Proof By Cases
Existence Proofs
Case Study: Tilings
Proof Techniques (II)
Haythem O. Ismail
c Haythem O. Ismail
Lecture 6
1 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Objectives
By the end of this lecture you will be able to
1
Construct proofs by cases.
2
Construct constructive and non-constructive existence proofs.
3
Construct uniqueness proofs.
c Haythem O. Ismail
Lecture 6
2 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Outline
1
Proof By Cases
2
Existence Proofs
3
Case Study: Tilings
c Haythem O. Ismail
Lecture 6
3 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Outline
1
Proof By Cases
2
Existence Proofs
3
Case Study: Tilings
c Haythem O. Ismail
Lecture 6
4 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
The Strategy
A proof by cases is a proof method used to prove conjunctions or
conditionals with disjunctive antecedents.
To prove p1 ∧ p2 ∧ · · · ∧ pn , prove p1 , p2 , . . . , and pn separately.
(The result follows by the rule of conjunction.)
To prove (p1 ∨ p2 ∨ · · · ∨ pn ) → q, prove p1 → q, p2 → q, . . .,
and pn → q. (The result follows by logical equivalence.)
However, this method is mostly used to prove general
conditionals p → q, where it is simpler to make use of a logical
equivalence p ≡ (p1 ∨ p2 ∨ · · · ∨ pn ).
c Haythem O. Ismail
Lecture 6
5 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
The Strategy
A proof by cases is a proof method used to prove conjunctions or
conditionals with disjunctive antecedents.
To prove p1 ∧ p2 ∧ · · · ∧ pn , prove p1 , p2 , . . . , and pn separately.
(The result follows by the rule of conjunction.)
To prove (p1 ∨ p2 ∨ · · · ∨ pn ) → q, prove p1 → q, p2 → q, . . .,
and pn → q. (The result follows by logical equivalence.)
However, this method is mostly used to prove general
conditionals p → q, where it is simpler to make use of a logical
equivalence p ≡ (p1 ∨ p2 ∨ · · · ∨ pn ).
c Haythem O. Ismail
Lecture 6
5 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
The Strategy
A proof by cases is a proof method used to prove conjunctions or
conditionals with disjunctive antecedents.
To prove p1 ∧ p2 ∧ · · · ∧ pn , prove p1 , p2 , . . . , and pn separately.
(The result follows by the rule of conjunction.)
To prove (p1 ∨ p2 ∨ · · · ∨ pn ) → q, prove p1 → q, p2 → q, . . .,
and pn → q. (The result follows by logical equivalence.)
However, this method is mostly used to prove general
conditionals p → q, where it is simpler to make use of a logical
equivalence p ≡ (p1 ∨ p2 ∨ · · · ∨ pn ).
c Haythem O. Ismail
Lecture 6
5 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
The Strategy
A proof by cases is a proof method used to prove conjunctions or
conditionals with disjunctive antecedents.
To prove p1 ∧ p2 ∧ · · · ∧ pn , prove p1 , p2 , . . . , and pn separately.
(The result follows by the rule of conjunction.)
To prove (p1 ∨ p2 ∨ · · · ∨ pn ) → q, prove p1 → q, p2 → q, . . .,
and pn → q. (The result follows by logical equivalence.)
However, this method is mostly used to prove general
conditionals p → q, where it is simpler to make use of a logical
equivalence p ≡ (p1 ∨ p2 ∨ · · · ∨ pn ).
c Haythem O. Ismail
Lecture 6
5 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
An Exhaustive Proof
Example
Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are
8 and 9.
First, note that 8 and 9 are indeed consecutive positive integers which are perfect powers
(23 and 32 , respectively).
Second, note that may systematically check all pairs of consecutive integers not
exceeding 100 making sure that only 8 and 9 are both perfect powers.
This is a lot of work though.
Instead, we may list all perfect powers which do not exceed 100 and check if there are
any consecutive pairs.
Thus,
Powers of 2:
Powers of 3:
Powers of 4:
Powers of 5:
Powers of 6:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
1, 8, 27, 64.
1, 16, 81.
1, 32.
1, 64.
Only consecutive integers are 8 and 9.
c Haythem O. Ismail
Lecture 6
6 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
An Exhaustive Proof
Example
Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are
8 and 9.
First, note that 8 and 9 are indeed consecutive positive integers which are perfect powers
(23 and 32 , respectively).
Second, note that may systematically check all pairs of consecutive integers not
exceeding 100 making sure that only 8 and 9 are both perfect powers.
This is a lot of work though.
Instead, we may list all perfect powers which do not exceed 100 and check if there are
any consecutive pairs.
Thus,
Powers of 2:
Powers of 3:
Powers of 4:
Powers of 5:
Powers of 6:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
1, 8, 27, 64.
1, 16, 81.
1, 32.
1, 64.
Only consecutive integers are 8 and 9.
c Haythem O. Ismail
Lecture 6
6 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
An Exhaustive Proof
Example
Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are
8 and 9.
First, note that 8 and 9 are indeed consecutive positive integers which are perfect powers
(23 and 32 , respectively).
Second, note that may systematically check all pairs of consecutive integers not
exceeding 100 making sure that only 8 and 9 are both perfect powers.
This is a lot of work though.
Instead, we may list all perfect powers which do not exceed 100 and check if there are
any consecutive pairs.
