Is $(-1)^{a+b} = (-1)^{a-b}$ true $\forall a,b \in \mathbb{Z}$ ?
My argument: $$ (-1)^{a-b} = \frac{(-1)^a}{(-1)^b} = \frac{(-1)^a}{(-1)^b}.\frac{(-1)^b}{(-1)^b} = \frac{(-1)^{a+b}}{(-1)^{2b}} = \frac{(-1)^{a+b}}{1} = (-1)^{a+b} $$
Any flaw?
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2First of all, usually $(-1)^a \neq -1^a$. Fix that. – adjan Feb 03 '16 at 14:15
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@adjan I think he works with $(-1)^{a-b}$. It seems to be right... – Piotr Benedysiuk Feb 03 '16 at 14:17
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@PiotrBenedysiuk yes, that was a notation advice. – adjan Feb 03 '16 at 14:17
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1@adjan corrected – Pranav Bisht Feb 03 '16 at 14:17
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1Make it shorter: $(-1)^{a-b}=(-1)^{a-b}(-1)^{2b}=(-1)^{a+b}$. (Or $a+b$ and $a-b$ have the same parity.) – Feb 03 '16 at 14:20
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Nope, no flaw. (apart from your lack of parentheses, that's confusing). We know that $(-1)^k$ is $1$ when $k$ is even, and $-1$ when $k$ is odd. And it's trivial that $a-b\equiv a+b\mod 2$.
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Well, not working in $\mathbb{Z}_2$ though. Or may be I am not understanding the connection? – Pranav Bisht Feb 03 '16 at 14:20
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I mean that $a,b$ are from $\mathbb{Z}$, not from $\mathbb{Z}_2$. Also, +1, -1 is not in modulo 2 – Pranav Bisht Feb 03 '16 at 14:26
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I know $a,b\in\mathbb{Z}$, but only their value $\mod 2$ matters for the value $(-1)^a$ and $(-1)^b$. And actually, $-1\equiv 1\mod 2$ – Feb 03 '16 at 14:29
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I see, since whole of $\mathbb{Z}$ can be partitioned into odd, even partition classes, $mod 2$. And $a-b$, $a+b$ have same parity in $mod 2$, therefore -1 raised to powers of same parity will give equal answer. – Pranav Bisht Feb 03 '16 at 14:37
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