Irrational Numbers - Mathematics 15: Lecture 8

Irrational NumbersMathematics 15: Lecture 8

Dan Sloughter

Furman University

October 3, 2006

Dan Sloughter (Furman University) Irrational Numbers October 3, 2006 1 / 13

Commensurability

I Line segments of length a and length b are commensurable if thereexists a line segment of length c and positive integers m and n suchthat a = mc and b = nc .

I Example: Line segments of lengths a = 3 and b = 5.5 arecommensurable since a = 6× 0.5 and b = 11× 0.5.

I Example: Line segments of lengths a = 37 and b = 2

5 arecommensurable since a = 15× 1

35 and b = 14× 135 .

I Note: two numbers a and b are commensurable if and only if ab = m

nfor some integers n and m.

I In particular, a rational number mn is commensurable with 1 (and only

rational numbers are commensurable with 1).


Commensurability





35 and b = 14× 135 .






Commensurability





35 and b = 14× 135 .






Commensurability





35 and b = 14× 135 .






Commensurability





35 and b = 14× 135 .






Incommensurables

I Consider a square with vertices at A, B, C , and D :A

BC

D

B1

C1

D1

P

I If the diagonal of the square is commensurable with the sides of thesquare, we could mark off a distance AP along the side AB such thatboth AB and AC are multiples of AP.

I Let B1 be the point on AC such that B1C = AB and let C1 be thepoint on AB such that B1C1 is perpendicular to AC .

I Then AB1 = B1C1 = C1B.


Incommensurables


BC

D

B1

C1

D1

P





Incommensurables


BC

D

B1

C1

D1

P





Incommensurables


BC

D

B1

C1

D1

P





Incommensurables (cont’d)

I Hence AB1 and AC1 are multiples of AP. That is, for the square withvertices at A, B1, C1, and D1, the diagonal AC1 and the side AB1 areboth multiples of AP.

I But the sides of this new square are less than half the length of thesides of the original square.

I We could continue to construct squares in this fashion until thelength of a side of the square was less than AP, contradicting theassumption that the sides are a multiple of AP.

I Hence we must conclude that the diagonal of a square is notcommensurable with a side of the square.




















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√

2.

I Hence, 1 and√

2 are not commensurable; that is,√

2 is not rational.

I The Pythagoreans had assumed that all lengths were commensurableand had used this assumption in many of their basic geometricarguments. Hence Tannery’s comment that this was “un vertiblescandale logique.”

I The Greeks found that they could work with such numbers as long asthey considered them as lengths of line segments, but neverdeveloped methods for working with them algebraically.





2.

I Hence, 1 and√


2 is not rational.







2.

I Hence, 1 and√


2 is not rational.







2.

I Hence, 1 and√


2 is not rational.







2.

I Hence, 1 and√


2 is not rational.




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 lengthof the given number and to cube a number was to construct a cube.


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 lengthof the given number and to cube a number was to construct a cube.


Example

I Recall: (a + b)2 = a2 + 2ab + b2

I The algebraic proof:

(a + b)2 = (a + b)(a + b)

= (a + b)a + (a + b)b

= a2 + ab + ab + b2

= a2 + 2ab + b2.

I Geometrically, the result follows from the picture:

a

a b

ab

b ab b

a 2

2


Example

I Recall: (a + b)2 = a2 + 2ab + b2


(a + b)2 = (a + b)(a + b)

= (a + b)a + (a + b)b

= a2 + ab + ab + b2

= a2 + 2ab + b2.


a

a b

ab

b ab b

a 2

2


Example

I Recall: (a + b)2 = a2 + 2ab + b2


(a + b)2 = (a + b)(a + b)

= (a + b)a + (a + b)b

= a2 + ab + ab + b2

= a2 + 2ab + b2.


a

a b

ab

b ab b

a 2

2


Theory of proportions

I Eudoxus (408? B.C. - 355? B.C.) restored the foundations of Greekmathematics 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 ofthe following is true:

I na > mb and nc > md ,I na = mb and nc = md ,I na < mb and nc < md .

I Note: if ab and c

d are rational numbers, this is saying that ab = c

d if forany rational number m

n , either ab > m

n and cd > m

n , or ab = m

n andcd = m

n , or ab < m

n and cd < m

n .




I na > mb and nc > md ,

I na = mb and nc = md ,I na < mb and nc < md .

