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Does there exist an infinite subset, $S$, of the real numbers such that both of the following are true:

  1. $\forall a,b\in S, \exists c\in S$ such that $a+b=c$
  2. $\forall c \in S, \exists a,b \in S$ such that $ab\neq c$

My suspicion is that such a set does not exist, but I'm not entirely sure. I know that the real numbers are closed under both addition and multiplication, but I'm not sure if this will help.

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    $\begingroup$ $\{n\pi\}_{n\in\mathbb{Z}}$ satisfies those conditions. (1) because $n\pi+m\pi=(n+m)\pi$. (2) because if $m\pi=r\pi s\pi$, then $\pi$ satisfies a polynomial with integer coefficients. $\endgroup$ – user530511 Feb 12 '18 at 4:09
  • $\begingroup$ With $\exists$ you can just take $\mathbf R$ in its entirety. For all $c \in \mathbf R$ take $a = c + 1$ and $b = 1$. $\endgroup$ – Trevor Gunn Feb 12 '18 at 4:13
  • $\begingroup$ No, I did mean $\exists$ for 2. $\endgroup$ – Mike Ainsel Feb 12 '18 at 4:14
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    $\begingroup$ Every infinite subset satisfies (2): if $c=0$ choose $a\ne0$, $b\ne0$; if $c\ne0$ choose $a=c$, $b\ne1$. $\endgroup$ – David Feb 12 '18 at 4:17

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