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I have been having an argument with a friend, and they claim $-8^0$ is $1$, and I claim that $-8^0$ is really $-(8^0)$ and is therefore $-1$. Who is right?

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4 Answers 4

Neither of you is right - or at least, you are both half right.. $$-8^0=-8^{1/2-1/2}=\frac{-8^{1/2}}{-8^{1/2}}$$ Now we compromise so that everyone gets an equal chance: $$\frac{-2\sqrt 2}{2\sqrt 2 i}={-1\over i}=i$$ So the answer is $i$. The imaginary unit. The enigma of the conundrums. The unspeakable number.

I have divulged this information at great risk to our agents in the field. Please use this remarkable knowledge with discretion, and always remember: use more parentheses.

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I just want to say i love this answer, for its simultaneous insanity and reason –  Asimov Apr 22 at 1:44
    
Um... please explain how you are not assuming the friend's approach from the second equality (in the first line) of this answer? –  anorton Apr 22 at 2:21
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@anorton: You're right, he really should've used the compromise approach there too, and declared that $-8^{1/2-1/2}=-\frac{-8^{1/2}}{-8^{1/2}}$, which finally leads to the conclusion that $-8^0 = -i$. But that's fine, since it's really just another branch of the same multivalued function. ;-) –  Ilmari Karonen Apr 22 at 3:20
    
This is not correct... There is only one solution to this. Any expression of the form $a^{\frac{b}{c}}$ has $c$ solutions in the complex plane... However, when $b = 0$, like in this case, this will always equal $1$ because $a^{\frac{b}{c}} = \left(a^{\frac{1}{c}}\right)^b$ (assuming $a^{\frac{1}{c}} \ne 0 \Leftrightarrow a \ne 0$). $\forall a^{\frac{1}{c}} \ne 0$, $\left(a^{\frac{1}{c}}\right)^b$ = 1. Since $8^{\frac{1}{c}} \ne 0$ for all $c \in \mathbb{R}$, this outcome is not possible. –  stonebrakermatt Apr 22 at 3:29
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Whoooooooooosh. sound of point going over heads –  blue Apr 22 at 21:31

Exponents take precedence over multiplication. Your friend is wrong.

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So by process of elimination, user144807 is correct. –  SimonT Apr 22 at 3:03
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It should be noted that this is simply a convention -- there's no mathematical reason why it would have to be so. Nonetheless, this convention is pretty much universally accepted, so that any mathematician would understand $-a^b$ as $-(a^b)$, not as $(-a)^b$. It does have the convenient advantage that it lets us write polynomials without having to stuff them full of parentheses whenever a coefficient happens to be negative. On the other hand, plenty of formulas involving alternating sums like $\sum_k (-1)^k f(k)$ would get simpler if it were the other way, so... –  Ilmari Karonen Apr 22 at 3:32
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@SimonT We would have to eliminate all other possibilities to use the process of elimination, and $-8^0$ could be anything. –  Potatoswatter Apr 22 at 4:32
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I've actually had someone once tell me their calculator was broken, showing me $-4^2 = -16$ on the screen and saying that it was wrong because "the sign doesn't matter when squaring" etc etc... I added the brackets and gave the calculator back, and he understood. –  Thomas Apr 22 at 7:45
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"precedence over multiplication" So you consider negation a kind of multiplication? One could have chosen a convention where $-a^b$ meant $(-a)^b$ and still have multiplication lower in precedence, such that $-ca^b$ still meant $-(c(a^b))$ or equivalently $(-c)(a^b)$, not $(-(ca))^b$ or $((-c)a)^b$. –  Jeppe Stig Nielsen Apr 22 at 21:05

In general, operations are done in this order:

  1. Brackets
  2. Exponents
  3. Division and Multiplication
  4. Addition and Subtraction

For example, $10 + 0 \times 5$ isn't read as "$10 + 0 = 10$, $10 \times 5 = 50$", it's read as "$0 \times 5 = 0$, $10 + 0 = 10$".

Your example is slightly trickier as you have to recognize what the minus sign really means. "$-n$" is essentially shorthand for $0-n$, and realizing that makes the answer clear.

Your equation is "really" $0-8^0$. Looking at the list from earlier, we see that exponents are done before subtraction, so that step comes next. $0-8^0$ becomes $0-1$, which we write as $-1$. You are correct.

Really, though, the question is less than ideal. Writing it as $-(8^0)$, as you say, would clear up a lot of problems. It's very easy to misread or misunderstand a problem and steps clearly were never taken to avoid that here.

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One might also say that $-n$ is shorthand for $(-1)n$, which of course does not change the answer. –  jpmc26 Apr 22 at 21:18
    
@jpm I can't think of any situations where $0-n \neq (-1)n$ (for $n \in \mathbb{R}$, at least) so having different ways to think about it is always good. :) –  undergroundmonorail Apr 22 at 22:10

You are correct: without parenthesis around the (−8), the formula −80 is equivalent to (−1) * (80).

You trying to determine whether the equation, −80 = x, like −23 = x, will result in x being positive or negative. 1 − 23 = −7 can be written as 1 + −23 = −7 or 1 + (−1) * (23) = 7. Clearly the negative addition of terms is separate from the result of the exponent.

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