^{2}= a

^{2}+ b

^{2}.

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 a

^{2}/b

^{2}= p and therefore a

^{2}= pb

^{2}.

(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}= pb

^{2}which means that p

^{2}a'

^{2}= pb

^{2}.

(5) Dividing both sides by p, we get: pa'

^{2}= b

^{2}.

(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

## 10 comments :

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

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

Cheers,

-Larry

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

Rob

Hi Rob,

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

Cheers,

-Larry

This in response to Corollary 4.1.

What if x and y are any numbers (i.e. the could be either irrational or rational)?

Hi Yo_Shi,

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

-Larry

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).

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)

-Larry

Lemma 3:should

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

be

if a,b are rational and α is irrational, and a+bα

=0, then a=b=0.In

Corollary 4.1ε 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.

Rob

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