Download Mathematical Proofs

Reading and Writing
Mathematical Proofs
Spring 2016
Lecture 1: Basic Proving Techniques I
Organization
The required stuff…
Organization
Lecturer:
Arthur van Goethem MSc, MF 4.144,
[email protected]
Web page:
Part of Data Structures
http://www.win.tue.nl/aga/teaching/2IL50/
Book:
Daniel Solow.
How to Read and Do Proofs (5th edition)
not mandatory
Schedule
Only 4 lectures
Prerequisites
Not covered
 Logical inferences
 Logical derivations (flag proofs)
 In other words: Logic and Set Theory (2IT60)
Covered
 Mathematical proofs in common English
 Common proof techniques (proof by contradiction, induction, etc.)
 Common errors
 Many proof examples
Today on:
Reading and Writing
Mathematical Proofs
Today on:
Basic Proving Techniques I
The proving cookbook…
Today:
“How to properly slap your sidekick.”
“Really poorly drawn circles!”
“Proving my hamsters are strange.”
What is a proof?
Why explanations are not proofs…
What is a proof?
A method for establishing truth
What establishes truth depends on context
Physics
Sufficient experimental
evidence
Courtroom
Admissible evidence
and witness testimony
Mathematical proof
Not a doubt possible!
What is a proof?
A form of communication
Proof must convince reader (not the writer!) of correctness
Proofs must be:
 Clearly written
 Should be easy to follow
 Very different from “proving process”
 Very precise
 No ambiguities!
 Leaving no doubts
What is a proof?
Clearly written
“We turn the solution into an array. We invert the graph and then it look
like the array. Because the solution can be an inverted graph and the
array is input, the solution exists.”
Very precise
“We use Depth-First-Search to find some nodes
that are too deep and somehow form the solution.”
Leaving no doubts
“In most cases it is the case that this conversion works. As we only
have the best cases we likely can always perform this conversion.”
Definition
Mathematical proof
A convincing argument for the reader to establish the correctness of a
mathematical statement without any doubt.
Definition
Mathematical proof
A convincing argument for the reader to establish the correctness of a
mathematical statement without any doubt.
Statement must be true or false
3+6
3+6=9
Definition
Mathematical proof
A convincing argument for the reader to establish the correctness of a
mathematical statement without any doubt.
In what format should a proof be?
Logical derivation
Good
 Very systematic
 Hard to make mistakes
Bad
 Not convenient for statements
not stated in logical formulas
 Emphasis on logical reasoning
→ detract from crux argument
 Hard to read
 Cumbersome
Common English
Theorem
If x is odd, then x2 is odd
Proof
Since x is odd, there exists a k ϵ ℤ
such that x = 2k + 1. Then,
x2 = (2k + 1)2 = 4k2 + 4k + 1 =
2(2k2 + 2k) + 1 = 2m+1.
As there exists a m ϵ ℤ such that
x2 = 2m + 1, x2 is odd. □
This is the kind of proof we expect
in Data Structures!
Good
 Short and to the point
 Easy to read
Bad
 Logical reasoning
somewhat hidden
 Natural language can be
ambiguous
Two Proof Formats
Theorem
If x is odd, then x2 is odd
Proof
Since x is odd, there exists a k ϵ ℤ
such that x = 2k + 1. Then,
x2 = (2k + 1)2 = 4k2 + 4k + 1 =
2(2k2 + 2k) + 1 = 2m+1.
As there exists a m ϵ ℤ such that
x2 = 2m + 1, x2 is odd. □
This is what you should write down
This should be in the back of your mind
Two Proof Formats
Theorem
If x is odd, then x2 is odd
Proof
Since x is odd, there exists a k ϵ ℤ
such that x = 2k + 1. Then,
x2 = (2k + 1)2 = 4k2 + 4k + 1 =
2(2k2 + 2k) + 1 = 2m+1.
