Linear Algebra, Theory And Applications, 2012a

F.3. TRANSCENDENTAL NUMBERS 455
Notethisshowsπ is irrational. If π = k/m where k, m are integers, then both iπ and
−iπ are roots of the polynomial with integer coefficients,
m 2 x 2 + k 2
which would require from what was just shown that
0 ≠2+e iπ + e −iπ
whichisnotthecasesincethesumontherightequals0.
The following corollary follows from this.
Corollary F.3.3 Let K and c i for i =1, ··· ,n be nonzero integers. For each k between 1
and n let {β (k) i
} m(k)
i=1
be the roots of a polynomial with integer coefficients,
where v k ,u k ≠0.Then
⎛
∑m 1
K + c 1
⎝
e β(1) j
j=1
Q k (x) ≡ v k x m k
+ ···+ u k
⎞ ⎛
∑m 2
⎠ + c 2
⎝
e β(2) j
j=1
⎞ ⎛
∑
⎠ + ···+ c n
m n
⎝
e β(n) j
j=1
⎞
⎠ ≠0.
Proof: Defining f k (x) andI k (s) as in Lemma F.3.2, it follows from Lemma F.3.2 that
for each k =1, ··· ,n,
∑
I k (β (k) i
) =
c k
m k
i=1
(
∑m k
K k + c k
deg(f k )
∑
−K k
j=0
i=1
f (j)
k
e β(k) i
) deg(fk )
∑
j=0
∑m k
(0) − c k
i=1
f (j)
k
(0)
deg(f k )
∑
j=0
f (j)
k
(β (k) i
)
This is exactly the same computation as in the beginning of that lemma except one adds
∑ deg(fk )
and subtracts K k j=0
f (j)
k
(0) rather than K ∑ deg(f k )
j=0
f (j)
k
(0) where the K k are chosen
such that their sum equals K. By Lemma F.3.2,
(
)
∑m k
∑m k ( )
c k I k (β (k) i
)= K k + c k e β(k) i v (m k−1)p
u p k + N kp
and so
i=1
c k
m k
−K k
(
v (m k−1)p
k
i=1
(
∑
∑m k
I k (β (k) i
)= K k + c k
i=1
)
u p k + N kp − c k N kp
′
i=1
e β(k) i
k
) ( )
v (m k−1)p
u p k + N kp
−K k v (m k−1)p
k
u p k + M kp
for some integer M k . By multiplying each Q k (x) by a suitable constant, it can be assumed
without loss of generality that all the v m k−1
k
u k are equal to a constant integer U. Then the
above equals
(
)
∑m k
∑m k
c k I k (β (k) i
)= K k + c k e β(k) i (U p + N k p)
i=1
i=1
k