What is $\varphi(0)$?

Question

$\varphi$ 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.

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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?

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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.

Answer 1

$\varphi(n)$ 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 $n=0$.