As there exists a m ϵ ℤ such that
x2 = 2m + 1, x2 is odd. □
Must be able to translate:
Logic ↔ English
Logic and Ambiguities
The stuff that is kind of doing that thing…
Logic?
Logic?
Logic?
Logic?
Logic?
Logic vs. English
Logic
English
P⋀Q
P and Q
Both P and Q
P⋁Q
P or Q
careful….
¬P
¬(P ⋁ Q)
not P
doesn’t hold, cannot
neither P nor Q
Logic vs. English
Logic
English
P⇒Q
P implies Q
If P, then Q
P⇔Q
P is equivalent to Q
P if and only if Q
P iff Q
For P ⇒ Q:
¬Q ⇒ ¬P (equivalent)
Q ⇒ P (not equivalent!)
contrapositive
converse
Ambiguities
Do you want beer or water?
Do you want beer or water or both? (or)
OR
Do you want either beer or water? (exclusive-or)
If you can solve any problem we pose, then you ace the course
If you can solve some problem we pose, then you ace the course
OR
If you can solve all problems we pose, then you ace the course
Always avoid ambiguities!
Detail
What to describe and what not
Definition
Mathematical proof
A convincing argument for the reader to establish the correctness of a
mathematical statement without any doubt.
How detailed should a proof be?
A Simple Proof
Theorem
Two distinct circles can have at most two intersection points.
A Simple Proof
Theorem
Two distinct circles can have at most two intersection points.
Proof
For the sake of contradiction, assume that two distinct circles C1
and C2 have (at least) three intersection points p1, p2, and p3. Note
that p1, p2, and p3 must lie on both C1 and C2. But, three points on a
circle uniquely define the circle. Thus, C1 = C2. Contradiction! □
A Simple Proof
Theorem
Two distinct circles can have at most two intersection points.
Proof
For the sake of contradiction, assume that two distinct circles C1
and C2 have (at least) three intersection points p1, p2, and p3. Note
that p1, p2, and p3 must lie on both C1 and C2. But, three points on a
circle uniquely define the circle. Thus, C1 = C2. Contradiction! □
A Simple Proof
Theorem
Two distinct circles can have at most two intersection points.
Proof
For the sake of contradiction, assume that two distinct circles C1
and C2 have (at least) three intersection points p1, p2, and p3. Note
that p1, p2, and p3 must lie on both C1 and C2. But then the center
of C1 (and C2) must be at the intersection of the bisector of p1 and
p2, and the bisector of p2 and p3. Since the bisectors are lines, and
two lines intersect only once, this uniquely defines the center (and
radius) of both C1 and C2.Thus, C1 = C2. Contradiction! □
A Simple Proof
Theorem
Two distinct circles can have at most two intersection points.
Proof
For the sake of contradiction, assume that two distinct circles C1
and C2 have (at least) three intersection points p1, p2, and p3. Note
that p1, p2, and p3 must lie on both C1 and C2. But then the center
of C1 (and C2) must be at the intersection of the bisector of p1 and
p2, and the bisector of p2 and p3. Since the bisectors are lines, and
two lines intersect only once, this uniquely defines the center (and
radius) of both C1 and C2.Thus, C1 = C2. Contradiction! □
Proof Detail
When is a proof detailed enough?
 What will the reader accept as true statements
 More detail results in “more obvious” statements…
 … but is also much longer
 A good proof must find the right balance
Axiomatic System
Axiom
 A simple statement assumed to be true
Which axioms are you allowed to use?
‘Anything’ from high school math (middelbare school)
Anything covered in class
If you are unsure, just ask!
Exception: If you are asked to prove a statement, the statement itself
can never be used as an axiom
Axioms in this Course
Exception:
If you are asked to prove a statement, the statement itself can
never be used as an axiom
Practice
What are the axioms?
Theorem
If x is odd, then x2 is odd
Proof
Since x is odd, there exists a k ϵ ℤ such that x = 2k + 1. Then,
x2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Thus, x2 is odd. □
Analogy
Proving process
Analogy
Written proof
Analogy
The Maze Analogy
entrance/premise
Proving process
 Find path between
entrance and exit
 Erratic, creative
Written proof
 Give instructions for
others to find path
 Stick to essentials
exit/goal
Proving Tips
Tips for writing proofs
1. State the proof techniques you’re using (see next lecture)
2.
