MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 1
Feb 8, 2012 (Week 2)
MATH4221 Euclidean and Non-Euclidean Geometries
Tutorial Note 1
TA:
CHENG, Kam Hang Henry
Office:
Email:
[email protected]
Phone Number: 2358 7468 / 9853 8144
Tutorial Session:
T1A
Wed 17:30 – 18:20 @ 1505 (25-26)
3215 (Lift 21)
Office Hours:
Wed 15:30 – 17:30
Tutorial Website:
http://ihome.ust.hk/~kero/
4221.html
Topics covered in week 1:
1. Euclid’s Elements and Five Postulates
2. Informal Logic
1. Euclid’s Elements and Five Postulates
What you need to know:
 The five postulates in Euclid’s Elements
 Diagrams: Uses and caveats
The axiomatic method for studying geometry starts from Euclid’s Elements, and every result in
this book was based on the five postulates Euclid stated in Book I.
Postulate 1 (Unique Segment):
Two distinct points determine a unique line segment.
Postulate 2 (Segment Extension): Given any line segment, we can extend it in either direction
as long as we wish.
Postulate 3 (Circle):
Given any point 𝑂 and any line segment 𝐴𝐵, there exists a
circle with center 𝑂 and radius equal to 𝐴𝐵.
Postulate 4 (Right Angles):
All right angles are equal to each other.
Postulate 5 (Parallel Postulate):
Given any line 𝑙 and any point 𝑃 not on 𝑙, there exists one
unique line 𝑚 that passes through 𝑃 and is parallel to 𝑙.
Euclid’s deductions also made use of some rules called common notions.
Common Notion 1: Two things equal to the same thing are equal (i.e. = is transitive).
Common Notion 2: Two pairs of equal things add up to give equal things (i.e. + preserves =).
Common Notion 3: When two equal parts are subtracted from two equal things, the
remainders are equal (i.e. − preserves =).
Common Notion 4: Things coincide with each other are equal.
Common Notion 5: The “whole” is greater than the “part”.
Page 1 of 8
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 1
Feb 8, 2012 (Week 2)
We have seen in the lecture Euclid’s Proposition I.1, which states that given a line segment 𝐴𝐵,
one can construct an equilateral triangle with 𝐴𝐵 as one of the edges. Let’s look at how this
helped Euclid to prove his Proposition I.2.
Example 1.1
(Proposition I.2, Elements) Show that given a line segment 𝐴𝐵 and a point 𝐶,
one can construct a line segment 𝐶𝐷 which equals 𝐴𝐵.
Euclid’s Proof (adapted):
There’s nothing to prove if 𝐴 = 𝐵, so we assume that 𝐴 and 𝐵 are distinct points.
Since 𝐴 and 𝐵 are distinct points, at least one of them is
different from 𝐶. Without loss of generality, we suppose
that 𝐴 and 𝐶 are distinct points.
By Postulate 1, there exists a unique line segment 𝐴𝐶.
By Proposition I.1, we can construct an equilateral triangle
𝑐2
𝐵
𝑐1
𝐴
𝐸
𝐺
𝐹
𝐶
Δ𝐴𝐶𝐸.
𝐷
By Postulate 3, we can construct a circle 𝑐1 with center 𝐴
𝐻
and radius 𝐴𝐵.
By Postulate 2, we can extend the line segment 𝐸𝐴 as long as we wish, say to 𝐸𝐹, so that the
line segment 𝑨𝑭 meets 𝒄𝟏 at some point 𝑮. (?)
Now by Postulate 3 again, we can construct a circle 𝑐2 with center 𝐸 and radius 𝐸𝐺.
By Postulate 2, we can extend the line segment 𝐸𝐶 as long as we wish, say to 𝐸𝐻, so that the
line segment 𝑬𝑯 meets 𝒄𝟐 at some point 𝑫. (?)
We claim that 𝐶𝐷 is our required line segment, i.e. 𝐶𝐷 = 𝐴𝐵.
