Download Proof Techniques 1 Vacuous Proof 2 Trivial Proof 3 Direct Proof 4

Proof Techniques
ICS141 (H1.6)
1
p
!
8/23/94
Vacuous Proof
q
is true if p is false. Thus, we can prove
)
p
q
by establishing that
is false.
p
Example 1
Theorem: If 2x is odd for some integer x, x2 + 1 is prime.
Proof: By the denition of odd and even integers, it is not the case that 2x is odd.
2
p
!
Trivial Proof
q
is true if q is true. Thus, we can prove
p
)
q
by establishing that
q
is true.
Example 2
Theorem: If x = p + 1 > 0 for some prime integer p, then 2x + 1 is odd.
Proof: By the denition of odd integers, 2x + 1 is obviously odd.
3
2
2
Direct Proof
Apply inference rules in order to deduce the truth of
q
from the truth of p.
Example 3
Theorem: If 6x + 9y = 101, then either x or y is not an integer.
Proof: Assume that 6x + 9y = 101. This can be rewritten as 3(2x + 3y ) = 101. But 101=3 is not an
integer. Hence 2x + 3y is not an integer. Therefore, either x or y must not be an integer.
2
Example 4
Theorem: If S is a set of one- and two-digit integers such that each of the digits 0 through 9 occurs
exactly once in the set S , then the sum of the elements of S is divisible by 9.
Proof: Assume that the hypothesis of the above assertion is true. Note that the digits 0 through
9 sum to 45. In any set S , some of the digits will occur in the 10's position and the reminder will
occur in the 1's position. Let k denote the sum of the digits which occur in the 10's position. Then
the sum of the elements of S can be expressed as follows.
10k + (45 0 k) = 9k + 45 = 9(k + 5)
Since it has a factor 9, the sum is divisible by 9 regardless of the value k.
4
2
Proof of the Contrapositive
Show that the contrapositive
: !:
q
p
of
p
!
q
is true, by using other proof techniques.
Example 5
A perfect number is an integer which is equal to the sum of all its divisors except the number itself.
For example, 6 = 1 + 2 + 3 and 28 = 1 + 2 + 4 + 7 + 14.
Theorem: A perfect number is not a prime number.
Proof: The contrapositive is that a prime number is not a perfect number. We use direct proof to
show the contrapositive. Suppose that k is a prime number. Then k 2 and k has exactly two
divisors: 1 and k. The sum of all its divisors except k itself is 1. Since k 2, k 6= 1. Thus k is not
a perfect number.
2
1
5
Proof by Contradiction
Assume that p ! q were false, i.e.,
r ^ :r from this assumption.
p
were true and
q
were false. Derive a contradiction such as
General Principle of Proof by Contradiction
Assume that p were false. Derive a contradiction from this assumption. Then we can conclude that
p is true.
Example 6
Theorem: There is no largest prime number.
Proof: Assume that there were a largest prime number, say K . Since all prime numbers are greater
than 1 and none of prime numbers are greater than K , there must be only a nite number of prime
numbers. Consider the product of all those prime numbers and call it j .
j
= 2 1 3 1 5 1 7 1 1 1K
If we divide j + 1 by any prime number between 2 and K , the remainder is 1. It means that j + 1
cannot be expressed as a product of any two integers other than (j + 1) and 1. Thus j + 1 must
be a prime number. Since j > K , j + 1 is a prime number greater than K . This contradicts the
assumption that K is the largest prime number. Therefore, there is no largest prime number.
2
6
Reduction to Absurdity
Assume that p ! q were false. That is, p is true and q is false. Derive :r for some assertion r which
is known to be true, by using inference rules.
Example 7
Theorem: If x2 + x 0 2 = 0, then x 6= 0.
2
2
2
Proof: Assume that x + x 0 2 = 0 and x = 0. Then x + x 0 2 = 0 + 0 0 2 = 02 must be equal to
0. It is obvious that 02 6= 0. Thus the assumption is not valid and hence the theorem is proved. 2
7
Proof by Cases
(p1 _ p2 _ 1 1 1 _ pn) ! q is equivalent to (p1 ! q) ^ (p2 ! q) ^ 1 1 1 ^ (pn
(1 i n).
!
q
). Show pi ) q for each
i
Example 8
Let max(a; b) be the operation to choose the maximum of two integers a and b.
Theorem: max(a; max(b; c)) = max(max(a; b); c)
Proof: For any three integers a, b and c, one of the following 6 cases must hold.
Case 1: a b c
Then max(b; c) = b and max(a; b) = a and max(max(a; b); c) = max(a; c) = a. Therefore the
equality holds.
Case 2: a c b
max(a; max(b; c)) = max(a; c) = a and max(max(a; b); c) = max(a; c) = a. Thus the equality
holds.
Case 3: b a c
max(a; max(b; c)) = max(a; b) = b and max(max(a; b); c) = max(b; c) = b. Thus the equality
holds.
2
Case 4: b c a
max(a; max(b; c)) = max(a; b) = b and max(max(a; b); c) = max(b; c) = b. Thus the equality
holds.
Case 5: c a b
max(a; max(b; c)) = max(a; c) = c and max(max(a; b); c) = max(a; c) = c. Thus the equality
holds.
Case 6: c b a
max(a; max(b; c)) = max(a; c) = c and max(max(a; b); c) = max(b; c) = c. Thus the equality
holds.
Since the equality holds in all cases, the equality holds.
3
2