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In Analysis this past week, we touched upon indeterminate forms which included some talk on undefined terms. One of the forms that got my attention was $0^0.$ Some of my colleagues argued that the best definition was $0^0=1$ since it made it easier to prove other theorems. The proof they gave me was this: $$1=(0+1)^1$$ $$= \binom{1}{0}\times0^1\times1^0+\binom{1}{1}\times0^0\times1^1$$ $$= 1\times0\times1+1\times0^0\times1$$ $$= 0^0.$$ Thus by Binomial Theorem and transitivity, $$1=0^0.$$

My problem with this is that it can easily be disproven by a counterexample (or so I think). Observe similarly, $$2=(0+2)^1$$ $$= \binom{2}{0}\times0^2\times2^0+\binom{2}{1}\times0^1\times2^1+\binom{2}{2}\times0^0\times2^2$$ $$= 1\times0\times1+2\times0\times2+1\times0^0\times4$$ $$= 0^0\times4.$$ Thus, $$2= 0^0\times4$$ $$\implies \frac{1}{2}=0^0.$$ But if $0^0=1$ then it follows that, $$2=4.$$ But this is a contradiction. Therefore $0^0\neq1.$

Would this work? Or did I do something wrong?

P.S. THIS IS NOT A DUPLICATE. I did check out the other link but the closest answer (which is not satisfying for me) was by celtschk. But celtschk said, "Therefore the definition $0^0=1$ is the most reasonable one" which I disagree and I have shown my reason above. If I made a mistake or am wrong, I hope someone will charitable point it out.

gen-ℤ ready to perish
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2 Answers2

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Zach Teitler pointed out the mistake. So it should actually be the following: $$2=(0+2)^1$$ $$= \binom{1}{0}\times0^1\times2^0+\binom{1}{1}\times0^0\times2^1$$ $$= 1\times0\times1+1\times0^0\times2$$ $$=0^0\times2.$$ Thus, $$2= 0^0\times2$$ $$\implies 1=0^0.$$

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You're doing Algebra, and indeterminate forms are from the realm of Analysis. This means that when one mentions $0^0$ in this context, it is never a real expression/variable which equals $0$, raised to the null power, which, strictly speaking, is meaningless, but an expression which gets smaller and smaller, raised to a power which itself gets smaller and smaller.

Bernard
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  • I disagree. It's perfectly fine to just define $0^0 = 1$. This is not problematic in the way $0/0$ is. This is convenient for algebra (as in the binomial theorem example) and we don't have to bring in any discussion of limits. But the definition will probably depend on the particular text you read. – Jair Taylor Oct 07 '17 at 22:46
  • @JairTaylor. "But the definition will probably depend on the particular context you read". No, I would say, the definition certainly depends on the context you read". Without any defined and meaningful context. $0^0$ cannot be defined to be $1$ and be done with it. – imranfat Oct 07 '17 at 22:54
  • Well, at some point this becomes a kind of sociological question. If it can cause confusion you should of course specify. But there are plenty of contexts where we tacitly assume $0^0 = 1$ and no one complains, like in any number of identities that would be false otherwise. If a particular book contains hundreds of these identities, I don't see what's wrong with clarifying on page 1 that $0^0 = 1$ thenceforth. – Jair Taylor Oct 07 '17 at 23:05
  • There's nothing wrong if it's specified from the very beginning. The most common context, nevertheless, is that of Analysis. – Bernard Oct 07 '17 at 23:09
  • @JairTaylor But isn't $0^0= 0^a \times 0^{-a} = 0^a/0^a$ which is what you said to be problematic? –  Oct 07 '17 at 23:12