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QUESTION

The arithmetic function $i$ is defined by
$i(n)=\left\{\begin{array}{cl}1&\textrm{if } n=1\\0&\textrm{if }
n>1\end{array}\right.$. Prove

\begin{description}

\item[(i)]
$i$ is multiplicative.

\item[(ii)]
$i*f=f*i=f$ for all arithmetic function $f$.

\item[(iii)]
If $f(1)\neq0$, we can find an arithmetic function $g$ satisfying
$f*g=g*f=i$.

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[Note: from questions 6 and 7, we can see that the set of all
arithmetic functions $f$ with $f(1)\neq0$ form an abelian group
under *. This allows us to apply results from group theory to the
study of arithmetic functions.]



ANSWER

\begin{description}

\item[(i)]
If gcd$(m,n)=1$, then either $m=n=1$, or at least one of $m$ and
$n$ is $>1$. In the latter case $mn>1$. Thus
$i(mn)=\left\{\begin{array}{cl}1&\textrm{ if } m=n=1\\0&\textrm{
otherwise}\end{array}\right.$

But if at least one of $m,n$ is $>1$, then at least one of $i(m),\
i(n)$ is equal to 0, and so their product is 1. Thus in all cases
$i(mn)=i(m)i(n)$ and $i$ is multiplicative.

\item[(ii)]
By part (ii) of question 6, $i*f=f*i$, so it will be enough to
proof $i*f=f$.

Now $i*f(n)=\sum_{d|n}i(d)f\left(\frac{n}{d}\right)$. But $i(d)=0$
unless $d=1$, so the only term contributing to this sum is the
first, so we get
$i*f(n)=i(1)f\left(\frac{n}{1}\right)=1.f(n)=f(n)$, This is true
for all values of $n$, so $i*f=f$, as required.

\item[(iii)]
Again, question 6(ii) tells us that we need only find $g$ such
that $f*g=i$. As $gŁ$ is an arithmetic function, to describe $g$
we need to specify its values on the natural numbers. We will find
$g$ by describing $g(1), g(2), g(3),\ldots$ and eventually getting
$g(n)$ in terms of the values already spacified for $g(k)$ with
$k<n$.

We first want $f*g(1)=i(1)=1$. Now
$(f*g)(1)=\sum_{d|1}f(d)g\left(\frac{1}{d}\right)$ and as the onlt
divisor of 1 is 1, this says $(f*g)(1)=f(1)g(1)$. Thus if we
define $g(1)=f(1)^{-1}$ (allowable as $f(1)\neg0$), we will have
$f*g(1)=i(1)$. Next we want $f*g(2)=i(2)=0$, We have
$f*g(2)=\sum_{d|2}f(d)g\left(\frac{2}{d}\right)=f(1)g(2)+f(2)g(1)$.
We have already defined $g(1)$, so we may define
$g(2)=\frac{-f(2)g(1)}{f(1)}=\frac{-f(2)}{f(1)^2}$. This ensures
that $f*g$ and $i$ agree at 1 and 2. Similarly we want
$f*g(3)=i(3)=0$, i.e. $f(1)g(3)+f(3)g(1)=0$, so again define
$g(3)=\frac{-f(3)g(1)}{f(1)}=\frac{-f(3)}{f(1)^2}$.To get
$f*g(4)=i(4)=0$ we need
$0=f*g(4)=\sum_{d_4}f(d)g\left(\frac{4}{d}\right)=f(1)g(4)+f(2)g(2)+f(4)g(1)$.
(as the divisors of 4 are 1, 2 and 4). This will give
$g(4)=\frac{-f(2)g(2)-f(4)g(1)}{f(1)}$, and as $g(1)$ and $g(2)$
are already defined this tells us how to define $g(4)$.

We continue in this way until $g(1),g(2),\ldots g(n-1)$ have all
been defined. To define $g(n)$ we note that we want
$f*g(n)=i(n)=0$, so we want
$0=\sum_{d|n}f(d)g\left(\frac{n}{d}\right)=
f(1)g(n)+\sum_{d>1,d|n}f(d)g\left(\frac{n}{d}\right)$. Now all the
terms in the sum $\sum_{d>1,d|n}f(d)g\left(\frac{n}{d}\right)$ are
already specified, so we may now define
$g(n)=\frac{-1}{f(1)}\sum_{d>1,d|n}f(d)g\left(\frac{n}{d}\right)$.

In this way the value of $g(n)$ is defined inductively for all
$n$, giving us an arithmetic function $g$ with the properties
required.

[It is worth noting that an arithmetic function is any function
whose domain in $N$, The functions $d, \sigma,\sigma_1$ etc, that
we've considered happen to take values in $N$, but this is not a
general requirement. If the function $f$ we start with here has
$f(1)\neq1$, the values of $g$ will not be integers, but this does
not matter.]

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