Timeline for Is $a^0 = 1$ for a nonzero, real number $a$, a theorem or an axiom?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 3, 2014 at 10:34 | comment | added | schnittstabil | You're right, one may start with a definition $n>=1$ and $a^{n+m}$ motivates $a^0:=1$, but at this moment you may want to extend the definition! It's neither a theorem, nor an axiom that $a^0=1$ holds. | |
| Nov 3, 2014 at 10:29 | comment | added | Joonas Ilmavirta | The definition that $a^n=a\times\cdots\times a$ (with $n$ copies of $a$) is natural for students if $n\geq1$. The empty product does not make that much sense for students that learn these things for the first time. They can observe that $a^{n+m}=a^na^m$, and this provides a good justification for letting $a^0=1$. In schools students don't necessarily start from definitions when they think. Universities are different. | |
| Nov 3, 2014 at 10:25 | comment | added | schnittstabil | The properties of powers are consequences of its definition, not the other way around. The meaning of $a^{n-m}$ for example can't be a good motivation of the definition. The motivation of exponentiation is simple: $a^5$ is an abbreviation of $a \cdot a\cdot a\cdot a\cdot a$ | |
| Nov 3, 2014 at 10:15 | comment | added | Joonas Ilmavirta | As I remarked under Philip Gibbs' answer, that is the only definition of $a^0$ for $a\neq0$ that is compatible with other properties of powers. Outside universities I would prefer to consider that as a consequence of the other power rules, although it may technically be a definition. | |
| Nov 3, 2014 at 10:12 | history | edited | schnittstabil | CC BY-SA 3.0 |
typos
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| Nov 3, 2014 at 9:52 | history | edited | schnittstabil | CC BY-SA 3.0 |
clarifications
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| Nov 3, 2014 at 9:00 | review | First posts | |||
| Nov 3, 2014 at 10:15 | |||||
| Nov 3, 2014 at 8:58 | history | answered | schnittstabil | CC BY-SA 3.0 |