First, since 𝐸𝐺 and 𝐸𝐷 are radii of the same circle 𝑐2 , 𝐸𝐺 = 𝐸𝐷 by the definition of radius.
By Common Notion 3, since 𝐸𝐷 − 𝐸𝐶 = 𝐸𝐺 − 𝐸𝐴, we have 𝐶𝐷 = 𝐴𝐺.
Next, since 𝐴𝐵 and 𝐴𝐺 are radii of the same circle 𝑐1, 𝐴𝐵 = 𝐴𝐺 by the definition of radius.
Finally by Common Notion 1, since 𝐶𝐷 and 𝐴𝐵 both equal 𝐴𝐺, we get 𝐶𝐷 = 𝐴𝐵.
∎
In fact, there are some notions that have not been rigorously established yet in the above
ancient Greek proof. Such notions include what it should mean by being equal, what it should
mean by adding up two line segments, why the circle must cut the segment if the segment is
long enough, and so on. These notions will be built up step-by-step in an axiomatic way in this
course. We will do every proof rigorously starting from the next section.
Remark:
Starting from next week, we will develop the whole theory from a more systematic
set of axioms than Euclid’s five postulates. We will base our work on that set of
axioms instead, so please pay attention on the different contexts.
Page 2 of 8
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 1
Feb 8, 2012 (Week 2)
One more thing we see from the previous proof is that a diagram is sometimes useful for
geometrical deductions. However, be careful not to fall into traps of badly drawn diagrams!
Apart from the false proof on that “all triangles are isosceles” given in the lecture, here is
another example of a false proof that results from a badly drawn diagram. Use your
high-school geometry in this example.
Example 1.2
(Lewis Carroll)
The following is a false proof on that “a right angle is equal to
an arbitrary obtuse angle”.
“Proof”:
Given a right angle and an arbitrary obtuse angle 𝛼, we draw a quadrilateral
𝐴𝐵𝐶𝐷 with ∠𝐷𝐴𝐵 = 90°, ∠𝐴𝐵𝐶 = 𝛼, and 𝐴𝐷 = 𝐵𝐶.
Next let 𝐸 and 𝐹 be the mid-points of 𝐴𝐵 and 𝐶𝐷
respectively, and draw the two perpendicular bisectors
𝐺
𝐴
𝐵
𝐸
of 𝐴𝐵 and of 𝐶𝐷.
𝐷
It can be easily shown that 𝐴𝐵 and 𝐶𝐷 are not parallel,
so their perpendicular bisectors must meet at some point 𝐺.
𝛼
𝐶
𝐹
Since ∆𝐴𝐸𝐺 ≅ Δ𝐵𝐸𝐺 and ∆𝐷𝐹𝐺 ≅ Δ𝐶𝐹𝐺 (SAS), we have 𝐴𝐺 = 𝐵𝐺 and
𝐷𝐺 = 𝐶𝐺 (corresponding sides of congruent triangles).
Thus ∆𝐴𝐷𝐺 ≅ Δ𝐵𝐶𝐺 (SSS).
Now we have ∠𝐺𝐴𝐷 = ∠𝐺𝐵𝐶 and ∠𝐺𝐴𝐸 = ∠𝐺𝐵𝐸 (corresponding angles of
congruent triangles), so we get ∠𝐷𝐴𝐵 = ∠𝐴𝐵𝐶 by subtracting these two
equalities. Therefore a right angle is equal to an arbitrary obtuse angle.
∎
Find out the flaw in the above false proof.
Solution:
The flaw is in the final step.
Subtracting the two equalities ∠𝐺𝐴𝐷 = ∠𝐺𝐵𝐶 and ∠𝐺𝐴𝐸 = ∠𝐺𝐵𝐸
in fact does not give ∠𝐷𝐴𝐵 = ∠𝐴𝐵𝐶, as shown in the more accurate
diagram on the right.
𝐺
𝐴
𝐷
𝐵
𝐸
𝛼
𝐹
𝐶
We should note that diagrams should not be necessary in any proof of geometry. The use of
diagrams is just to help us think, but not to mislead us!
