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What if we define the number next to $0$ on the real number line to be special like $0$, but not quite $0$. Is there some work done in this direction? Can someone point me towards it because I am extremely curious. But for now let's consider the number next to $0$ to be $j$ which is special abiding the following rules -

  • $\frac{j}{x} = j$ for $x \gt 1$
  • $\frac{j}{x} = 0$ for $x = 0$
  • $\frac{x}{\infty} = j$ for $x \gt 0$
  • $j + j = 2j$

Of course, some of the special properties could be different in the refined system. However, I wonder how we can prove the nonexistence of such a number. Or why doesn't such a defined system already exist?

Not sure if it's already been proven to be non-existent, but, how can we prove it's non-existence. My question also expands to the fact that, since we define the system and it's rules how can we say for certain that even with the allowance of bending those rules we cannot prove it's existence. I guess the idea here is to not think of the number next $0$ like a number but a special number/representation, for example it's extremely similar to $\infty$ just like how $\infty$ is not a number but a representation. Same questions for proving it's existence as well.

mathlander
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gonerogue
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    Something vaguely related would be the idea of the hyperreal numbers (an extension of the real numbers), studied a lot in the field of nonstandard analysis. Perhaps something to look into? – PrincessEev Dec 20 '22 at 04:23
  • We can at least show that no real number can do the job. If $\epsilon>0$ is given , $\frac{\epsilon}{2}$ is smaller and still positive. So, you need to create an exotic object like an infinitesimal. The challenge is to establish rules that fit to the rules to the real numbers. The above comment could be helpful. – Peter Dec 20 '22 at 07:25
  • this is not possible, as every $x>0$, for $x\in \mathbb{R}$, represents a limit ordinal, so AFAIK there is no way to define the successor of zero in $(0,1)$ – Masacroso Dec 20 '22 at 09:16

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