[lotharbot]

> "x * f(x) = x cannot be equivalent to f(x) = x/x"

Sure it's equivalent, over a domain not including x=0. This does not break algebra any more than, say, restricting the domain of the square root (when working in the reals) to non-negative numbers. We work in restricted domains in mathematics all the time.

> " f'(x) = 2x/x"

f'(x) = lim (h->0) [2(x+h)-2x]/h. Since h is approaching (and therefore not equal to) zero, there is no problem. Any appearance of 0/0 in the problem is a result of an attempted (but unsuccessful, that is, indeterminate) evaluation -- it's not actually 0/0, it's 2h/h where h is close to but not equal to 0.

We don't need to define 0/0=1 in order to have either algebra or calculus work. We choose to define 0/0=1 in certain circumstances which make certain calculations go more smoothly, and we choose not to define 0/0 in other circumstances where it's either unnecessary or potentially misleading.

[einhverfr]

> This does not break algebra any more than, say, restricting the domain of the square root (when working in the reals) to non-negative numbers.

Sure it does, because if that is the case, you restrict your domain when you divide by a variable expression. If you divide both sides by x-1, then you effectively rule out 1 from the domain.

That's the problem.

Now this is not the same as 0/0. The point is that 0/0 is only undefined when it persists after simplification and only because you can't define a relationship between the two zeros.

I.e. 0/0 is undefined because 2x/x, 52x/x, and x^2/x give you three different answers as x->0. That doesn't mean that every function which has not been reduced and can transiently evauate to 0/0 is treated as non-continuous.

Regarding derivatives, this highlights the problem because to solve the first derivative of a variable to a simple exponent, you multiply by the exponent and divide by the variable (x^2 becomes 2x, 2x becomes 2, and so forth). What this means is that you may be dealing with a limit but the limit defines a function, which is something like 2x^2/x for the derivative of x^2 and 2x/x for the second derivative.

Unless you allow simplification before determining whether the function is continuous, these things don't make sense. If you allow reduction first, then x/x^2 is undefined where x = 0, but x/x is not, because you can reduce it to 1 before applying any further logic. Both may appear to evaluate to 0/0 however.

There are a huge number of things that seem to break in algebra and calculus if one treats x/x as non-continuous and undefined. The simpler solution is to allow reduction to 1 before determining that it is undefined. (of course 52x/x would reduce to 52 instead, again showing why 0/0 is oversimplifying the problem).

[lotharbot]

> "you restrict your domain when you divide by a variable expression"

Why is this a problem?

Whenever you perform an operation that has a restricted domain, you restrict your domain. This may result in an actual "not defined at x=1" result, or simply "the value at x=1 is found through an alternative method" result.

> "0/0 is only undefined when it persists after simplification"

When you're working in the context of limits, it wasn't an actual 0/0 to begin with; it was near-0/near-0, which is perfectly OK to simplify. The limit defines a function that already has a restricted domain -- h->0 means h is not actually zero. The expression naively evaluating to 0/0 simply tells you that you need to do more work to properly evaluate it -- 0/0 is not the actual result.

Note that using the limit to find the derivative gives you a function that you'd like to be continuous in x, but the divide-by-zero is in h. Consider f(x)=x^2. The derivative is

lim h->0 [(x+h)^2 - x^2 ] /h

lim h->0 [ x^2 + 2xh + h^2 - x^2 ] /h

lim h->0 [ 2xh + h^2 ] / h

lim h->0 [2x + h] * h/h

since h does NOT equal zero, we can treat h/h=1, and the limit trivially collapses to 2x. Note that we never had the variable x in the denominator of our fraction; we never placed a restriction on x or suggested anything about a discontinuity relative to x. We only restricted h, which was already restricted by the limit itself.

[ndeine]

The expression 52x/x has a restricted domain as well (x /= 0). But that's normally not what you mean when you write it. It isn't often you really care about expressions like 52x/x; they are generally just intermediate steps in getting to a real solution.

For example: I have done a lot of work on some equation that is interesting to me, and finally I have reduced it to 5+yx=52x+5. Now obviously the rules of algebra let me subtract 5 from each side and be left with yx=52x, and this subtraction also has no effect on the domains for which our variables may be defined. All is well.

But dividing out the x is what we are concerned with now. Surely y=52 is a solution to the equation - why can this not be true for all values of x?

Well, for nonzero x we have y=52 and nobody will complain. For x=0, though, solving for y is problematic. Note that if x=0, y could be 1, or 33, or any number. If there is some function f such that y=f(x), then it follows that f(x) holds a unique value y for each input of x/=0, but for x=0 we cannot know what y might be; this is what we mean by undefined. Thus we say the domain of f(x) is the set of all real numbers x, such that x is not equal to zero.

If you have been told otherwise, or even gotten away with doing algebra or calculus under the assumption that the domain of our function f may include zero, you are taking a mathematical shortcut rather than performing formal analysis. It is not calculus nor algebra that is broken by saying f is undefined for x=0, but rather your (albeit practically useful) misconception of these systems.

I'll finish with some formal rules of algebra, to hammer this in:

- [P6] Existence of a multiplicative identity: a * 1 = 1 * a = a ; 1 /= 0.

- [P7] Existence of multiplicative inverses: a * a^(-1) = a^(-1) * a = 1, for a /= 0.

These are taken from page 9 of Spivak's Calculus, 3rd edition. He goes on to build the foundations of all of calculus from rules like these. Surely he would not present this as a fundamental axiom of his system, only to immediately (and silently) reject it and build a flawed calculus instead!

Indeed, on pg. 41, when defining functions, Spivak later writes (emphasis his):

> It is usually understood that a definition such as "k(x) = (1/x) + 1/(x-1), x /= 0, 1" can be shortened to "k(x) = (1/x) + 1/(x-1)"; in other words, unless the domain is explicitly restricted further, it is understood to consist of all numbers for which the definition makes any sense at all.

[tomp]

> Sure it does, because if that is the case, you restrict your domain when you divide by a variable expression. If you divide both sides by x-1, then you effectively rule out 1 from the domain. > That's the problem.

That's the problem that a mathematician must handle. The solution to the equation

  x * f(x) = x

is very simple: x is either 0 or such that f(x) = 1.