# Is my definition of division by zero problematic? [closed]

I am a complete amateur, but I've attempted to define division by zero. This it's usually classified as undefined. After all, if 1/0 = b, where b is some non-zero number, then b times 0 would have to equal 1, and there exists no real number that, when multiplied by zero, equals 1. Therefore, it was evident that my definition would necessitate the creation of a new kind of number. This number, which I call Z, is defined as Z=1/0. Z is thus the reciprocal of 0.
Playing with this new number for awhile less me to discover that 2/0=2Z, and so on where n/0=nZ. This leads to a whole set of numbers, Z numbers, similar to imaginary numbers. After all, imaginary numbers were created to define something that was formerly undefined, namely, a number whose square was -1. This leads to the whole set of imaginary / complex numbers. Already I've discovered interesting facts about this number Z, such as the square of Z is equal to Z, since 1/0 times 1/0 equals 1/0.
My question is, is there some way in which my new set of numbers to define division by zero falls apart under closer scrutiny? I have yet to discover one, but I'm very new to mathematical exploration. I apologize for not being familiar with the Math font used to write equations.

## closed as unclear what you're asking by Peter, Mostafa Ayaz, Isaac Browne, José Carlos Santos, NamasteJul 31 '18 at 0:07

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• I think you quickly run into problems with your definition if you want any of the standard rules of arithmetic to apply. For example, if $Z^2 = Z$ as you say, then we should also have $Z^2 \times 0^2 = Z \times 0^2 = (Z \times 0) \times 0 = 1 \times 0 = 0$, but on the other hand $Z^2 \times 0^2 = (Z \times 0)^2 = 1^2 = 1$. – Gregory J. Puleo Jul 30 '18 at 4:14
• This doesn't seem any different from the symbol $\infty$ any entails the same difficulties. What is $Z-Z?$ or $Z/Z?$ – saulspatz Jul 30 '18 at 4:15
• @GregoryJ.Puleo Since $0=-0$ and pretending basic algebra stands $\,(-1)/0=1/(-0)=1/0=Z\,$. – dxiv Jul 30 '18 at 4:25
• You would lose associativity since $2 = 0\times(2\times Z) \ne (0\times 2)\times Z = 1$. – Rahul Jul 30 '18 at 4:30
• Why are you trying to divide by zero ? – Rene Schipperus Jul 30 '18 at 4:49

With your definition of Z, we have $1/Z= 1/(1/0) =0$
Now we get $$2/Z = 2(0)=0 =1/Z$$ Multiply by $Z$ and we get $1=2$
Similarly you can prove $m=n$ for any two integers which we do not approve of.