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Irrational Numbers
Mathematics 15: Lecture 8
Dan Sloughter
Furman University
October 3, 2006
Dan Sloughter (Furman University)
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October 3, 2006
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Commensurability
I
Line segments of length a and length b are commensurable if there
exists a line segment of length c and positive integers m and n such
that a = mc and b = nc.
Dan Sloughter (Furman University)
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Commensurability
I
Line segments of length a and length b are commensurable if there
exists a line segment of length c and positive integers m and n such
that a = mc and b = nc.
I
Example: Line segments of lengths a = 3 and b = 5.5 are
commensurable since a = 6 × 0.5 and b = 11 × 0.5.
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Commensurability
I
Line segments of length a and length b are commensurable if there
exists a line segment of length c and positive integers m and n such
that a = mc and b = nc.
I
Example: Line segments of lengths a = 3 and b = 5.5 are
commensurable since a = 6 × 0.5 and b = 11 × 0.5.
I
Example: Line segments of lengths a = 73 and b = 25 are
1
1
commensurable since a = 15 × 35
and b = 14 × 35
.
Dan Sloughter (Furman University)
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Commensurability
I
Line segments of length a and length b are commensurable if there
exists a line segment of length c and positive integers m and n such
that a = mc and b = nc.
I
Example: Line segments of lengths a = 3 and b = 5.5 are
commensurable since a = 6 × 0.5 and b = 11 × 0.5.
I
Example: Line segments of lengths a = 73 and b = 25 are
1
1
commensurable since a = 15 × 35
and b = 14 × 35
.
I
Note: two numbers a and b are commensurable if and only if
for some integers n and m.
Dan Sloughter (Furman University)
Irrational Numbers
a
b
October 3, 2006
=
m
n
2 / 13
Commensurability
I
Line segments of length a and length b are commensurable if there
exists a line segment of length c and positive integers m and n such
that a = mc and b = nc.
I
Example: Line segments of lengths a = 3 and b = 5.5 are
commensurable since a = 6 × 0.5 and b = 11 × 0.5.
I
Example: Line segments of lengths a = 73 and b = 25 are
1
1
commensurable since a = 15 × 35
and b = 14 × 35
.
I
Note: two numbers a and b are commensurable if and only if
for some integers n and m.
I
In particular, a rational number m
n is commensurable with 1 (and only
rational numbers are commensurable with 1).
Dan Sloughter (Furman University)
Irrational Numbers
a
b
October 3, 2006
=
m
n
2 / 13
Incommensurables
I
Consider a square with vertices at A, B, C , and D :
D
A
P
B1
D1
C1
C
Dan Sloughter (Furman University)
B
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October 3, 2006
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Incommensurables
I
Consider a square with vertices at A, B, C , and D :
D
A
P
B1
D1
C1
C
I
B
If the diagonal of the square is commensurable with the sides of the
square, we could mark off a distance AP along the side AB such that
both AB and AC are multiples of AP.
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
3 / 13
Incommensurables
I
Consider a square with vertices at A, B, C , and D :
D
A
P
B1
D1
C1
C
B
I
If the diagonal of the square is commensurable with the sides of the
square, we could mark off a distance AP along the side AB such that
both AB and AC are multiples of AP.
I
Let B1 be the point on AC such that B1 C = AB and let C1 be the
point on AB such that B1 C1 is perpendicular to AC .
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Incommensurables
I
Consider a square with vertices at A, B, C , and D :
D
A
P
B1
D1
C1
C
B
I
If the diagonal of the square is commensurable with the sides of the
square, we could mark off a distance AP along the side AB such that
both AB and AC are multiples of AP.
I
Let B1 be the point on AC such that B1 C = AB and let C1 be the
point on AB such that B1 C1 is perpendicular to AC .
I
Then AB1 = B1 C1 = C1 B.
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Incommensurables (cont’d)
I
Hence AB1 and AC1 are multiples of AP. That is, for the square with
vertices at A, B1 , C1 , and D1 , the diagonal AC1 and the side AB1 are
both multiples of AP.
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Incommensurables (cont’d)
I
Hence AB1 and AC1 are multiples of AP. That is, for the square with
vertices at A, B1 , C1 , and D1 , the diagonal AC1 and the side AB1 are
both multiples of AP.
I
But the sides of this new square are less than half the length of the
sides of the original square.
Dan Sloughter (Furman University)
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October 3, 2006
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Incommensurables (cont’d)
I
Hence AB1 and AC1 are multiples of AP. That is, for the square with
vertices at A, B1 , C1 , and D1 , the diagonal AC1 and the side AB1 are
both multiples of AP.