Thus,
Powers of 2:
Powers of 3:
Powers of 4:
Powers of 5:
Powers of 6:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
1, 8, 27, 64.
1, 16, 81.
1, 32.
1, 64.
Only consecutive integers are 8 and 9.
c Haythem O. Ismail
Lecture 6
6 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
An Exhaustive Proof
Example
Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are
8 and 9.
First, note that 8 and 9 are indeed consecutive positive integers which are perfect powers
(23 and 32 , respectively).
Second, note that may systematically check all pairs of consecutive integers not
exceeding 100 making sure that only 8 and 9 are both perfect powers.
This is a lot of work though.
Instead, we may list all perfect powers which do not exceed 100 and check if there are
any consecutive pairs.
Thus,
Powers of 2:
Powers of 3:
Powers of 4:
Powers of 5:
Powers of 6:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
1, 8, 27, 64.
1, 16, 81.
1, 32.
1, 64.
Only consecutive integers are 8 and 9.
c Haythem O. Ismail
Lecture 6
6 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
An Exhaustive Proof
Example
Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are
8 and 9.
First, note that 8 and 9 are indeed consecutive positive integers which are perfect powers
(23 and 32 , respectively).
Second, note that may systematically check all pairs of consecutive integers not
exceeding 100 making sure that only 8 and 9 are both perfect powers.
This is a lot of work though.
Instead, we may list all perfect powers which do not exceed 100 and check if there are
any consecutive pairs.
Thus,
Powers of 2:
Powers of 3:
Powers of 4:
Powers of 5:
Powers of 6:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
1, 8, 27, 64.
1, 16, 81.
1, 32.
1, 64.
Only consecutive integers are 8 and 9.
c Haythem O. Ismail
Lecture 6
6 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
An Exhaustive Proof
Example
Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are
8 and 9.
First, note that 8 and 9 are indeed consecutive positive integers which are perfect powers
(23 and 32 , respectively).
Second, note that may systematically check all pairs of consecutive integers not
exceeding 100 making sure that only 8 and 9 are both perfect powers.
This is a lot of work though.
Instead, we may list all perfect powers which do not exceed 100 and check if there are
any consecutive pairs.
Thus,
Powers of 2:
Powers of 3:
Powers of 4:
Powers of 5:
Powers of 6:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
1, 8, 27, 64.
1, 16, 81.
1, 32.
1, 64.
Only consecutive integers are 8 and 9.
c Haythem O. Ismail
Lecture 6
6 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
An Exhaustive Proof
Example
Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are
8 and 9.
First, note that 8 and 9 are indeed consecutive positive integers which are perfect powers
(23 and 32 , respectively).
Second, note that may systematically check all pairs of consecutive integers not
exceeding 100 making sure that only 8 and 9 are both perfect powers.
This is a lot of work though.
Instead, we may list all perfect powers which do not exceed 100 and check if there are
any consecutive pairs.
Thus,
Powers of 2:
Powers of 3:
Powers of 4:
Powers of 5:
Powers of 6:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
1, 8, 27, 64.
1, 16, 81.
1, 32.
1, 64.
Only consecutive integers are 8 and 9.
c Haythem O. Ismail
Lecture 6
6 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
An Exhaustive Proof
Example
Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are
8 and 9.
First, note that 8 and 9 are indeed consecutive positive integers which are perfect powers
(23 and 32 , respectively).
Second, note that may systematically check all pairs of consecutive integers not
exceeding 100 making sure that only 8 and 9 are both perfect powers.
This is a lot of work though.
Instead, we may list all perfect powers which do not exceed 100 and check if there are
any consecutive pairs.
Thus,
Powers of 2:
Powers of 3:
Powers of 4:
Powers of 5:
Powers of 6:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
1, 8, 27, 64.
1, 16, 81.
1, 32.
1, 64.
Only consecutive integers are 8 and 9.
c Haythem O. Ismail
Lecture 6
6 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
An Exhaustive Proof
Example
Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are
8 and 9.
First, note that 8 and 9 are indeed consecutive positive integers which are perfect powers
(23 and 32 , respectively).
Second, note that may systematically check all pairs of consecutive integers not
exceeding 100 making sure that only 8 and 9 are both perfect powers.
This is a lot of work though.
Instead, we may list all perfect powers which do not exceed 100 and check if there are
any consecutive pairs.
Thus,
Powers of 2:
Powers of 3:
Powers of 4:
Powers of 5:
Powers of 6:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
1, 8, 27, 64.
1, 16, 81.
1, 32.
1, 64.
Only consecutive integers are 8 and 9.
c Haythem O. Ismail
Lecture 6
6 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
An Exhaustive Proof
Example
Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are
8 and 9.
First, note that 8 and 9 are indeed consecutive positive integers which are perfect powers
(23 and 32 , respectively).
Second, note that may systematically check all pairs of consecutive integers not
exceeding 100 making sure that only 8 and 9 are both perfect powers.
This is a lot of work though.
Instead, we may list all perfect powers which do not exceed 100 and check if there are
any consecutive pairs.
Thus,
Powers of 2:
Powers of 3:
Powers of 4:
Powers of 5:
Powers of 6:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
1, 8, 27, 64.
1, 16, 81.
1, 32.
1, 64.
