What kind of number is . . . ?
Feryal Alayont
alayontf@gvsu.edu
March 13, 2007
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
1 / 25
Kinds of numbers
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
2 / 25
Kinds of numbers
• Natural numbers: 1, 2, 3, 4, . . .
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
2 / 25
Kinds of numbers
• Natural numbers: 1, 2, 3, 4, . . .
• Integers: . . . , -2, -1, 0, 1, 2, . . .
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
2 / 25
Kinds of numbers
• Natural numbers: 1, 2, 3, 4, . . .
• Integers: . . . , -2, -1, 0, 1, 2, . . .
• Rational numbers: . . ., 0, 1, 1/2, 2, 1/3, 3, 1/4, 2/3, . . .
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
2 / 25
Kinds of numbers
• Natural numbers: 1, 2, 3, 4, . . .
• Integers: . . . , -2, -1, 0, 1, 2, . . .
• Rational numbers: . . ., 0, 1, 1/2, 2, 1/3, 3, 1/4, 2/3, . . .
• And the rest . . .
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
2 / 25
The rest
An irrational number is a real number which is not rational, i.e. it cannot
be expressed as
m
n
F. Alayont (GVSU)
where both m and n are integers.
What kind of number is . . . ?
March 13, 2007
3 / 25
The rest
An irrational number is a real number which is not rational, i.e. it cannot
be expressed as
m
n
where both m and n are integers.
Please write down a few irrational numbers that come to your mind.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
3 / 25
Irrationality of
√
2
Proved around 500 B.C. by Pythagoreans.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
4 / 25
Irrationality of
√
2
Proved around 500 B.C. by Pythagoreans.
proof: Assume
√
F. Alayont (GVSU)
2=
m
n
in reduced form.
What kind of number is . . . ?
March 13, 2007
4 / 25
Irrationality of
√
2
Proved around 500 B.C. by Pythagoreans.
proof: Assume
√
2=
m
n
in reduced form.
2n2 = m2
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
4 / 25
Irrationality of
√
2
Proved around 500 B.C. by Pythagoreans.
proof: Assume
√
2=
m
n
in reduced form.
2n2 = m2
F. Alayont (GVSU)
→
m = 2k
What kind of number is . . . ?
March 13, 2007
4 / 25
Irrationality of
√
2
Proved around 500 B.C. by Pythagoreans.
proof: Assume
√
2=
m
n
in reduced form.
2n2 = m2
→
m = 2k
n2 = 2k 2
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
4 / 25
Irrationality of
√
2
Proved around 500 B.C. by Pythagoreans.
proof: Assume
√
F. Alayont (GVSU)
2=
m
n
in reduced form.
2n2 = m2
→
m = 2k
n2 = 2k 2
→
n = 2l
What kind of number is . . . ?
March 13, 2007
4 / 25
Irrationality of
√
2
Proved around 500 B.C. by Pythagoreans.
proof: Assume
√
2=
m
n
in reduced form.
2n2 = m2
→
m = 2k
n2 = 2k 2
→
n = 2l
Contradiction.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
4 / 25
Other roots
For an integer a, a1/n is not a rational number unless a is an nth power
of an integer.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
5 / 25
Other roots
For an integer a, a1/n is not a rational number unless a is an nth power
of an integer.
More irrational numbers:
√ √ √
3, 5, 6, . . . ,
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
5 / 25
Other roots
For an integer a, a1/n is not a rational number unless a is an nth power
of an integer.
More irrational numbers:
√ √ √
√
√
√
3
3
3
3, 5, 6, . . . , 2, 3, 4, . . . ,
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
5 / 25
Other roots
For an integer a, a1/n is not a rational number unless a is an nth power
of an integer.
More irrational numbers:
√ √ √
√
√
√
√
√
√
3
3
4
4
3
4
3, 5, 6, . . . , 2, 3, 4, . . . , 2, 3, 5, . . .
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
5 / 25
Making new irrational numbers
Theorem: If x is rational and y is irrational, then x + y and xy (x 6= 0)
are irrational.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
6 / 25
Making new irrational numbers
Theorem: If x is rational and y is irrational, then x + y and xy (x 6= 0)
are irrational.
proof: Assume x is rational, y is irrational and x + y is rational.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
6 / 25
Making new irrational numbers
Theorem: If x is rational and y is irrational, then x + y and xy (x 6= 0)
are irrational.
proof: Assume x is rational, y is irrational and x + y is rational.