I
But the sides of this new square are less than half the length of the
sides of the original square.
I
We could continue to construct squares in this fashion until the
length of a side of the square was less than AP, contradicting the
assumption that the sides are a multiple of AP.
Dan Sloughter (Furman University)
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October 3, 2006
4 / 13
Incommensurables (cont’d)
I
Hence AB1 and AC1 are multiples of AP. That is, for the square with
vertices at A, B1 , C1 , and D1 , the diagonal AC1 and the side AB1 are
both multiples of AP.
I
But the sides of this new square are less than half the length of the
sides of the original square.
I
We could continue to construct squares in this fashion until the
length of a side of the square was less than AP, contradicting the
assumption that the sides are a multiple of AP.
I
Hence we must conclude that the diagonal of a square is not
commensurable with a side of the square.
Dan Sloughter (Furman University)
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The square root of 2
I
Consider a square with sides of length 1.
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The square root of 2
I
Consider a square with sides of length 1.
I
By the Pythagorean theorem, the diagonal of the square is
Dan Sloughter (Furman University)
Irrational Numbers
√
2.
October 3, 2006
5 / 13
The square root of 2
I
I
I
Consider a square with sides of length 1.
√
By the Pythagorean theorem, the diagonal of the square is 2.
√
√
Hence, 1 and 2 are not commensurable; that is, 2 is not rational.
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
5 / 13
The square root of 2
I
I
I
I
Consider a square with sides of length 1.
√
By the Pythagorean theorem, the diagonal of the square is 2.
√
√
Hence, 1 and 2 are not commensurable; that is, 2 is not rational.
The Pythagoreans had assumed that all lengths were commensurable
and had used this assumption in many of their basic geometric
arguments. Hence Tannery’s comment that this was “un vértible
scandale logique.”
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
5 / 13
The square root of 2
I
I
I
Consider a square with sides of length 1.
√
By the Pythagorean theorem, the diagonal of the square is 2.
√
√
Hence, 1 and 2 are not commensurable; that is, 2 is not rational.
I
The Pythagoreans had assumed that all lengths were commensurable
and had used this assumption in many of their basic geometric
arguments. Hence Tannery’s comment that this was “un vértible
scandale logique.”
I
The Greeks found that they could work with such numbers as long as
they considered them as lengths of line segments, but never
developed methods for working with them algebraically.
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October 3, 2006
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The geometry of numbers
I
The Greeks thought of numbers as the lengths of line segments.
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The geometry of numbers
I
The Greeks thought of numbers as the lengths of line segments.
I
To square a number was to construct a square with sides of the length
of the given number and to cube a number was to construct a cube.
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Example
I
Recall: (a + b)2 = a2 + 2ab + b 2
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Example
I
Recall: (a + b)2 = a2 + 2ab + b 2
I
The algebraic proof:
(a + b)2 = (a + b)(a + b)
= (a + b)a + (a + b)b
= a2 + ab + ab + b 2
= a2 + 2ab + b 2 .
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
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Example
I
Recall: (a + b)2 = a2 + 2ab + b 2
I
The algebraic proof:
(a + b)2 = (a + b)(a + b)
= (a + b)a + (a + b)b
= a2 + ab + ab + b 2
= a2 + 2ab + b 2 .
I
Geometrically, the result follows from the picture:
b
a
Dan Sloughter (Furman University)
2
ab
b
a2
ab
a
b
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Theory of proportions
I
Eudoxus (408? B.C. - 355? B.C.) restored the foundations of Greek
mathematics with his definition for the equality of two proportions:
ratios a : b and c : d are equal if for any two integers n and m one of
the following is true:
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
8 / 13
Theory of proportions
I
Eudoxus (408? B.C. - 355? B.C.) restored the foundations of Greek
mathematics with his definition for the equality of two proportions:
ratios a : b and c : d are equal if for any two integers n and m one of
the following is true:
I
na > mb and nc > md,
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
8 / 13
Theory of proportions
I
Eudoxus (408? B.C. - 355? B.C.) restored the foundations of Greek
mathematics with his definition for the equality of two proportions:
ratios a : b and c : d are equal if for any two integers n and m one of
the following is true:
I
I
na > mb and nc > md,
na = mb and nc = md,
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
8 / 13
Theory of proportions
I
Eudoxus (408? B.C. - 355? B.C.) restored the foundations of Greek
mathematics with his definition for the equality of two proportions:
ratios a : b and c : d are equal if for any two integers n and m one of
the following is true:
I
I
I
na > mb and nc > md,
na = mb and nc = md,
na < mb and nc < md.