Page 3 of 8
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 1
Feb 8, 2012 (Week 2)
2. Informal Logic
What you need to know:
 Sentential logic, the five logical connectives:
 Predicate logic, quantifiers: ∀, ∃
 Logic rules
 The six “legal” types of justification
 Contrapositive, Proof by contradiction
 Proof by cases
not, and, or, →, ↔
To improve the system of common notions in Euclid’s Elements, we introduce the subject of
informal logic. Logic is the basis of all arguments in our subsequent proofs in geometry.
First of all, there should be no unstated assumptions in any proof. We should justify each step
in the proof clearly. The “legal” justifications include only the following six types:
 Statements in the hypothesis
 Axioms
 Theorems or proved results
 Definitions
 Previous steps in the argument
 Logic rules
Here are some logic rules about negation that can be used in justifying the steps in our proofs:
(Double Negation)
not(not 𝑆) = 𝑆.
(de Morgan’s Law)
not(𝑆1 and 𝑆2 ) = (not 𝑆1 ) or (not 𝑆2 ).
(de Morgan’s Law)
not(𝑆1 or 𝑆2 ) = (not 𝑆1 ) and (not 𝑆2 ).
(Universal Negation)
(Existential Negation)
Example 2.1
not(∀𝑥)(𝑆(𝑥)) = (∃𝑥)(not 𝑆(𝑥)).
not(∃𝑥)(𝑆(𝑥)) = (∀𝑥)(not 𝑆(𝑥)).
Show that not(𝑃 → 𝑄) = 𝑃 and (not 𝑄).
Proof:
not(𝑃 → 𝑄) = not(not 𝑃 or 𝑄)
= not(not 𝑃) and (not 𝑄)
= 𝑃 and (not 𝑄)
(By definition of →)
(de Morgan’s Law)
(Double Negation)
∎
Page 4 of 8
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Example 2.2
Proof:
Tutorial Note 1
Feb 8, 2012 (Week 2)
Show that not(∃𝑥)(𝜙(𝑥) and 𝜓(𝑥)) = (∀𝑥)(𝜙(𝑥) → not 𝜓(𝑥)).
not(∃𝑥)(𝜙(𝑥) and 𝜓(𝑥))
= not(∃𝑥)(𝜙(𝑥) and not(not 𝜓(𝑥)))
= not(∃𝑥)(not(𝜙(𝑥) → not 𝜓(𝑥)))
= (∀𝑥)(not(not(𝜙(𝑥) → not 𝜓(𝑥))))
= (∀𝑥)(𝜙(𝑥) → not 𝜓(𝑥))
(Double Negation)
(By the result from Example 2.1)
(Existential Negation)
(Double Negation)
∎
Here are some other logic rules about implication that are useful in this course:
(Modus Ponens)
If 𝑃 is true and (𝑃 ⇒ 𝑄), then 𝑄 is true.
(Modus Tollendo Tollens)
If (𝑃 ⇒ 𝑄) and not 𝑄 is true, then not 𝑃 is true.
(Modus Tollendo Ponens)
If (𝑃 or 𝑄) and not 𝑃 are true, then 𝑄 is true.
(Hypothetical Syllogism)
If (𝑃 ⇒ 𝑄) and (𝑄 ⇒ 𝑅), then (𝑃 ⇒ 𝑅).
If (𝑃 and 𝑄) is true, then 𝑃 is true. If (𝑃 and 𝑄) is true,
then 𝑄 is true.
(not 𝑄 → not 𝑃) = (𝑃 → 𝑄)
(Contrapositive)
The names of the above rules are just for your reference. They are not required in your work
in this course.
(Conjuction Elimination)
Remark:
The truth value of a statement depends on the axioms of the corresponding theory.
For example, you will see later in this course that the statement “the angle sum of a
triangle is 180°” is true in Euclidean geometry, but is false in hyperbolic geometry.
Example 2.3
(Rubin 2.F4 adapted) Suppose that the statements (𝑂 → 𝑉) , (𝐺 → 𝑀) ,
(𝐺 → (𝑂 or not 𝑀)), (𝐺 or not 𝑀) and (𝑀 or not 𝑉) are all true. Deduce
that (𝑂 ⇔ 𝐺).
