# Thread: Irrational root theorm question

1. ## Irrational root theorm question

Going by the Irrational Root Theorem:

Let a and b be rational numbers and let √b be an irrational number.
If a + √v=b is a root of a polynomial equation with rational coefficients then the conjugate a - √b also is a root.

Here's my question:

what if its 3^√3? The conjugate would be -3^√3 would it not? but when put into the equation x^3 - 3 = 0. 3^√3 works but not its conjugate. I don't understand why.

2. Originally Posted by lax600
Going by the Irrational Root Theorem:

Let a and b be rational numbers and let √b be an irrational number.
If a + √v=b is a root of a polynomial equation with rational coefficients then the conjugate a - √b also is a root.

Here's my question:

what if its 3^√3? The conjugate would be -3^√3 would it not? but when put into the equation x^3 - 3 = 0. 3^√3 works but not its conjugate. I don't understand why.
$\displaystyle 3^{\sqrt{3}}$ is not a solution to $\displaystyle x^3 - 3 = 0$

3. but WHY?

4. Originally Posted by lax600
but WHY?
plug it in:

$\displaystyle (3^{\sqrt 3})^3 - 3 = 3^{3 \sqrt 3} - 3 \ne 0$

(by the way, i've never heard of the irrational root theorem)

5. Going by the Irrational Root Theorem:

Let a and b be rational numbers and let √b be an irrational number.
If a + √v=b is a root of a polynomial equation with rational coefficients then the conjugate a - √b also is a root.

Here's my question:

what if its 3^√3? The conjugate would be -3^√3 would it not? but when put into the equation x^3 - 3 = 0. 3^√3 works but not its conjugate. I don't understand why.
Jhevon showed you why it doesn't work. I may add that $\displaystyle 3^{\sqrt{3}}$ doesn't fit your theorem. Would that be $\displaystyle a$? or $\displaystyle b$? But even in these case, where is the other term? You cannot apply the theorem in such an example. That's why it doesn't have to work! And it doesn't. Never heard of that theorem, but seems nice.

6. Originally Posted by arbolis
Jhevon showed you why it doesn't work. I may add that $\displaystyle 3^{\sqrt{3}}$ doesn't fit your theorem. Would that be $\displaystyle a$? or $\displaystyle b$? But even in these case, where is the other term? You cannot apply the theorem in such an example. That's why it doesn't have to work! And it doesn't. Never heard of that theorem, but seems nice.
i don't think it works in general, unless there's something we are missing, what's that v about?

counter-example: take the example given, $\displaystyle x^3 - 3 = 0$

then $\displaystyle \sqrt 3 = \underbrace{0}_a + \underbrace{\sqrt 3}_b$ is a root, while $\displaystyle 0 - \sqrt{3}$ isn't

7. i don't think it works in general, unless there's something we are missing, what's that v about?

counter-example: take the example given,

then is a root, while isn't
I didn't check the theorem well enough! Sorry. But maybe there's something we are missing, we should investigate.

8. i always wondered if there was an irrational roots theorem...

9. Originally Posted by Jonboy
i always wondered if there was an irrational roots theorem...
It is actually the irrational conjugate roots theorem