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I got into a big argument with my teacher about this. I am saying that it is undefined because every time I work it out, I end up getting $\frac{0}{0}$ which I know to be undefined.

Git Gud
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6thYearWizard
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    What do you get for $0!$ when you work it out? (Hint: $0!$ is not something you can "work out", it's given by a definition.) – David C. Ullrich Dec 20 '15 at 18:41
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    Maybe this will help $0!:=1$ – meiji163 Dec 20 '15 at 18:41
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    But it is not undefined. Why wouldn't it be not undefined? (perhaps check that triple negative in the title). – Darth Geek Dec 20 '15 at 18:42
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    Triple negations are discouraged. – Thomas Dec 20 '15 at 19:01
  • There shouldn't have been an argument. Either the teacher should have said that $0! = 1$ or the student should have listened and realized s/he was assuming $0! = 0$. Someone wasn't saying something or someone wasn't listening. Now as to why $0! = 1$ which is counter-intuitive (have sympathy; it is counterintuitive) is another story altogether. – fleablood Dec 20 '15 at 19:12
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    "Triple negations are discouraged". At least they aren't encouraged to not be avoided. – fleablood Dec 20 '15 at 19:14
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    Can you explain how you got to $\frac00$? That would mean you somehow got $0!^0 = 0$, which is not true since $0!=1$, and $1^0=1$... – 5xum Dec 20 '15 at 19:16
  • It seems easy to see that the empty sum is zero (the additive identity). The empty product is $1$ - the multiplicative identity. If you assume that all terms of the product are positive, you can convert one to the other using logs/exponentials, if that halls anyone to get why it makes sense. – Mark Bennet Dec 20 '15 at 19:32

2 Answers2

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Conventionally, $0!=1$ and $x^0=1~\forall x\neq0$. From this we obtain \begin{align} \frac{0^{0!}}{0!^0} = \frac{0^1}{1^0} = \frac{0}{1} = 0\,. \end{align} Hope this helps.

BoisterousLemma
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Hi there, you're wrong - you should apologise to your teacher

I don't know why this question has gotten so much attention, the OP needs to learn to read and not panic! There is no problem.

$0!=1$ the denominator of $0!^0$ is $1^0$ which is $1$

The numerator is $0^{0!}=0^1=0$

The only problem you could have is if you had $x^0$ with $x=0$, which I consider undefined.

Next.

Why is $0!= 0$?

There is something I was taught as "the axiom of choice" but it isn't, it has a different name, it states that:

Given a decision with m outcomes, and another with n outcomes regardless of the first decision then the number of ways to decide is exactly: $$mn$$

If you have $n$ things and you have to choose an ordering, you have $n$ choices for the first thing, $n-1$ for the second, $n-3$ for the third.... so forth. This is where the $n!$ definition comes from

How many ways are there to arrange a collection of 0 objects? One way.

There is only one way I can present you with nothing.

Alec Teal
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    I'm not sure a student should apologize for challenging a teacher. Even if the student is wrong. Now, if the student is pig-headed and never listens is one thing but to simply challenge a teacher is another. – fleablood Dec 20 '15 at 19:50
  • Thanks for sharing @fleablood – Alec Teal Dec 20 '15 at 19:57