Proving $\sqrt3 + \sqrt[3]{2}$ to be irrational

Question

In a test I tried to solve recently I came across the following question:

Prove $$\sqrt3 + \sqrt[3]{2}$$ is irrational

I tried proving it by saying it is equal to some rational number $$\sqrt3 + \sqrt[3]{2} = \frac{m}{n}$$ and reaching a contradiction. I tried squaring / cubing both sides and that didnt work. So how do I go about solving this problem?

Answer 1

It might be useful to first prove the following two statements - the sum of a rational and irrational number is irrational, and the product of a rational and irrational number is irrational. You can then move the $\sqrt 3$ and get:

$\sqrt[3] 2=\frac{m}{n}-\sqrt 3$

$2=(\frac{m}{n}-\sqrt 3)^3$

After expanding the term at the RHS, the two lemmas you proved might come in handy.

Answer 2

It's possible to find a polynomial that has $\sqrt{3}+\sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots. $$\left((x-\sqrt{3})^3-2\right)\left((x+\sqrt{3})^3-2\right)=0$$ $$\left(x^3-3\sqrt{3}x^2+9x-3\sqrt{3}-2\right)\left(x^3+3\sqrt{3}x^2+9x+3\sqrt{3}-2\right)=0$$ $$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$ By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $\sqrt{3}+\sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $\pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $\pm 23$ the $x^6$ term is much larger than everything else combined.

This is not the easy way, of course.

Answer 3

Suppose $\sqrt 3 + \sqrt[3]2$ were rational; that is,

$\sqrt 3 + \sqrt[3]2 = r \in \Bbb Q: \tag 1$

then

$\sqrt[3]2 = r - \sqrt 3; \tag 2$

we cube:

$2 = r^3 - 3r^2\sqrt 3 + 3r(\sqrt 3)^2 - (\sqrt 3)^3, \tag 3$

or

$2 = r^3 - 3r^2 \sqrt 3 + 9r - 3\sqrt 3, \tag 4$

or

$2 = r^3 + 9r - (3r^2 + 3)\sqrt 3, \tag 5$

or

$\sqrt 3 = \dfrac{2 - r^3 -9r}{3r^2 + 3} \in \Bbb Q, \tag 6$

which contradicts the fact that

$\sqrt 3 \notin \Bbb Q; \tag 7$

therefore, (1) is false. $OE\Delta$.

Answer 4

Let $\sqrt3+\sqrt[3]2=r\in\mathbb Q$.

Thus, $$2=(r-\sqrt3)^3$$ or $$2=r^3-3\sqrt3r^2+9r-3\sqrt3$$ or $$\sqrt3=\frac{r^3+9r-2}{3(r^2+1)}\in\mathbb Q,$$ which is a contradiction.

Answer 5

Squaring is simpler than cubing. ;-)

Let $u=\sqrt[3]{2}+\sqrt{3}$. Then $\sqrt{3}=u-\sqrt[3]{2}$ and therefore $$ (u^2-3)-2u\sqrt[3]{2}+\sqrt[3]{4}=0 $$ If $u\in\mathbb{Q}$, this contradicts $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ being a basis of $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$, which stems from $x^3-2$ being irreducible over $\mathbb{Q}$ (Eisenstein) and standard facts on simple algebraic extensions.

Even simpler: if $u$ is rational, then $\mathbb{Q}(\sqrt{3})$ is a subfield of $\mathbb{Q}(\sqrt[3]{2})$, which is ruled out by the dimension theorem.