# Degree of this number

1. Mar 6, 2005

### T-O7

Does anyone know how to find the degree over Q of this number:
$$\sqrt{3} + \sqrt[3]{4}$$

2. Mar 6, 2005

### Zurtex

I don't know what the question is that you're asking so I'm probably being really stupid here, but I assume it would help if you had that expressed as the rote of some polynomial equation with integer coefficients:

$$x = \sqrt{3} + \sqrt[3]{4}$$

$$x - \sqrt{3} = \sqrt[3]{4}$$

$$x^3 - 3\sqrt{3}x^2 + 9x - 3\sqrt{3} = 4$$

$$x^3 + 9x - 4 = 3\sqrt{3}x^2 + 3\sqrt{3}$$

$$x^6 + 6x^4 - 8x^3 + 81x^2 - 72x - 16 = 27x^4 + 54x^2 + 27$$

$$x^6 - 21x^4 - 8x^3 + 27x^2 - 72x - 43 = 0$$

That help at all? (still probably worth checking my steps haha)

Last edited: Mar 6, 2005
3. Mar 6, 2005

### Hurkyl

Staff Emeritus
Or, if you'd like to do it a different way, just keep raising your number to powers until you get a linearly dependant set. (Treating each distinct irrational number as a basis vector)

For example, for a = √2 + √3:

a^0 = 1
a^1 = √2 + √3
a^2 = 5 + 2 √6
a^3 = 11 √2 + 9 √3
a^4 = 49 + 20 √6

So, Q(a) is a vector space over Q, with basis vectors 1, √2, √3, and √6. We have 5 different vectors, so 1, a, a^2, a^3, and a^4 form a linearly dependant set...

4. Mar 6, 2005

### HallsofIvy

Staff Emeritus
Of course, the fact that you have to go to 5th power to get a dependent set means that the degree is 4?

(And notice that Hurkyl was using $\sqrt{2}+\sqrt{3}$, NOT the number of the original question. I like Zurtex's method: $\sqrt{2}+ ^3\sqrt{3}$ is algebraic of order 6 and so must have degree 6 over Q.

5. Mar 6, 2005

### Hurkyl

Staff Emeritus
Zurtex's approach is certainly easier to execute, but sometimes it's nonobvious how to manipulate things so that radicals don't proliferate. For example, if the cube roots of both 2 and 4 are in the number, when you cube to get rid of the cube root of 2, the cube root of 4 will just introduce more cube roots of 2.

6. Mar 6, 2005

### T-O7

Great, thanks a lot guys!
Following Zurtex's method, i calculated:
$$x^6-9x^4-8x^3+27x^2-72x-11=0$$ for $$x=\sqrt{3}+\sqrt[3]{4}$$.
(There was a little error with zurtex's calculation i think)
I managed to prove that this was irreducible tediously......but (this might be a dumb question) how did you know x was algebraic of order 6 to begin with? That would save me a lot of trouble heh

7. Mar 7, 2005

### Hurkyl

Staff Emeritus
Because $[\mathbb{Q}(\sqrt{3}) : \mathbb{Q}] = 2$ and $[\mathbb{Q}(\sqrt[3]{4}}) : \mathbb{Q}] = 3$. Therefore, both 2 and 3 must divide $[\mathbb{Q}(\sqrt{3}, \sqrt[3]{4}}) : \mathbb{Q}]$.

This strongly suggests that the degree of your number must be 6... but more work would be needed to prove it.

Oh, and a bit of shamelessness. If you used my approach to come up with a polynomial your number satisfies, you can also use it directly to prove that it is the minimum polynomial.

For example, for √2 + √3, I just have to prove that {1, a, a^2, a^3} is linearly independent, which is a fairly straightforward task.