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.
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.