Then y = (x + y ) − x is rational too.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
6 / 25
Making new irrational numbers
Theorem: If x is rational and y is irrational, then x + y and xy (x 6= 0)
are irrational.
proof: Assume x is rational, y is irrational and x + y is rational.
Then y = (x + y ) − x is rational too.
Contradiction.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
6 / 25
Making new irrational numbers
Theorem: If x is rational and y is irrational, then x + y and xy (x 6= 0)
are irrational.
proof: Assume x is rational, y is irrational and x + y is rational.
Then y = (x + y ) − x is rational too.
Contradiction.
So: 1 +
√
2,
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
6 / 25
Making new irrational numbers
Theorem: If x is rational and y is irrational, then x + y and xy (x 6= 0)
are irrational.
proof: Assume x is rational, y is irrational and x + y is rational.
Then y = (x + y ) − x is rational too.
Contradiction.
So: 1 +
√
2, 2 −
F. Alayont (GVSU)
√
3
3,
What kind of number is . . . ?
March 13, 2007
6 / 25
Making new irrational numbers
Theorem: If x is rational and y is irrational, then x + y and xy (x 6= 0)
are irrational.
proof: Assume x is rational, y is irrational and x + y is rational.
Then y = (x + y ) − x is rational too.
Contradiction.
So: 1 +
√
2, 2 −
F. Alayont (GVSU)
√
3
3,
5
3
√
7,
What kind of number is . . . ?
March 13, 2007
6 / 25
Making new irrational numbers
Theorem: If x is rational and y is irrational, then x + y and xy (x 6= 0)
are irrational.
proof: Assume x is rational, y is irrational and x + y is rational.
Then y = (x + y ) − x is rational too.
Contradiction.
So: 1 +
√
2, 2 −
F. Alayont (GVSU)
√
3
3,
5
3
√
√
7, 3 + 4 3 5 etc. are all irrational.
What kind of number is . . . ?
March 13, 2007
6 / 25
Making new irrational numbers
Theorem: If x is rational, then x 2 , x 3 , x 4 , . . . are rational.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
7 / 25
Making new irrational numbers
Theorem: If x is rational, then x 2 , x 3 , x 4 , . . . are rational.
Theorem: (Equivalent) If x is irrational, then
irrational.
F. Alayont (GVSU)
What kind of number is . . . ?
√
x,
√
3
x,
√
4
x, . . . are
March 13, 2007
7 / 25
Making new irrational numbers
Theorem: If x is rational, then x 2 , x 3 , x 4 , . . . are rational.
Theorem: (Equivalent) If x is irrational, then
irrational.
So:
p√
√
x,
√
3
x,
√
4
x, . . . are
2,
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
7 / 25
Making new irrational numbers
Theorem: If x is rational, then x 2 , x 3 , x 4 , . . . are rational.
Theorem: (Equivalent) If x is irrational, then
irrational.
So:
√
x,
√
3
x,
√
4
x, . . . are
p√ p
√
3
2, 1 + 2,
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
7 / 25
Making new irrational numbers
Theorem: If x is rational, then x 2 , x 3 , x 4 , . . . are rational.
Theorem: (Equivalent) If x is irrational, then
irrational.
So:
√
x,
√
3
x,
√
4
x, . . . are
p√ p
√ q √
3
2, 1 + 2, 53 4 5 + 3 ,
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
7 / 25
Making new irrational numbers
Theorem: If x is rational, then x 2 , x 3 , x 4 , . . . are rational.
Theorem: (Equivalent) If x is irrational, then
irrational.
So:
√
x,
√
3
x,
√
4
x, . . . are
q
p√ p
√ q √
√
3
2, 1 + 2, 53 4 5 + 3 , 2 + 3 7 + 32 5 etc. are all irrational.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
7 / 25
Making new irrational numbers
Theorem: If x is rational, then x 2 , x 3 , x 4 , . . . are rational.
Theorem: (Equivalent) If x is irrational, then
irrational.
So:
√
x,
√
3
x,
√
4
x, . . . are
q
p√ p
√ q √
√
3
2, 1 + 2, 53 4 5 + 3 , 2 + 3 7 + 32 5 etc. are all irrational.
