Download Math Structures Indirect Proof Examples By Contradiction Thm 4.6.1

Math Structures
Indirect Proof Examples
By Contradiction
Thm 4.6.1
There is no greatest integer.
Proof
Assume the opposite. In other words, assume that there is a greatest integer x. We must show a
contradiction. Let k  x 1 . Then k is an integer but k  x which contradicts the assumption
that x is the greatest integer.
4.6: 6
There is no greatest negative real number.
Proof
Assume the opposite. In other words, assume there is a greatest negative real number x. We
must show a contradiction. Choose y  x 2 . Then y is a negative real number. Since
x
x   0 , we have x  y . Hence y is a negative real number greater than x. This contradicts
2
the assumption that x is the greatest negative real number.
4.6: 3
For all integers n, 3n  2 is not divisible by 3.
Proof
Assume the opposite. In other words, assume 3n  2 is divisible by 3. We must show a
contradiction. By assumption, 3n  2  3k for some integer k. Hence, using basic algebra,
3k  3n  2
3n  2
2
k
 n
3
3
2
k n 
3
The left hand side of this equation is an integer since it’s the difference of integers, but the right
hand side is not an integer. Hence, a contradiction.
Thm 4.6.2
There is no integer that is both even and odd.
Proof
Assume the opposite. In other words, suppose there is an integer n that is both even and odd. We
must show a contradiction. By definition, we can find integers k and j such that n  2k and
n  2 j  1 . Since n  n , we must have 2k  2 j  1 . So, by elementary algebra,
2k  2 j  1
1
2
The left hand side of this equation is an integer since it’s the difference of integers, but the right
hand side is not an integer. Hence, a contradiction.
k j
By Contraposition
4.6: 19
If a product of two positive real numbers is greater than 100, then at least one of the
numbers is greater than 10.
Proof
We will prove the contrapositive. In other words, we’ll prove “If two numbers are both less than
or equal to 10, then their product is less than or equal to 100.” So, let n and m be integers such
that n  10 and m  10 . We must show n  m  100 . But by basic algebra,
n  m  10 10  100
Lemma 4.6.4
For all integers n, if n 2 is even then n is even.
Proof
We will prove the contrapositive. In other words, we'll prove "If n is odd, then n 2 is odd". So,
let n be an odd integer. We must show that n 2 is odd. Since n is odd, we can choose an integer
k such that n  2k 1 . Hence,
2
n 2   2k  1
 4k 2  4k  1
 2  2k 2  2k   1
 2t  1
where t   2k  2k  is an integer. Therefore, by definition, n 2 is odd.
2