## Sunday, January 01, 2006

### Irrational Numbers

Irrational numbers are numbers that cannot be formed by the ratio of two integers. In Ancient Greece, the mathematician and philosopher Pythagoras created a secret society whose sole goal was to study the universe in terms of numbers. Pythagorus had discovered that musical harmonies result from perfect ratios between string sizes and became convinced that all the universe could be studied in terms of these perfect ratios. For example, the most famous discovery of his group was the Pythagorean Theorem which states that the hypotenuse (the diagonal) of a right triangle is equal to the sum of the squares of its sides. In other words: if c is the length of the hypotenus, a,b are the length of the other sides, then c2 = a2 + b2.

It turns out that the Pythagorean Theorem is itself a proof for the existence of irrational numbers. For example, if a=1, b = 1, then c = √2. Here in lies one of the most famous irrational numbers.

Theorem: if p is a prime then √p is irrational.

(1) Assume that there exists two values a,b such that: a/b = √p and such that a,b are the lowest positive fraction such that gcd(a,b)=1.

NOTE: We know that in fractions, if a,b had any common factors, then we could divide them off and still maintain the same ratio.

(2) Then a2/b2 = p and therefore a2 = pb2.

(3) So, we see that p divides a. So there exists a value a' such that a=pa'. [This is true by Euclid's Lemma since p is a prime and it divides either a or a]

(4) So, we get (pa')2 = pb2 which means that p2a'2 = pb2.

(5) Dividing both sides by p, we get: pa'2 = b2.

(6) But now we see that p divides b (again by Euclid's Lemma). This is a contradiction since a,b do not have any common divisors so we can reject our initial assumption.

QED

Lemma 1: if α is irrational, a,b are rational, and b ≠ 0, then a + bα is irrational

(1) Let x = a + bα

(2) bα = x - a

(3) α = (x - a)/b

(4) Now, this proves that x is not rational. Since if x is rational, then (x-a)/b would be rational, but this is impossible since (x-a)/b = α which is irrational.

QED

Lemma 2: if α is irrational, then 1/α is irrational

(1) Let y = 1/α

(2) Then, α = 1/y.

(3) This proves that y is not rational. If it were, then 1/y would be rational but it isn't since 1/y = α which is irrational.

QED

Lemma 3: if a,b are rational and α is irrational, and a + bα is rational, then a=b=0.

(1) By Lemma 1 above, if a + bα is rational, then b = 0.

(2) Since a + bα = 0 we know that a = 0 - bα = 0 - 0 = 0

QED

Lemma 4: For any positive real number ε, there exists a positive irrational number that is less than ε

Proof:

(1) Let p be a prime such that p is greater than 1/ε [We can make this assumption based on Euclid's Theorem about infinite primes, see here]

(2) By the Theorem above, we know that p is an irrational number.

(3) Now, if based on step #1, we know that:

1/√p is less than ε

(4) We also know that 1/√p is an irrational number from Lemma 2 above.

QED

Corollary 4.1: For any two distinct rational numbers, there exists an irrational number that is in between.

Proof:

(1) Let x,y be two distinct rational numbers where x is greater than y.

(2) We can see that if ε is any positive number less than x - y, then, y + ε lies in between x and y.

(3) By Lemma 4 above, we know that there exists an irrational number α that is less than ε.

(4) Further, we know that y + α is also irrational [by Lemma 1 above] and we know that y + α lies in between x and y.

QED