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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
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
• 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”
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.
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.
Let ABC be a triangle where angle ABC equals angle ACB
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.
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.
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
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
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 }
We could also have the set of all “non-teacups”
Which contains itself
Russell’s Paradox
Call the set of all sets that are not members of
themselves “R.”
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.
A book about all books which don’t refer to themselves
Books which refer to themselves
Books which don’t refer to
Principles of mathematics
“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
• Theory of Types.
Russell’s Paradox and the Theory of
• All men in a town must be clean
• Men must either shave
themselves or be shaved by the
• 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
• Don’t let anyone shave
• 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 =∅
11.11 and 11.35 just let you split a property into two and prove
it separately
"Conclusion (1), with propositions ∗11.11
and ∗11.35, implies that if
∃ 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 =∅
And if
∃ 𝑥, 𝑦 and 𝛼 = {𝑥}, 𝛽 = {𝑦} ⇒ 𝛼 ∪ 𝛽 = 2 ≡ 𝛼 ∩ 𝛽 = ∅
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
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 )