Only consecutive integers are 8 and 9.
c Haythem O. Ismail
Lecture 6
6 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
An Exhaustive Proof
Example
Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are
8 and 9.
First, note that 8 and 9 are indeed consecutive positive integers which are perfect powers
(23 and 32 , respectively).
Second, note that may systematically check all pairs of consecutive integers not
exceeding 100 making sure that only 8 and 9 are both perfect powers.
This is a lot of work though.
Instead, we may list all perfect powers which do not exceed 100 and check if there are
any consecutive pairs.
Thus,
Powers of 2:
Powers of 3:
Powers of 4:
Powers of 5:
Powers of 6:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
1, 8, 27, 64.
1, 16, 81.
1, 32.
1, 64.
Only consecutive integers are 8 and 9.
c Haythem O. Ismail
Lecture 6
6 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Integers and Their Squares
Example
Prove that if n is an integer, then n2 ≥ n.
Proof
First, note that “n is an integer” ≡ n = 0 ∨ n ≤ −1 ∨ n ≥ 1.
We consider each case.
n = 0. It is indeed the case that 02 = 0 ≥ 0.
n ≤ −1. Since n < 0 and n2 > 0, it follows that n2 ≥ n.
n ≥ 1. We know that n > 1. Multiplying both sides by the
positive integer n, we get n2 > n.
c Haythem O. Ismail
Lecture 6
7 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Integers and Their Squares
Example
Prove that if n is an integer, then n2 ≥ n.
Proof
First, note that “n is an integer” ≡ n = 0 ∨ n ≤ −1 ∨ n ≥ 1.
We consider each case.
n = 0. It is indeed the case that 02 = 0 ≥ 0.
n ≤ −1. Since n < 0 and n2 > 0, it follows that n2 ≥ n.
n ≥ 1. We know that n > 1. Multiplying both sides by the
positive integer n, we get n2 > n.
c Haythem O. Ismail
Lecture 6
7 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Integers and Their Squares
Example
Prove that if n is an integer, then n2 ≥ n.
Proof
First, note that “n is an integer” ≡ n = 0 ∨ n ≤ −1 ∨ n ≥ 1.
We consider each case.
n = 0. It is indeed the case that 02 = 0 ≥ 0.
n ≤ −1. Since n < 0 and n2 > 0, it follows that n2 ≥ n.
n ≥ 1. We know that n > 1. Multiplying both sides by the
positive integer n, we get n2 > n.
c Haythem O. Ismail
Lecture 6
7 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Integers and Their Squares
Example
Prove that if n is an integer, then n2 ≥ n.
Proof
First, note that “n is an integer” ≡ n = 0 ∨ n ≤ −1 ∨ n ≥ 1.
We consider each case.
n = 0. It is indeed the case that 02 = 0 ≥ 0.
n ≤ −1. Since n < 0 and n2 > 0, it follows that n2 ≥ n.
n ≥ 1. We know that n > 1. Multiplying both sides by the
positive integer n, we get n2 > n.
c Haythem O. Ismail
Lecture 6
7 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Integers and Their Squares
Example
Prove that if n is an integer, then n2 ≥ n.
Proof
First, note that “n is an integer” ≡ n = 0 ∨ n ≤ −1 ∨ n ≥ 1.
We consider each case.
n = 0. It is indeed the case that 02 = 0 ≥ 0.
n ≤ −1. Since n < 0 and n2 > 0, it follows that n2 ≥ n.
n ≥ 1. We know that n > 1. Multiplying both sides by the
positive integer n, we get n2 > n.
c Haythem O. Ismail
Lecture 6
7 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Integers and Their Squares
Example
Prove that if n is an integer, then n2 ≥ n.
Proof
First, note that “n is an integer” ≡ n = 0 ∨ n ≤ −1 ∨ n ≥ 1.
We consider each case.
n = 0. It is indeed the case that 02 = 0 ≥ 0.
n ≤ −1. Since n < 0 and n2 > 0, it follows that n2 ≥ n.
n ≥ 1. We know that n > 1. Multiplying both sides by the
positive integer n, we get n2 > n.
c Haythem O. Ismail
Lecture 6
7 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Absolute Values
Example
Prove that |xy| = |x||y| for real numbers x and y.
Proof
First, note that “x and y are real numbers”
[x, y ≥ 0] ∨ [x, y < 0] ∨ [x and y have different signs] .
We consider each case.
Case 1. In this case, |xy| = xy = x × y = |x||y|.
Case 2. In this case, |xy| = xy = (−x) × (−y) = |x||y|.
Case 3. In this case, we assume without loss of generality
that x < 0 and y ≥ 0. Hence,
|xy| = −xy = (−x) × y = |x||y|
c Haythem O. Ismail
Lecture 6
8 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Absolute Values
Example
Prove that |xy| = |x||y| for real numbers x and y.
Proof
First, note that “x and y are real numbers”
[x, y ≥ 0] ∨ [x, y < 0] ∨ [x and y have different signs] .
We consider each case.
Case 1. In this case, |xy| = xy = x × y = |x||y|.
Case 2. In this case, |xy| = xy = (−x) × (−y) = |x||y|.
Case 3. In this case, we assume without loss of generality
that x < 0 and y ≥ 0. Hence,
|xy| = −xy = (−x) × y = |x||y|
c Haythem O. Ismail
Lecture 6
8 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Absolute Values
Example
Prove that |xy| = |x||y| for real numbers x and y.
