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I am looking for solutions for $x^2 = 0$ where $x \neq 0$.

I understand this has no solutions in familiar number systems (by the zero-factor property).

I've googled on numerous occasions, but the search results are always related to simple math. The closest I've come to finding a solution is from the wikipedia entry for the zero product rule, which gives examples of number systems where the product of the additive identity does not necessarily equal itself.

Does anyone know of number systems with solutions to this equation? Could anyone point me to educational resources related to those number systems?

Jafego
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    It really depends on what you mean by "number system" since the definition of a field precludes this outright, but certain rings do have this property for nonzero $x$. – Derek Allums May 16 '22 at 09:03
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    For matrices, you can find matrices that are nilpotent with degree 2, but I don't see how you can have a number system with such solution. We do have Infinitesimal calculus where terms such as $x^2$ and higher are regarded as zero, but I don't think the number itself is regarded that. For example you can read John Bell's book: https://www.amazon.com/Primer-Infinitesimal-Analysis-John-Bell/dp/0521887186/ref=sr_1_2?crid=DN4UJJI8Q5CC&keywords=john+bell+infinitesimal+calculus&qid=1652691816&sprefix=john+bell+infinitesimal+calculus%2Caps%2C259&sr=8-2 – MathematicalPhysicist May 16 '22 at 09:04
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    Adding to the above comments: an element in a ring satisfying $x^n=0$ for some $n$ is called nilpotent. There are common examples of rings with nilpotent elements of varying degrees (the degree of a nilpotent element is the smallest $n$ for which the above is satisfied) – Mor A. May 16 '22 at 09:05
  • A nice large class of examples is "the integers modulo $n$, where $n$ is not squarefree". – Izaak van Dongen May 16 '22 at 09:07
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    I'd suggest looking up Dual numbers. – Oria May 16 '22 at 09:13
  • See dual numbers for universal examples. – Bill Dubuque May 16 '22 at 11:03

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Consider the commutative ring $\Bbb Z/25\Bbb Z$. Then $x^2=0$ has five solutions, namely $$ x=0,5,10,15,20. $$ The ring $\Bbb Z/n\Bbb Z$ is a field if and only if $n$ is prime. The equation $x^2=0$ has a nontrivial solution if $n$ is not squarefree.

Dietrich Burde
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Let $n\geq2$ be an integer, and consider the ring $\mathbb{Z}_{n^2}$ (the integers modulo $n^2$). Then clearly $n<n^2$, so $n$ is not the zero element in $\mathbb{Z}_{n^2}$, but

$$n^2\equiv0\pmod{n^2},$$

now $n^2$ is zero in $\mathbb{Z}_{n^2}$.

Lorago
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