Keep a linear flow
 Proving process different from written proof
3.
Describe every step clearly in words
4.
Don’t use complicated notation
5.
Make sure your axioms are actually “obvious”
 What is obvious to you may not be obvious to the reader
6.
Finish your proof
 Connect everything with what you were trying to prove
Practice
Practice Proofs
Theorem (handshaking lemma)
For any graph G=(V, E) the number of vertices with odd degree is
even.
Practice Proofs
Theorem (handshaking lemma)
Consider a group of people shaking hands with each other (but not
necessarily with everyone). The number of people that shook
hands with an odd number of people is even.
Practice Proofs
Theorem (handshaking lemma)
For any graph G=(V, E) the number of vertices with odd degree is
even.
Proof
Consider the sum of the degrees of all vertices: Σv d(v).
This sum counts every edge exactly twice.
Thus the sum is 2*|E|, which is even.
Vertices with an even degree do not affect the parity of this sum.
Hence, if V’ is the set of vertices with odd degree,
then Σv ϵ V’ d(v) must also be even.
Since the sum of an odd number of odd numbers is odd,
the sum Σv ϵ V’ d(v) is even if and only if |V’| is even. □
Practice Proofs
Theorem (Pythagoras)
For a right triangle T with side lengths x, y, and z (x ≤ y ≤ z), it holds
that x2 + y2 = z2.
Practice Proofs
Theorem (Pythagoras)
For a right triangle T with side lengths x, y, and z (x ≤ y ≤ z), it holds
that x2 + y2 = z2.
Proof
Seems tricky… Let’s try to prove something simpler. What if x = y?
Then the area of the triangle is x2/2. We need to prove that 2x2 = z2.
So 4 triangles should together have area z2. How about this
configuration?
z
z
x
y
z
z
z
Practice Proofs
Theorem (Pythagoras)
For a right triangle T with side lengths x, y, and z (x ≤ y ≤ z), it holds
that x2 + y2 = z2.
Proof
Let’s return to the general case. What happens if we again try to
form a square with side length z? We get a hole in the middle which
seems to be a square. We may have enough for a proof here…
z
z
x
z
z
y
z
Practice Proofs
Theorem (Pythagoras)
For a right triangle T with side lengths x, y, and z (x ≤ y ≤ z), it holds
that x2 + y2 = z2.
Proof
Consider the configuration of four copies of T shown in the figure.
Note that, for a right triangle, the sum of the non-right angles is 90°,
since the sum of all angles is 180°. Hence, the outer shape is a
square with side length z and has area z2. The
z
area of a single triangle is xy/2. Because T is a
right triangle, the middle hole is also a square
with side length (y - x) and area (y - x)2. We
z
z
thus obtain: z2 = 2xy + (y - x)2 = x2 + y2. □
z
Basic Proving Techniques
Proving 101…
Overview
Basic Proving Techniques
Overview
Basic Proving Techniques
1.
Forward-backward method
2.
Mathematical induction
3.
Case analysis
4.
Proof by contradiction
Forward-Backward Method
How to get from A to B and B to A…
Forward-Backward Method
The most basic approach
Logically combine axioms, definitions, and earlier theorems (forward)
Simplify the goal (backward)
FACTS
This should always be your default approach
GOAL
Usage
When to use?
Generally used for statements of the form: If P then Q
Premise
Reason forward from the premise
Reason backward from the goal
Goal
Basic Example
Theorem
If my hamsters do excessive exercise, I will be tired in the morning.
Proof
My hamsters do excessive exercise
They are running in their exercise wheel
The exercise wheel is making noise
Something is keeping me awake
I do not get a good night’s sleep
I am tired in the morning
Example
Theorem
If 0 ≤ 𝑥 ≤ 2, then – 𝑥3 + 4𝑥 + 1 > 0
Proof
Perhaps first plot the function – 𝑥3 + 4𝑥 + 1.
