Lecture 3 – February 17, 2003
Mar 30, 2015
Lecture 3 – February 17, 2003
Chapter 3
Elementary Number Theory and Methods of Proof
Section 3.1
Direct Proof and Counterexample I: Introduction
Definitions A definition gives meaning to a term. A non-primitive term is defined using previously
defined terms. A primitive term is undefined. Example
A function f : R R is increasing if f(x) f(y) whenever x y.
Previously defined terms: function, real numbers, greater than.
Definitions Definitions are not theorems. Example
Def: A number n is a perfect square if n = k2 for some integer k.
Now suppose t is a perfect square. Then t = k2 for some integer k. Is this the “error of the converse”?
Definitions are automatically “if and only if,” even though they don’t say so.
Proofs A proof is an argument leading from a
hypothesis to a conclusion in which each step is so simple that its validity is beyond doubt.
That is a subjective judgment – what is simple to one person may not be so simple to another.
Types of Proofs Proving universal statements
Prove something is true in every instance Proving existential statements
Prove something is true in at least one instance Disproving universal statements
Prove something is false in at least one instance Disproving existential statements
Prove something is false in every instance
Proving Universal Statements The statement is generally of the form
x D, P(x) Q(x) Use the method of generalizing from the
generic particular. Select an arbitrary x in D (generic particular). Assume that P(x) is true (hypothesis). Argue that Q(x) is true (conclusion).
Theorem: The sum of two consecutive triangle numbers is a perfect square. Definition: Let n be a positive integer. The nth
triangle number Tn is the number n(n + 1)/2.
Definition: Let n be a positive integer. The nth perfect square Sn is the number n2.
Example: Direct Proof
Proof of Theorem Proof:
Let n be a positive integer. Tn + Tn + 1 = n(n + 1)/2 + (n + 1)(n + 2)/2
= (n2 + n + n2 + 3n + 2)/2= (2n2 + 4n + 2)/2= (n + 1)2
= Sn + 1. Therefore, Tn + Tn + 1 = Sn + 1 for all n 1.
Example: Direct Proof Theorem: If x, y R, then
x2 + y2 2xy. Incorrect proof:
x2 + y2 2xy. x2 – 2xy + y2 0. (x – y)2 0, which is known to be true.
What is wrong?
Lecture 2 – Feb 19, 2003
Proving Existential Statements Proofs of existential statements are also
called existence proofs. Two types of existence proofs
Constructive Construct the object. Prove that it has the necessary properties.
Non-constructive Argue indirectly that the object must exist.
Example: Constructive Proof Theorem: Given a segment AB, there is a
midpoint M of AB. Proof:
A B
M
C
Justification Argue by SAS that triangles ACM and BCM
are congruent and that AM = MB.
A B
M
C
Example: Constructive Proof Theorem: The equation
x2 – 7y2 = 1.
has a solution in positive integers. Proof:
Let x = 8 and y = 3. Then 82 – 732 = 64 – 63 = 1.
Example: Constructive Proof Theorem: The equation
x2 – 67y2 = 1.
has a solution in positive integers. Proof: ?
Example: Non-Constructive Proof Theorem: There exists x R such that
x5 – 3x + 1 = 0. Proof:
Let f(x) = x5 – 3x + 1. f(1) = –1 < 0 and f(2) = 27 > 0. f(x) is a continuous function. By the Intermediate Value Theorem, there
exists x [1, 2] such that f(x) = 0.
Disproving Universal Statements Construct an instance for which the
statement is false. Also called proof by counterexample.
Example: Proof by Counterexample Disprove the conjecture (Fermat): All
integers of the form 22n + 1, for n 1, are prime.
(Dis)proof: Let n = 5. 225 + 1 = 4294967297. 4294967297 = 6416700417.
Example: Proof by Counterexample Disprove the statement: If a function is
continuous at a point, then it is differentiable at that point.
(Dis)proof: Let f(x) = |x| and consider the point x = 0. f(x) is continuous at 0. f(x) is not differentiable at 0.
