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There is a chapter in my math book about pascals triangle. In one of the problems you are supposed to prove that:

$$\sum_{i=0}^n \binom{n}{i} = 2^n$$ where $n$ and $i$ are natural numbers.

My first instinct was to just use the binomial formula:

$$(a+b)^n= \sum_{i=0}^n \binom{n}{i}a^{n-i}b^{i}$$

and let $(a+b)^n = (2+0)^n$ which gives you:

$$(2+0)^n= \sum_{i=0}^n \binom{n}{i}2^{n-i}0^{i} = \binom{n}{0}2^{n-0} 0^0+0 $$ because when $i\ge1$ the rest of the sum will be zero. The thing is that I don't know is if I am allowed to use that $0^0=1$. My math book does not have solutions so I don't know how the author indented the proof should be done. I then went on wikipedia and read that:

Zero to the power of zero, denoted by $0^0$, is a mathematical expression that is either defined as 1 or left undefined, depending on context. In algebra and combinatorics, one typically defines $0^0$ = 1.

So I guess I am allowed? My casio calculator does not allow me to do $0^0=1$ which made me more uncertain. Another way to the proof possibly could have been done is by letting: $(a+b)^n=(1+1)^n$ which would avoid using $0^0=1$. But I have not been able to complete the proof using this method yet.

volticus
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  • Thank you for the response @José Carlos Santos. They dont discuss if you are allowed to use 0^0 =1 in this proof. So it does not answer my question fully. – volticus Nov 25 '23 at 11:44
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    I have answered your question yourself, when you stated that, in the context of combinatorics, $0^0=1$. Anyway, I fail to understand the relevance that this has to to, since all you get from $2^n=(2+0)^n$ is that$$2^n=\binom n02^n=2^n.$$ – José Carlos Santos Nov 25 '23 at 11:47
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    @JoséCarlosSantos, ahh. Thank you, I see now that I did not prove anything. – volticus Nov 25 '23 at 13:23

1 Answers1

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Let's rephrase your question slightly -- to set you on the right path. You shouldn't think of "$0^0 = 1$" as something you're allowed to do: it is a definition, and in principle, you can define it to be whatever works for you. Of course, you should check that it makes sense with what you're trying to actually do with it. In this case, it's about the binomial formula. So we can just check: is the binomial formula for $(a + b)^n$ still true if we let $b = 0$, and we let $0^0 = 1$? $$ a^n = (a + 0)^n \stackrel{?}{=} \sum_{i = 0}^n \binom ni a^i 0^{n-i} = \binom n n a^n = a^n. $$ So indeed, it seems like it does work! In this case using the definition $0^0 = 1$ is perfectly fine. (More generally I think you'll find that $0^0 = 1$ is the correct definition in almost all cases. You can't make $(x, y) \mapsto x^y$ continuous at $(0, 0)$ regardless.)

However, this won't help you solve your textbook problem. The other way you suggest at the end of your question should be more fruitful.

Mees de Vries
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