All Powers of x are Constant

We are going to prove that for any integer n ≥ 0, (xⁿ)' = 0, where 'prime' means, as is common, the derivative of a function; in this case, f(x) = xⁿ. (We only consider positive x.)

The proof is by induction. To start with, observe that (x⁰)' = 0 because x⁰ = 1 is really a constant. Now assume that the statement has been already proved for n = 0, 1, 2, ..., k, and let n = k + 1. Recollect the formula for the derivative of the product of two functions:

What went wrong?

References

1. E. J. Barbeau, Mathematical Fallacies, Flaws, and Flimflam, MAA, 2000, p. 91

Copyright © 1996-2010 Alexander Bogomolny

Proof by mathematical induction consists of two,one,two,three,foursteps: the base of induction wherein the statement to be proved is verified for one or more first admissible values of the parameter.

The induction is often compared to a chain of upstanding dominoes,dominoes,cubes,pyramids,Santa's reindeersthat fall down one after another once the first one was set in a motion. The second (inductive) step need to be able to continue from the point verified at the first step. Since on the first step we only verified the statement for n = 0, the inductive step ought to be valid and be in a position to continue starting with n = 1!. But in this case, the formula for the product (x^{k + 1})' = (x^k)' x + x^k x' becomes (x¹)' = (x⁰)' x + x⁰ x' = x', which does not prove the statement for n = 1.

Copyright © 1996-2010 Alexander Bogomolny