Proof
First, note that “x and y are real numbers”
[x, y ≥ 0] ∨ [x, y < 0] ∨ [x and y have different signs] .
We consider each case.
Case 1. In this case, |xy| = xy = x × y = |x||y|.
Case 2. In this case, |xy| = xy = (−x) × (−y) = |x||y|.
Case 3. In this case, we assume without loss of generality
that x < 0 and y ≥ 0. Hence,
|xy| = −xy = (−x) × y = |x||y|
c Haythem O. Ismail
Lecture 6
8 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Absolute Values
Example
Prove that |xy| = |x||y| for real numbers x and y.
Proof
First, note that “x and y are real numbers”
[x, y ≥ 0] ∨ [x, y < 0] ∨ [x and y have different signs] .
We consider each case.
Case 1. In this case, |xy| = xy = x × y = |x||y|.
Case 2. In this case, |xy| = xy = (−x) × (−y) = |x||y|.
Case 3. In this case, we assume without loss of generality
that x < 0 and y ≥ 0. Hence,
|xy| = −xy = (−x) × y = |x||y|
c Haythem O. Ismail
Lecture 6
8 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Absolute Values
Example
Prove that |xy| = |x||y| for real numbers x and y.
Proof
First, note that “x and y are real numbers”
[x, y ≥ 0] ∨ [x, y < 0] ∨ [x and y have different signs] .
We consider each case.
Case 1. In this case, |xy| = xy = x × y = |x||y|.
Case 2. In this case, |xy| = xy = (−x) × (−y) = |x||y|.
Case 3. In this case, we assume without loss of generality
that x < 0 and y ≥ 0. Hence,
|xy| = −xy = (−x) × y = |x||y|
c Haythem O. Ismail
Lecture 6
8 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Absolute Values
Example
Prove that |xy| = |x||y| for real numbers x and y.
Proof
First, note that “x and y are real numbers”
[x, y ≥ 0] ∨ [x, y < 0] ∨ [x and y have different signs] .
We consider each case.
Case 1. In this case, |xy| = xy = x × y = |x||y|.
Case 2. In this case, |xy| = xy = (−x) × (−y) = |x||y|.
Case 3. In this case, we assume without loss of generality
that x < 0 and y ≥ 0. Hence,
|xy| = −xy = (−x) × y = |x||y|
c Haythem O. Ismail
Lecture 6
8 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Outline
1
Proof By Cases
2
Existence Proofs
3
Case Study: Tilings
c Haythem O. Ismail
Lecture 6
9 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive and Nonconstructive Proofs
Existence proofs are proofs of statements of the form ∃x[P(x)].
We need to prove that objects with particular properties exist.
Existence proofs fall in two major classes:
Constructive proofs, in which we describe a particular object for
which the property holds.
Nonconstructive proofs, in which we do not describe any
particular object but prove the statement in some
other way.
c Haythem O. Ismail
Lecture 6
10 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive and Nonconstructive Proofs
Existence proofs are proofs of statements of the form ∃x[P(x)].
We need to prove that objects with particular properties exist.
Existence proofs fall in two major classes:
Constructive proofs, in which we describe a particular object for
which the property holds.
Nonconstructive proofs, in which we do not describe any
particular object but prove the statement in some
other way.
c Haythem O. Ismail
Lecture 6
10 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive and Nonconstructive Proofs
Existence proofs are proofs of statements of the form ∃x[P(x)].
We need to prove that objects with particular properties exist.
Existence proofs fall in two major classes:
Constructive proofs, in which we describe a particular object for
which the property holds.
Nonconstructive proofs, in which we do not describe any
particular object but prove the statement in some
other way.
c Haythem O. Ismail
Lecture 6
10 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive and Nonconstructive Proofs
Existence proofs are proofs of statements of the form ∃x[P(x)].
We need to prove that objects with particular properties exist.
Existence proofs fall in two major classes:
Constructive proofs, in which we describe a particular object for
which the property holds.
Nonconstructive proofs, in which we do not describe any
particular object but prove the statement in some
other way.
c Haythem O. Ismail
Lecture 6
10 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive and Nonconstructive Proofs
Existence proofs are proofs of statements of the form ∃x[P(x)].
We need to prove that objects with particular properties exist.
Existence proofs fall in two major classes:
Constructive proofs, in which we describe a particular object for
which the property holds.
Nonconstructive proofs, in which we do not describe any
particular object but prove the statement in some
other way.
c Haythem O. Ismail
Lecture 6
10 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive Proofs
Example
Prove that there are 100 consecutive positive integers that are not
perfect squares.
Proof.
Consider the sequence of integers 2501, 2502, . . . , 2600.
These are 100 consecutive positive integers.
To show that none of them is a perfect square we note that
1
2
If m > n, then m2 > n2 for positive integers m and n.
502 = 2500 and 512 = 2601.
Given the above, there are no perfect squares among
2501, 2502, . . . , 2600.
c Haythem O. Ismail
Lecture 6
11 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive Proofs
Example
Prove that there are 100 consecutive positive integers that are not
perfect squares.
Proof.
Consider the sequence of integers 2501, 2502, . . . , 2600.
These are 100 consecutive positive integers.
