Prove Irrationality of \sqrt{3}, \sqrt{5}, \sqrt{6} and 2^1/3, 3^1/3

[zooxanthellae]

Homework Statement

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.

Homework Equations

Can't think of any, really.

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?

 

[Mark44]

Homework Statement

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.

What you meant to say but your LaTeX script wasn't quite right was this:
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.

Homework Equations

Can't think of any, really.

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?

 

[zooxanthellae]

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

 

[Mark44]

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.

 

[zooxanthellae]

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.

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"]

 

[Mark44]

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.

 

[zooxanthellae]

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.

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.

 

[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.

 

[zooxanthellae]

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.

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.

 

[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?