⇐ Indirect Proofs | Existence Proofs ⇒ |

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.

Claim:There are an infinite number of prime numbers.

Proof:Suppose not. Then, there is a finite number of prime numbers and we can letp_{1},p_{2}, ...,pbe all the prime numbers. We define_{n}Q=p_{1}p_{2}...p+ 1._{n}If

Qis prime, thenQis a prime number not on the listp_{1},p_{2}, ...,p_{n}becauseQis larger than everyp. That contradicts the assumption that_{i}p_{1}, ...,pcontain all of the prime numbers._{n}On the other hand, suppose that

Qis composite. Letqbe a prime factor ofQ. Note thatQis not divisible by any of thep. Why? because the product_{i}p_{1}p_{2}...pis a multiple of_{n}pand the next multiple of_{i}pis_{i}p_{1}p_{2}...p+_{n}p. Since_{i}p_{1}p_{2}...p<_{n}Q<p_{1}p_{2}...p+_{n}p,_{i}Qcannot be a multiple ofp. That means the prime factor_{i}qofQdoes not appear on the list of prime numbersp_{1},p_{2}, ...psince_{n}Qis a multiple ofq. 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

Claim:The graph below is not 3-colorable.

Proof:Suppose by contradiction that the graphis3-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 vertexabe red. Verticesbanddmust be colored different colors that are not red. Let these two colors be, respectively, blue and green. Now, vertexeis adjacent to verticesaandd, so it cannot be colored red or green. Thus, vertexemust be colored blue. Similarly,cmust be green,gmust be red and we arrive at the following partial coloring:

Note that vertex

fis adjacent to verticesb,dandg, which are, respectively, blue, green and red. Thus, vertexfcannot 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

Claim:√2 is an irrational number.

Proof:Suppose not. Then we can express √2 as a ratio of two integersnandmsuch that the greatest common divisor ofnandmis 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 =Squaring both sides gives us:n/m.2 =Thus,n^{2}/m^{2}

2m^{2}=n^{2}.n^{2}is an even number. So,nitself must also be an even number. (Ifnwere odd,n^{2}would also be odd.) Therefore,n= 2kfor some integerkand we have2So,m^{2}=n^{2}

2m^{2}= (2k)^{2}

2m^{2}= 4k^{2}

m^{2}= 2k^{2}.m^{2}is also even, which in turn implies thatmis even. Therefore, bothnandmare divisible by 2. This contradicts our assumption that gcd(n,m) = 1. Thus, our initial assumption that √2 is rational must be false.QED

⇐ Indirect Proofs | Existence Proofs ⇒ |

Last Modified: 15 Feb 2008 17:29:25 EST by Richard Chang to Spring 2008 CMSC 203 Section Homepage