To show that none of them is a perfect square we note that
1
2
If m > n, then m2 > n2 for positive integers m and n.
502 = 2500 and 512 = 2601.
Given the above, there are no perfect squares among
2501, 2502, . . . , 2600.
c Haythem O. Ismail
Lecture 6
11 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive Proofs
Example
Prove that there are 100 consecutive positive integers that are not
perfect squares.
Proof.
Consider the sequence of integers 2501, 2502, . . . , 2600.
These are 100 consecutive positive integers.
To show that none of them is a perfect square we note that
1
2
If m > n, then m2 > n2 for positive integers m and n.
502 = 2500 and 512 = 2601.
Given the above, there are no perfect squares among
2501, 2502, . . . , 2600.
c Haythem O. Ismail
Lecture 6
11 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive Proofs
Example
Prove that there are 100 consecutive positive integers that are not
perfect squares.
Proof.
Consider the sequence of integers 2501, 2502, . . . , 2600.
These are 100 consecutive positive integers.
To show that none of them is a perfect square we note that
1
2
If m > n, then m2 > n2 for positive integers m and n.
502 = 2500 and 512 = 2601.
Given the above, there are no perfect squares among
2501, 2502, . . . , 2600.
c Haythem O. Ismail
Lecture 6
11 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive Proofs
Example
Prove that there are 100 consecutive positive integers that are not
perfect squares.
Proof.
Consider the sequence of integers 2501, 2502, . . . , 2600.
These are 100 consecutive positive integers.
To show that none of them is a perfect square we note that
1
2
If m > n, then m2 > n2 for positive integers m and n.
502 = 2500 and 512 = 2601.
Given the above, there are no perfect squares among
2501, 2502, . . . , 2600.
c Haythem O. Ismail
Lecture 6
11 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive Proofs
Example
Prove that there are 100 consecutive positive integers that are not
perfect squares.
Proof.
Consider the sequence of integers 2501, 2502, . . . , 2600.
These are 100 consecutive positive integers.
To show that none of them is a perfect square we note that
1
2
If m > n, then m2 > n2 for positive integers m and n.
502 = 2500 and 512 = 2601.
Given the above, there are no perfect squares among
2501, 2502, . . . , 2600.
c Haythem O. Ismail
Lecture 6
11 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Constructive Proofs
Example
Prove that there are 100 consecutive positive integers that are not
perfect squares.
Proof.
Consider the sequence of integers 2501, 2502, . . . , 2600.
These are 100 consecutive positive integers.
To show that none of them is a perfect square we note that
1
2
If m > n, then m2 > n2 for positive integers m and n.
502 = 2500 and 512 = 2601.
Given the above, there are no perfect squares among
2501, 2502, . . . , 2600.
c Haythem O. Ismail
Lecture 6
11 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Nonconstructive Proof
Example
Prove that there are irrational numbers x and y such that xy is
rational.
Proof.
√
Recall that 2 is irrational.
√ √2
Consider the number 2 .
We have√two cases:
√
1
2
√
2
2√ is rational. In this case, we are done. (With x = y =
2
2
√
2.)
is irrational. In this case, we proceed as follows.
√ √2
√
y = 2, which
2 and
are both irrational.
√ √
√ 2 2 √ √2·√2 √ 2
y
Hence, x = ( 2 ) = 2
= 2 = 2, which is
rational.
Let x =
c Haythem O. Ismail
Lecture 6
12 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Nonconstructive Proof
Example
Prove that there are irrational numbers x and y such that xy is
rational.
Proof.
√
Recall that 2 is irrational.
√ √2
Consider the number 2 .
We have√two cases:
√
1
2
√
2
2√ is rational. In this case, we are done. (With x = y =
2
2
√
2.)
is irrational. In this case, we proceed as follows.
√ √2
√
y = 2, which
2 and
are both irrational.
√ √
√ 2 2 √ √2·√2 √ 2
y
Hence, x = ( 2 ) = 2
= 2 = 2, which is
rational.
Let x =
c Haythem O. Ismail
Lecture 6
12 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Nonconstructive Proof
Example
Prove that there are irrational numbers x and y such that xy is
rational.
Proof.
√
Recall that 2 is irrational.
√ √2
Consider the number 2 .
We have√two cases:
√
1
2
√
2
2√ is rational. In this case, we are done. (With x = y =
2
2
√
2.)
is irrational. In this case, we proceed as follows.
√ √2
√
y = 2, which
2 and
are both irrational.
√ √
√ 2 2 √ √2·√2 √ 2
y
Hence, x = ( 2 ) = 2
= 2 = 2, which is
rational.
Let x =
c Haythem O. Ismail
Lecture 6
12 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Nonconstructive Proof
Example
Prove that there are irrational numbers x and y such that xy is
rational.
Proof.
√
Recall that 2 is irrational.
√ √2
Consider the number 2 .
We have√two cases:
√
1
2
√
2
2√ is rational. In this case, we are done. (With x = y =
2
2
√
2.)
is irrational. In this case, we proceed as follows.
√ √2
√
y = 2, which
2 and
are both irrational.
√ √
√ 2 2 √ √2·√2 √ 2
y
Hence, x = ( 2 ) = 2
= 2 = 2, which is
rational.
Let x =
c Haythem O. Ismail
Lecture 6
12 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Nonconstructive Proof
Example
Prove that there are irrational numbers x and y such that xy is
rational.
