Here is another reason, albeit a slightly more advanced one.

When dealing with real numbers, we can talk about them as a field.   One requirement to be a field is that there has to be multiplicative inverses (division), except for the additive identity (0).  In other words, every real number has a multiplicative inverse besides 0.

So when you say x / 0, what you are really saying is, "x times the multiplicative inverse of 0".  But 0 doesn't have a multiplicative inverse!  So the statement makes no sense at all!

What do you mean by unclear and clear?

In a field, the additive identity can't have a multiplicative inverse.  0 is the additive identity of the field of real numbers, so saying it has a multiplicative inverse "/0" makes the reals not a field.

Multiplicative inverse of x: what you multiply by x to get 1. (=1/x or x-¹)
Additive inverse of x: what you add to x to get 0. (=-x)

For anyone who is interested:

Prove that the additive identity can not have a multiplicative inverse in a field.

Proof: This proof is by contradiction.  Let F be a field, and assume that the addititive identity, 0, has a multiplicative inverse.  That is, 0a = a0 = 1, for some a ∈ F.  For F to be a field, it must hold that F is a ring as well.  Thus, the distributive property holds (as it must in all rings):

a(b + c) = ab + ac and (a + b)c = ac + bc

But consider the equation 0 + 0 = 0, which must hold because 0 is the additive identity.  Applying some x ∈ F to both sides we get x(0 + 0) = x(0).  And by the distributive property, x0 + x0 = x0.  Since the field must be a group under addition, every element must have an additive inverse, and this includes x0.  So x0 + x0 + (x0)-¹ = x0 + (x0)-¹, and so x0 + (0) = 0 which means x0 = 0.

The same argument can be used to show that 0x = 0.  Thus, any element multiplied by 0 is 0.  But field must also be an integral domain, which must have a non-zero multiplicative identity (called the unity).  Thus, it must be that 0 ≠ 1.  Contradiction.

∴ It must be that the additive identity does not have a multiplicative inverse.  QED.

So x0 + x0 + (x0)-¹ = x0 + (x0)-¹, did you mean x0 + x0 - x0 = x0 - x0 ?

Yes, exactly!  Subtraction is definited as adding the inverse of a number.  Just like division is defined as multiplying the inverse of a number.

Over time, I am being slowly affected by you Ricky.

I guess that's a good thing.  But don't take everything I say is fact.  Remember, I make mistakes too, and trust me, I've been known to make some great ones.

so it seems, unless you start believing in infinity/2 and that whole realm of varying sizes of infinity

There are varying sizes of infinity, if you talk about it in the right context.  For example, the set of natural numbers is infinitely large.  But it is still smaller than the set of real numbers.  And weirdly enough, even though the natural numbers are a proper subset of the integers, the integers and natural numbers are the same size.

It also seems to me that 0/0 = anything and everything.

let a be some real number,

a * 0 = 0, a = 0/0

thus 0/0 = any and every real number.  Not really a legitimate proof but its fun to pretend. :-)

Yeah I know, like I said its fun to pretend.