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.
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, $$2= 0^0\times2$$ $$\implies 1=0^0.$$