this post was submitted on 20 Nov 2024
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A number divided by zero equals infinity.
Except if it's zero then (so 0/0) it is either undefined or any number IIRC.
If you plot out any number divided by x, as x approaches 0 the answer goes towards Infinity, yes.
When it reaches zero it ceases to be a number.
Every number divided by 0 is "undefined", and it is not undefined because we can't describe it, it is undefined because it does not exist, because you cannot divide things by 0.
Funny that you posted this in a dbzer0.com community (dbzer0 = device by zero).
You might say that if you don't divide a number the number remains itself. But you are saying to divide a number by not dividing the number.
That mathematical process does not work. One or the other must be true for the operation to happen.
You might be saying that an infinite amount of nothing can go into any something, but that is also not true. For there to be nothing, there cannot be something.
Zero is not a number in and of itself save for when it is literally the descriptor of the lack of the existence of a quantity.
Trying to divide a number by zero is like trying to divide existence by non-existence. If existence exists, then there is no non-existence to divide it with.
Therefore you cannot mathematically compute how much non-existence there is in existence.
No. The standard field (that is, a ring where both operations are abelian groups) on the complex numbers doesn't have a multiplicative inverse of 0; rings can't have a multiplicative inverse for the additive identity. You can create an algebra with a ring as a sub-algebra with such, but it will no longer be a ring. My preferred method is to impose such an algebra on the one-point compactification of the Complex Numbers, where the single added point is denoted as "Ω".
I started this project when I was 12, and when I could show that the results were self-consistent this was what I had settled on:
let z be a complex number that is not otherwise specified by the following equations. Note: the complex numbers contain the Real numbers, and so the following equations apply to the them as well.
0Ω=Ω0=1
z+Ω=Ω+z=zΩ=Ωz=Ω=ΩΩ
Ω-Ω=0. Ω-Ω=Ω+(-Ω)=Ω+(-1Ω)=Ω+Ω=0
The algebra described above is not associative. That is to say, (AB)C does not always equal A(BC).
Addendum: despite what my earlier statement implied, there is exactly 1 ring for which the additive identity has a multiplicative inverse: the trivial ring, which has only 1 element (that I will label as 0).
The operations are a such: 0+0=0=0*0. Note: this is also the only ring for which the additive and multiplicative identities are the same element.
I was originally going to mention it, but I didn't want to make my comment more complicated than in needed to be. Then I realized that the way I phrased it was technically inaccurate, and so this addendum exists.