# Irrational numbers

What is a proof that a number is irrational.
For instance, how do we know the PI goes on forever without a pattern?

CRGreathouse
Homework Helper
Proofs of irrationality and transcendental-ness are typically very difficult. I don't know an easy proof for pi offhand.

http://en.wikipedia.org/wiki/Square_root_of_2" [Broken]

One proof of the number's irrationality is the following proof by infinite descent. It is also a proof by contradiction, which means the proposition is proved by assuming that the opposite of the proposition is true and showing that this assumption is false, thereby implying that the proposition must be true.

1. Assume that $$\sqrt{2}$$ is a rational number, meaning that there exists an integer a and an integer b such that $$a / b = \sqrt{2}$$.
2. Then $$\sqrt{2}$$ can be written as an irreducible fraction $$a / b$$ such that $$a$$ and $$b$$ are coprime integers and $$(a / b)^2 = 2$$.
3. It follows that $$a^2 / b^2 = 2$$ and $$a^2 = 2 b^2$$. ($$(a / b)^n = a^n / b^n$$)
4. Therefore $$a^2$$ is even because it is equal to $$2 b^2$$. ($$2 b^2$$ is necessarily even because it is 2 times another whole number; that is what "even" means.)
5. It follows that $$a$$ must be even as (squares of odd integers are also odd, referring to b) or (only even numbers have even squares, referring to a).
6. Because a is even, there exists an integer $$k$$ that fulfills: $$a = 2k$$.
7. Substituting $$2k$$ from (6) for a in the second equation of (3): $$2b^2 = (2k)^2$$ is equivalent to $$2b^2 = 4k^2$$ is equivalent to $$b2 = 2k^2$$.
8. Because $$2k^2$$ is divisible by two and therefore even, and because $$2k^2 = b^2$$, it follows that $$b^2$$ is also even which means that $$b$$ is even.
9. By (5) and (8) $$a$$ and $$b$$ are both even, which contradicts that $$a / b$$ is irreducible as stated in (2).

Since there is a contradiction, the assumption (1) that $$\sqrt{2}$$ is a rational number must be false. The opposite is proven: $$\sqrt{2}$$ is irrational.

Proofs of transcendental-ness are not as easy.

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also, irrationality and transcendental-ness are not the same thing.

one usually implies the other but not always.

HallsofIvy
Homework Helper
??? "Transcendental" always implies "irrational".

All rational numbers are algebraic (of order 1).

HallsofIvy
Homework Helper
There is a very interesting theorem that says

"Let c be a positive real number. If there exist a function f, continuous on [0, c] and positive on (0, c), such that f and all of its iterated anti-derivatives can be taken to be integer valued at 0 and c, then c is irrational."

If f(x)= sin(x), then all anti-derivatives can be taken (by taking the constant of integration to be 0) as sin(x), -sin(x), cos(x), or -cos(x). The values of those at 0 and $\pi$ are 0, 1, or -1, all integers. Therefore, $\pi$ is irrational.

Transcendental" always implies "irrational

that's true yeah.

http://en.wikipedia.org/wiki/Square_root_of_2" [Broken]

One proof of the number's irrationality is the following proof by infinite descent. It is also a proof by contradiction, which means the proposition is proved by assuming that the opposite of the proposition is true and showing that this assumption is false, thereby implying that the proposition must be true.

1. Assume that $$\sqrt{2}$$ is a rational number, meaning that there exists an integer a and an integer b such that $$a / b = \sqrt{2}$$.
2. Then $$\sqrt{2}$$ can be written as an irreducible fraction $$a / b$$ such that $$a$$ and $$b$$ are coprime integers and $$(a / b)^2 = 2$$.
3. It follows that $$a^2 / b^2 = 2$$ and $$a^2 = 2 b^2$$. ($$(a / b)^n = a^n / b^n$$)
4. Therefore $$a^2$$ is even because it is equal to $$2 b^2$$. ($$2 b^2$$ is necessarily even because it is 2 times another whole number; that is what "even" means.)
5. It follows that $$a$$ must be even as (squares of odd integers are also odd, referring to b) or (only even numbers have even squares, referring to a).
6. Because a is even, there exists an integer $$k$$ that fulfills: $$a = 2k$$.
7. Substituting $$2k$$ from (6) for a in the second equation of (3): $$2b^2 = (2k)^2$$ is equivalent to $$2b^2 = 4k^2$$ is equivalent to $$b2 = 2k^2$$.
8. Because $$2k^2$$ is divisible by two and therefore even, and because $$2k^2 = b^2$$, it follows that $$b^2$$ is also even which means that $$b$$ is even.
9. By (5) and (8) $$a$$ and $$b$$ are both even, which contradicts that $$a / b$$ is irreducible as stated in (2).

Since there is a contradiction, the assumption (1) that $$\sqrt{2}$$ is a rational number must be false. The opposite is proven: $$\sqrt{2}$$ is irrational.

Proofs of transcendental-ness are not as easy.
That is a flipping awesome proof!

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DaveC426913
Gold Member
I followed it all except this:
2. Then $$\sqrt{2}$$ can be written as an irreducible fraction $$a / b$$
How does it follow from 1. that it must be irreducible?

think what it's supposed to mean is that if it's rational, it can be written as a fraction.

any fraction can be 'reduced' to it's simplest form (that you can't divide top and bottom by 2/3/.../whatever...)

copy and paste from wikipedia so don't blame me for any inaccuracies.

DaveC426913
Gold Member
think what it's supposed to mean is that if it's rational, it can be written as a fraction.

any fraction can be 'reduced' to it's simplest form (that you can't divide top and bottom by 2/3/.../whatever...)
Right. That's obvious now. Thanks.

There is a very interesting theorem that says

"Let c be a positive real number. If there exist a function f, continuous on [0, c] and positive on (0, c), such that f and all of its iterated anti-derivatives can be taken to be integer valued at 0 and c, then c is irrational."

If f(x)= sin(x), then all anti-derivatives can be taken (by taking the constant of integration to be 0) as sin(x), -sin(x), cos(x), or -cos(x). The values of those at 0 and $\pi$ are 0, 1, or -1, all integers. Therefore, $\pi$ is irrational.

Is the reverse also true?

CRGreathouse