Bertrand Russell 18th May 1872 – 2nd February 1970

Early history

Born in 1872 into one of the most prestigious families in Britain. Both parents died before he was five years old Raised by a staunchly Presbyterian Grandmother Educated by various tutors but introduced to Euclid’s works by his brother. (video) Went to Cambridge in 1890

His problems with Euclid’s Elements (1902)

• Proposition 1 – “an assumption not noticed by Euclid because of the dangerous habit of using a figure”

• Proposition 4 – “a tissue of nonsense”

• Proposition 6 – missing an axiom

• Proposition 7 – “so thoroughly fallacious that Euclid would have done better not to have attempted a proof”

http://www-history.mcs.st-and.ac.uk/Extras/Russell_Euclid.html

http://www-history.mcs.st-and.ac.uk/Extras/Russell_Euclid.htmlhttp://www-history.mcs.st-and.ac.uk/Extras/Russell_Euclid.htmlhttp://www-history.mcs.st-and.ac.uk/Extras/Russell_Euclid.htmlhttp://www-history.mcs.st-and.ac.uk/Extras/Russell_Euclid.htmlhttp://www-history.mcs.st-and.ac.uk/Extras/Russell_Euclid.htmlhttp://www-history.mcs.st-and.ac.uk/Extras/Russell_Euclid.html

Proposition 1 It is required to construct an equilateral triangle Describe the circle BCD with centre A and radius AB. Again describe the circle ACE with centre B and radius BA Join the straight lines CA and CB from the point C at which the circles cut one another to the points A and B. Now, since the point A is the centre of the circle CDB then AC equals AB. Since the point B is the centre of the circle CAE therefore BC equals BA But AC was proved equal to AB therefore each of the straight lines AC and BC equals AB And things which equal the same thing also equal one another, therefore AC also equals BC Therefore the three straight lines AC, AB and BC equal one another. Therefore the triangle ABC is equilateral and it has been constructed on the given finite straight line AB.

Proposition 4 If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides.

A

B C

D

E F

AB and AC equal DE and DF respectively and angle BAC equals EDF

Superimpose ABC on DEF and all points, lines and angles coincide.

Proposition 6 If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.

I say that the side AB also equals the side AC.

If AB does not equal AC, then one of them is greater.

Let AB be greater. Cut off DB from AB the greater equal to AC the less, and

join DC. I.4

C.N.5

Since DB equals AC, and BC is common, therefore the two

sides DB and BC equal the two sides AC and CB respectively, and the

angle DBC equals the angle ACB.

Therefore the base DC equals the base AB, and the triangle

DBC equals the triangle ACB, the less equals the greater, which is absurd.

Therefore AB is not unequal to AC, it therefore equals it.

Let ABC be a triangle where angle ABC equals angle ACB

http://aleph0.clarku.edu/~djoyce/elements/bookI/propI4.htmlhttp://aleph0.clarku.edu/~djoyce/elements/bookI/cn.html

Proposition 7 Given two straight lines constructed from the ends of a straight line and meeting in a point, there cannot be constructed from the ends of the same straight line, and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each equal to that from the same end.

“Many more general criticisms might be passed on Euclid's methods, and on his conception of Geometry; but the above definite fallacies seem sufficient to show that the value of his work as a masterpiece of logic has been very grossly exaggerated.”

B Russell, The Teaching of Euclid, The Mathematical Gazette 2 (33) (1902), 165-167.

Russell’s Paradox 1902: Gottlob Frege - Grundgesetze der Arithmetik (Basics of Arithmetic) Volume II

We could also have the set of all “non-teacups”

Which contains itself

For any formula 𝑃(𝑥) where 𝑥 can be anything then there exists a set {𝑥: 𝑃 𝑥 } In other words if T is the property of being a teacup then the set of all teacups S is

𝑆 = {𝑥: 𝑇 𝑥 }

Call the set of all sets that are not members of themselves “R.”

Russell’s Paradox

If R is a member of itself, then by definition it must not be a member of itself.

Similarly, if R is not a member of itself, then by definition it must be a member of itself.