Disproving Existential Statements These can be among the most difficult of all
proofs. Famous examples
There is no formula “in radicals” for the general solution of a 5th degree polynomial.
There is no solution in positive integers of the equation
xn + yn = zn.
Example: Disproving an Existential Statement Theorem: There is no solution in integers to the
equation
x2 – y2 = 101010 + 2. Proof:
A perfect square divided by 4 has remainder 0 or 1. Therefore, x2 – y2 divided by 4 has remainder 0, 1, or 3. However, 101010 + 2 divided by 4 has remainder 2. Therefore, x2 – y2 101010 + 2 for any integers x and y.
Section 3.2
Direct Proof and Counterexample II: Rational Numbers
Rational Numbers A rational number is a number that equals
the quotient of two integers. Let Q denote the set of rational numbers. An irrational number is a number that is
not rational. We will assume, for the time being, that
there exist irrational numbers.
Direct Proof Theorem: The sum of two rational numbers
is rational. Proof:
Let r = a/b and s = c/d be rational. Then r + s = (ad + bc)/bd, which is rational.
Proof by Counterexample Disprove: The sum of two irrationals is
irrational. Counterexample:
Let α be irrational. Then –α is irrational. α + (–α) = 0, which is rational.
Direct Proof Theorem: Between every two distinct
rationals, there is a rational. Proof:
Let r, s Q. Assume that r < s. Let t = (r + s)/2. Then t Q. We must show that r < t < s.
Proof continued Given: r < s. Add r: 2r < r + s. Divide by 2: r < (r + s)/2 = t. Given: r < s. Add s: r + s < 2s. Divide by 2: t = (r + s)/2 < s. Therefore, r < t < s.
Lecture 3 – Feb 20, 2003
Other Theorems Theorem: Between every two distinct
irrationals there is a rational. Proof: ? Theorem: Between every two distinct
irrationals there is an irrational. Proof: ?
An Interesting Question Why are the last two theorems so hard to
prove? Because they involve “negative”
hypotheses and “negative” conclusions.
Positive and Negative Statements A positive statement asserts the existence of
a number. A negative statement asserts the
nonexistence of a number. It is much easier to use a positive
hypothesis than a negative hypothesis. It is much easier to prove a positive
conclusion than a negative conclusion.
Positive and Negative Statements “r is rational” is a positive statement.
It asserts the existence of integers a and b such that r = a/b.
“α is irrational” is a negative statement. It asserts the nonexistence of integers a and b
such that α = a/b. Is there a “positive” characterization of
irrational numbers?
Section 3.3
Direct Proof and Counterexample III: Divisibility
Divisibility Definition: An integer a divides an integer
b if a 0 and there exists an integer c such that
ac = b. Write a | b to indicate that a divides b. Divisibility is a positive property.
Units Definition: An integer u is a unit if u | 1. This is a positive property.
Why? The only units are 1 and –1.
Composite Numbers Definition: An integer n is composite if
there exist non-units a and b such that
n = ab. A composite number factors in a non-trivial
way. Is this a positive property?
What about the non-units?
Prime Numbers Definition: An integer p is prime if p is not
a unit and p is not composite. A prime number factors only in a trivial
way. This is a negative property. Prime numbers: 2, 3, 5, 7, 11, …
Example: Direct Proof Theorem: If u and v are units, then uv is a unit. Proof:
Let u and v be units. There exist integers r and s such that ur = 1 and vs = 1. Therefore, (ur)(vs) = 1. Rearrange: (uv)(rs) = 1. Therefore, uv is a unit.
Example: Direct Proof Theorem: Let a and b be integers. If a | b and
b | a, then a/b and b/a are units. Proof:
Let a and b be integers. Suppose a | b and b | a. There exist integers c and d such that ac = b and bd = a. Therefore, acd = bd = a. Therefore, cd = 1. Thus, c and d are units.
Corollary: If a | b and b | a, then a = b or a = –b.
Example: Direct Proof Theorem: Let a, b, c be integers. If a | b and
b | a + c, then a | c. Proof:
Let a, b, and c be integers. Suppose a | b and b | a + c. There exist integers d and e such that ad = b and
be = a + c. Substitute: (ad)e = a + c. Rearrange: a(de – 1) = c. Therefore, a | c.
