 + 4i^2}}2=\frac 12i(\sqrt{5} -1)$. So $0 \le d < 1$. $i = 0 = d = e$ is a trivial answer. $i = 1; d = \frac 12(\sqrt{5} -1)=0.61803398874989484820458683436564$ works. if $i \ge 2$ then $d \ge \sqrt{5} -1 > 1$ so that is not possible. We must have $e = d + i = \frac 12(\sqrt{5} -1) + 1 = 1.61803398874989484820458683436564$ or $e = 0$. $\endgroup$ – fleablood Aug 14 '16 at 19:17
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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?

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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 $d=r-1$.

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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.

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  • $\begingroup$ Nice hint! But I'd say rather than assume $i=1$ (we could also assume $i = 2$ or $i=3$ or whatever) we could solve for $x = d/i$. Then we have to have $i = d/x < 1/x$. Which as it turns out has only one possible value for $i$. $\endgroup$ – fleablood Aug 14 '16 at 19:43

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