I Note: if ab and c



n , either ab > m

n and cd > m

n , or ab = m

n andcd = m

n , or ab < m

n and cd < m

n .




I na > mb and nc > md ,I na = mb and nc = md ,

I na < mb and nc < md .

I Note: if ab and c



n , either ab > m

n and cd > m

n , or ab = m

n andcd = m

n , or ab < m

n and cd < m

n .





I Note: if ab and c



n , either ab > m

n and cd > m

n , or ab = m

n andcd = m

n , or ab < m

n and cd < m

n .





I Note: if ab and c



n , either ab > m

n and cd > m

n , or ab = m

n andcd = m

n , or ab < m

n and cd < m

n .


Theory of proportions (cont’d)

I Note: if ab is not a rational number, then the second option is not

possible.

I In that case, the ratio a : b splits the rational numbers into two sets:

I those for which na > mb, that is, ab > m

nI those for which na < mb, that is, a

b < mn .




possible.I In that case, the ratio a : b splits the rational numbers into two sets:



b < mn .






n

I those for which na < mb, that is, ab < m

n .







b < mn .


Dedekind cuts

I Richard Dedekind (1831 - 1916) provides the first logically completedefinition of irrational numbers in 1872.

I Motivation: the identification of real numbers with points on a lineI (A1,A2) is a Dedekind cut if A1 and A2 are subsets of the rational

numbers Q such that

I neither A1 nor A2 is emptyI every rational number is in exactly one of A1 and A2

I if a1 is in A1 and a2 is in A2, then a1 < a2.


Dedekind cuts


I Motivation: the identification of real numbers with points on a line

I (A1,A2) is a Dedekind cut if A1 and A2 are subsets of the rationalnumbers Q such that




Dedekind cuts



numbers Q such that




Dedekind cuts



numbers Q such thatI neither A1 nor A2 is empty

I every rational number is in exactly one of A1 and A2



Dedekind cuts



numbers Q such thatI neither A1 nor A2 is emptyI every rational number is in exactly one of A1 and A2



Dedekind cuts



numbers Q such thatI neither A1 nor A2 is emptyI every rational number is in exactly one of A1 and A2



Dedekind cuts (cont’d)

I Given a Dedekind cut (A1A2), then one of the following must hold

I A1 has a largest element and A2 does not have a smallest elementI A2 has a smallest element and A1 does not have a largest elementI A1 does not have a largest element and A2 does not have a smallest

element.

I In the first two cases, the Dedekind cut is determined by somerational number q:

I either A1 is the set of all rational numbers x with x ≤ q and A2 is theset of all rational numbers x with x > q,

I or A1 is the set of all rational numbers x with x < q and A2 is the setof all rational numbers x with x ≥ q.

I The third case defines an irrational number.

I Example:√

2 is defined by letting A1 be the set of all negativerational numbers along with all nonnegative rational numbers q forwhich q2 < 2 and A2 is the set of all positive rational numbers forwhich q2 > 2.



I Given a Dedekind cut (A1A2), then one of the following must holdI A1 has a largest element and A2 does not have a smallest element

I A2 has a smallest element and A1 does not have a largest elementI A1 does not have a largest element and A2 does not have a smallest

element.





I Example:√




I Given a Dedekind cut (A1A2), then one of the following must holdI A1 has a largest element and A2 does not have a smallest elementI A2 has a smallest element and A1 does not have a largest element

I A1 does not have a largest element and A2 does not have a smallestelement.





I Example:√




I Given a Dedekind cut (A1A2), then one of the following must holdI A1 has a largest element and A2 does not have a smallest elementI A2 has a smallest element and A1 does not have a largest elementI A1 does not have a largest element and A2 does not have a smallest

element.





I Example:√





element.





I Example:√





element.





I Example:√





element.





I Example:√





element.





I Example:√





element.





I Example:√



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, suchthat

√2 = m

n , that is

2 =m2

n2.

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 = 4c2. Explain why this implies that nis an even number.

c. Explain why (a) and (b) imply a contradiction of our original assumptionthat

√2 is rational.

2. Adapt the proof of the previous problem that√

2 is irrational to showthat

√3 is irrational.


Problems (cont’d)

3. In class we showed how the algebraic identity

(a + b)2 = a2 + 2ab + b2

could be illustrated geometrically. Illustrate each of the following in asimilar fashion:

a. a(b + c) = ab + acb. (a− b)2 = a2 − 2ab + b2.


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Irrational Numbers - Mathematics 15: Lecture 8