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I am currently working in my Discrete math class with elementary number theory and methods of proof. I have been given the problem $-a^n = (-a)^n$. According to the professor and the book this property is true for some integers and false for others integers. For example: Let $a=1$. Then, $-1^n$ = $(-1)^n$. Wouldn't that be true? But How can I show an example where this property is false for other integers?

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Try varying $n$ instead of $a$ – mrf Feb 19 '13 at 23:48
$-1^n = (-1)^n$, right... – Kaster Feb 19 '13 at 23:52
Yes, there is something wrong if you think that $-1^n=(-1)^n$ in general. The former is the opposite of $1^n=1$ so it is always $-1$, while $(-1)^n$ is the product of $(-1)$ with itself $n$ times. What does this give when $n=1, 2, 3$...? – 1015 Feb 20 '13 at 0:02
To be clear, the usual notation rules mean that $-a^n$ = $-(a^n)$. – Ben Millwood Feb 20 '13 at 0:08
up vote 4 down vote accepted

Yes, let a = 1. Then we can NOT say $-1^n = (-1)^n$ for all n. For example take n = 2. Well, $-1^2 = -1$, but $(-1)^2 = 1 \neq -1$. Now consider n = 3. This time it works: $-1^3 = (-1)^3 = -1$. Therefore have shown that $-a^n = (-a)^n$ only holds for some combinations of a and n.

To be more precise we can say that the equation only holds for odd $n$ (assuming n is an integer, to avoid imaginary numbers). It turns out that whether this equation holds is actually independent of a. Choose any $a$ you'd like and the equation will only hold for odd $n$.

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Is it independent of $a$? Let $n = 2$, $a = 0$. Then $0 = -0^2 = (-0)^2 = 0$. – Stahl Apr 20 '13 at 16:19
Yes, 0 is the exception. This is because the only solution to -a = a is a = 0. For all other a, though, the generalization holds. – daniel.wright Apr 21 '13 at 17:55

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