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$\varphi$ is Euler's totient function. My question is:

  • When/if $\varphi$ is defined at $0$, what is it usually defined as?

  • Is there a "most natural" or more commonly accepted definition of $\varphi(0)$?

This is a soft question, because it's of course a matter of convention.

Here are three possible definitions, with some justification.

Definition 1: $\boldsymbol{\varphi(0) = 2}$. Note that for $n \ge 1$, $\varphi(n) = \left| \left( \mathbb{Z} / n \mathbb{Z} \right)^* \right|$, the number of units mod $n$. Plugging in $n = 0$, we get $$ \varphi(0) = \left| \left( \mathbb{Z} / 0 \mathbb{Z} \right)^* \right| = \left| \left( \mathbb{Z} \right)^* \right| = \left| \{-1, 1\} \right| = 2. $$

Definition 2: $\boldsymbol{\varphi(0) = 0}$. If $a \mid b$, then $\varphi(a) \mid \varphi(b)$. To preserve this property for $b = 0$, we need that $\varphi(a) \mid \varphi(0)$ for all $a$, implying $\varphi(0) = 0$.

WolframAlpha returns $\varphi(0) = 0$. However, the Wolfram Math World page explains:

By convention, $\phi(0)=1$, although the Wolfram Language defines EulerPhi[0] equal to 0 for consistency with its FactorInteger[0] command.

which gives us

Definition 3: $\boldsymbol{\varphi(0) = 1}$. Does anyone know the reason for this convention?

More observations

  • Any choice of $\varphi(0)$ is consistent with the multiplicativity of $\varphi$.

  • $\varphi(mn) = \varphi(m) \varphi(n) \frac{d}{\varphi(d)}$, where $d = \gcd(m,n)$, implies $\varphi(0) = 0$. This supports definition 2.

  • If $\varphi(n) = \sum_{ab = n} a \mu(b)$, then note that when $ab = 0$, $a = 0$ or $b = 0$, and since $\mu(0) = 0$, $a \mu(b) = 0$. So this is $\sum 0 = 0$, again agreeing with definition 2.

Caleb Stanford
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1 Answers1

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$\varphi(n)$ is the number of positive integers not larger than $n$ that are coprime to $n$.

There are no positive integers not larger than $0$, so by definition, if we were to define $\varphi(0)$, we would want it to equal $0$. This agrees with $$\varphi(n)=n\prod_{\text{prime} \ p\lvert n} \left(1-\frac{1}{p} \right), $$ where the product runs over all the primes, when $n=0$. We may indeed note that we cannot deduce from this the infinitude of the primes, since in any finite case the product is smaller than $1$, and is thus nullified by $n=0$.

Vincenzo Oliva
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  • (+1) Another good argument for definition 2--the product form of $\varphi(n)$ is particularly convincing. But then I'm still puzzled by Wolfram Math World's "By convention" statement. – Caleb Stanford Aug 23 '15 at 01:40
  • Maple also give us the value $0$. – GEdgar Aug 23 '15 at 01:44
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    But Pete Clark raised an important point: defining $\varphi(0)$ is completely useless. There is never any reason you need such a definition. This is different from the case of defining $n!$ at $n = 0$. – KCd Aug 23 '15 at 02:45
  • @KCd: Well, we define $1$ not to be prime just for convenience, don't we? In particular, it should be because we want to avoid saying "except for" in statements of various prime theorems, especially the Fundamental Theorem of Arithmetic. The situation here is similar, I think. – Vincenzo Oliva Aug 23 '15 at 19:39
  • @6005: What's your opinion? – Vincenzo Oliva Aug 23 '15 at 19:39
  • Yes, it has to be a matter of convenience, but $1$ being not prime is much more useful, I am not sure now if the value $\varphi(0)$ is useful at all. – Caleb Stanford Aug 23 '15 at 19:59
  • There is no purpose in defining $\varphi(0)$, or certainly no important purpose. I have never seen any reason to need a definition of $\varphi(0)$. This is quite different from the case of deciding whether 1 should or should not be a prime number. – KCd Aug 24 '15 at 11:09