> The relationship f(x)=x/x is only defined for x /= 0 and thus is not equivalent to 1.
If that's the case then:
x * f(x) = x cannot be equivalent to f(x) = x/x which breaks algebra in pretty fundamental ways (since the former would certainly be defined for 0 but the latter would not).
The second major problem is that it also breaks calculus. Let's start with a straight line: f(x) = 2x.
Now let's take the first derivative of this: f'(x) = 2x/x.
Does the line at the point where x = 0 have a slope or not? If this is discontinuous, then you have also broken calculus.
This gets as to why 0/0 is undefined, namely because when you cannot express it as a limit, and have no idea how both zeros are derived (and hence what they mean) you cannot give a specific number. You can come up with equations which for some value reduce to 0/0 but whose limits range from negative infinity all the way to positive infinity. But that doesnt mean that x/x is undefined where x = 0. x/x reduces to 1. Always. Anything else breaks higher mathematics generally.
If that's the case then:
x * f(x) = x cannot be equivalent to f(x) = x/x which breaks algebra in pretty fundamental ways (since the former would certainly be defined for 0 but the latter would not).
The second major problem is that it also breaks calculus. Let's start with a straight line: f(x) = 2x.
Now let's take the first derivative of this: f'(x) = 2x/x.
Does the line at the point where x = 0 have a slope or not? If this is discontinuous, then you have also broken calculus.
This gets as to why 0/0 is undefined, namely because when you cannot express it as a limit, and have no idea how both zeros are derived (and hence what they mean) you cannot give a specific number. You can come up with equations which for some value reduce to 0/0 but whose limits range from negative infinity all the way to positive infinity. But that doesnt mean that x/x is undefined where x = 0. x/x reduces to 1. Always. Anything else breaks higher mathematics generally.