And so is:
11
+
5
F. Alayont (GVSU)
r
7
1 2
+
2 5
q
4+
What kind of number is . . . ?
√
3
March 13, 2007
7 / 25
Irrationality of e
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
8 / 25
Irrationality of e
Euler: e is irrational, using continued fractions, 1737.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
8 / 25
Irrationality of e
Euler: e is irrational, using continued fractions, 1737.
Fourier: a more elementary proof using the series expansion, 1815.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
8 / 25
Irrationality of e
proof: Taylor expansion of e x : e x = 1 + x +
F. Alayont (GVSU)
x2
2!
What kind of number is . . . ?
+
x3
3!
+ ... +
xk
k!
+ ...
March 13, 2007
9 / 25
Irrationality of e
proof: Taylor expansion of e x : e x = 1 + x +
So: e = 1 + 1 +
F. Alayont (GVSU)
1
2!
+ ... +
1
k!
x2
2!
+
x3
3!
+ ... +
xk
k!
+ ...
+ ...
What kind of number is . . . ?
March 13, 2007
9 / 25
Irrationality of e
proof: Taylor expansion of e x : e x = 1 + x +
So: e = 1 + 1 +
Assume e =
m
n,
1
2!
+ ... +
1
k!
x2
2!
+
x3
3!
+ ... +
xk
k!
+ ...
+ ...
m and n integers
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
9 / 25
Irrationality of e
proof: Taylor expansion of e x : e x = 1 + x +
So: e = 1 + 1 +
1
2!
+ ... +
1
k!
x2
2!
+
x3
3!
+ ... +
xk
k!
+ ...
+ ...
Assume e = m
n , m and n integers
1
1
n! e − 1 + 1 + 2!
+ . . . + n!
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
9 / 25
Irrationality of e
proof: Taylor expansion of e x : e x = 1 + x +
So: e = 1 + 1 +
1
2!
+ ... +
1
k!
x2
2!
+
x3
3!
+ ... +
xk
k!
+ ...
+ ...
Assume e = m
n , m and n integers
1
1
is a positive integer, call it A
n! e − 1 + 1 + 2!
+ . . . + n!
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
9 / 25
Irrationality of e
proof: Taylor expansion of e x : e x = 1 + x +
So: e = 1 + 1 +
1
2!
+ ... +
1
k!
x2
2!
+
x3
3!
+ ... +
xk
k!
+ ...
+ ...
Assume e = m
n , m and n integers
1
1
is a positive integer, call it A
n! e − 1 + 1 + 2!
+ . . . + n!
A = n!
F. Alayont (GVSU)
1
1
1
+
+
+ ...
(n + 1)! (n + 2)! (n + 3)!
What kind of number is . . . ?
March 13, 2007
9 / 25
Irrationality of e
proof: Taylor expansion of e x : e x = 1 + x +
So: e = 1 + 1 +
1
2!
+ ... +
1
k!
x2
2!
+
x3
3!
+ ... +
xk
k!
+ ...
+ ...
Assume e = m
n , m and n integers
1
1
is a positive integer, call it A
n! e − 1 + 1 + 2!
+ . . . + n!
A = n!
=
1
1
1
+
+
+ ...
(n + 1)! (n + 2)! (n + 3)!
1
1
1
+
+
+ ...
n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3)
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
9 / 25
Irrationality of e
proof: Taylor expansion of e x : e x = 1 + x +
So: e = 1 + 1 +
1
2!
+ ... +
1
k!
x2
2!
+
x3
3!
+ ... +
xk
k!
+ ...
+ ...
Assume e = m
n , m and n integers
1
1
is a positive integer, call it A
n! e − 1 + 1 + 2!
+ . . . + n!
A = n!
=
1
1
1
+
+
+ ...
(n + 1)! (n + 2)! (n + 3)!
1
1
1
+
+
+ ...
n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3)
<
F. Alayont (GVSU)
1
1
1
1
+
+ ... =
+
n + 1 (n + 1)2 (n + 1)3
n
What kind of number is . . . ?
March 13, 2007
9 / 25
Irrationality of e
proof: Taylor expansion of e x : e x = 1 + x +
So: e = 1 + 1 +
1
2!
