# Irrationality Proofs

1. Jun 23, 2010

### zooxanthellae

1. The problem statement, all variables and given/known data

a) Prove that $$\sqrt{3}, \sqrt{5}, \sqrt{6}$$ are irrational. Hint: To treat $$\sqrt{3}$$, for example, use the fact that every integer is of the form $$3n$$ or $$3n + 1$$ or $$3n + 2$$. Why doesn't this proof work for $$\sqrt{4}$$?

b) Prove that $$2 ^ 1/3$$ and $$3 ^ 1/3$$ are irrational.

2. Relevant equations

Can't think of any, really.

3. The attempt at a solution

a) I futzed around with fractions involving $$3n$$ and $$3n + 1$$ and so forth but I never really got anywhere. I figure that the proof must have something to do with 3, or $$\sqrt{3}^2$$, but I can't tell exactly what.

b) I actually got something here: I reasoned that $$\sqrt[3]{2} = (\sqrt[3]{2}^2)\sqrt{2}$$. Since we know that $$\sqrt{2}$$ is irrational, and that (rational) x (irrational) is irrational unless one of the terms is $$0$$, and that (rational) x (rational) is rational, we can conclude that it is impossible for $$\sqrt[3]{2}$$ to be rational because if it was, then a rational number ($$\sqrt[3]{2}^2$$) multiplied by an irrational number ($$\sqrt{2}$$) would equal a rational number, which here cannot be possible. Is this a proof by contradiction?

Last edited: Jun 23, 2010
2. Jun 23, 2010

### Staff: Mentor

Prove that $$2 ^ {1/3}$$ and $$3 ^ {1/3}$$ are irrational.

If your exponent inside a [ tex] tag takes two or more characters, surround them with braces - { }.

You can also use cube roots, using \sqrt[3]{ ... }, like this:
Prove that $$\sqrt[3]{2}$$ and $$\sqrt[3]{3}$$ are irrational.

3. Jun 23, 2010

### zooxanthellae

Sorry about that. I went back and fixed the $$\sqrt[3]{2}$$ in the second half but forgot about the first!

4. Jun 23, 2010

### Staff: Mentor

To prove that, say, sqrt(2) is irrational, assume the opposite: that it is rational and so can be written as a/b, with no factors in common in a and b.

sqrt(2) = a/b ==> 2 = a2/b2

You should arrive at a contradiction, which means that your assumption that sqrt(2) is rational must not be true.

5. Jun 23, 2010

### zooxanthellae

Oh yes, I know that (Spivak actually provides what is essentially the same proof). I'm just looking to see if my reasoning is correct [b) in "The attempt at a solution"]

6. Jun 23, 2010

### Staff: Mentor

In b, you have $$\sqrt[3]{2} = (\sqrt[3]{2}^2)\sqrt{2}$$
Why would you think this? The left side is about 1.4422; the right side is about 2.2449.
I don't get what you're trying to do here.

7. Jun 23, 2010

### zooxanthellae

Oh dear. It's an arithmetic error that I overlooked (I multiplied two of the exponents when I should have added them).

I guess there's a reason Spivak's proof looked so much longer than mine!

Should probably turn attention back to a), then.

8. Jun 24, 2010

### Gib Z

This doesn't use the hint, but you could generalize the proof in Spivak for sqrt 2 to any non-perfect square. Keep an eye out for which step its crucial that its not a perfect square.

9. Jun 24, 2010

### zooxanthellae

The step that most sticks out to me is the one in which Spivak asserts that $$p^2 = 2q^2$$. If 2 were some perfect square, then the equation would look like $$p^2 = (aq)^2$$ where $$a$$ is some integer, which would allow one to solve for $$p/q$$. But I'm puzzled as to how I could generalize that proof, since it seems to rest on the fact that the numbers involved are even, which wouldn't work for 3.

In other words, I'd end up with $$p^2 = 3q^2$$. All this seems to tell me is that $$p$$ and $$q$$ are both even or both odd. But whereas Spivak's proof for 2 relied on $$p$$ and $$q$$ being even, and so having a common factor of 2, that doesn't seem to be possible with $$p$$ and $$q$$ being odd.

10. Jun 24, 2010

### Count Iblis

You can also look at the factorization in prime numbers. If you have:

sqrt[p] = r/s

where p is a prime number and r and s are integers that have no divisirs in common, then you have:

p = r^2/s^2 ---------->

r^2 = p s^2

If you were to factor both sides and count (by multiplicity) how many prime factors there are on each side, what would you find?