# I I think I solved divisions by zero

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1. Apr 16, 2017

### IDK10

First of all, I didn't know where to put this in general math or differential equations.

Let's start with the basic, x/0 = α. Where α is every number and decimal number from -∞ to +∞, by rearranging, we get x = 0α, x= 0, therefore only 0/0 = α. Now, we can integrate this with graphs.

Take the equations:
y=0/0,
and y=x/0, and the variant y=(x+a)/0

y=0/0:
If y=2 then there will be a horizontal line going through the points (x, 2), where x is any number on the x-axis. However, 0/0 = α, and therefore y=α, and as this means there will be infinite jorizontal lines, the entire graph will be full.

y=x/0:
Now we are dealing with a change in x. By rearrangin, we get x=0, therefore a graph of y=x/0 ≡ x=0. Another way of proving this, is that when x is greater than, or less than 0. It won't work, for example y=1/0, 0y=1, 0=1, but 0≠1, but if it is replaced with 0, y=0/0, we get the infinite vertical line from before, but is trapped at x=0 because of the change in x making other lines impossible.

y=(x+a)/0:
This time, by rearranging, we get x=-a. By using what we said before, it works the same way. For example, y=(x-4)/0. If x = 4, then we get y=0/0, therefore we get an infinite vertical line at x=4, but mot anyother line because if x/0=0 (where x≠0), it won't work.

y=x/0, is the same as y= x*1/0, and by differentiation we get dy/dx = 1/0, but 1/0 is impossible since, by rearranging, we get from 1/0=x to 1=0, but 1≠0, yet there is a gradient, otherwise it wouldn't be travelling upwards, the graphs and differentiation contradict each other.

The gradient of y=(x+a)/0 will be the same, as nothing would change to the line, apart from its translation across the x-axis. Proof:
y=(x+a)/0
x - dy/dx = 1
a - dy/dx = 0
dy/dx = (1+0)0 = 1/0.

2. Apr 16, 2017

### Stephen Tashi

State what you are trying to do. "Solved divisions by zero" doesn't say anything specific.

3. Apr 16, 2017

### Staff: Mentor

... there is no problem to solve. $0$ when it is used alongside addition and multiplication, denotes the neutral element of addition, which doesn't belong to the multiplicative structure consider simultaneously. Thus it is simply impossible to think about its inverse. If you add it to the multiplicative structure, then you consequently invent an entire new arithmetic, which you would have to define in the first place. Given the usual structure, contradictions cannot be avoided. Thus a combination of both is doomed to fail.

Beside that we don't discuss personal theories on PF.