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
8 / 13
Theory of proportions
I
Eudoxus (408? B.C. - 355? B.C.) restored the foundations of Greek
mathematics with his definition for the equality of two proportions:
ratios a : b and c : d are equal if for any two integers n and m one of
the following is true:
I
I
I
I
na > mb and nc > md,
na = mb and nc = md,
na < mb and nc < md.
Note: if ba and dc are rational numbers, this is saying that ba = dc if for
a
m
c
m
a
m
any rational number m
n , either b > n and d > n , or b = n and
c
m
a
m
c
m
d = n , or b < n and d < n .
Dan Sloughter (Furman University)
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October 3, 2006
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Theory of proportions (cont’d)
I
Note: if ba is not a rational number, then the second option is not
possible.
Dan Sloughter (Furman University)
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Theory of proportions (cont’d)
I
I
Note: if ba is not a rational number, then the second option is not
possible.
In that case, the ratio a : b splits the rational numbers into two sets:
Dan Sloughter (Furman University)
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October 3, 2006
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Theory of proportions (cont’d)
I
I
Note: if ba is not a rational number, then the second option is not
possible.
In that case, the ratio a : b splits the rational numbers into two sets:
I
those for which na > mb, that is,
Dan Sloughter (Furman University)
a
b
>
Irrational Numbers
m
n
October 3, 2006
9 / 13
Theory of proportions (cont’d)
I
I
Note: if ba is not a rational number, then the second option is not
possible.
In that case, the ratio a : b splits the rational numbers into two sets:
I
I
those for which na > mb, that is,
those for which na < mb, that is,
Dan Sloughter (Furman University)
a
b
a
b
>
<
Irrational Numbers
m
n
m
n.
October 3, 2006
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Dedekind cuts
I
Richard Dedekind (1831 - 1916) provides the first logically complete
definition of irrational numbers in 1872.
Dan Sloughter (Furman University)
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October 3, 2006
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Dedekind cuts
I
Richard Dedekind (1831 - 1916) provides the first logically complete
definition of irrational numbers in 1872.
I
Motivation: the identification of real numbers with points on a line
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
10 / 13
Dedekind cuts
I
Richard Dedekind (1831 - 1916) provides the first logically complete
definition of irrational numbers in 1872.
I
Motivation: the identification of real numbers with points on a line
(A1 , A2 ) is a Dedekind cut if A1 and A2 are subsets of the rational
numbers Q such that
I
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
10 / 13
Dedekind cuts
I
Richard Dedekind (1831 - 1916) provides the first logically complete
definition of irrational numbers in 1872.
I
Motivation: the identification of real numbers with points on a line
(A1 , A2 ) is a Dedekind cut if A1 and A2 are subsets of the rational
numbers Q such that
I
I
neither A1 nor A2 is empty
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
10 / 13
Dedekind cuts
I
Richard Dedekind (1831 - 1916) provides the first logically complete
definition of irrational numbers in 1872.
I
Motivation: the identification of real numbers with points on a line
(A1 , A2 ) is a Dedekind cut if A1 and A2 are subsets of the rational
numbers Q such that
I
I
I
neither A1 nor A2 is empty
every rational number is in exactly one of A1 and A2
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
10 / 13
Dedekind cuts
I
Richard Dedekind (1831 - 1916) provides the first logically complete
definition of irrational numbers in 1872.
I
Motivation: the identification of real numbers with points on a line
(A1 , A2 ) is a Dedekind cut if A1 and A2 are subsets of the rational
numbers Q such that
I
I
I
I
neither A1 nor A2 is empty
every rational number is in exactly one of A1 and A2
if a1 is in A1 and a2 is in A2 , then a1 < a2 .
Dan Sloughter (Furman University)
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October 3, 2006
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Dedekind cuts (cont’d)
I
Given a Dedekind cut (A1 A2 ), then one of the following must hold
Dan Sloughter (Furman University)
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Dedekind cuts (cont’d)
I
Given a Dedekind cut (A1 A2 ), then one of the following must hold
I
A1 has a largest element and A2 does not have a smallest element
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
11 / 13
Dedekind cuts (cont’d)
I
Given a Dedekind cut (A1 A2 ), then one of the following must hold
I
I
A1 has a largest element and A2 does not have a smallest element
A2 has a smallest element and A1 does not have a largest element
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
11 / 13
Dedekind cuts (cont’d)
I
Given a Dedekind cut (A1 A2 ), then one of the following must hold
I
I
I
A1 has a largest element and A2 does not have a smallest element
A2 has a smallest element and A1 does not have a largest element
A1 does not have a largest element and A2 does not have a smallest
element.