Proof.
√
Recall that 2 is irrational.
√ √2
Consider the number 2 .
We have√two cases:
√
1
2
√
2
2√ is rational. In this case, we are done. (With x = y =
2
2
√
2.)
is irrational. In this case, we proceed as follows.
√ √2
√
y = 2, which
2 and
are both irrational.
√ √
√ 2 2 √ √2·√2 √ 2
y
Hence, x = ( 2 ) = 2
= 2 = 2, which is
rational.
Let x =
c Haythem O. Ismail
Lecture 6
12 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Nonconstructive Proof
Example
Prove that there are irrational numbers x and y such that xy is
rational.
Proof.
√
Recall that 2 is irrational.
√ √2
Consider the number 2 .
We have√two cases:
√
1
2
√
2
2√ is rational. In this case, we are done. (With x = y =
2
2
√
2.)
is irrational. In this case, we proceed as follows.
√ √2
√
y = 2, which
2 and
are both irrational.
√ √
√ 2 2 √ √2·√2 √ 2
y
Hence, x = ( 2 ) = 2
= 2 = 2, which is
rational.
Let x =
c Haythem O. Ismail
Lecture 6
12 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Nonconstructive Proof
Example
Prove that there are irrational numbers x and y such that xy is
rational.
Proof.
√
Recall that 2 is irrational.
√ √2
Consider the number 2 .
We have√two cases:
√
1
2
√
2
2√ is rational. In this case, we are done. (With x = y =
2
2
√
2.)
is irrational. In this case, we proceed as follows.
√ √2
√
y = 2, which
2 and
are both irrational.
√ √
√ 2 2 √ √2·√2 √ 2
y
Hence, x = ( 2 ) = 2
= 2 = 2, which is
rational.
Let x =
c Haythem O. Ismail
Lecture 6
12 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Nonconstructive Proof
Example
Prove that there are irrational numbers x and y such that xy is
rational.
Proof.
√
Recall that 2 is irrational.
√ √2
Consider the number 2 .
We have√two cases:
√
1
2
√
2
2√ is rational. In this case, we are done. (With x = y =
2
2
√
2.)
is irrational. In this case, we proceed as follows.
√ √2
√
y = 2, which
2 and
are both irrational.
√ √
√ 2 2 √ √2·√2 √ 2
y
Hence, x = ( 2 ) = 2
= 2 = 2, which is
rational.
Let x =
c Haythem O. Ismail
Lecture 6
12 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp!
Example
Chomp is a two-player game.
Cookies are laid out on a rectangular m × n grid.
The top-left cookie is poisonous.
The two players take turns making moves.
In each move, a player eats a cookie together with all cookies
below and to the right of it.
The loser is the one who eats the poisonous cookie.
Does the first player has a winning strategy?
c Haythem O. Ismail
Lecture 6
13 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp!
Example
Chomp is a two-player game.
Cookies are laid out on a rectangular m × n grid.
The top-left cookie is poisonous.
The two players take turns making moves.
In each move, a player eats a cookie together with all cookies
below and to the right of it.
The loser is the one who eats the poisonous cookie.
Does the first player has a winning strategy?
c Haythem O. Ismail
Lecture 6
13 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp!
Example
Chomp is a two-player game.
Cookies are laid out on a rectangular m × n grid.
The top-left cookie is poisonous.
The two players take turns making moves.
In each move, a player eats a cookie together with all cookies
below and to the right of it.
The loser is the one who eats the poisonous cookie.
Does the first player has a winning strategy?
c Haythem O. Ismail
Lecture 6
13 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp!
Example
Chomp is a two-player game.
Cookies are laid out on a rectangular m × n grid.
The top-left cookie is poisonous.
The two players take turns making moves.
In each move, a player eats a cookie together with all cookies
below and to the right of it.
The loser is the one who eats the poisonous cookie.
Does the first player has a winning strategy?
c Haythem O. Ismail
Lecture 6
13 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp!
Example
Chomp is a two-player game.
Cookies are laid out on a rectangular m × n grid.
The top-left cookie is poisonous.
The two players take turns making moves.
In each move, a player eats a cookie together with all cookies
below and to the right of it.
The loser is the one who eats the poisonous cookie.
Does the first player has a winning strategy?
c Haythem O. Ismail
Lecture 6
13 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp!
Example
Chomp is a two-player game.
Cookies are laid out on a rectangular m × n grid.
The top-left cookie is poisonous.
The two players take turns making moves.
In each move, a player eats a cookie together with all cookies
below and to the right of it.
The loser is the one who eats the poisonous cookie.
Does the first player has a winning strategy?
c Haythem O. Ismail
Lecture 6
13 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp!
Example
Chomp is a two-player game.
Cookies are laid out on a rectangular m × n grid.
The top-left cookie is poisonous.
The two players take turns making moves.
In each move, a player eats a cookie together with all cookies
below and to the right of it.
The loser is the one who eats the poisonous cookie.
Does the first player has a winning strategy?
c Haythem O. Ismail
Lecture 6
13 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp: Nonconstructive Proof
Example
Proof.
Note the following:
1 The game ends after at most mn steps.
2 There must be a loser and, hence, a winner.
Consider the first player starting by eating the bottom-right cookie.
