Chapter 14 Cardinality Rules586
Rule(Inclusion-Exclusion-II).
ˇˇ
ˇˇ
ˇ
[n
iD 1
Si
ˇˇ
ˇˇ
ˇ
D
X
;¤If1;:::;ng
. 1/jIjC^1
ˇˇ
ˇˇ
ˇ
\
i 2 I
Si
ˇˇ
ˇˇ
ˇ
(14.6)
A proof of these rules using just highschool algebra is given in Problem 14.52.
14.9.5 Computing Euler’s Function
We can also use Inclusion-Exclusion to derive the explicit formula for Euler’s func-
tion claimed in Corollary 8.10.11: if the prime factorization ofnisp 1 e^1 pemmfor
distinct primespi, then
.n/Dn
Ym
iD 1
1
1
pi
: (14.7)
To begin, letSbe the set of integers inŒ0::n/that arenotrelatively prime ton.
So.n/Dn jSj. Next, letCabe the set of integers inŒ0::n/that are divisible by
a:
CaWWDfk 2 Œ0::n/j ajkg:
So the integers inSare precisely the integers inŒ0::n/that are divisible by at least
one of thepi’s. Namely,
SD
[m
iD 1
Cpi: (14.8)
We’ll be able to find the size of this union using Inclusion-Exclusion because the
intersections of theCpi’s are easy to count. For example,Cp\Cq\Cris the set
of integers inŒ0::n/that are divisible by each ofp,qandr. But since thep;q;r
are distinct primes, being divisible by each of them is the same as being divisible
by their product. Now ifkis a positive divisor ofn, then there are exactlyn=k
multiples ofkinŒ0::n/. So exactlyn=pqrof the integers inŒ0::n/are divisible by
all three primesp,q,r. In other words,
jCp\Cq\CrjD
n
pqr
:
This reasoning extends to arbitrary intersections ofCp’s, namely,
ˇ
ˇˇ
ˇˇ
ˇ
\
j 2 I
Cpj
ˇ
ˇˇ
ˇˇ
ˇ
D
n
Q
j 2 Ipj