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I have been working on proving that $(-1)^{n+1}$ = $(-1)^{(n-1)}$ by mathematical induction. Is this the correct proof?

Formula:

$(-1)^{n+1} = (-1)^{n-1}$

Chose $n = 2$. $-1 = -1$. True for $n = 2$.

Assumption:

$(-1)^{n+1} = (-1)^{n-1}$

Case n+1:

$(-1)^{(n+1)+1} = (-1)^{(n+1)-1}$

$(-1)^{n+2} = (-1)^{n}$

Divide by $-1$ on both sides.

$(-1)^{n+1} = (-1)^{n - 1}$

This is the same as the assumed formula, so the statement is proven.

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    When you do an induction, it is critically important to state what you're assuming and what you're trying to prove. In Case $n+1$ are you assuming that the statement is true for $n+1$ or trying to prove it's true for $n+1$? – 2017-02-28
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    @MichaelBurr - thank you. I have edited my post to reflect this. – 2017-02-28
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    @ΘΣΦGenSan Just because induction isn't needed, doesn't mean that it isn't a good problem to use to practice induction. – 2017-02-28
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    I understand. No worries.... – 2017-02-28

2 Answers 2

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Your proof is not quite correct.

First, you made a typo, it is $(-1)^{n+1}$ and not $(-1)^n$ at the beginning.

Then, you can not start from $(-1)^{(n+1)+1}=(-1)^{(n+1)-1}$ because you don't know at first this is true.

Instead, you have to do the reasoning in the other direction, starting from what you know:

$$(-1)^{n+1}=(-1)^{n-1}$$

and multiply by $(-1)$ to get the result for the case $n+1$.

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    Thank you for the reply. I edited my post to reflect @michael-burr's comment. Is the edited version correct? I think it reflects my original intention more clearly. – 2017-02-28
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    @Daniel It is better, but no, you still can assume that what you are trying to proof is correct as the beginning of the proof. – 2017-02-28
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    Thank you - I originally thought that is how mathematical induction works. I.E., you take statement A, prove it works for some integer, then prove it works for statement A with (n+1). If that's the case, then statement A is true. Is this incorrect? – 2017-02-28
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    @Daniel This is correct. Take a statement $P(n)$. You prove that $P(0)$ is true. Then you prove that **if** $P(n)$ is true, then $P(n+1)$ is also true. – 2017-02-28
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    @Daniel But in your case you are assuming that $P(n+1)$ is true, which you can not do since this is what you are aiming to prove. – 2017-02-28
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    So in this case, my impression is that $P(n) = (-1)^{n+1} = (-1)^{n-1}$. $P(n+1) = (-1)^{(n+1)+1} = (-1)^{(n+1)-1}$. I assumed that the first one was true, then proved the second one based on this assumption. Did I go wrong somewhere in this process? – 2017-02-28
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    @Daniel You are right. This is what you should do :) – 2017-02-28
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    Perfect - thanks for the help! – 2017-02-28
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Your assumption really should be:

Assume $(-1)^{n+1}=(-1)^{n-1}$ for all numbers up to and including $n$.

Then when you finish your induction proof, you must realize that have really proven it only for integers 2 and above (as your base case is 2--you could have chosen 0 or any other integer for your base, nothing special about 2 here). You could make your base case lower, but to prove this using induction for all integers, you have to prove things in the other direction, for number flowing off in the direction below your base case (1,0,-1,-2,-3,...), perhaps with a separate induction.

It's much more complicated than the simple way of showing this identity true for all integers in the comments above.