# All Powers of *x* are Constant

We are going to prove that for any integer *n ≥ 0, (x ^{n})' = 0*, where 'prime' means, as is common, the derivative of a function; in this case,

*f(x) = x*.

^{n}*x*.)

The proof is by induction. To start with, observe that *(x ^{0})' = 0*

*x*

^{0}= 1*n = 0, 1, 2, ..., k*,

*n = k + 1*.

(fg)' = f ' g + fg'.

Take *f(x) = x ^{k}* and

*g(x) = x.*Then

*(fg)(x) = x*is the function whose derivative we need to calculate:

^{k + 1}(x^{k + 1})' = (x^{k})' x + x^{k} x' = 0,

because both derivatives above vanish according to the inductive assumption.

### References

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

|Contact| |Front page| |Contents| |Algebra| |Fallacies|

Copyright © 1996-2018 Alexander Bogomolny

### What Went Wrong?

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*,*n = 1!*.*(x ^{k + 1})' = (x^{k})' x + x^{k} x'*

*(x*,

^{1})' = (x^{0})' x + x^{0}x' = x'*n = 1*.

|Contact| |Front page| |Contents| |Algebra| |Fallacies|

Copyright © 1996-2018 Alexander Bogomolny

65110472 |