+ ... +
1
k!
x2
2!
+
x3
3!
+ ... +
xk
k!
+ ...
+ ...
Assume e = m
n , m and n integers
1
1
is a positive integer, call it A
n! e − 1 + 1 + 2!
+ . . . + n!
A = n!
=
1
1
1
+
+
+ ...
(n + 1)! (n + 2)! (n + 3)!
1
1
1
+
+
+ ...
n + 1 (n + 1)(n + 2) (n + 1)(n + 2)(n + 3)
<
1
1
1
1
+
+ ... =
+
n + 1 (n + 1)2 (n + 1)3
n
Contradiction.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
9 / 25
Irrationality of π
Lambert: π is irrational, using continued fractions, 1761.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
10 / 25
Irrationality of π
Lambert: π is irrational, using continued fractions, 1761.
Niven: a relatively simple proof following an idea of Hermite, using
integrals, 1947
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
10 / 25
Irrationality of π
Niven’s polynomials: P(x) =
F. Alayont (GVSU)
x n (1−x)n
n!
What kind of number is . . . ?
March 13, 2007
11 / 25
Irrationality of π
Niven’s polynomials: P(x) =
x n (1−x)n
n!
example: For n = 3, P(x) =
1
3
3! (x
F. Alayont (GVSU)
− 3x 4 + 3x 5 − x 6 ).
What kind of number is . . . ?
March 13, 2007
11 / 25
Irrationality of π
Niven’s polynomials: P(x) =
x n (1−x)n
n!
example: For n = 3, P(x) =
1
3
3! (x
− 3x 4 + 3x 5 − x 6 ).
Properties: 1) P(x) is a polynomial of degree 2n.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
11 / 25
Irrationality of π
Niven’s polynomials: P(x) =
x n (1−x)n
n!
example: For n = 3, P(x) =
1
3
3! (x
− 3x 4 + 3x 5 − x 6 ).
Properties: 1) P(x) is a polynomial of degree 2n.
2) For 0 < x < 1, 0 < P(x) <
F. Alayont (GVSU)
1
n! .
What kind of number is . . . ?
March 13, 2007
11 / 25
Irrationality of π
Niven’s polynomials: P(x) =
x n (1−x)n
n!
example: For n = 3, P(x) =
1
3
3! (x
− 3x 4 + 3x 5 − x 6 ).
Properties: 1) P(x) is a polynomial of degree 2n.
2) For 0 < x < 1, 0 < P(x) <
1
n! .
3) Any kth derivative of P(x) yields an integer for x = 0 and x = 1.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
11 / 25
Irrationality of π
Theorem: π 2 is irrational.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
12 / 25
Irrationality of π
Theorem: π 2 is irrational.
proof: Assume π 2 = ba , a and b integers.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
12 / 25
Irrationality of π
Theorem: π 2 is irrational.
proof: Assume π 2 = ba , a and b integers.
R1
n
n
.
Let N = an 0 P(x)π sin πx dx , P(x) = x (1−x)
n!
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
12 / 25
Irrationality of π
Theorem: π 2 is irrational.
proof: Assume π 2 = ba , a and b integers.
R1
n
n
.
Let N = an 0 P(x)π sin πx dx , P(x) = x (1−x)
n!
N
(by parts)
=
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
12 / 25
Irrationality of π
Theorem: π 2 is irrational.
proof: Assume π 2 = ba , a and b integers.
R1
n
n
.
Let N = an 0 P(x)π sin πx dx , P(x) = x (1−x)
n!
(by parts)
N =
an − cos πx P(x) −
F. Alayont (GVSU)
P 00 (x)
π2
+ . . . + sin πx
P 0 (x)
π
What kind of number is . . . ?
−
P (3) (x)
π3
+ ...
1
0
March 13, 2007
12 / 25
Irrationality of π
Theorem: π 2 is irrational.
proof: Assume π 2 = ba , a and b integers.
R1
n
n
.
Let N = an 0 P(x)π sin πx dx , P(x) = x (1−x)
n!
(by parts)
N =
an − cos πx P(x) −
=
−an
P 00 (x)
π2
+ . . . + sin πx
cos πx P(x) −
F. Alayont (GVSU)
P 00 (x)
π2
P 0 (x)
π
+ ... ±
What kind of number is . . . ?
