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Find the mistake in the following proof that purports to show that every nonnegative integer power of every nonzero real number is 1.

Let $r$ be any nonzero real number and let the property $P(n)$ be the equation "$r^n=1$".

Basis

The property is true for $n=0$ by definition of the zeroth power.

Hypothesis

Let $k>0$ be an integer and suppose that $r^i=1$ for all integers $i$ with $0\leq i <k.$

Now

$r^k = r^{(k-1)+(k-1)-(k-2)}$

$ = \frac{r^{(k-1)} \cdot r^{(k-1)}}{r^{(k-2)}}$ by laws of exponent

$= \frac{1* 1}{1}=1$ by hypothesis

Thus, $r^k=1 $ as was to be shown.

Since we have proved the basis step and inductive step, we conclude that $r^n =1$ for all integers $n\geq1$.


I know it's obviously wrong, but the logic looks so convincing. Nonetheless, my guess is that it is misusing the laws of exponents; however, I don't know how to explain the deception of the proof in thorough detail.So, how is it wrong, and where?

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    $\begingroup$ What happens when $k=1$? $\endgroup$ – Michael Burr Aug 17 '15 at 1:58
  • $\begingroup$ @MichaelBurr I know that you could uses when n=2 as a counter example, but the goal of this problem is to find where the mistake is. $\endgroup$ – user261954 Aug 17 '15 at 2:01
  • $\begingroup$ To rephrase @MihirSinghal's answer: $k=0$ does not imply $k=1$ in your inductive argument and the reason is exactly as Alex noted. $\endgroup$ – Cameron Williams Aug 17 '15 at 2:14
  • $\begingroup$ If the result were true when $k=1$, then the induction would work. If you try to prove the case where $k=1$ using the induction, then you'll find that you don't have the right assumptions to use the inductive step. $\endgroup$ – Michael Burr Aug 17 '15 at 2:24
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You need two base cases ($n=0$ and $n=1$), since you're using $r^{k-2}=1$. The $n=1$ case is clearly false, so the proof is wrong.

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  • $\begingroup$ So in strong induction we have to check for two base cases? Because in my text book it says let a and b be fixed integers with $"a\leq b"$ Is it possible to have a=b, as in this "proof" a=b=1. Basis: P(a), P(a+1),...,P(b) are true. $\endgroup$ – user261954 Aug 17 '15 at 2:07
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    $\begingroup$ When $n=1$, you cannot use this proof because $n-2=-1$ is not in the range of consideration. You do not always need two base cases for strong induction, but this subtlety in the proof you give here requires it. $\endgroup$ – Alex S Aug 17 '15 at 2:11
  • $\begingroup$ This is just repeating Alex's comment in different words, but I hope it might help anyway. Your induction hypothesis was, as you said, $r^i=1$ for integers $i$ in the range $0\leq i<k$. You then applied this for two values of $i$, namely $k-1$ and $k-2$. So you need $0\leq k-1<k$ and $0\leq k-2<k$. For both of those to be true, you need $2\leq k$. So your induction step won't work when $k=1$. Specifically, when $k=1$, then the value $k-2$ of $i$ that you want to use in the induction hypothesis is not in the available range $0\leq i<k$ because it's $-1$. $\endgroup$ – Andreas Blass Aug 17 '15 at 3:27

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