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.


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.


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.


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


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


(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.


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


(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.



Anonymous said...

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

In its current form this lemma is incorrect.
It should be:

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

Larry Freeman said...

Thanks for the comment. I agree with your suggestion about making the lemma clearer -- I've updated the blog entry.



Scouse Rob said...

In Lemma 1, doesn't b have to be rational as well?


Larry Freeman said...

Hi Rob,

I've updated the blog to make it clearer that b must also be rational.



Yo_shi said...

This in response to Corollary 4.1.
What if x and y are any numbers (i.e. the could be either irrational or rational)?

Larry Freeman said...

Hi Yo_Shi,

You are right. There is always an irrational number between any two real numbers.


Anonymous said...

Is corollary 4 indeed rigorous?

We have established that x < y+ε < y, and that we can find an irrational number α such that α < ε. Hence we can say that x < y+α, where y+α is irrational. But we cannot assert that y+α < y. It may, for all we know, that x < y < y+α.

In fact, based on the crude way that we pick our irrational number (i.e. the reciprocal of the square root of a rational number), we can let x = 1, y = 9/10, and find that there is in fact no such α that lies in the open interval (x, y).

Larry Freeman said...

Hi Anonymous,

To be clear, we are assuming that x > y so that y < y + ε < x.

If ε is a positive number less than x-y (which is positive), it follows that:

y + ε > y (since ε is positive)

x > y + ε (since ε is less than x - y)


Scouse Rob said...

Lemma 3:


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


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

Scouse Rob said...

In Corollary 4.1

ε is any positive number less than x-y.

How do you prove the existence of ε?

You need a Lemma saying that for any positive number there is a smaller positive number.

You can use Lemma 4 to get an irrational number ε that is less than x-y.

But then there is no need to use Lemma 4 again to get an α that is less than ε..

x+ε would then be a sufficient irrational number between x and y.