P (3) (x)
π3
1
P (2n) (x)
2n
π
0
−
+ ...
1
0
March 13, 2007
12 / 25
Irrationality of π
Theorem: π 2 is irrational.
proof: Assume π 2 = ba , a and b integers.
R1
n
n
.
Let N = an 0 P(x)π sin πx dx , P(x) = x (1−x)
n!
(by parts)
N =
an − cos πx P(x) −
=
−an
P 00 (x)
π2
+ . . . + sin πx
cos πx P(x) −
P 00 (x)
π2
P 0 (x)
π
+ ... ±
an π12n
P (3) (x)
π3
1
P (2n) (x)
2n
π
0
−
+ ...
1
0
is an integer and the derivative values are integers, so N is an
integer.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
12 / 25
Irrationality of π
0<N=a
n
Z
1
P(x)π sin πx dx <
0
F. Alayont (GVSU)
What kind of number is . . . ?
πan
n!
March 13, 2007
13 / 25
Irrationality of π
0<N=a
n
Z
1
P(x)π sin πx dx <
0
N is an integer but we can make
F. Alayont (GVSU)
πan
n!
πan
n!
small by choosing a large n.
What kind of number is . . . ?
March 13, 2007
13 / 25
Irrationality of π
0<N=a
n
Z
1
P(x)π sin πx dx <
0
N is an integer but we can make
πan
n!
πan
n!
small by choosing a large n.
Contradiction.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
13 / 25
Irrationality of π
0<N=a
n
Z
1
P(x)π sin πx dx <
0
N is an integer but we can make
πan
n!
πan
n!
small by choosing a large n.
Contradiction.
Corollary: π is irrational.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
13 / 25
Making more irrational numbers
Theorem: (?) If x and y are irrational numbers, then x + y is irrational.
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
14 / 25
Making more irrational numbers
Theorem: (?) If x and y are irrational numbers, then x + y is irrational.
Sigh... not true: x = π and y = −π
F. Alayont (GVSU)
What kind of number is . . . ?
March 13, 2007
14 / 25
Making more irrational numbers
Theorem: (?) If x and y are irrational numbers, then x + y is irrational.
Sigh... not true: x = π and y = −π
What about π +
F. Alayont (GVSU)
√
2? Is this irrational?
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March 13, 2007
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Algebraic numbers
A complex number is called an algebraic number if it is a root of
a polynomial with rational coefficients.
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Algebraic numbers
A complex number is called an algebraic number if it is a root of
a polynomial with rational coefficients.
example:
√
2 is algebraic:
F. Alayont (GVSU)
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March 13, 2007
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Algebraic numbers
A complex number is called an algebraic number if it is a root of
a polynomial with rational coefficients.
example:
√
2 is algebraic: p(x) = x 2 − 2
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March 13, 2007
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Algebraic numbers
A complex number is called an algebraic number if it is a root of
a polynomial with rational coefficients.
example:
√
2 is algebraic: p(x) = x 2 − 2
example: Any rational number is algebraic:
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March 13, 2007
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Algebraic numbers
A complex number is called an algebraic number if it is a root of
a polynomial with rational coefficients.
example:
√
2 is algebraic: p(x) = x 2 − 2
example: Any rational number is algebraic: x =
p(x) = x − m
n
F. Alayont (GVSU)
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m
n
is the root of
March 13, 2007
15 / 25
Algebraic numbers
A complex number is called an algebraic number if it is a root of
a polynomial with rational coefficients.
example:
√
2 is algebraic: p(x) = x 2 − 2
example: Any rational number is algebraic: x =
p(x) = x − m
n
example: 1 +
√
F. Alayont (GVSU)
m
n
is the root of
2 is algebraic:
What kind of number is . . . ?
March 13, 2007
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Algebraic numbers
A complex number is called an algebraic number if it is a root of
a polynomial with rational coefficients.
example:
√
2 is algebraic: p(x) = x 2 − 2
example: Any rational number is algebraic: x =
p(x) = x − m
n
example: 1 +
√
F. Alayont (GVSU)
m
n
is the root of
2 is algebraic: p(x) = x 2 − 2x − 1
What kind of number is . . . ?
