Is $(1-p)^0$ equal to $1$ or $-1$? [closed]

Context:

I would like to proof that $C^2_0 p^0 (1-p)^2 + C^2_1 p^1 (1-p) + C^2_2 p^2 (1-p)^0 = 0$. When I do derivation for the equation, I stopped in the following point: $(1-p)^2 + 2p(1-p) + p^2$

I would like to find out the output of the first term, $(1-p)^0$, is it 1 or -1?

Ibrahim

closed as off-topic by Aqua, Martin Sleziak, B. Goddard, AugSB, GNUSupporter 8964民主女神 地下教會Jan 9 '18 at 19:08

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• I have added the context. Would be possible to remove the on hold? –  Ibrahim EL-Sanosi Jan 11 '18 at 12:48
• Hi Ibrahim, I voted to reopen, but you will get a better answer if you explain (in the question, not in comments) why you think it might be -1. – Matthew Towers Jan 11 '18 at 13:47

$$x^0=1\space(\forall)\space x\in\Bbb{R}^*$$

I did say $\Bbb{R}^*$, because $0^0$ is not unanimously agreed upon or might be context-dependent.

• $0^0$ is less than context dependent, it can only be expressed as the limit of some function of x and thus defined as whatever that limit is. It’s exactly like $\frac00$. – Marcus Aurelius Jan 9 '18 at 17:33
• @MarcusAurelius: There is a lot of discussion of this topic on this site. Often, $0^0$ is defined to be $1$ with the caveat that $x^y$ is not continuous at $(0,0)$. – robjohn Jan 9 '18 at 17:44

For every $x$, $x^0$ is $1$ except for $x=\infty$.

• Yes, it makes sense, it is 1, because p is probability and p is always <=1. –  Ibrahim EL-Sanosi Jan 9 '18 at 17:14

For all $1-p>0$ it's $1$, otherwise it's not defined.

• Why is it undefined for negative numbers? – John Doe Jan 9 '18 at 17:08
• $$1=k^n\cdot k^{-n}=k^0$$This doesn't have any specific requirement on the sign of $k$. – John Doe Jan 9 '18 at 17:17
• After you have defined a branch of the $(\cdot)^\frac12$ function, you'd get $$i\cdot(-i)=1$$ – John Doe Jan 9 '18 at 17:21
• $(-1)^0$ is also 1 – Plexus Jan 9 '18 at 17:25
• Well, if you wish to square root negative numbers, of course we will end up in $\Bbb C$. In any case, I'm sure there are other ways to 'prove' that $(-k)^0=1$, other than the argument I used above - it was the first I thought of. The way I learned it, it was taken to be the definition that $k^0=1$ for any non-zero $k$, and then that it was convenient (for example in probability, and binomial expansions) to define $0^0=1$ also. – John Doe Jan 9 '18 at 17:28