Smallest around 𝑥 = 0 and 𝑥 = 2.
But −03 + 4 ∗ 0 + 1 = 1, and
−23 + 4 ∗ 2 + 1 = 1.
It seems that – 𝑥3 + 4𝑥 + 1 ≥ 1 for 0 ≤ 𝑥 ≤ 2.
Simplify goal: −𝑥3 + 4𝑥 ≥ 0 for 0 ≤ 𝑥 ≤ 2.
Example
Theorem
If 0 ≤ 𝑥 ≤ 2, then – 𝑥3 + 4𝑥 + 1 > 0
Proof
It seems that – 𝑥3 + 4𝑥 + 1 ≥ 1 for 0 ≤ 𝑥 ≤ 2.
Simplify goal: −𝑥3 + 4𝑥 ≥ 0 for 0 ≤ 𝑥 ≤ 2.
Let’s simplify further: −𝑥3 + 4𝑥 = 𝑥 (4 – 𝑥2) ≥ 0
Also, 𝑥 (4 – 𝑥2) ≥ 0 if 𝑥 ≥ 0 and 4 – 𝑥2 ≥ 0.
I think we’re ready for the proof…
Example
Theorem
If 0 ≤ 𝑥 ≤ 2, then – 𝑥3 + 4𝑥 + 1 > 0
Proof
We first show that – 𝑥3 + 4𝑥 ≥ 0.
We can rewrite the function as :
– 𝑥3 + 4𝑥 = 𝑥 (4 – 𝑥2) = 𝑥 (2 – 𝑥) (2 + 𝑥).
Since 𝑥, (2 − 𝑥), and (2 + 𝑥) are all nonnegative for 0 ≤ 𝑥 ≤ 2.
It follows that 𝑥 (2 – 𝑥) (2 + 𝑥) = – 𝑥3 + 4𝑥 ≥ 0 for 0 ≤ 𝑥 ≤ 2.
By adding 1 to both sides we obtain – 𝑥3 + 4𝑥 + 1 ≥ 1 > 0. □
Another Example
Theorem
If f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n)f2(n) = O(g1(n)g2(n))
Proof
What does f(n) = O(g(n)) mean again?
There exist positive constants c and n0 such that f(n) ≤ c g(n)
for all n ≥ n0.
Another Example
Theorem
If f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n)f2(n) = O(g1(n)g2(n))
Proof
By definition there exist positive constants c’ and n0’ such that
f1(n) ≤ c’ g1(n) for all n ≥ n0’. Similarly, there exist positive constants
c’’ and n0’’ such that f2(n) ≤ c’’ g2(n) for all n ≥ n0’’. We need to
show that there exist positive constants c and n0 such that
f1(n)f2(n) ≤ c g1(n)g2(n) for all n ≥ n0.
We must be very careful with variables! Use different names!
Another Example
Theorem
If f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n)f2(n) = O(g1(n)g2(n))
Proof
By definition there exist positive constants c’ and n0’ such that
f1(n) ≤ c’ g1(n) for all n ≥ n0’. Similarly, there exist positive constants
c’’ and n0’’ such that f2(n) ≤ c’’ g2(n) for all n ≥ n0’’. We need to
show that there exist positive constants c and n0 such that
f1(n)f2(n) ≤ c g1(n)g2(n) for all n ≥ n0.
Another Example
Theorem
If f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n)f2(n) = O(g1(n)g2(n))
Proof
By definition there exist positive constants c’ and n0’ such that
f1(n) ≤ c’ g1(n) for all n ≥ n0’. Similarly, there exist positive constants
c’’ and n0’’ such that f2(n) ≤ c’’ g2(n) for all n ≥ n0’’. We need to
show that there exist positive constants c and n0 such that
f1(n)f2(n) ≤ c g1(n)g2(n) for all n ≥ n0.