We have one of two possibilities:
1 This move is the first move of a winning strategy, in which case we are done.
2 This move is not the first move of a winning strategy of the first player.
Thus, the first move of the second player is the beginning of a winning
strategy.
Hence, instead of eating the bottom-right cookie, the first player could have
made the same move of the second player and secured the game.
c Haythem O. Ismail
Lecture 6
14 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp: Nonconstructive Proof
Example
Proof.
Note the following:
1 The game ends after at most mn steps.
2 There must be a loser and, hence, a winner.
Consider the first player starting by eating the bottom-right cookie.
We have one of two possibilities:
1 This move is the first move of a winning strategy, in which case we are done.
2 This move is not the first move of a winning strategy of the first player.
Thus, the first move of the second player is the beginning of a winning
strategy.
Hence, instead of eating the bottom-right cookie, the first player could have
made the same move of the second player and secured the game.
c Haythem O. Ismail
Lecture 6
14 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp: Nonconstructive Proof
Example
Proof.
Note the following:
1 The game ends after at most mn steps.
2 There must be a loser and, hence, a winner.
Consider the first player starting by eating the bottom-right cookie.
We have one of two possibilities:
1 This move is the first move of a winning strategy, in which case we are done.
2 This move is not the first move of a winning strategy of the first player.
Thus, the first move of the second player is the beginning of a winning
strategy.
Hence, instead of eating the bottom-right cookie, the first player could have
made the same move of the second player and secured the game.
c Haythem O. Ismail
Lecture 6
14 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp: Nonconstructive Proof
Example
Proof.
Note the following:
1 The game ends after at most mn steps.
2 There must be a loser and, hence, a winner.
Consider the first player starting by eating the bottom-right cookie.
We have one of two possibilities:
1 This move is the first move of a winning strategy, in which case we are done.
2 This move is not the first move of a winning strategy of the first player.
Thus, the first move of the second player is the beginning of a winning
strategy.
Hence, instead of eating the bottom-right cookie, the first player could have
made the same move of the second player and secured the game.
c Haythem O. Ismail
Lecture 6
14 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp: Nonconstructive Proof
Example
Proof.
Note the following:
1 The game ends after at most mn steps.
2 There must be a loser and, hence, a winner.
Consider the first player starting by eating the bottom-right cookie.
We have one of two possibilities:
1 This move is the first move of a winning strategy, in which case we are done.
2 This move is not the first move of a winning strategy of the first player.
Thus, the first move of the second player is the beginning of a winning
strategy.
Hence, instead of eating the bottom-right cookie, the first player could have
made the same move of the second player and secured the game.
c Haythem O. Ismail
Lecture 6
14 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp: Nonconstructive Proof
Example
Proof.
Note the following:
1 The game ends after at most mn steps.
2 There must be a loser and, hence, a winner.
Consider the first player starting by eating the bottom-right cookie.
We have one of two possibilities:
1 This move is the first move of a winning strategy, in which case we are done.
2 This move is not the first move of a winning strategy of the first player.
Thus, the first move of the second player is the beginning of a winning
strategy.
Hence, instead of eating the bottom-right cookie, the first player could have
made the same move of the second player and secured the game.
c Haythem O. Ismail
Lecture 6
14 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp: Nonconstructive Proof
Example
Proof.
Note the following:
1 The game ends after at most mn steps.
2 There must be a loser and, hence, a winner.
Consider the first player starting by eating the bottom-right cookie.
We have one of two possibilities:
1 This move is the first move of a winning strategy, in which case we are done.
2 This move is not the first move of a winning strategy of the first player.
Thus, the first move of the second player is the beginning of a winning
strategy.
Hence, instead of eating the bottom-right cookie, the first player could have
made the same move of the second player and secured the game.
c Haythem O. Ismail
Lecture 6
14 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp: Nonconstructive Proof
Example
Proof.
Note the following:
1 The game ends after at most mn steps.
2 There must be a loser and, hence, a winner.
Consider the first player starting by eating the bottom-right cookie.
We have one of two possibilities:
1 This move is the first move of a winning strategy, in which case we are done.
2 This move is not the first move of a winning strategy of the first player.
Thus, the first move of the second player is the beginning of a winning
strategy.
Hence, instead of eating the bottom-right cookie, the first player could have
made the same move of the second player and secured the game.
c Haythem O. Ismail
Lecture 6
14 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Chomp: Nonconstructive Proof
Example
Proof.
Note the following:
1 The game ends after at most mn steps.
2 There must be a loser and, hence, a winner.
Consider the first player starting by eating the bottom-right cookie.
We have one of two possibilities:
1 This move is the first move of a winning strategy, in which case we are done.
2 This move is not the first move of a winning strategy of the first player.
Thus, the first move of the second player is the beginning of a winning
strategy.
Hence, instead of eating the bottom-right cookie, the first player could have
made the same move of the second player and secured the game.
c Haythem O. Ismail
Lecture 6
14 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Uniqueness
Uniqueness proofs are used to prove statements that there is a
unique object having some property.
Recall that these statements have the form
∃x[P(x) ∧ ∀y[P(y) → y = x]].
c Haythem O. Ismail
Lecture 6
15 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Uniqueness of Multiplicative Inverse
Example
Prove that for a 6= 0, there is a unique real number b such that
ab = 1.
Proof.
