Zero to the negative power.

Question

The zero to the negative power is undefined. Am I right?

@Dr.SonnhardGraubner, what is the difference betweenundefined and not defined? — trafalgarLaww, Nov 25 '17 at 16:37
@Dr.SonnhardGraubner, as far as I know "not defined" and "undefined" are synonymous. It appears from your comment that you are making a distinction between the two expressions? Perhaps not. — G Tony Jacobs, Nov 25 '17 at 16:44
you asked me what is the difference between undefined and not defined — Dr. Sonnhard Graubner, Nov 25 '17 at 16:47
@Dr.SonnhardGraubner, I do not know the answer. By the way, this was the reason I asked. — trafalgarLaww, Nov 25 '17 at 16:48
i have also no answer, but i will ask someone in the university — Dr. Sonnhard Graubner, Nov 25 '17 at 16:49
@Dr.SonnhardGraubner , do not bother. I think that there is no difference. — trafalgarLaww, Nov 25 '17 at 16:51

score 3 · Accepted Answer · answered Nov 25 '17 at 16:40

Traditionally, $x^{-k}$ denotes the multiplicative inverse of $x$, $x^{-1}$, raised to the $k^{th}$ power. For instance, in $\mathbf{R}$, we have that the inverse of $2$ is $\frac{1}{2}$ so that $2^{-1}=\frac{1}{2}$. Thus, $2^{-k}=(\frac{1}{2})^k=\frac{1}{2^k}.$ In the case of $0$, there is no multiplicative inverse in $\mathbf{R}$ (or in any field, for that matter), so that the symbol $0^{-1}$ makes no sense. For instance, can you find a real number $r$ so that $r\cdot 0=1$? The answer is no.

In short, $0^{-1}$ does not make sense without further interpretation.

score 1 · Answer 2 · answered Nov 25 '17 at 16:39

1

Indeed, raising $0$ to a negative power implies division by zero, which is undefined. Explicitly, for $a>0$: $$0^{-a}=\frac1{0^a}=\frac10.$$

answered Nov 25 '17 at 16:39

G Tony Jacobs

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Zero to the negative power.

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