To establish this, we need to find suitable values for c and n0.
Another Example
Theorem
If f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n)f2(n) = O(g1(n)g2(n))
Proof
By definition there exist positive constants c’ and n0’ such that
f1(n) ≤ c’ g1(n) for all n ≥ n0’. Similarly, there exist positive constants
c’’ and n0’’ such that f2(n) ≤ c’’ g2(n) for all n ≥ n0’’. We need to
show that there exist positive constants c and n0 such that
f1(n)f2(n) ≤ c g1(n)g2(n) for all n ≥ n0.
To establish this, we need to find suitable values for c and n0.
We already are given constants c’, c’’, n0’, and n0’’.
Another Example
Theorem
If f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n)f2(n) = O(g1(n)g2(n))
Proof
By definition there exist positive constants c’ and n0’ such that
f1(n) ≤ c’ g1(n) for all n ≥ n0’. Similarly, there exist positive constants
c’’ and n0’’ such that f2(n) ≤ c’’ g2(n) for all n ≥ n0’’. We need to
show that there exist positive constants c and n0 such that
f1(n)f2(n) ≤ c g1(n)g2(n) for all n ≥ n0.
To establish this, we need to find suitable values for c and n0.
We already are given constants c’, c’’, n0’, and n0’’.
Note that f1(n)f2(n) ≤ c’g1(n) c’’g2(n) = c’ c’’ g1(n)g2(n) ,
Another Example
Theorem
If f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n)f2(n) = O(g1(n)g2(n))
Proof
By definition there exist positive constants c’ and n0’ such that
f1(n) ≤ c’ g1(n) for all n ≥ n0’. Similarly, there exist positive constants
c’’ and n0’’ such that f2(n) ≤ c’’ g2(n) for all n ≥ n0’’. We need to
show that there exist positive constants c and n0 such that
f1(n)f2(n) ≤ c g1(n)g2(n) for all n ≥ n0.
To establish this, we need to find suitable values for c and n0.
We already are given constants c’, c’’, n0’, and n0’’.
Note that f1(n)f2(n) ≤ c’g1(n) c’’g2(n) = c’ c’’ g1(n)g2(n) ,
but only if n ≥ n0’ and n ≥ n0’’ .
So n ≥ n0 should imply n ≥ n0’ and n ≥ n0’’ .
Another Example
Theorem
If f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n)f2(n) = O(g1(n)g2(n))
Proof
By definition there exist positive constants c’ and n0’ such that
f1(n) ≤ c’ g1(n) for all n ≥ n0’. Similarly, there exist positive constants
c’’ and n0’’ such that f2(n) ≤ c’’ g2(n) for all n ≥ n0’’. We need to
show that there exist positive constants c and n0 such that
f1(n)f2(n) ≤ c g1(n)g2(n) for all n ≥ n0.
Let n0 = max(n0’, n0’’) and c = c’ c’’. Then, for all n ≥ n0 (which
implies n ≥ n0’ and n ≥ n0’’), f1(n)f2(n) ≤ c’ g1(n) c’’ g2(n) = c’ c’’ g1(n)
g2(n) = c g1(n) g2(n).
We have f1(n)f2(n) ≤ c g1(n) g2(n) for all n ≥ n0.
Thus, f1(n)f2(n) = O(g1(n)g2(n)). □
Another Example
Theorem
If f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n)f2(n) = O(g1(n)g2(n))
Proof
By definition there exist positive constants c’ and n0’ such that
f1(n) ≤ c’ g1(n) for all n ≥ n0’. Similarly, there exist positive constants
c’’ and n0’’ such that f2(n) ≤ c’’ g2(n) for all n ≥ n0’’. We need to
show that there exist positive constants c and n0 such that
f1(n)f2(n) ≤ c g1(n)g2(n) for all n ≥ n0.
