I think you break algebra if you start treating y = x^2/x as distinct from y = x, and the both the former and x/x have the same problem with x=0.
Edit:
As has occurred to me, you might actually have a point, in that x/x may be a continuous function, but so is 0^x, which gives you a different limit. So it may actually determine whether in context it makes sense to treat 0^0 as x/x or whether it makes sense to treat it as a limit of 0^x as x -> 0. This may be application dependent.
Edit2:
Ouch. 0^x can't be defined for negative numbers, so I am back to siding with 1 for reasons that any other answer breaks algebra.
Because if you can't simplify, you can't treat them as the same, which is a fundamental premise of algebra.
I.e. the promise/premise of algebra is that if you take an equation, it remains equally valid when you add, subtract, multiply, or divide both sides by the same expression. If you treat these as distinct, then every division operation where you divide by a variable expression reduces the domain of possible answers.
This could have some very nasty corner cases where a division operation removes from the domain the answer you are looking for, and therefore renders an equation which could otherwise be solved undefined.
Hence my simple observation that in both algebra and calculus, we treat simplify equations before determining if they are continuous or not.
Here's a trivial example:
Solve for f(x) in the following equation (x + 2) * f(x) = x^2 + 4x + 4, where x = -2.
The problem you run into is that your first step is to simplify, so you divide both sides by x + 2, and therefore you end up with a division by 0 unless you are allowed to simplify before evaluating. Otherwise, you eliminate -2 from your domain and the answer is undefined. (The answer should be 0).
> "the promise/premise of algebra is that if you take an equation, it remains equally valid when you add, subtract, multiply, or divide both sides by the same expression"
No.
The promise/premise of algebra is that an equation remains equally valid when you perform valid algebraic operations on two equal quantities.
Division is defined as the inverse of multiplication. Multiplying by zero has no inverse, and thus, it is not valid to divide by zero. If you are in a circumstance where you want to divide to solve an equation, you must check to make sure you're not dividing by zero (and write a special case for any circumstance in which you might otherwise have divided by zero.)
> "Solve for f(x) in the following equation (x + 2) * f(x) = x^2 + 4x + 4, where x = -2."
The way you've written it, f(x) is undefined for x=-2. Any value I select for f(2) makes that expression true. Try it -- if f(2)=813, 0 * 813 = 4-8+4. That solution works.
You might prefer to define f(2)=0 because it makes f(x) continuous. But you have to define that value separately; you can't solve for it from the equation, because dividing by (x+2) is not a valid operation for x=-2. You could, alternatively, choose to define f(2)=7 or f(2)=-318 or any other value. Because you've written an equation that allows it.
That's the thing about mathematics. It's very carefully defined and very precise. Your inclination is to try to simplify first, but you're simplifying in a way that is not valid for x=-2 and thereby accidentally making an incorrect statement (that f(2) "should" be zero.) If you simplify properly, you'll see what ubercow said is true -- the solution is f(x)=x+2 except at x=-2, where f(-2) can be any number and therefore you have a continuum of possible solutions.
I don't really understand what you mean. You can never divide by zero. You can't treat the functions as the same because they are not the same function. It's not a fundamental premise of algebra that you should be able to.
The corner cases aren't nasty, you have to take care of them every time. Like with your example, the solution is f(x)=x+2, except where x=-2 where f(-2) can be any number. There are a continuum of solutions to the equation each with a different value of f at x=-2.
Edit:
As has occurred to me, you might actually have a point, in that x/x may be a continuous function, but so is 0^x, which gives you a different limit. So it may actually determine whether in context it makes sense to treat 0^0 as x/x or whether it makes sense to treat it as a limit of 0^x as x -> 0. This may be application dependent.
Edit2:
Ouch. 0^x can't be defined for negative numbers, so I am back to siding with 1 for reasons that any other answer breaks algebra.