CMPS 260
Examples of Proofs

Here are a few examples of (informal) mathematical proofs that use approaches other than mathematical induction, which is covered in a separate document. Among the techniques employed are Assume the Antecedant, proof by contradiction, and proving the contrapositive. The Pigeonhole Principle is also illustrated.

The Assume the Antecedant approach is relevant when proving an implication A ⟹ B. (Here, A is the antecedant and B is the consequent.) What one does is to assume that A is true and, making use of that assumption, to prove that B is also true.

One might identify two forms of proof by contradiction. In one, A is proved by showing ¬A ⟹ A (which is equivalent to A). In the other form, A is proved by showing ¬A ⟹ ¬B, where B is something already known to be true.

Typically, such proofs follow the Assume the Antecedant scheme. With respect to the first form, you would assume ¬A to be true and then proceed to show that A follows from that assumption. With respect to the second form, you would assume ¬A to be true and then proceed to draw a conclusion from it that is contradictory to a known fact/theorem.

The contrapositive of an implication A ⟹ B is ¬B ⟹ ¬A. They are equivalent to each other. In the proof by contrapostive approach, one proves an implication by proving its contrapositive.

The Pigeonhole Principle says that if you have n items and you distribute them among m "containers", then, assuming that n > m, some container must have more than one item in it.


Example 1:

Prove that there are infinitely many prime numbers (i.e., for any prime number, there exists a prime number that is greater).

Proof: (by contradiction)
Assume, contrary to what is to be proved, that there are only finitely many prime numbers, and let the set of all prime numbers be {P1, P2, ... Pn}.

Now let K = P1 · P2 · ... · Pn + 1 (i.e., the product of all the primes, plus 1). Clearly, K is greater than one and is not divisible by any of the Pi's. (Dividing K by any Pi yields a remainder of 1.) But every positive integer greater than one has at least one prime divisor! Hence, we have a contradiction! ■


Example 2:

Recall that a free tree is a connected and acyclic graph.

Lemma: Every free tree having at least two nodes has at least one node of degree one (i.e., an endpoint). (The degree of a node is the number of edges to which it is incident.)

Proof: (by contradiction) Suppose, contrary to what is to be proved, that there exists a free tree T having at least two nodes and in which every node has degree two or more. Start at an arbitrary node in T and trace out a path.

At each step, traverse an edge from the current node to another node, without using any edge for a second time. (In other words, arrive at each node via one edge and leave it via a different edge.) This is possible due to every node having degree at least two. Because the number of nodes is finite, at some point along the path a node will be visited for a second time. Therefore, T has a cycle, which contradicts the assumption that it is a tree. ∎


Example 3

Prove that, if there is a path connecting nodes u and v in a graph, the length of the shortest such path (i.e., the number of edge crossings) is fewer than the number of nodes in the graph.

Proof: (by contradiction) Suppose, contrary to what is to be proved, that G is a graph having m nodes and having nodes u and v such that the shortest path between u and v is of length n≥m. Let such a path be (u1, u2, ..., un+1), with u = u1 and v = un+1. Because n≥m, the Pigeonhole Principle1 tells us that some node must appear in this path twice. More precisely, there must exist i and j, with 1≤i<j≤n+1, such that ui = uj. This means that this path includes a cycle going from ui back to itself. If we remove the edges of that cycle from the path, we get a shorter path connecting u and v (namely (u1, u2, ..., ui, uj+1, ..., un+1)), which contradicts the assumption. ■


Example 4

(Note: This example is under construction!)

Letting k, m, and n be integers, prove that if k is not a divisor of mn, then k is neither a divisor of m nor a divisor of n. Formalizing this, taking its contrapositive, and simplifying, we get:

    ¬(k|mn)  ⟹  ¬(k|m) ∧ ¬(k|n)

 =    < contrapositive: p ⇒ q  ≡  ¬q ⇒ ¬p >

    ¬(¬(k|m) ∧ ¬(k|n))  ⟹  ¬¬(k|mn) 

 =    < DeMorgan >

    ¬¬(k|m) ∨ ¬¬(k|n)  ⟹  ¬¬(k|mn) 

 =    < Double Negation (three times) >

    k|m ∨ k|n  ⟹  k|mn

What remains is to show that if k is a divisor of either m or n, then it is a divisor of mn. But that should be obvious.

For the reader who is in need of a more complete argument, we can continue by showing this implication using the Assume the Antecdant approach. That is, we assume that either k is a divisor of m or k is a divisor of n, and from that assumption we show that k is a divisor of mn:

Case 1: k is a divisor of m, which is to say that, for some integer j, m = kj. Then mn = (kj)n = k(jn), which is to say that k is a divisor of mn.

Case 2: k is a divisor of n. Similar logic to case 1.


Example 5

Prove that √2 is irrational.

In support of the proof, a number-theoretic lemma and two corollaries are stated first.

Lemma: Let a and b be positive integers and suppose that their prime factorizations are as follows:

a = 2i2 · 3i3 · 5i5 · 7i7 · 11i11 · ...
b = 2j2 · 3j3 · 5j5 · 7j7 · 11j11 · ...

Then the prime factorization of a·b is a·b = 2i2+j2 · 3i3+j3 · 5i5+j5 · 7i7+j7 · 11i11+j11 · ...

Proof: Elementary number theory

Corollary 1: Let a be a positive integer and suppose that its prime factorization is a = 2i2 · 3i3 · 5i5 · 7i7 · 11i11 · ...

Then the prime factorization of a2 is a2 = 22i2 · 32i3 · 52i5 · 72i7 · 112i11 · ...

Corollary 2: Let a be a positive integer and suppose that a2 is even (i.e., 2|a2). Then a is even (i.e., 2|a).

Now we go ahead with the proof that √2 is irrational.

Proof: (by contradiction) Assume, contrary to what is to be proved, that √2 is rational, which is to say that there exist positive integers k and m, whose greatest common divisor is 1, such that √2 = k/m.

    √2 = k/m

⇒     < a = b ⇒ a2 = b2 >

    (√2)2 = (k/m)2

=      < (√a)2 = a >

    2 = (k/m)2

=      < (a/b)c = ac/bc >

    2 = k2/m2

=      < a = b/c ≡ ac = b >

    2m2 = k2

⇒    < 2a = b ⇒ b is even; thus, k2 is even.
        By Corollary 2, k is even, so let k = 2r >

    2m2 = (2r)2

=      < (ab)c = acbc; 22 = 4 >

    2m2 = 4r2

=      < dividing both sides by 2 >

    m2 = 2r2

⇒   < a = 2b ⇒ a is even >

    m2 is even (i.e., 2|m2)

⇒    < Corollary 2 >

    m is even

During the course of the proof, we found that both k and m must be even. But this contradicts the assumption that their GCD is 1. ■