A number theory problem

Question

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)?

@wellwisher: please flag the questions you claim are cheating and list a source and I think the moderators will remove it! — , Mar 05 '13 at 15:48
I don't think the number theory tag is appropriate, removing it. — Marc van Leeuwen, Apr 15 '13 at 10:18

score 3 · Answer 1 · answered Mar 03 '13 at 17:15

3

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) $\theta \in S$.

answered Mar 03 '13 at 17:15

You need one more as $1$ doesn't count. But it's not hard to find – Ross Millikan Mar 03 '13 at 20:09
@RossMillikan, I needed to show 1 is in S before I could show theta=0.754877... in S – Mar 03 '13 at 20:25

score 2 · Answer 2 · edited Apr 15 '13 at 10:14

2

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$~1$

edited Apr 15 '13 at 10:14

Marc van Leeuwen

115,048

answered Mar 03 '13 at 17:14

Ross Millikan

374,822

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 03 '13 at 17:51
4

$-1$ 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 03 '13 at 19:47

A number theory problem

2 Answers2