A note on notation:

If

are arithmetic functions, then

is the Dirichlet product thereof, such that

.

**Theorem:** *If both and are multiplicative, then is multiplicative.* *Proof:* We shall prove by contradiction. Suppose

is not multiplicative. Let

. Since

is not multiplicative, there exist

such that

. We choose

as small as possible. If

, then

so

. Since

,

is not multiplicative, a contradiction. If

, we have

for all

and

. Now,

.

Since

,

. Therefore

is not multiplicative, a contradiction.

Okay, here's my problem with this "proof": Why in the world do we have " for all and " if we are assuming that is *not* multiplicative? As far as I can see, such a statement might be true or false for some . Yet the author of the proof takes it to be true. Why?

Thanks!