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alexandræ · that time i got chatgpt to divide by zero
that time i got chatgpt to divide by zerojanuary 13th, 2024
happy new year, hope y'all had nice holidays ! to put an end to all this 2023 stuff, let me tell y'all about something i did around five months ago — i just didn't think it was worth a blog article back then, but it's kinda funny in retrospect. so, i tried to formalize some kind of division by zero in such a way that makes all numbers invertible, while losing as few properties as possible. the way i did it was by taking a field \(K\) and just define \(K_{-1}:=K+\{0,0^{-1}\}\) where \(0^{-1}\) is defined using these properties :
- \(0^{-1}\cdot0:=1=:0\cdot0^{-1}.\) similarly, \((0^{-1})^{-1}:=0.\)
- multiplication is commutative\(\,:\,\forall\,x,y\in K_{-1},\,x\cdot y=y\cdot x.\)
- for \(x>0,\) akin to how \(0^x=0\), we have \(0^{-x}:=0^{-1}.\)
- for \(x\ne0,\) \(x\cdot0^{-1}:=0^{-1}=:0^{-1}\cdot x.\)
- there is \((K,\cdot)\)-distributivity over \((K_{-1},+),\) which implies that for \(x,y\in K,\) \(x\cdot(y+0^{-1})=\left\{\begin{array}{ll}xy+0^{-1}&x\ne0\\1&x=0\end{array}\right.\)
- for \(x,y\in K,\) \((x+0^{-1})\cdot(y+0^{-1})=xy+0^{-1}.\)
- for \(x,x'\in K_{-1},\) \(x'/x=x'x^{-1}.\)
- for \(x\in K,\) \((x+0^{-1})^{-1}=0.\)
sure, multiplication is not associative, nor is it distributive over addition. still, as long as you're careful with what you're doing, tackling division by zero this way is very much doable. there are some odd properties we can easily show ; notably, \(0^{-1}=-0^{-1}\) so \(0^{-1}+0^{-1}=0,\) even though \(2\cdot0^{-1}=0^{-1}\ne0\ldots\) yeah. chatgpt got involved when i asked of it to compute \((x+0^{-1})^n.\) i thought i'd solved this, but when i asked chatgpt, it actually found \((x+0^{-1})^n=x^n+0^{-1},\) which was different from my result (that i got using newton's binomial formula, which clearly didn't work here...), and it's by examining its proof that i realised i was wrong and it had successfully figured it out. here is its proof, translated from french :
we can prove this identity by induction on \(n.\) for \(n=1,\) we have \((x+0^{-1})^1=x^1+0^{-1}\) which is trivial. now, suppose that \((x+0^{-1})^n=x^n+0^{-1}\) for some \(n>0.\) let's show \((x+0^{-1})^{n+1}=x^{n+1}+0^{-1}.\) indeed, we have
$$\begin{array}{rrl}(x+0^{-1})^{n+1}&=&(x+0^{-1})^n(x+0^{-1})\\&=&(x^n+0^{-1})(x+0^{-1})\\&[=&\cdots]\\&=&x^{n+1}+0^{-1}\end{array}$$
where we used the properties of \(K_{-1}\) to justify each step. thus, by induction, we showed that \((x+0^{-1})^n=x^n+0^{-1}\) for all \(x\in K\) and \(n>0\).
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alexandræ
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