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
11 / 13
Dedekind cuts (cont’d)
I
Given a Dedekind cut (A1 A2 ), then one of the following must hold
I
I
I
I
A1 has a largest element and A2 does not have a smallest element
A2 has a smallest element and A1 does not have a largest element
A1 does not have a largest element and A2 does not have a smallest
element.
In the first two cases, the Dedekind cut is determined by some
rational number q:
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
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Dedekind cuts (cont’d)
I
Given a Dedekind cut (A1 A2 ), then one of the following must hold
I
I
I
I
A1 has a largest element and A2 does not have a smallest element
A2 has a smallest element and A1 does not have a largest element
A1 does not have a largest element and A2 does not have a smallest
element.
In the first two cases, the Dedekind cut is determined by some
rational number q:
I
either A1 is the set of all rational numbers x with x ≤ q and A2 is the
set of all rational numbers x with x > q,
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
11 / 13
Dedekind cuts (cont’d)
I
Given a Dedekind cut (A1 A2 ), then one of the following must hold
I
I
I
I
A1 has a largest element and A2 does not have a smallest element
A2 has a smallest element and A1 does not have a largest element
A1 does not have a largest element and A2 does not have a smallest
element.
In the first two cases, the Dedekind cut is determined by some
rational number q:
I
I
either A1 is the set of all rational numbers x with x ≤ q and A2 is the
set of all rational numbers x with x > q,
or A1 is the set of all rational numbers x with x < q and A2 is the set
of all rational numbers x with x ≥ q.
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
11 / 13
Dedekind cuts (cont’d)
I
Given a Dedekind cut (A1 A2 ), then one of the following must hold
I
I
I
I
In the first two cases, the Dedekind cut is determined by some
rational number q:
I
I
I
A1 has a largest element and A2 does not have a smallest element
A2 has a smallest element and A1 does not have a largest element
A1 does not have a largest element and A2 does not have a smallest
element.
either A1 is the set of all rational numbers x with x ≤ q and A2 is the
set of all rational numbers x with x > q,
or A1 is the set of all rational numbers x with x < q and A2 is the set
of all rational numbers x with x ≥ q.
The third case defines an irrational number.
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
11 / 13
Dedekind cuts (cont’d)
I
Given a Dedekind cut (A1 A2 ), then one of the following must hold
I
I
I
I
In the first two cases, the Dedekind cut is determined by some
rational number q:
I
I
I
I
A1 has a largest element and A2 does not have a smallest element
A2 has a smallest element and A1 does not have a largest element
A1 does not have a largest element and A2 does not have a smallest
element.
either A1 is the set of all rational numbers x with x ≤ q and A2 is the
set of all rational numbers x with x > q,
or A1 is the set of all rational numbers x with x < q and A2 is the set
of all rational numbers x with x ≥ q.
The third case defines an irrational number.
√
Example: 2 is defined by letting A1 be the set of all negative
rational numbers along with all nonnegative rational numbers q for
which q 2 < 2 and A2 is the set of all positive rational numbers for
which q 2 > 2.
Dan Sloughter (Furman University)
Irrational Numbers
October 3, 2006
11 / 13
Problems
√
√
1. Here is another proof that 2 is not rational: Suppose 2 is rational.
Then√there exist integers m and n, having no common divisors, such
that 2 = m
n , that is
m2
2= 2.
n
Hence m2 = 2n2 .
a. Explain why we may conclude that m is an even number.
b. Since m is an even number, we may write m = 2c for some integer c.
Then we have 2n2 = m2 = (2c)2 = 4c 2 . Explain why this implies that n
is an even number.
c. Explain
√ why (a) and (b) imply a contradiction of our original assumption
that 2 is rational.
2. Adapt√the proof of the previous problem that
that 3 is irrational.
Dan Sloughter (Furman University)
Irrational Numbers
√
2 is irrational to show
October 3, 2006
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Problems (cont’d)
3. In class we showed how the algebraic identity
(a + b)2 = a2 + 2ab + b 2
could be illustrated geometrically. Illustrate each of the following in a
similar fashion:
a. a(b + c) = ab + ac
b. (a − b)2 = a2 − 2ab + b 2 .
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