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

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?

## 5 Answers

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.

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.

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. $$OE\Delta$$.

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.

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.