March 13, 2007
15 / 25
Algebraic numbers
A complex number is called an algebraic number if it is a root of
a polynomial with rational coefficients.
example:
√
2 is algebraic: p(x) = x 2 − 2
example: Any rational number is algebraic: x =
p(x) = x − m
n
example: 1 +
√
m
n
is the root of
2 is algebraic: p(x) = x 2 − 2x − 1
example: i is algebraic
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March 13, 2007
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Algebraic numbers
There are algebraic numbers which are not expressible by radicals, Abel,
1824
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Closure property of addition and multiplication
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Closure property of addition and multiplication
Closure property of algebraic numbers: if we add or multiply two algebraic
numbers we get another algebraic number.
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Closure property of addition and multiplication
Closure property of algebraic numbers: if we add or multiply two algebraic
numbers we get another algebraic number.
√
√
So: 2 3 2 + 32 7,
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Closure property of addition and multiplication
Closure property of algebraic numbers: if we add or multiply two algebraic
numbers we get another algebraic number.
√
√
√
√
√ √
So: 2 3 2 + 32 7, 1 + 2 + 3 3 + 5 7 3 130 etc. are algebraic.
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Kinds of numbers again
• Natural numbers: 1, 2, 3, 4, . . .
• Integers: . . . , -2, -1, 0, 1, 2, . . .
• Rational numbers: . . ., 0, 1, 1/2, 2, 1/3, 3, 1/4, 2/3, . . .
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Kinds of numbers again
• Natural numbers: 1, 2, 3, 4, . . .
• Integers: . . . , -2, -1, 0, 1, 2, . . .
• Rational numbers: . . ., 0, 1, 1/2, 2, 1/3, 3, 1/4, 2/3, . . .
• Algebraic numbers:
F. Alayont (GVSU)
√
2,
√
3
5, 1 +
√
√
3 + 2 3 13, . . .
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March 13, 2007
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Kinds of numbers again
• Natural numbers: 1, 2, 3, 4, . . .
• Integers: . . . , -2, -1, 0, 1, 2, . . .
• Rational numbers: . . ., 0, 1, 1/2, 2, 1/3, 3, 1/4, 2/3, . . .
• Algebraic numbers:
√
2,
√
3
5, 1 +
√
√
3 + 2 3 13, . . .
• And the rest . . .
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The rest
A transcendental number is a number which is not algebraic, i.e. it is
not the root of a polynomial with rational coefficients.
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The rest
A transcendental number is a number which is not algebraic, i.e. it is
not the root of a polynomial with rational coefficients.
e and π are both transcendental:
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The rest
A transcendental number is a number which is not algebraic, i.e. it is
not the root of a polynomial with rational coefficients.
e and π are both transcendental:
e transcendental conjectured by Legendre, 1794; proved by Hermite, 1873
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The rest
A transcendental number is a number which is not algebraic, i.e. it is
not the root of a polynomial with rational coefficients.
e and π are both transcendental:
e transcendental conjectured by Legendre, 1794; proved by Hermite, 1873
π transcendental conjectured by Euler, 1755; proved by Lindemann, 1882
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Making more more irrational numbers
Theorem: If x is algebraic and y is transcendental, then x + y and xy
(x 6= 0) are transcendental, and hence irrational.
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March 13, 2007
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Making more more irrational numbers
Theorem: If x is algebraic and y is transcendental, then x + y and xy
(x 6= 0) are transcendental, and hence irrational.
So: π +
√
2,
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Making more more irrational numbers
Theorem: If x is algebraic and y is transcendental, then x + y and xy
(x 6= 0) are transcendental, and hence irrational.
So: π +
√
2, e −
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p
3
1+
√
5,
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March 13, 2007
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Making more more irrational numbers
Theorem: If x is algebraic and y is transcendental, then x + y and xy
(x 6= 0) are transcendental, and hence irrational.
q
p
√
√
√
√
3
So: π + 2, e − 1 + 5, π(1 + 7) − 3 2 − 53 11, etc. are all
transcendental, and hence irrational.
F. Alayont (GVSU)
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March 13, 2007
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Making more more irrational numbers
Theorem: If x is algebraic, then
F. Alayont (GVSU)
√
x is algebraic.
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March 13, 2007
21 / 25
Making more more irrational numbers
Theorem: If x is algebraic, then
√
x is algebraic.
