#### UMBC CMSC203, Discrete Structures, Fall 2010

In a proof by contradiction, we begin by assuming that the claim does not hold. Using this premise, we make a series of arguments and show that a contradiction is reached. That is, to prove a proposition p, we show that ¬p ⇒ FALSE. The, contrapositive of that implication is TRUE ⇒ p. since TRUE is indeed true, we can conclude that p holds.

A proof by contradiction is similar to an indirect proof. However, indirect proofs are used with implications, whereas, a proof by contradiction can be used in a more general setting.

Example 1:
Claim: There are an infinite number of prime numbers.

Proof: Suppose not. Then, there is a finite number of prime numbers and we can let p1, p2, ..., pn be all the prime numbers. We define
Q = p1 p2 ... pn + 1.

If Q is prime, then Q is a prime number not on the list p1, p2, ..., pn because Q is larger than every pi. That contradicts the assumption that p1, ..., pn contain all of the prime numbers.

On the other hand, suppose that Q is composite. Let q be a prime factor of Q. Note that Q is not divisible by any of the pi. Why? because the product p1 p2 ... pn is a multiple of pi and the next multiple of pi is p1 p2 ... pn + pi. Since

p1 p2 ... pn  <  Q  <  p1 p2 ... pn + pi,
Q cannot be a multiple of pi. That means the prime factor q of Q does not appear on the list of prime numbers p1, p2, ... pn since Q is a multiple of q. Thus, we also have a contradiction in this case.

In either case, we found a prime number not on the list of prime numbers. Thus, we can conclude that our premise that there is a finite number of prime numbers must be false. Thus, there must be an infinite number of primes.

QED

Example 2:
Claim: The graph below is not 3-colorable.

Proof: Suppose by contradiction that the graph is 3-colorable using 3 colors we'll call red, green and blue. Then, every vertex in the graph is colored red, green or blue. Without loss of generality, let the color of vertex a be red. Vertices b and d must be colored different colors that are not red. Let these two colors be, respectively, blue and green. Now, vertex e is adjacent to vertices a and d, so it cannot be colored red or green. Thus, vertex e must be colored blue. Similarly, c must be green, g must be red and we arrive at the following partial coloring:

Note that vertex f is adjacent to vertices b, d and g, which are, respectively, blue, green and red. Thus, vertex f cannot be colored blue, green or red. This contradicts our previous conclusion that every vertex is colored red, green or blue. Thus, our initial assumption that the graph is 3-colorable must be false.

QED

Example 3:
Claim:2 is an irrational number.

Proof: Suppose not. Then we can express √2 as a ratio of two integers n and m such that the greatest common divisor of n and m is 1 (i.e., gcd(n,m) = 1). [This argument relies on the fact we learned in elementary school about reducing fractions to lowest terms.] Thus, we have
2 = n/m.
Squaring both sides gives us:
2 = n2/m2
2 m2 = n2.
Thus, n2 is an even number. So, n itself must also be an even number. (If n were odd, n2 would also be odd.) Therefore, n = 2k for some integer k and we have
2 m2 = n2
2 m2 = (2 k)2
2 m2 = 4 k2
m2 = 2 k2.
So, m2 is also even, which in turn implies that m is even. Therefore, both n and m are divisible by 2. This contradicts our assumption that gcd(n,m) = 1. Thus, our initial assumption that √2 is rational must be false.
QED