In this text I would like to prove that 0 divided by itself does not equal any complex number, with the exception of 0 and ?, although ? is not actually considered to be a number.

The proof I will be using is called proof by contradiction.

Let's say that the result of 0/0 is equal to a number we will represent as "k".

Let a equal b.

a = b
a * a = a * b
a^(2) = ab
a^(2) - b^(2) = ab - b^(2)
(a - b)(a + b) = b(a - b)
((a - b)(a + b) / (a - b)) = (b(a - b) / (a - b))
((a - b) / (a - b)) * (a + b) = ((a - b) / (a - b)) * b
(0 / 0) * (a + b) = (0 / 0) * b
k * (a + b) = k * b
k * (b + b) = k * b
k * 2b = k * 1b
(k * 2 * b / b) = (k * 1 * b / b)
k * 2 * (b / b) = k * 1 * (b / b)
k * 2 * 1 = k * 1 * 1
2 * k = 1 * k
2k = 1k

We have now proved that k is not equal to any complex number (excluding 0 and ?), since none of them multiplied by 2 are equal to themselves.

2 * 6 ? 1 * 6
2 * 5i ? 1 * 5i
2 * 4 + 7i ? 2 * 4 + 7i

The only numbers that k could equal are 0 and ? (even though infinity is not technically a number). This is because they have the following property:

x * 0 = y * 0 (True, even if x ? y)
x * ? = y * ? (True, even if x ? y. Not certain that this one is true when either x or y equal 0)

Because of the property mentioned above if k was equal to 0 or ? then the equation 2k = 1k would not be impossible.

2k = 1k
2 * 0 = 2 * 0
0 = 0

or

2k = 1k
2 * ? = 1 * ?
? = ?

Of course, k could be a new, undiscovered number. That new number though would have to have the property mentioned above.