Fake Proof: $a^n = 1$ f or all nonnegative integers $n$

Question

Find the flaw with the following "proof" that $a^n = 1$ for all nonnegative integers $n$, whenever $a$ is a nonzero real number.

Basis Step: $a^0 = 1$ is true by definitino of $a^0$.

Inductive Step: Assume that $a^j = 1$ for all nonnegative integers $j$ with $j \leq k$. Then note that $$ a^{k+1} = \frac{a^k \cdot a^k}{a^{k-1}} = \frac{1 \cdot 1}{1} = 1 $$

Where is the flaw in the above proof? My answer is that the basis step needs to iterate nonnegative integers beyond $0$, like $1$, $2$, etc.

Any thoughts?

Answer 1

$a^1=\cfrac {a^0\cdot a^0}{a^{-1}}$ and your hypothesis doesn't cover the case of $a^{-1}$.

Answer 2

Clearly the induction fails on the first step, since when you only have $k = 0$, you cannot make use of $a^{k-1} = 1$.

Answer 3

The inductive step requires $k\geq1$, because of the denominator (you are using that $a^{-1}=1$, which forces $a=1$). But you want to start your induction at $k=0$. So the base step and the inductive step are not connected.

Answer 4

2 things wrong:

1)You can not assume anything more in your induction step than was shown in the base step.

In here you are assuming that k-1 is also nonnegative. That is not the case for k = 0 so we can not assume that.

2) Even more basic. When you assume a base step is true for k, you can not assume you have shown it is true for all j <= k. You can only assume it is true for k and only k. This is important if the inductive step requires not just k but k- 1, then your base step must not merely show for k = 1 (or zero) but for k = 0 (or negative 1) as well. Or equivalently for k=1 and k = 2.

If you do show for both k =1 and k=2, your induction is sound.

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This reminds me of the proof that all the marbles in a bag of marbles are the same color.

Base case: n = 1. All the marbles in a bag of just one marble are same color. (Because there is only 1 marble so there can't be more than 1 color.)

Inductive case: Assume all the marbles in a bag of n marbles are the same color. Take a bag of n + 1 marble. Take a marble out. You now have a bag of n marbles. Therefore they are all the same color. Remove one of these marbles. Put the original marble you removed back in. You have a bag of n marbles so they are all the same color. No return the 2nd marble. It was the same color so now all the marbles in the bag of n+1 marbles are the same color.

Therefore by induction all the marbles in any size bag of marbles are the same color.