Wolfram MathWorld defines the Euler Totient function as follows:

The totient function phi(n), also called Euler's totient function, is defined    
as the number of positive integers <=n that are relatively prime to (i.e., do 
not contain any factor in common with) n, where 1 is counted as being 
relatively prime to all numbers

By this definition would $\phi(0)=0$?

  • $\begingroup$ Is there any context for which you would ever need the totient function to be defined for 0? $\endgroup$ – Alan Apr 8 '15 at 23:37
  • $\begingroup$ None than curiosity. $\endgroup$ – terminex9 Apr 8 '15 at 23:38
  • 3
    $\begingroup$ Also consider this property: if $d\mid n$ then $\phi(d)\mid\phi(n).$, and notice that every integer divides $0.$ $\endgroup$ – awllower Apr 8 '15 at 23:41
  • $\begingroup$ Why do you think it would need a definition or want one? You could theoretically call it $2$ since $\mathbb Z/0\mathbb Z$ has two units, I suppose. $\endgroup$ – Thomas Andrews Apr 9 '15 at 0:01

Yes, because we must conclude that there are zero positive integers less than or equal to zero. We do not even have to consider the remainder of the criteria or the fact that $1$ is relatively prime to all numbers.

  • $\begingroup$ Okay. I was just curious if there was a different number that was accepted condition or if the totient function limits n to natural numbers. $\endgroup$ – terminex9 Apr 8 '15 at 23:38
  • $\begingroup$ I don't see how you would make use of the fact that $\phi(0) = 0$ but based on the definition you gave it is the only sensible answer. I'm sure the input is restricted to the positive integers in some definitions. $\endgroup$ – user157227 Apr 8 '15 at 23:40
  • $\begingroup$ @terminex9 He only based his answer on your definition, since you asked: "by this definition would $\phi(0)=0$?", and the answer is of course yes, because there are no positive integers less than $n$. He didn't consider anything else outside your definition, which could very well be incomplete. Indeed, the same link you took your definition from adds the exception $\phi (0)=1$. $\endgroup$ – user26486 Apr 30 '15 at 15:23

If you read a little further down in the MathWorld entry, you find this:

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

However, no reference is given for the convention. It's a rare formula that calls out for $\phi(0)$ to be included. The Book of Numbers, by Conway and Guy, offers one possibility. On page 156, they write:

...the $n$th Farey series has length

$$1+\phi(1)+\phi(2)+\phi(3)+\cdots+\phi(n-1)+\phi(n)$$ (the initial $1$ comes from the fact that we count both $0/1$ and $1/1$).

Note, however, that Conway and Guy do not write $\phi(0)$ for the initial $1$.

  • $\begingroup$ Perhaps sometimes $\phi(0)$ is thought of as an empty product and other times as an empty sum? Just a thought. $\endgroup$ – user157227 Apr 9 '15 at 1:16

In general, we don't often need to define $\phi(n)$ for $n<1$.

For $n=1$, we define $\phi(1)=1$ because we want $\phi$ to be multiplicative, and because it lets us deduce that $\sum_{d\mid n} \phi(d) = n$. The whole reason Wolfram defines it with $\leq n$ rather than $<n$ is just for that case of $n=1$. Essentially, if $\rho$ is any multiplicative function, then $\rho(1)$ is a "empty product," and thus needs to be $1$.

There isn't much value in defining $\phi(0)$. Sometimes, we just leave things undefined. In broader contexts, like algebraic number fields, you can define $\phi(I)$ where $I$ is any ideal in a commutative ring, and $\phi(I)$ is the number of units in $R/I$. That yields a definition of $\phi$ that is more natural and you get $\phi(0)=\phi(0\mathbb Z)=2$. But I don't think that value is useful.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.