Proof:
We need to show (𝑂 ⇒ 𝐺) and (𝐺 ⇒ 𝑂):
(i) Since (𝑀 or not 𝑉) is true, we have (𝑉 ⇒ 𝑀) by definition of →.
Since (𝐺 or not 𝑀) is true, we have (𝑀 ⇒ 𝐺) by definition of → again.
From (𝑂 ⇒ 𝑉), (𝑉 ⇒ 𝑀) and (𝑀 ⇒ 𝐺), we have (𝑂 ⇒ 𝐺) by Hypothetical Syllogism.
(ii) Now to show that (𝐺 ⇒ 𝑂), we first suppose that 𝐺 is true.
Then since 𝐺 and (𝐺 → 𝑀) are true, 𝑀 is true by Modus Ponens.
Since 𝐺 and (𝐺 → (𝑂 or not 𝑀)) are true, (𝑂 or not 𝑀) is true by Modus Ponens.
Since 𝑀 and (𝑂 or not 𝑀) are true, 𝑂 is true by Modus Tollendo Ponens.
Thus (𝐺 ⇒ 𝑂).
Now we have both (𝑂 ⇒ 𝐺) and (𝐺 ⇒ 𝑂), so (𝑂 ⇔ 𝐺).
Page 5 of 8
∎
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Example 2.4
Tutorial Note 1
Feb 8, 2012 (Week 2)
(Rubin 8.A2 adapted) Suppose that (∀𝑥)(𝑃(𝑥) → (not 𝑄(𝑥) or not 𝑅(𝑥)))
and (∃𝑥)(𝑄(𝑥) and 𝑅(𝑥)) are both true. Show that (∃𝑥)(not 𝑃(𝑥)) is true.
Proof:
Since (∃𝑥)(𝑄(𝑥) and 𝑅(𝑥)) is true, we can substitute 𝑥 with some particular 𝑢 to get
(𝑄(𝑢) and 𝑅(𝑢)) is true.
By de Morgan’s Law, not(not 𝑄(𝑢) or not 𝑅(𝑢)) is also true.
But since (∀𝑥)(𝑃(𝑥) → (not 𝑄(𝑥) or not 𝑅(𝑥))) is true, 𝑃(𝑢) → (not 𝑄(𝑢) or not 𝑅(𝑢)) is
also true.
Now both not(not 𝑄(𝑢) or not 𝑅(𝑢)) and 𝑃(𝑢) → (not 𝑄(𝑢) or not 𝑅(𝑢)) are true, so
not 𝑃(𝑢) is true by Modus Tollendo Tollens.
This implies that (∃𝑥)(not 𝑃(𝑥)) is true.
∎
Sometimes we do proofs indirectly, for example by contradiction or by proving the
contrapositive.
Example 2.5
Suppose that (𝐴 → (𝐵 → 𝐶)) , (𝐶 → not 𝐷) , (not 𝐸 → 𝐷) and (𝐴 and 𝐵)
are all true. Deduce that 𝐸 is true.
Proof:
[Proof by contradiction]
We first suppose on the contrary that not 𝐸 is true, and then deduce a contradiction:
(i) Since not 𝐸 and (not 𝐸 → 𝐷) are true, 𝐷 is true by Modus Ponens.
Since 𝐷 and (𝐶 → not 𝐷) are true, not 𝐶 is true by Modus Tollendo Tollens.
(ii) Since (𝐴 and 𝐵) is true by hypothesis, 𝐴 and 𝐵 are both true by Conjunction Elimination.
Since 𝐴 and (𝐴 → (𝐵 → 𝐶)) are true, (𝐵 → 𝐶) is true by Modus Ponens.
Since 𝐵 and (𝐵 → 𝐶) are true, 𝐶 is true by Modus Ponens.
Now both 𝐶 and not 𝐶 are true, which gives a contradiction.
Therefore 𝐸 is true.