First, note that a ·
1
a
= 1.
Hence, there is a number b(= 1/a) such that ab = 1.
We now need to prove the uniqueness of this number.
Suppose that ac = ab = 1.
Hence, ac = ab.
Since, a 6= 0, we can divide both sides by a to get c = b.
c Haythem O. Ismail
Lecture 6
16 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Uniqueness of Multiplicative Inverse
Example
Prove that for a 6= 0, there is a unique real number b such that
ab = 1.
Proof.
First, note that a ·
1
a
= 1.
Hence, there is a number b(= 1/a) such that ab = 1.
We now need to prove the uniqueness of this number.
Suppose that ac = ab = 1.
Hence, ac = ab.
Since, a 6= 0, we can divide both sides by a to get c = b.
c Haythem O. Ismail
Lecture 6
16 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Uniqueness of Multiplicative Inverse
Example
Prove that for a 6= 0, there is a unique real number b such that
ab = 1.
Proof.
First, note that a ·
1
a
= 1.
Hence, there is a number b(= 1/a) such that ab = 1.
We now need to prove the uniqueness of this number.
Suppose that ac = ab = 1.
Hence, ac = ab.
Since, a 6= 0, we can divide both sides by a to get c = b.
c Haythem O. Ismail
Lecture 6
16 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Uniqueness of Multiplicative Inverse
Example
Prove that for a 6= 0, there is a unique real number b such that
ab = 1.
Proof.
First, note that a ·
1
a
= 1.
Hence, there is a number b(= 1/a) such that ab = 1.
We now need to prove the uniqueness of this number.
Suppose that ac = ab = 1.
Hence, ac = ab.
Since, a 6= 0, we can divide both sides by a to get c = b.
c Haythem O. Ismail
Lecture 6
16 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Uniqueness of Multiplicative Inverse
Example
Prove that for a 6= 0, there is a unique real number b such that
ab = 1.
Proof.
First, note that a ·
1
a
= 1.
Hence, there is a number b(= 1/a) such that ab = 1.
We now need to prove the uniqueness of this number.
Suppose that ac = ab = 1.
Hence, ac = ab.
Since, a 6= 0, we can divide both sides by a to get c = b.
c Haythem O. Ismail
Lecture 6
16 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Uniqueness of Multiplicative Inverse
Example
Prove that for a 6= 0, there is a unique real number b such that
ab = 1.
Proof.
First, note that a ·
1
a
= 1.
Hence, there is a number b(= 1/a) such that ab = 1.
We now need to prove the uniqueness of this number.
Suppose that ac = ab = 1.
Hence, ac = ab.
Since, a 6= 0, we can divide both sides by a to get c = b.
c Haythem O. Ismail
Lecture 6
16 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Uniqueness of Multiplicative Inverse
Example
Prove that for a 6= 0, there is a unique real number b such that
ab = 1.
Proof.
First, note that a ·
1
a
= 1.
Hence, there is a number b(= 1/a) such that ab = 1.
We now need to prove the uniqueness of this number.
Suppose that ac = ab = 1.
Hence, ac = ab.
Since, a 6= 0, we can divide both sides by a to get c = b.
c Haythem O. Ismail
Lecture 6
16 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Outline
1
Proof By Cases
2
Existence Proofs
3
Case Study: Tilings
c Haythem O. Ismail
Lecture 6
17 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Tiling Problems
Tiling problems consist of grids of squares of various sizes.
A checkerboard uses a square grid. A standard checkerboard has
8 rows and 8 columns.
A domino is a rectangular piece that is one square by two
squares.
A board is tiled by dominoes when all its squares are covered
with no overlapping dominoes and no dominoes overhanging the
board.
c Haythem O. Ismail
Lecture 6
18 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Tiling Problems
Tiling problems consist of grids of squares of various sizes.
A checkerboard uses a square grid. A standard checkerboard has
8 rows and 8 columns.
A domino is a rectangular piece that is one square by two
squares.
A board is tiled by dominoes when all its squares are covered
with no overlapping dominoes and no dominoes overhanging the
board.
c Haythem O. Ismail
Lecture 6
18 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Tiling Problems
Tiling problems consist of grids of squares of various sizes.
A checkerboard uses a square grid. A standard checkerboard has
8 rows and 8 columns.
A domino is a rectangular piece that is one square by two
squares.
A board is tiled by dominoes when all its squares are covered
with no overlapping dominoes and no dominoes overhanging the
board.
c Haythem O. Ismail
Lecture 6
18 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Tiling Problems
Tiling problems consist of grids of squares of various sizes.
A checkerboard uses a square grid. A standard checkerboard has
8 rows and 8 columns.
A domino is a rectangular piece that is one square by two
squares.
A board is tiled by dominoes when all its squares are covered
with no overlapping dominoes and no dominoes overhanging the
board.
c Haythem O. Ismail
Lecture 6
18 / 19
Proof By Cases
Existence Proofs
Case Study: Tilings
Exercises
Example
Prove or disprove.
1
A standard checkerboard can be tiled using dominoes.
2
The board resulting from removing the top-left square from the
standard checkerboard can be tiled using dominoes.
3
The board resulting from removing both the top-left and
bottom-right squares from the standard checkerboard can be tiled
using dominoes.
c Haythem O. Ismail
Lecture 6
19 / 19