Let n0 = max(n0’, n0’’) and c = c’ c’’. Then, for all n ≥ n0 (which
implies n ≥ n0’ and n ≥ n0’’), f1(n)f2(n) ≤ c’ g1(n) c’’ g2(n) = c’ c’’ g1(n)
g2(n) = c g1(n) g2(n).
We have f1(n)f2(n) ≤ c g1(n) g2(n) for all n ≥ n0.
Thus, f1(n)f2(n) = O(g1(n)g2(n)). □
Mathematical Induction
An introduction to induction…
Recurring structures
Many algorithms and (data)structures have recurrences
sum = 0
 Loop
for i = 1 to A.length
do sum = sum + A[i]
 List
 Tree
The Idea
The Base Idea of Induction
P(1)
P(2)
P(3)
…
P(n)
Base Case
One (or more) very simple cases that we can trivially proof.
Induction Hypothesis
The statement that we want to prove (for any 𝑛).
Step
Prove that if the statement holds for a small instance, it must also hold for
a larger instance.
Usage
When to use?
 Whenever you need to prove something is true for all values of 𝑛.
 (or all values ≥ 𝑥)
 Infinite possibilities!
 When there is a clear structure in the problem (e.g., trees)
 We will talk more about this later
Basic Example
Theorem
If 𝑛 dominos are placed in a row and I push the first; they all fall.
Proof:
We use induction on 𝑛.
Base Case (𝑛 = 1):
If there is only 1 domino, it must also be the first.
I will push the first over, so trivially they all fall.
Basic Example
Proof:
We use induction on 𝑛.
Base Case (𝑛 = 1):
If there is only 1 domino, it must also be the first.
I will push the first over, so trivially they all fall and the IH holds.
Induction Hypothesis:
If 𝑛 dominos are placed in a row and I push the first; they all fall.
Step:
Assume the IH holds for 𝑛 dominos.
If there were 𝑛 + 1 dominos in a row, the first 𝑛 form a row of length 𝑛.
By IH the first 𝑛 dominos will all fall. As all 𝑛 dominos fall, so must the
𝑛th domino. If the 𝑛th domino falls, then it will tip over the 𝑛 + 1th. The
first 𝑛 dominos fall over and the 𝑛 + 1th domino also falls over.
So all 𝑛 + 1 dominos fall over. Thus, the IH holds.
Basic Example
Proof:
We use induction on 𝑛.
Base Case (𝑛 = 1):
If there is only 1 domino, it must also be the first.
I will push the first over, so trivially they all fall and the IH holds.
Induction Hypothesis:
If 𝑛 dominos are placed in a row and I push the first; they all fall.
Step:
Assume the IH holds for 𝑛 dominos.
If there were 𝑛 + 1 dominos in a row, the first 𝑛 form a row of length 𝑛.
By IH the first 𝑛 dominos will all fall. As all 𝑛 dominos fall, so must the
𝑛th domino. If the 𝑛th domino falls, then it will tip over the 𝑛 + 1th. The
first 𝑛 dominos fall over and the 𝑛 + 1th domino also falls over.
So all 𝑛 + 1 dominos fall over. Thus, the IH holds.
Example
Theorem
For all positive integers 𝑛, 3𝑛 − 1 is even.
Proof:
We use induction on 𝑛.
Base Case (n = 1): 31 − 1 = 2, which is indeed even.
IH: 3𝑛 − 1 is even.
Step (n >= 1):
Assume that 3𝑛 − 1 is even. (IH)
We need to show that 3𝑛+1 − 1 is even.
We have: 3𝑛+1 − 1 = 3 ∗ 3𝑛 − 1 = (2 ∗ 3𝑛 ) + (3𝑛 − 1).
A multiplication with an even number is always even (2 ∗ 3𝑛 ).
By IH, (3𝑛 − 1) is also even. The sum of two even numbers is also even.
Thus, 3𝑛+1 − 1 must be even. The IH holds. □
Practice 1
Theorem
For all positive integers 𝑛,
𝑛
𝑘=1 𝑘
= 𝑛(𝑛 + 1)/2
Practice 1
Proof:
We use induction on 𝑛.
