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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
$f*g=g*f=i$.
\end{description}
[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
$k1,d|n}f(d)g\left(\frac{n}{d}\right)$. 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.]
\end{description}
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