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Let $S$ be a set of real numbers satisfying the following conditions:

i. $0$ is in $S$.

ii. Whenever $x$ is in $S$ then $2^x+3^x$ is in S.

iii. Whenever $x^2+x^3$ is in $S$ then $x$ is in $S$.

How can I prove that $S$ contains at least two distinct numbers between $0$ and $1$, i.e., (0, 1)?

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By between do you mean elements in $(0,1)$ or in $[0,1]$? – Oliver E. Anderson Mar 3 '13 at 17:13
@OliverE.Anderson (0, 1) – Somebody Mar 3 '13 at 17:14
Sumac problem ..its actually cheating :P – user65141 Mar 5 '13 at 15:44
@wellwisher: please flag the questions you claim are cheating and list a source and I think the moderators will remove it! – user58512 Mar 5 '13 at 15:48
@wellwisher: This SUMaC? – Did Mar 5 '13 at 16:22

by (i) $0 \in S$

by (ii) $2^0 + 3^0 = 2 \in S$

now $2 = 1^3 + 1^2$ so by (iii) $1 \in S$

now $1 = \theta^2 + \theta^3$ where $\theta = 0.7548776662\ldots$ so by (iii) $\theta \in S$.

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You need one more as $1$ doesn't count. But it's not hard to find – Ross Millikan Mar 3 '13 at 20:09
@RossMillikan, I needed to show 1 is in S before I could show theta=0.754877... in S – user58512 Mar 3 '13 at 20:25

Use ii to show $2$ is in $S$. Then solve $x^3+x^2-2=0$ to.get two more elements, one of which is $-1$ so ii.gets you $5/6$. Solve$x^3+x^2=1$ for another.

Added: if you replace condition i by: there is a positive number in $S$, then you can use iii alone because $x^3+x^2$ is monotonic. Given $y$ in $S$ there is $y'\lt \sqrt[3]y$ in $S$ and you can keep going down until you are below$~1$

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You don't have to solve the equation, just show there is a root in $(0,1)$ and it is not $5/6$ – Ross Millikan Mar 3 '13 at 17:51
$-1$ is not a root of $x^3+x^2-2$. But it is a root of $x^3+x^2$, and so it is in $S$ because of (i) and (iii) immediately. – Greg Martin Mar 3 '13 at 19:47

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