√
proof: If x satisfies ax 2 + bx + c = 0, then x satisfies ax 4 + bx 2 + c = 0.
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March 13, 2007
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Making more more irrational numbers
Theorem: If x is algebraic, then
√
x is algebraic.
√
proof: If x satisfies ax 2 + bx + c = 0, then x satisfies ax 4 + bx 2 + c = 0.
Theorem: If x is algebraic, then
F. Alayont (GVSU)
√
n
x is algebraic.
What kind of number is . . . ?
March 13, 2007
21 / 25
Making more more irrational numbers
Theorem: If x is algebraic, then
√
x is algebraic.
√
proof: If x satisfies ax 2 + bx + c = 0, then x satisfies ax 4 + bx 2 + c = 0.
Theorem: If x is algebraic, then
√
n
x is algebraic.
Theorem: (Equivalent) If x is transcendental, then x 2 , x 3 , x 4 , . . . are
transcendental.
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March 13, 2007
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Making more more irrational numbers
Theorem: If x is algebraic, then
√
x is algebraic.
√
proof: If x satisfies ax 2 + bx + c = 0, then x satisfies ax 4 + bx 2 + c = 0.
Theorem: If x is algebraic, then
√
n
x is algebraic.
Theorem: (Equivalent) If x is transcendental, then x 2 , x 3 , x 4 , . . . are
transcendental.
So: π 2 , e 5 ,
√
2 + π 2 , etc. are all irrational.
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More transcendence results
Theorem: (Hermite, Lindemann) 1) If α is an algebraic number not equal
to 0 or 1, then log α is transcendental.
2) If α is a non-zero algebraic number, then e α is transcendental.
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More transcendence results
Theorem: (Hermite, Lindemann) 1) If α is an algebraic number not equal
to 0 or 1, then log α is transcendental.
2) If α is a non-zero algebraic number, then e α is transcendental.
Transcendence of π: If π were algebraic, then iπ is algebraic and e iπ = −1
would have to be transcendental.
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Hilbert’s 7th problem
Theorem: (Gel’fond, Schneider, 1934) If α is an algebraic number not
equal to 0 or 1 and β is a non-rational algebraic number, then αβ is
transcendental.
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March 13, 2007
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Hilbert’s 7th problem
Theorem: (Gel’fond, Schneider, 1934) If α is an algebraic number not
equal to 0 or 1 and β is a non-rational algebraic number, then αβ is
transcendental.
√
So: 3
3,
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Hilbert’s 7th problem
Theorem: (Gel’fond, Schneider, 1934) If α is an algebraic number not
equal to 0 or 1 and β is a non-rational algebraic number, then αβ is
transcendental.
√
So: 3
3,
21+
√
2,
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March 13, 2007
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Hilbert’s 7th problem
Theorem: (Gel’fond, Schneider, 1934) If α is an algebraic number not
equal to 0 or 1 and β is a non-rational algebraic number, then αβ is
transcendental.
√
So: 3
3,
21+
√
2,
F. Alayont (GVSU)
√
√
2
2
, etc. are all transcendental.
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March 13, 2007
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Hilbert’s 7th problem
Theorem: (Gel’fond, Schneider, 1934) If α is an algebraic number not
equal to 0 or 1 and β is a non-rational algebraic number, then αβ is
transcendental.
√
So: 3
3,
21+
√
2,
√
√
2
2
, etc. are all transcendental.
Also: e π is transcendental.
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March 13, 2007
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Hilbert’s 7th problem
Theorem: (Gel’fond, Schneider, 1934) If α is an algebraic number not
equal to 0 or 1 and β is a non-rational algebraic number, then αβ is
transcendental.
√
So: 3
3,
21+
√
2,
√
√
2
2
, etc. are all transcendental.
Also: e π is transcendental.
If e π were algebraic, then (e π )i = e iπ = −1 would have to be
transcendental.
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March 13, 2007
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Conclusion
MTH 210 conjecture: π + e is irrational.
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Conclusion
MTH 210 conjecture: π + e is irrational.
Open problems:
Are π + e, π/e, π e , 2e , ln(π) irrational?
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March 13, 2007
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References
Aigner, M., Ziegler, G., Proofs from the book, 2000
Ribenboim, P., My Numbers, My Friends, 2000
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