Lecture 12 – Feb 20, 2003
Section 3.4
Direct Proof and Counterexample IV: Division into Cases and the
Quotient-Remainder Theorem
The Quotient-Remainder Theorem Theorem: Let n and d be integers, d 0.
Then there exist unique integers q and r such that
n = qd + r
and
0 r < d. q is the quotient and r is the remainder.
Example: Proof by Cases Theorem: For any integer n, n3 – n is a
multiple of 6. Proof:
Divide n by 6 to get q and r: n = 6q + r, where 0 r < 6.
Substitute: n3 – n = (6q + r)3 – (6q + r). Expand and rearrange:
n3 – n = 6(36q3 + 18q2r + 3qr2 – q) + (r3 – r).
Proof continued Therefore, 6 | (n3 – n) if and only if 6 | (r3 – r). Consider the 6 possible cases:
Case 1: r = 0. r3 – r = 03 – 0 = 0 = 60. Case 2: r = 1. r3 – r = 13 – 1 = 0 = 60. Case 3: r = 2. r3 – r = 23 – 2 = 6 = 61. Case 4: r = 3. r3 – r = 33 – 3 = 24 = 64. Case 5: r = 4. r3 – r = 43 – 4 = 60 = 610. Case 6: r = 5. r3 – r = 53 – 5 = 120 = 620.
Proof continued In every case, 6 | (r3 – r).
Therefore, 6 | (r3 – r) in general. Therefore, 6 | (n3 – n) for all integers n.
Section 3.5
Direct Proof and Counterexample V: Floor and Ceiling
The Floor Function Let x be a real number. The floor of x,
denoted x, is the integer n such that
n x < n + 1. If x is an integer, then x = x. If x is not an integer, then x is the first integer
such that x < x.
The Ceiling Function The ceiling of x, denoted x, is the integer n
such that
n – 1 < x n. If x is an integer, then x = x. If x is not an integer, then x is the first integer
such that x > x.
Example: Direct Proof Theorem: Let x and y be real numbers.
Thenx + y x + y < x + y + 1.
Proof (1st inequality): By definition, x x and y y. Therefore, x + y x + y.
Proof (2nd inequality): By definition, x + y < x + y + 1.
Exercise: The Ceiling Function Theorem: Let x and y be real numbers.
Then
x + y – 1 < x + y x + y. Proof: Exercise
Questions Is – –x = x true for all real numbers x?
Proof: ? Is x – 1 < x x true for all real numbers x?
Proof: ? Is 2x + 2y = x + y + x + y true for all
real numbers x and y? Proof: ?
An Interesting Theorem Theorem: Let x be a positive real number. Then x
is irrational if and only if the two sequences
1 + x, 2 + 2x, 3 + 3x, …and
1 + 1/x, 2 + 2/x, 3 + 3/x, …together contain every positive integer exactly once.
Proof: ?
Lecture 13 – Feb 20, 2003
Section 3.6
Indirect Argument: Contradiction and Contraposition
Form of Proof by Contraposition Theorem: p q. This is logically equivalent to q p. Outline of the proof of the theorem:
Assume q. Prove p. Conclude that p q.
This is a direct proof of the contrapositive.
Benefit of Proof by Contraposition If p and q are negative statements, then p
and q are positive statements. We may be able to give a direct proof that
q p.
Example: Proof by Contraposition Theorem: The sum of a rational and an
irrational is irrational. Restate the theorem: Let r be a rational
number and let α be a number. If α is irrational, then r + α is irrational.
Restate again: Let r be a rational number and let α be a number. If r + α is rational, then α is rational.
The Proof Proof:
Let r be rational and α be a number. Suppose that r + α is rational. Let s = r + α. Then α = s – r, which is rational. Therefore, if r + α is rational, then α is rational. It follows that if α is irrational, then r + α is
irrational.
