Proofs in Mathematics

In this section, we discuss the proofs of a number of interesting results in mathematics.

The main goal is to see how mathematical proofs are constructed. But first, we need some terminology.

The natural numbers are the positive integers or whole numbers, 1, 2, 3, 4, 5, ….

The integers are the set consisting of the natural numbers, zero (0), and the negatives of the natural numbers: … , -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ….

The rational numbers are all the numbers which are quotients of integers (except that we do not allow any quotients in which the denominator is zero).

The word “rational" comes from “ratio”.

We begin with a famous problem. It is said that when the ancient Greeks discovered this fact, they sacrificed two hundred oxen to celebrate:

The square-root of 2 is “irrational”, i.e., not a rational number.

Proof: Suppose √2 is a rational number, that is, it is possible to write √2 = a/b, for some natural numbers a and b. As usual, we can assume that this fraction a/b is in lowest terms, which amounts to saying that a and b have no common divisors other than 1. Squaring both sides of the equation gives

2 =

or 2 b2 = a2

Now the left side of this equation is an even number, so the right side must also be even. But this forces a to be even, because a2 is odd if a is odd.

But if a is even, i.e., is divisible by 2, then a2 must be divisible by 4. After all, if a = 2k, then a2 = (2k)2 = 2k(2k) = 4k2 , which is clearly divisible by 4.

Now if 2 b2 is divisible by 4, then b2 must be divisible by 2 and, as before, we conclude that b is even. But this means a and b are both even, which contradicts our assumption that they have no common divisors.

= 4k2

b2 = 2k2

The fact that we arrived at a contradiction means it must have been wrong to make the assumptions we did. In this case, this means it was wrong to assume that the square-root of 2 is rational.

We conclude that it is irrational, completing the proof.

QED

This is an example of what mathematicians call “proof by contradiction”. You assume that the result you want is false and from that derive a contradiction. Then you can conclude that the result must be true.

Hilbert's Hotel. David Hilbert (1862 — 1943) was a celebrated German mathematician.

A number of important concepts in mathematics are named after him,

but there is also a more frivolous example, known as Hilbert's Hotel.

This establishment has an infinite number of rooms, numbered 1, 2, 3, 4, 5, . . . .

One day, there is a convention, and every room is full, but then a large group of additional guests arrives.

“No problem", says the clerk. He simply asks each guest to move to the room with double his or her current number.

All the guests move to the even-numbered rooms, freeing up all the odd-numbered rooms for the newcomers.

This story can be summarized by saying that there are just as many even numbers as there are numbers altogether. This seems peculiar to us, because usually if we take part of something, we expect it to be smaller.

The point, of course, is that that logic does not apply to infinite sets.

A set is called “countable" if it has the same number of elements as the natural numbers 1, 2, 3, 4, 5, . . . .We have just seen that the even numbers constitute a countable set.The set of integers . . . , -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, . . . is another (the negative numbers . . . , 5, -4, -3, -2, -1 clearly correspond to the natural numbers, but then they also correspond to the even numbers).We need to convince ourselves that the remaining numbers 0, 1, 2, 3, 4, 5, . . . . correspond to the odd numbers.

But this is not hard: if we double each number and add 1, we get 1, 3, 5, 7, 9, 11, . . . ).

A little more surprising is the set of rational numbers a/b . We shall now see that therationals are countable.

Actually, since we have seen that combining two countable sets gives a countable set (like the even and odd numbers combining to make all the natural numbers), it is enough to see that the set of positive rational numbers is countable.

We begin by making a table.

Eliminate any duplications.

1 2 3 4 5 6 7

1 1/1=1 2 3 4 5 6 7

2 1/2 2/2=1 3/2 4/2=2 5/2 6/2=3 7/2

3 1/3 2/3 3/3=1 4/3 5/3 6/3=2 7/3

4 1/4 2/4=1/2 3/4 4/4=1 5/4 6/4=3/2 7/4

5 1/5 2/5 3/5 4/5 5/5=1 6/5 7/5

numerator

Den

omin

ator

Then enumerate the positive rational numbers in a zigzag pattern.

1 2 3 4 5 6 7

1 1/1=1 2 3 4 5 6 7

2 1/2 2/2=1 3/2 4/2=2 5/2 6/2=3 7/2

3 1/3 2/3 3/3=1 4/3 5/3 6/3=2 7/3

4 1/4 2/4=1/2 3/4 4/4=1 5/4 6/4=3/2 7/4

5 1/5 2/5 3/5 4/5 5/5=1 6/5 7/5

numerator

Den

omin

ator

We find that there are countably many positive rational numbers.

A number n > 1 is “prime" if the only natural numbers which divide it are 1 and n.

The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

But 4 and 6 are not prime, since each is divisible by 2. Likewise, 15 is divisible by 3 and 5.

It is natural to wonder if the prime numbers come to an end, or if there are infinitely many.

We need to recall the principle of “prime factorization”, which says that any number N can be written as a product of prime numbers: N = p1p2p3 … pm;where p1, p2, p3, . . . , pm are prime numbers, possibly with repetitions.Moreover, this factorization is unique, apart from the order in which the factors are written.

So, for example,15 = 3 . 5, 8 = 2 . 2 . 2, 36 = 2 . 2 . 3 . 3.

Now suppose that there were only finitely many prime numbers, and label themp1, p2, p3, . . . , pm . Consider the numberM = (p1

. p2 . p3 . . . . . pm ) + 1.Clearly M is not divisible by any of the primes p1, p2, p3, . . . , pm , because if you divide M by any of them, there is always a remainder of 1.

But since we are assuming these are the only primes, this contradicts the principle of prime factorization.This contradiction proves that there must be infinitely many primes. QED

Johann Carl Friedrich Gauss (1777 –1855)