Base case (𝑛 = 1):
𝑛
𝑘=1 𝑘
1
𝑘=1 𝑘
=
= 1 =
1 1+1
2
=
𝑛 𝑛+1
2
.
As both values equate to the same the IH holds.
IH: 𝑛𝑘=1 𝑘 = 𝑛(𝑛 + 1)/2
Step (𝑛 ≥ 1):
Suppose that 𝑛𝑘=1 𝑘 = 𝑛(𝑛 + 1)/2. (IH)
We need to show that 𝑛+1
𝑘=1 𝑘 = (𝑛 + 1)(𝑛 + 2)/2.
We have:
𝑛+1
𝑘=1 𝑘
=
=
𝑛
𝑘=1 𝑘
𝑛 𝑛+1 +2(𝑛+1)
2
+ 𝑛+1 =
=
𝑛 𝑛+1
2
+ 𝑛 + 1 (by IH)
𝑛+1 (𝑛+2)
.
2
Thus the IH holds and it follows by induction that
𝑛
𝑘=1 𝑘 = 𝑛(𝑛 + 1)/2 for all positive integers 𝑛. □
Practice 2
Theorem
For every integer 𝑛 ≥ 5, 2𝑛 > 𝑛2
Practice 2
Proof:
We use induction on 𝑛.
Base case (𝑛 = 5): 2𝑛 = 25 = 32 > 25 = 52 = 𝑛2
IH: 2𝑛 > 𝑛2
Step (𝑛 ≥ 5):
Suppose that 2𝑛 > 𝑛2 (IH).
We need to show that 2𝑛+1 > (𝑛 + 1)2 .
We have:
2𝑛+1 = 2 ∗ 2𝑛 > 2 ∗ 𝑛2 (by IH)
So it is sufficient to show that 2 ∗ 𝑛2 ≥ (𝑛 + 1)2 = 𝑛2 + 2𝑛 + 1 for
𝑛 ≥ 5. This can be simplified to 𝑛2 – 2𝑛 – 1 ≥ 0 or (𝑛 − 1)2 ≥ 2.
This is clearly true for 𝑛 ≥ 5.
So it follows by induction that 2𝑛 > 𝑛2 for 𝑛 ≥ 5. □
Loop Invariant
An introduction to proving loops…
Loop Invariant
sum = 0
for i = 1 to A.length
do sum = sum + A[i]
What do we want?
How do we prove something is true at the end?
What do we really know?
Loop Invariant
sum = 0
𝑖=1
𝑖=1
for i = 1 to A.length
do sum = sum + A[i]
What do we really know?
At the start:
Sum = 0
= Sum first 0 elements
Loop Invariant
sum = 0
𝑖=2
𝑖=2
for i = 1 to A.length
do sum = sum + A[i]
What do we really know?
At the start:
Sum = 0
After first iteration:
= Sum first 0 elements
Sum = 0 + 𝐴[1]
= Sum first 1 element
Loop Invariant
sum = 0
𝑖=3
for i = 1 to A.length
do sum = sum + A[i]
What do we really know?
At the start:
Sum = 0
After first iteration:
= Sum first 0 elements
Sum = 0 + 𝐴[1]
= Sum
first 1 element
Not an
Axiom!!
After second iteration:
Sum = 0 + 𝐴 1 + 𝐴[2]
= Sum first 2 elements
Loop Invariant
sum = 0
𝑖=3
for i = 1 to A.length
do sum = sum + A[i]
What do we really know?
At the start:
Sum = 0
After first iteration:
= Sum first 0 elements
Sum = 0 + 𝐴[1]
After second iteration:
= Sum first 1 element
Sum = 0 + 𝐴 1 + 𝐴[2]
= Sum first 2 elements
Loop Invariant
Loop Invariants replicate the chain of logical derivations
At the start:
Sum = 0
After first iteration:
= Sum first 0 elements
Sum = 0 + 𝐴[1]
After second iteration:
= Sum first 1 element
Sum = 0 + 𝐴 1 + 𝐴[2]
= Sum first 2 elements
To prove a claim is true at the end, we show…
 …it is true at the start
 …if it is true at the start of a random iteration 𝑖,
it is still true at the start of the next iteration 𝑖 + 1
 …the claim is true at the end of the loop.
