Download pdf file

More Examples of Proofs
Contradiction Proofs
Definition: A prime number is an integer greater than 1 which is divisible
only by 1 and itself. Ex: 2, 5, 11 are primes; 6, 15, 100 are not primes.
There are an infinite number of primes.
Pf: BWOC suppose that there are only a finite number of primes.
Let p1, p2, ..., pk be the complete list of all prime numbers.
Consider the integer N = p1p2...pk + 1.
N is not a prime because it is larger than every element in the list.
Suppose the integer d, with 1 < d < N, divides N.
Let p be a prime divisor of d.
p divides N since d divides N.
Thus, p divides N – p1p2....pk since p is in the list of primes.
Therefore, p divides 1. →←
Thus, there are an infinite number of primes.
Quiz #2
If all the integers of A are either even or squares, then either
there is an even integer in A or all the integers of A are squares.
(∀x ∈ A)(P(x) ∨ Q(x)) ⇒ (∃ x ∈ A)P(x) ∨ (∀ x ∈ A)Q(x)
~ (A ⇒ B) ≡ A ∧ ~B
(∀x ∈ A)(P(x) ∨ Q(x)) ∧ (∀ x ∈ A) ~P(x) ∧ (∃ x ∈ A) ~Q(x)
All the integers of A are either even or squares, and (but) none of
the integers in A are even and there is a non-square in A.
unless
When used in an English sentence the meaning can be ambiguous.
The “usual” interpretation of A unless B is :
~B ⇒ A ≡ ~(~B ∧ ~A) ≡ B ∨ A.
Thus,
Tuition will increase unless Referenda C & D pass
does not “usually” imply that tuition will not go up if C & D pass,
i.e.,
B ⇒ ~A ≡ ~(B ∧ ~~A) ≡ ~B ∨ ~A.
However,
I will fix your car unless you don't pay me.
Would have the stronger interpretation of ~B ⇔ A.
Contradiction Proofs
There do not exist three consecutive natural numbers
such that the cube of the largest is the sum of the cubes
of the smaller numbers.
Pf. Look at (n+2)3 = (n+1)3 + n3
n3 + 6n2 + 12n + 8 = n3 + 3n2 + 3n + 1 + n3
-1
3
5 6
0 = n3 - 3n2 - 9n - 7 = f(n)
f'(n) = 3n2 - 6n - 9 = 3(n-3)(n+1)
f(-1) = -2, f(3) = -34
f(5) = -2, f(6) = 47
the only root of f(n) is non-integral (between 5 and 6).
→←
Contradiction Proofs
There are no prime numbers a,b,c such that c3 = a3 + b3.
Pf. At least one of a, b is even.
May assume b = 2.
8 = c3 - a3 = (c - a)(c2 + ca + a2)
but each term in the sum is greater than or equal to 4 .
→←
Contradiction Proofs
If a,b,c are integers such that a2 + b2 = c2, show that
at least one of a or b is even.
Pf: Suppose that both a and b are odd.
There exist integers k and m so that a = 2k + 1 and b = 2m + 1.
a2 + b2 = (2k+1)2 + (2m+1)2 = 4(k2+m2) + 4(k+m) + 2.
hmmmmmmmm ???????
Try again.
Pf: Suppose that both a and b are odd.
a2 and b2 are odd, so c2 is even.
Since c2 is even, c is even.
There is an integer n so that c = 2n and so c2 = 4n2.
hmmmmmmmmmmm ???????
Contradiction Proofs
If a,b,c are integers such that a2 + b2 = c2, show that
at least one of a or b is even.
Pf: Suppose that both a and b are odd.
There exist integers k and m so that a = 2k + 1 and b = 2m + 1.
a2 + b2 = (2k+1)2 + (2m+1)2 = 4(k2+m2) + 4(k+m) + 2.
Thus, c2 = 2[2(k2+m2 + k+m) + 1], which, since k2+m2 + k+m is
an integer, is two times an odd number.
Since a2 and b2 are odd, c2 is even.
Since c2 is even, c is even.
There is an integer n so that c = 2n and so c2 = 4n2.
So, c2 = 2(2n2) which is two times an even number. →←
Case Method
The case method is based on the tautology:
[(P ∨ Q) ⇒ R ] ⇔ [(P ⇒ R) ∧ (Q ⇒ R)]
If the hypothesis of an implication can be broken up into
special cases (which cover all possibilities) and each
special case implies the conclusion, then the implication is
true.
The special cases may arise naturally from the form of the
hypothesis or a definition used in the hypothesis. Other
times, the special cases may be more "forced" – imposed
because you see how to easily prove the statement under
additional conditions.
Case Method
Prove that every natural number n is either a prime, a
perfect square or divides (n-1)!
Pf: If n = 1, then it is a perfect square, so we may assume n > 1.
Any natural number is either a prime or not a prime.
If n > 1 is not a prime, then there are integers a and b with n = ab
and 1 < a,b < n.
Case I: a = b
If a = b then n = a2, so n is a perfect square.
Case II: a ≠ b
Since a < n, a| (n-1)!.
Since b < n and b ≠ a, b | (n-1)!/a
So, ab | (n-1)!
Thus, n is either a prime, a perfect square or divides (n-1)!
Pigeon-Hole Principle
The Pigeon-Hole Principle is the formal statement of a
common sense idea that we are all aware of.
If there are 7 pigeons and 6 pigeon holes and all the
pigeons are in a pigeon hole, then some pigeon hole must
have at least 2 pigeons in it!
- or If there are 13 pieces of paper all of which are in one of
3 drawers in a desk, then some drawer must have at least 5
pieces of paper.
Pigeon-Hole Principle
Pigeon-Hole Principle : If kn + 1 objects are distributed
amongst n sets, one of the sets must contain at least
k + 1 objects.
Pf: If no set contains at least k+1 objects, then each of the
n sets has at most k objects. Thus, the total number of
objects is at most nk. →←
Problem
Let there be 9 points in 3-space with integer coordinates.
Show that there is a pair of these points whose line
segment contains an interior point whose coordinates are
integers.
Outline of Proof:
Points in 3-space have 3 coordinates, (a,b,c).
Integer coordinates are either odd or even.
There are 8 odd-even patterns of integer coordinates in 3-space.
Since there are 9 points, at least two must have the same pattern
by the Pigeon-Hole Principle.
The midpoint of the line segment joining two points with the
same pattern has integer coordinates.
Special Assignment 1 : Fill in the missing details of this proof.