6Real numbers
IA Numbers and Sets
6.6 Algebraic numbers
Rational numbers are “nice”, because they can be written as fractions. Irrational
numbers are bad. However, some irrational numbers are worse than others. We
can further classify some irrational numbers as being transcendental.
Definition
(Algebraic and transcendental numbers)
.
An algebraic number is a
root of a polynomial with integer coefficients (or rational coefficients). A number
is transcendental if it is not algebraic.
Proposition. All rational numbers are algebraic.
Proof. Let x =
p
q
, then x is a root of qx − p = 0.
Example.
√
2 is irrational but algebraic since it is a root of x
2
− 2 = 0.
So do transcendental numbers exist?
Theorem. (Liouville 1851; Nonexaminable) L is transcendental, where
L =
∞
X
n=1
1
10
n!
= 0.11000100 ···
with 1s in the factorial positions.
Proof.
Suppose instead that
f
(
L
) = 0 where
f
(
x
) =
a
k
x
k
+
a
k−1
x
k−1
+
···
+
a
0
,
where a
i
∈ Z, a
k
6= 0.
For any rational p/q, we have
f
p
q
= a
k
p
q
k
+ ···+ a
0
=
integer
q
k
.
So if p/q is not a root of f, then f(p/q) ≥ q
−k
.
For any m, we can write L = first m terms + rest of the terms = s + t.
Now consider f(s) = f(L) − f(s) (since f (L) = 0). We have
f(L) − f (s) =
X
a
i
(L
i
− s
i
)
≤
X
a
i
(L
i
− s
i
)
=
X
a
i
(L − s)(L
i−1
+ ··· + s
i−1
)
≤
X
a
i
(L − s)i,
= (L − s)
X
ia
i

= tC
with C =
P
ia
i
.
Writing
s
as a fraction, its denominator is at most 10
m!
. So
f
(
s
)
 ≥
10
−k×m!
.
Combining with the above, we have tC ≥ 10
−k×m!
.
We can bound t by
t =
∞
X
j=m+1
10
−j!
≤
∞
X
`=(m+1)!
10
−`
=
10
9
10
−(m+1)!
.
So (10
C/
9)10
−(m+1)!
≥
10
−k×m!
. Pick
m ∈ N
so that
m > k
and 10
m!
>
10C
9
.
This is always possible since both
k
and 10
C/
9 are constants. Then the inequality
gives 10
−(m+1)
≥ 10
−(k+1)
, which is a contradiction since m > k.
Theorem. (Hermite 1873) e is transcendental.
Theorem. (Lindermann 1882) π is transcendental.