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I'm not a professional mathematician (nor anything remotely close to that), but for sometimes I've wondered about this.

Would it be possible to magicate some strange mathematical concept (lets call it $ S $), if that's even the correct word, such that:

$$ S \times 0 = 1 $$

Or in other words:

$$ S = \frac{1}{0} $$

I'm imagining something which at first looks absurd, such as the imaginary unit $ i = \sqrt{-1} $, which nonetheless works very well in mathematics and in the real word.

My sparse math skills can't take much further than some basic and sometimes bizarre properties such as the fact that $ S $ seems to be completely indiferente to addition/subtraction, division and exponentiation for the natural numbers:

$$ n \in \mathbb{N} $$

$$ S + n = \frac{1}{0} + n = \frac{1 + n \times 0}{0 \times 1} = \frac{1}{0} = S \Rightarrow S + n = S $$

$$ S - n = \frac{1}{0} - n = \frac{1 - n \times 0}{0 \times 1} = \frac{1}{0} = S \Rightarrow S - n = S $$

$$ \frac{S}{n} = \frac{1}{0} \times \frac{1}{n} = \frac{1 \times 1}{0 \times n} = \frac{1}{0} \Rightarrow \frac{S}{n} = S $$

$$ S^n = \frac{1^n}{0^n} = \frac{1}{0} = S \Rightarrow S^n = S $$

For multiplication it's a bit more complicated, but the result is the same and it doesn't violate any of the above (I think). Illustrating for $ n = 2 $:

$$ 2 \times S = S + S = \frac{1}{0} + \frac{1}{0} = \frac{1 \times 0 + 1 \times 0}{0^2} = \frac{1/S}{1/S} = \frac{S}{S} \Rightarrow S = \frac{S}{S} - S = S \Bigl(\frac{1}{S} - 1\Big) = S (0 - 1) = S \times n = S \Rightarrow S \times n = S$$

Would this idea break math? I feel it's the most likely case, but I'm not skilled enough to pursue this much further than the above mentioned basics. Anyway, for me it's an interesting thing to think about, and I'd love the support of the community to come up with ways in which this idea fails or violates it's own rule-set in some manner.

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    If S=0=1 there is no problem. x=0=1 for every x, and x+y=0=1 and xy=0=1 for any value. It's just very very boring. – fleablood Sep 27 '17 at 20:35
  • I'm also not an expert in algebra, I think anyway that your first propositions ($S+n=S$, $S^n=S$) are consistent, but that the computation of $S+S$ has a flaw. – Tom-Tom Sep 27 '17 at 20:36
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    Actually https://math.stackexchange.com/q/26445/29335 probably answers the question completely, that or any of the $\aleph_0$ duplicates linked to it. – rschwieb Sep 27 '17 at 20:37
  • You have two options. The extended reals include $\infty $. But it is not a ring and doesn't follow algebra. (Which doesn't "break" anything.) Or we can have a trivial ring where there is only one element in existence. Which also doesn't "break" anything. – fleablood Sep 27 '17 at 21:07

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