Mathematical Writing - Vivaldi Franco 2014
Mishandling Functions
Forms of Argument
Many things can go wrong when we deal with functions. For instance, we may apply a function to arguments outside its domain (e.g., forming the logarithm of zero), or invert a function incorrectly (e.g., taking the wrong sign of a square root).
We begin with a proof from calculus which isn’t correct, even if it captures the essence of the argument. The chain rule for differentiating the composition of two functions states that if and are differentiable, then
SLOPPY PROOF. Let be a real number. We compute
Letting tend to zero gives the desired result.
The right-hand side is obtained by multiplying and dividing by . But this quantity may well be zero for non-zero (e.g., if is constant), which invalidates the argument. A valid proof requires more delicacy.
PROOF. Let . Since is differentiable at , we write
where is a continuous function with . Specialising to , we find
As tend to zero, the last term tends to zero, being the product of a function that tends to zero and a function that tends to a finite limit (because is differentiable). This gives the desired result.
Mishandling a non-invertible function may have unpredictable consequences:
WRONG THEOREM. Every negative real number is positive.
WRONG PROOF. Let be a negative real number. Then
as required.
What’s wrong with this proof? The function is not invertible, and to invert it we must restrict the domain. If we restrict it to the range and take the positive sign for the square root, then the functions and are inverse of each other: . However, if is negative, these functions are inverse of each other only if we take the negative sign of the square root.
Thus the error occurred in the last equality
where, having squared a negative quantity, we mistakenly choose the positive sign of the square root. Exercise 7.3.2 deals with a more general instance of the same phenomenon.