what are the consequences of using ${-0}$ in exponentiation?

Question

mainly in regards to $10^{-0}$, $-10^{-0}$, $0^{-0}$, $-0^0$ and $-0^{-0}$

i have have tried wolframalpha but i can't get it to interpret ${-0}$ as ${-0}$, it just interprets it as ${0}$

Answer 1

As noted in the comments, the reason Wolfram Alpha interprets $-0$ as $0$ is because $-0$ is $0$. They are one and the same.

$0$ is the identity element of addition: the number such that for any other number $a$, $a + 0 = 0$. There can be only one identity element. Because if $e$ also has this property, then $e + 0 = 0$ by $0$ being an identity, and $e + 0 = e$ by $e$ being an identity. So $e = e + 0 = 0$. They are the same number.

For any number $a$, the number $-a$ is by definition the number such that $a + (-a) = 0$. Again, for a given $a$, there is can be only one such number. For if $a + b = 0$ and $a + c = 0$, then $$c = 0 + c =(b + a) + c = b + (a + c) = b + 0 = b$$So every number has a unique opposite.

But $0 + 0 = 0$, which means that $0$ is the unique number which when added to $0$ results in $0$. By the definition of "$-a$" given above, that means, $-0 = 0$. They are the exact same number.

So

• $10^{-0} = 10^0 = 1$.
• $-10^{-0} = -(10^{-0}) = -(1) = - 1$, because exponentiation has precedence over negation as operators.
• $(-10)^{-0} = (-10)^0 = 1$ because any real number raised to the power of $0$ is $1$ (Okay, some people have a snit about defining $0^0 = 1$, since $0^x = 0$ for all $x > 0$. But I won't get into why they are so, so wrong about that here.)
• $0^{-0} = 0^0 = 1$
• $(-0)^0 = 0^0 = 1$
• $(-0)^{-0} = 0^0 = 1$