Download Direct Proof More Examples Contraposition

Spring 2011
Proof Methods
Direct Proof
Prove: If an integer is divisible by 6, then twice the integer is divisible by 4.
Proof: Let x be an integer divisible by 6. Then x = 6k for some integer k by the definition of divisibility.
So 2x = 2(6k) = 12k. 12 is divisible by 4. Therefore 12k = 4(3k). Since k is an integer, 3k must
also be an integer. Thus by the definition of divisibility, 2x = 12k is divisible by 4.
More Examples
1. Prove or disprove: If 2n + 3 is odd then the number n is odd.
Disprove by counter example: Let n = 2. 2(2) + 3 = 7. In this case 2n+3 is odd but n is even
(not odd).
2. Prove or disprove: If 3n + 2 is odd then the number n is odd.
Proof: (Direct) If 3n + 2 is odd then 3n has to be odd. If 3n was even then you would be able to
factor a 2 from 3n + 2 which would prove that the sum 3n + 2 is even and we know that it is not.
Since 3n is odd then n has to be odd. Once again, if n were even then we would be able to factor
a 2 out of the expression 3n making then entire expression even and we know it is not. Therefore
we know that n must be odd.
3. Prove or disprove: If x + y = 7 then either x or y must be even.
Proof: (Exhaustive) Since we are given that x and y are positive integers, there are only 6 values
for x and y that sum to 7. We can make a table of all of these values and show that one must be
even
x
y
x+y
1 6 7
y is even
2 5 7
x is even
3 4 7
y is even
4 3 7
x is even
5 2 7
y is even
6 1 7
x is even
Contraposition
1. Prove: If n2 is odd then n is odd.
Proof: To prove “If n2 is odd then n is odd”, we will prove the contraposition: If n is even, then n2
is even.
If n is even, then n = 2k for some integer k. n2 = n  n = (2k)  (2k) = 2(2k2). Since k and 2 are
both integers, 2k2 is an integer. Therefore 2(2k2) is even. Therefore n2 is even.
2. Prove: If x is positive then so is x + 1.
Proof: Contraposition: If x + 1  0 then x  0.
If x + 1  0 then by subtracting both sides by 1, we have x  -1. Since -1 < 0, x  0.
Spring 2011
Proof Methods
3. Prove: If the product of 2 integers is not divisible by n then neither integer is.
Proof: Contraposition: If either integer x or y is divisible by 2 then their product xy is divisible by
2.
Without loss of generality, assume that x is divisible by 2. Then by the definition of divisibility
we know that x can be written as x = 2k for some integer k. Then the product xy can be written as
xy = 2ky. We know that k and y are both integers, therefore ky is also an integer. Therefore the
product 2(ky) is divisible by 2 which means that the product xy is also divisible by 2.
Contradiction
1. Prove: If 3n + 2 is odd then n is odd.
Proof: Assume n is even. Then n = 2k for some integer k. This implies that the expression
3n + 2 can be rewritten as follows:
3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1).
Since 3n + 2 = 2(3k + 1) we know that 3n + 2 is even. However, we are given that 3n + 2 is odd.
This is a contradiction. Therefore our assumption that n is even must be incorrect. Therefore
when 3n + 2 is odd, n must be odd.
2. Prove: No integer is both even and odd.
Proof: Assume there exists some integer x that is both even and odd. If x is even then it can be
written as x = 2a for some integer a. If x is odd then it can be written as x = 2b + 1 for some
integer b. Since x equals both, we must have that 2a = 2b + 1. Rewriting this we get:
2a = 2b + 1
2a – 2b = 1
2(a – b) = 1
a–b=½
Since a and b are both integers, a – b must be an integer. However, we have shown that a – b =
½ which is not an integer. Therefore we have a contraction. This means that our assumption that
an integer can be both even and odd is false. Therefore no integer can be both even and odd.
3. Prove: If the sum of two primes is prime, then one of the primes must be 2.
Proof: Let x and y be two prime numbers such that x + y is prime and both x and y > 2. Since x
and y are prime numbers not equal to 2, both x and y are odd. The sum of two odd numbers is
even. Therefore x + y is even. If x + y is even then x + y = 2k for some integer k. Since x and y
are both greater than 2, their sum x + y > 2. Therefore 2 is a divsor of x + y and x + y  2.
Therefore x + y is not prime. This is a contradiction since we are told that x + y is prime.
Therefore our assumption that x and y > 2 is incorrect. If the sum of two primes is prime then one
of the primes must be 2.