Loop Invariant
Loop Invariants replicate the chain of logical derivations
At the start:
Sum = 0
After first iteration:
= Sum first 0 elements
Sum = 0 + 𝐴[1]
After second iteration:
= Sum first 1 element
Sum = 0 + 𝐴 1 + 𝐴[2]
= Sum first 2 elements
To prove a statement is true at the end, we need..
 Invariant
(What remains true)
 Initialization
(Starting conditions)
 Maintenance
(Making sure it remains true)
 Termination
(Ending conditions)
Basic Example
sum = 0
for i = 1 to A.length
do sum = sum + A[i]
Invariant
At the start of iteration 𝑖, 𝑠𝑢𝑚 contains the sum of 𝐴[1. . 𝑖 − 1].
Initialization
At the start of the loop 𝑖 = 1 and 𝑠𝑢𝑚 = 0.
For the loop invariant to hold, 𝑠𝑢𝑚 must contains the sum of 𝐴 1. . 0 = ∅.
The sum of no elements is trivially 0.
So 𝑠𝑢𝑚 is correctly set to 0.
Basic Example
sum = 0
for i = 1 to A.length
do sum = sum + A[i]
Invariant
At the start of iteration 𝑖, 𝑠𝑢𝑚 contains the sum of 𝐴[1. . 𝑖 − 1].
Maintenance
At the start of iteration 𝑖, by the loop invariant 𝑠𝑢𝑚 contains the sum of 𝐴[1. . 𝑖 −
1]. In iteration 𝑖, sum is increased by 𝐴[𝑖]. So 𝑠𝑢𝑚 is the sum of elements
𝐴[1. . 𝑖 − 1] + 𝐴[𝑖] = the sum of elements 𝐴 1. . 𝑖 = 𝐴[1. . 𝑖 + 1 − 1]. Thus the
invariant will be maintained.
Basic Example
sum = 0
for i = 1 to A.length
do sum = sum + A[i]
Invariant
At the start of iteration 𝑖, 𝑠𝑢𝑚 contains the sum of 𝐴[1. . 𝑖 − 1].
Maintenance
At the start of iteration 𝑖, by the loop invariant 𝑠𝑢𝑚 contains the sum of 𝐴[1. . 𝑖 −
1]. In iteration 𝑖, sum is increased by 𝐴[𝑖]. So 𝑠𝑢𝑚 is the sum of elements
𝐴[1. . 𝑖 − 1] + 𝐴[𝑖] = the sum of elements 𝐴 1. . 𝑖 = 𝐴[1. . 𝑖 + 1 − 1]. Thus the
invariant will be maintained.
Basic Example
sum = 0
for i = 1 to A.length
do sum = sum + A[i]
Invariant
At the start of iteration 𝑖, 𝑠𝑢𝑚 contains the sum of 𝐴 1. . 𝑖 − 1 .
Termination
The loop terminates when 𝑖 > 𝐴. 𝑙𝑒𝑛𝑔𝑡ℎ, so 𝑖 = 𝐴. 𝑙𝑒𝑛𝑔𝑡ℎ + 1.
By the loop invariant we know that 𝑠𝑢𝑚 contains the sum
of 𝐴 1. . (𝐴. 𝑙𝑒𝑛𝑔𝑡ℎ + 1) − 1 = 𝐴[1. . 𝐴. 𝑙𝑒𝑛𝑔𝑡ℎ].
This is exactly what we wanted to compute.
Next time on:
Reading and Writing
Mathematical Proofs
Next time!
“Proving Teletubbies are unlikable!”
“How to prove you are cool (and own a Ferrari).”
“How to store way too many pigeons.”