Example: Proof by Contraposition Theorem: If u is a unit and p is prime, then
up is prime. Restatement: Let u be unit and p be an
integer. If p is a prime, then up is a prime. 2nd Restatement: Let u be unit and p be an
integer. If up is not a prime, then p is not a prime.
Proof continued Proof:
Let u be a unit and p an integer. There is an integer v such that uv = 1. Suppose up is not prime. Two possibilities:
up is a unit. up is composite.
Proof continued (Case 1) Case 1: up is a unit.
Then (up)v is a unit. However, (up)v = (uv)p = p. Therefore, p is a unit. Therefore, p is not a prime.
Proof continued (Case 2) Case 2: up is composite.
There exist non-units b and c such that up = bc. Then p = (uv)p = (up)v = (bc)v = (bv)c. bv and c are non-units. Therefore, p is composite. Therefore, p is not a prime.
Proof concluded In both cases p is not a prime. Therefore p is not a prime in general. Therefore, if p is prime, then up is prime.
Form of Proof by Contradiction Theorem: p q. Outline of the proof of the theorem :
Assume (p q). This is equivalent to assuming p q. Derive a contradiction, i.e., conclude r r for
some statement r. Conclude that p q.
Benefit of Proof by Contradiction The statement r may be any statement
whatsoever because any contradiction
r r
will suffice.
Contradiction vs. Contraposition Sometimes a proof by contradiction “becomes” a
proof by contraposition. Here is how it happens.
Assume (p q), i.e., p q. Prove p. Cite the contradiction p p. Conclude that p q.
Is this proof by contradiction or by contraposition? Proof by contraposition is preferred.
Lecture 14 – Feb 24, 2003
Useful Fact Theorem: An integer p is prime if and only
if, for all integers a and b, if p | ab, then p | a or p | b.
In symbols, p is prime if and only ifa, b Z, (p | ab p | a p | b)
This is a positive characterization of primes. It may allow a direct proof rather than a
proof by contradiction or contraposition.
Direct Proof Theorem: If u is a unit and p is prime, then
up is prime. Proof:
Let u be a unit and p a prime. There is an integer v such that uv = 1. Let a and b be integers and suppose that up | ab. There exists an integer c such that upc = ab. Therefore, p | ab.
Proof concluded Thus, p | a or p | b, since p is prime. Case 1: p | a.
Then there exists an integer d such that pd = a. Then (up)(dv) = (uv)(pd) = a. Therefore, up | a.
Case 2: p | b. Similar to Case 1.
Therefore, up | a or up | b. Therefore, up is prime.
Section 3.7
Two Classical Theorems
Classical Theorem #1 Theorem: 2 is irrational. Proof (Euclid):
Suppose 2 is rational. There exist integers a and b such that 2 = a/b. (WOLOG) Assume that a and b are relatively
prime. Square: 2b2 = a2. Therefore, 2 | a2 and so 2 | a.
Proof concluded Substitute 2c for a: 2b2 = 4c2. Simplify: b2 = 2c2. Therefore, 2 | b2 and so 2 | b. This contradicts the assumption that a and b are
relatively prime. Therefore, 2 is irrational.
Classical Theorem #2 Theorem: The set of prime numbers is infinite. Proof:
Suppose there are only finitely many primes. Let {p1, …,pn} be a complete list of the primes. Let k = (p1 … pn) + 1. k 2, yet pi does not divide k for any i. This is a contradiction. Therefore, there are infinitely many primes.
Euclid’s proof.
Example: Constructive Existence Proof Theorem: Between any two distinct
irrationals there is a rational and an irrational.
Proof: Let α and β be irrational numbers with α < β. Then β – α > 0. Choose an integer n such that n(β – α) > 1. Then 1/n < β – α.
Proof continued Let m = nβ – 1. Then
m < nβ m + 1. Then m/n < β and nβ – 1 m. Then α < β – 1/n = (nβ – 1)/n m/n. Therefore, α < m/n < β.
Proof concluded Choose an integer k such that
k(β – m/n) > 2. Divide by k:
β – m/n > 2/k. Then β > m/n + 2/k. Therefore,
α < m/n < m/n + 2/k < β.