Question

Consider the following proposed proof, which claims to show that every nonnegative integer power of every nonzero real number is 1.
Let r be any nonzero real number, and let P(n) be the equation r ^n = 1.
Show that P(0) is true : P(0) is true because r^0 = 1 by definition of zeroth power.
Show that for every integer k >= 0, if P(i) is true for each integer i from 0 through k, then P(k + 1) is also true:
Let k be any integer with k >= 0 and suppose that r^ i = 1 for each integer i from 0 through k. This is the inductive hypothesis. We must show that r^ k + 1 = 1. Now
r^ k + 1 = r^ k + k - (k - 1) because k + k - (k - 1) = k + 1
r ^k * r^ k / r^ k - 1 by the laws of exponents
1*1 / 1 by inductive hypothesis
=1
Thus r^ k + 1 = 1 [as was to be shown].
[Since we have proved the basis step and the inductive step of the strong mathematical induction, we conclude that the given statement is true.]
Question:
a) Identify the error(s) in the above "proof."

Answer 1

Answer:

The statement is not true. There is an error in the following step:

$\frac{r^{k} * r^{k} }{r^{(k-1)} } = \frac{1 * 1}{1}$

Step-by-step explanation:

Here is the complete proof:

$r^{k + 1} = r^{k + k - (k - 1)}$  (because k + k - (k - 1) = k + 1)

$= \frac{r^{k} * r^{k} }{r^{(k-1)} }$ (By laws of exponent)

$= \frac{r^{k} * r^{k} }{r^{k} * r^{-1} } }$

$= \frac{r^{k} * r^{k} * r^{1} }{r^{k}} }$ (Since the power of r is a negative integer i.e. -1 it will shift to the numerator)

$= r^{k} * r^{1}$  (We get this equation after canceling out the similar terms in numerator and denominator in previous step)

$= r^{k + 1} \neq 1$ (Provided r > 1 or r < -1)