consider this, 0=0, 2 x0 =0 , 1x0=0 rhs=lhs, hence 1x0=2x0, ergo 1=2 , however 1=! 2?

You skipped the step where you divide both sides by zero. There is a reason that zero is not in the domain of possible divisions.

The division function is simply not defined in a manner that allows for that.

Edit: It isn't that the value is unknown. Every function needs a definition, even a very simple function like addition. That definition tells you what domain of input values are allowed, what range of output values result, and how the inputs are mapped onto the outputs. For division, the function simply isn't defined when the denominator is zero because there is no definition that provides internally consistent results. It could have been a function like factorials where the normal definition doesn't tell you what to do with zero, but it's consistent when 0!=1, so that's how it's defined. But it isn't like that. The function is left undefined.