Divide by 0 != universe explode

Post inspired by: The Last Denominator

I just want to let people know that there is no way that any (true) theorem about the field of real numbers (or any field) has a division by zero. In order to get to the concluding statement would require an element that is 1/0 (which I will call the inverse of zero) to be summoned. Since this element does not exist in a field other than the zero ring (in which case 0 = 1), it cannot be used. The step taken is invalid.

Now that I have shown that the usage of zero's inverse is invalid, I will show that no true statement can imply a theorem that involves zero's inverse. By the axioms of inference that logicians, philosophers and mathematicians have developed, A implies B is equivalent to (NOT A) OR B. Therefore since B = conclusion with zero's inverse, if A is true, then the implication is false. And so we conclude that only false statements can imply the existence of zero's inverse (with the exception of the extremely uninteresting case of the zero ring).

To those of you who do not know to what I refer to as a ring and field, here is the break down (first we need the concept of a group):
A group G is a set of elements with a very limited amount of structure imposed on them under one operation (call it *):
1) There exists an identity element, 1, such that 1*x = x for each element x in G.
2) There exists an inverse element 1/x such that 1/x * x = 1 for each x in G.
3) For two elements in G, x and y, x*y is also in G.
4) For three elements in G, x,y and z, (x*y)*z = x*(y*z). In other words, * is an associative operation.

A ring R is simply a set of elements that have a certain structure under two operations defined among the elements (call them + and *):
1) R is an abelian group under + (abelian just means that for x,y in R, x + y = y + x. In other words, + is commutative).
2) For two elements x, y in R, x*y is in R.
3) The distributive laws hold: x,y,z in R imply x(y + z) = xy + xz, and (x + y)z = xz + yz.

A field F is a special kind of ring that really has a lot of great qualities that allow us to prove a lot of cool things:
1) F is a ring.
2) All elements of F except the identity of + form an abelian group under *.

The "except the identity of +" is the key term. Since we commonly call the identity of +, 0, this means that 0 is not required to have an inverse (with regards to *) in F. We know that the real numbers form a field because every element other than 0 has a multiplicative inverse (namely 1/x), and every element has an additive inverse (-x), and you can't add or multiply two numbers and get an answer that isn't a real number. Since 0*x = 0 for any x in the reals, 0 does not have an inverse z such that 0*z = 1. Thus 0 does not have an inverse in the real numbers.

Okay? Now you can be sure that if you get a division by 0 that either the number is too small for your calculator to keep the correct precision and it just rounded to 0, you started off with a false statement, or you had an incorrect deductive step.