Here’s a “correct” statement on the uniqueness of zero:
There exists a number 0 in the set of real numbers such that for all real numbers x we have: 0x=x0=0. Assume we have another number in the set of real numbers, say that number is a, such that for all x in the set of real numbers ax=xa=0. Then we have 0a=0 and we have 0a=a therefore 0=0*a=a and 0=a hence 0 is unique.
For why you can’t divide by zero:
For all x,y,z in the set of real numbers we have x(y+z)=xy +xz. Now, assume for the sake of contradiction that 0/0 in the set of real numbers exists, let a denote this number.
Consider 10=1(0+0)=10+10. then we have 1*(0/0)=1*(0/0) + 1*(0/0) or 1a=1a + 1*a since a is a real number we have a/a=1 so 1=1 + 1 which can not be. Hence there does not exist a in the set of real numbers such that a=0/0.
On a side note showing that -2 has not real square root is easy but showing that the square root of 2 is a real number is not so easy. It is not sufficient to state that the square of the square root of 2 is real thus the square root of two must be real, because certainly -2 is real and it’s square root is not. If you wanted to rigorously show that every positive real number has a real square root you actually need a rather involved proof. I don’t really remember exactly how it would be shown but you should be able to show it using an infinite series of rational numbers Xm/Yn with m,n real numbers, and you want to construct the series in such a manner that the limit is the square root of two. You actually don’t need the concept of a limit to show it you only need something a little weaker, the concept that every set bounded from has a least upper bound.
But as I said I don’t remember it exactly, analysis is not my strong point in math. If anybody else actually knows how it’s shown I would like to see it if you have the time.