Books which refer to themselves Books which don’t refer to themselves

A book about all books which don’t refer to themselves

Principles of mathematics (1903)

“In spite of its five hundred pages the book is much too short.” - G.H. Hardy

“[It] turned out to be a crude and rather immature draft of the subsequent work [Principia Mathematica]” – Russell (1959)

Principia Mathematica

• Was an attempt to describe a set of symbolic axioms and rules from which all mathematics was derived.

• Unrestricted creation of sets is no longer allowed.

• Theory of Types.

Russell’s Paradox and the Theory of Types

• All men in a town must be clean shaven

• Men must either shave themselves or be shaved by the barber

• The (male) barber is only allowed to shave those who don’t shave themselves.

• What does the barber do?

Russell’s Paradox and the Theory of Types

• Don’t let anyone shave themselves

• Form several castes in society

• Caste 1 shaves Caste 2

• Caste 2 shaves Caste 3 etc.

• Caste 1 remains unshaven

A proof from Principia Mathematica Volume I, page 372

Dot notation

Dots are effectively brackets

1 + 2 × 3 + 4 × (5 + 6)

1 + 2. × .3 + 4.× 5 + 6

1 + 2 × 3 + 4

1 + 2. × 3 ∶ + 4

Proof of 1+1

( 𝛼, 𝑏 ∈ 1 ⇒ 𝛼 ∩ 𝛽 = ∅ ≡ 𝛼 ∪ 𝛽 ∈ 2) )

1 and 2 have been clearly defined by this stage

( 𝛼, 𝑏 ∈ 1 ⇒ 𝛼 ∩ 𝛽 = ∅ ≡ 𝛼 ∪ 𝛽 ∈ 2)

This is simply a statement of a previous proposition that has already been proved

𝛼 = 𝑥 , 𝛽 = 𝑦 ⇒ 𝛼 ∪ 𝛽 ∈ 2 ≡ 𝑥 ≠ 𝑦

And by another previously proved theorem (if x and y are different then…)

𝑥 ∩ 𝑦 = ∅

And by another previously proved theorem (the two generic sets 𝛼 and 𝛽)

𝛼 ∩ 𝛽 = ∅

"Conclusion (1), with propositions ∗11.11 and ∗11.35, implies that if

11.11 and 11.35 just let you split a property into two and prove it separately

∃ 𝑥, 𝑦 and 𝛼 = {𝑥}, 𝛽 = {𝑦} ⇒ 𝛼 ∪ 𝛽 ∈ 2 ≡ 𝛼 ∩ 𝛽 = ∅

These two theorems just say that you can prove things separately and then put them back together to prove the whole

Putting it all together ( 𝛼, 𝑏 ∈ 1 ⇒ 𝛼 ∩ 𝛽 = ∅ ≡ 𝛼 ∪ 𝛽 ∈ 2)

𝛼 = 𝑥 , 𝛽 = 𝑦 ⇒ 𝛼 ∪ 𝛽 ∈ 2 ≡ 𝑥 ≠ 𝑦

≡ 𝑥 ∩ 𝑦 = ∅

≡ 𝛼 ∩ 𝛽 = ∅

∃ 𝑥, 𝑦 and 𝛼 = {𝑥}, 𝛽 = {𝑦} ⇒ 𝛼 ∪ 𝛽 = 2 ≡ 𝛼 ∩ 𝛽 = ∅

And if

(1)

(2)

By (1) and (2) the proposition has been demonstrated

Actual proof of 1+1=2

Vol II, page 86

Gödel's incompleteness theorem

For any system of consistent axioms, there will always be statements about the natural numbers that are true but unprovable in the system. Any system like this cannot demonstrate its own consistency

(1931)

Russell’s life after PM Spent time in prison for opposing WW1 Moved to America – lost his position at the University of Chicago over his opinions on sexual morality Eventually agreed that conflict opposing Nazi Germany was necessary during WW2 Survived a plane crash (1948) Spent post war period opposing nuclear armament Nobel prize for Literature in 1950 Jailed for 7 days (age 89) for his part in an anti-nuclear demonstration Died of Influenza in 1970

Further “reading”

Closing message from Bertrand Russell

Clip from https://www.youtube.com/watch?v=1bZv3pSaLtY )

https://www.youtube.com/watch?v=1bZv3pSaLtYhttps://www.youtube.com/watch?v=1bZv3pSaLtY