∎
Alternative proof:
[Direct proof]
Since (𝐴 and 𝐵) is true by hypothesis, 𝐴 and 𝐵 are both true by Conjunction Elimination.
Since 𝐴 and (𝐴 → (𝐵 → 𝐶)) are both true, (𝐵 → 𝐶) is true by Modus Ponens.
Since 𝐵 and (𝐵 → 𝐶) are true, 𝐶 is also true by Modus Ponens.
Since 𝐶 and (𝐶 → not 𝐷) are true, not 𝐷 is true by Modus Ponens.
Finally since not 𝐷 and (not 𝐸 → 𝐷) are true, 𝐸 is true by Modus Tollendo Tollens.
∎
Page 6 of 8
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 1
Feb 8, 2012 (Week 2)
Finally, it is also “legal” to construct proofs by cases.
(Law of Excluded Middles)
(Proof by cases)
Example 2.6
During any proof, we can insert the step (𝑃 or not 𝑃), which is
always true.
If (𝑆1 or 𝑆2 or ⋯ or 𝑆𝑛 ) is true and (𝑆𝑖 ⇒ 𝐶) for each
1 ≤ 𝑖 ≤ 𝑛, then 𝐶 is also true.
(Rubin 2.H2 adapted) Suppose that (𝐴 or 𝐵), (𝐶 → not 𝐴), (𝐵 → 𝐷) and
(𝐶 → not 𝐷) are true. Show that not 𝐶 is true.
Proof:
We prove that not 𝐶 is true by cases:
(i) If 𝐴 is true, then since (𝐶 → not 𝐴) is true, not 𝐶 is true by Modus Tollendo Tollens.
(ii) If not 𝐴 is true, then since (𝐴 or 𝐵) is true, 𝐵 is true by Modus Tollendo Ponens.
Since 𝐵 and (𝐵 → 𝐷) are true, 𝐷 is true by Modus Ponens.
Since 𝐷 and (𝐶 → not 𝐷) are true, not 𝐶 is true by Modus Tollendo Tollens.
Now not 𝐶 is true is both cases, so it is true.
Page 7 of 8
∎
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 1
Feb 8, 2012 (Week 2)
Exercise
1. (Greenberg 2.2, 3) State the negations of Euclid’s Postulate 4 and Postulate 5.
2. (Greenberg 2.17) Which of the followings are valid arguments?
Hint:
Do not focus on the meaning or truth value of the statements, just focus on the logic.
(a) No frogs are poetical; some ducks are unpoetical. Hence, some ducks are not frogs.
(b) Gold is heavy; nothing but gold will silence him. Hence, nothing light will silence him.
(c) All lions are fierce; some lions do not drink coffee. Hence some creatures that drink coffee
are not fierce.
(d) Some pillows are soft; no pokers are soft. Hence, some pokers are not pillows.
3. (Rubin 2.F8, H4, 8.A4 adapted) Construct proofs for the following:
(a) Suppose that ((𝑀 or 𝑉) → 𝐺), (𝐺 → 𝑉) and (𝑂 → (not 𝐽 or not 𝑉)) are all true. Show
that (𝑂 → (𝑀 → not 𝐽)) is true.
(b) Suppose that ((𝐴 → 𝐵) or (𝐴 and not 𝐶)), 𝐴, (𝐶 → not 𝐵) and (not 𝐶 → 𝐵) are all true.
Show that not 𝐶 is true.
(c) If (∀𝑥)(𝑃(𝑥) → 𝑄(𝑥)) , (∃𝑥)(𝑅(𝑥) and not 𝑄(𝑥)) and (∀𝑥) (𝑅(𝑥) → (𝑃(𝑥) or 𝑆(𝑥)))
are all true, show that (∃𝑥)(𝑅(𝑥) and 𝑆(𝑥)) is true.
4. Show that (∃𝑥)(∀𝑦)(𝜙(𝑥, 𝑦)) ⇒ (∀𝑦)(∃𝑥)(𝜙(𝑥, 𝑦)). Is the converse true?
Page 8 of 8