Prove that $2\sqrt 3+3\sqrt[3] 2-1$ is irrational
My attempt:
$$k=2\sqrt 3+3\sqrt[3] 2-1$$
Suppose $k\in \mathbb Q$, then $k-1\in \mathbb Q$.
$$2\sqrt 3+3\sqrt[3] 2=p/q$$
I'm stuck here and don't know how to procced. I tried to do this:
$$\sqrt 3=\frac{p/q-3\sqrt[3] 2}{2}$$
contradiction, but I'm not at all sure about that. How should I proceed?
On
After messy algebra*, we get that $k$ is a root of $x^6+6 x^5-21 x^4-232 x^3-93 x^2-3486 x-2411$.
The rational root theorem tells us that $k$ either is an integer or is irrational.
If $k$ were an integer, then $k$ would be a multiple of $2411$, because $2411$ is prime. But $0 < k < 9$ and $k$ so cannot be a multiple of $2411$.
Therefore, $k$ is irrational.
(*) or after asking WA.
On
Assuming $$\frac{p}{q}=2\sqrt{3}+3\sqrt[3]{2}$$ You will have : $$54 = (\frac{p}{q} - 2\sqrt{3})^3$$ $$= \frac{p^3}{q^3}-6\frac{p^2}{q^2}\sqrt{3}+18\frac{p}{q}-24\sqrt{3}$$ So finally : $$54 -\frac{p^3}{q^3}-18\frac{p}{q}= -6\frac{p^2}{q^2}\sqrt{3}-24\sqrt{3} = -6 (\frac{p^2}{q^2}+4)\sqrt{3}$$ which is impossible because $\sqrt{3}$ is irrational and $54 -\frac{p^3}{q^3}-18\frac{p}{q}$ and $-6 (\frac{p^2}{q^2}+4)$ are both rational
On
From your equation we obtain $$\sqrt 3=\frac{p}{2q}-\frac32\sqrt[3]2$$
Take to the power of $2$ and set $u=\frac32\sqrt[3]2$ to obtain $$3=\frac{p^2}{4q^2}+u^2-\frac pqu$$ This equaion has rational roots only if $\sqrt{\Delta}$ is rational where $$\Delta=(-\frac pq)^2-4 \times 1 \times (\frac{p^2}{4q^2}-3)=12 $$
On
I suppose it is established that $\sqrt 3$ and $\sqrt[3]2$ are irrational.
If $2\sqrt 3 +3\sqrt[3]2-1$ were rational, $\sqrt[3]2$ would be an element of the quadratic field $\mathbf Q(\sqrt3)$, which has degree $2$ over $\mathbf Q$, and $\mathbf Q(\sqrt[3]2)$ would be one of its subfields. Unfortunately,$[\mathbf Q(\sqrt[3]2):\mathbf Q]=3$.
On
This is an elementary but very good question. We have $$k = 2\sqrt{3} + 3\sqrt[3]{2} - 1$$ so that $$\sqrt[3]{2} = \frac{k + 1 - 2\sqrt{3}}{3}$$ and if $k$ is rational then it means that that $\sqrt[3]{2}$ is a quadratic irrationality i.e. it is a root of a quadratic equation with rational coefficients.
Hardy proves in "A Course of Pure Mathematics" that $\sqrt[3]{2}$ is not the root of such a quadratic equation. I show his method here. Let us suppose on the contrary that $\sqrt[3]{2}$ is a root of the equation $$ax^{2} + bx + c = 0\tag{1}$$ where $a \neq 0 \neq c$ and $a, b, c$ are integers. Now $\sqrt[3]{2}$ is also a root of $$x^{3} - 2 = 0\tag{2}$$ Multiplying $(1)$ by $2$ we get $$2ax^{2} + 2bx + 2c = 0$$and using $(2)$ we get $$2ax^{2} + 2bx + cx^{3} = 0$$ or $$cx^{2} + 2ax + 2b = 0\tag{3}$$ Multiply $(1)$ by $c$ and $(3)$ by $a$ and subtract to get $$(bc - 2a^{2})x + c^{2} - 2ab = 0$$ Now $x = \sqrt[3]{2}$ is irrational and hence the above equation is possible only when $c^{2} - 2ab = 0$ and $bc - 2a^{2} = 0$. This means that $c^{2} = 2ab$ and $bc = 2a^{2}$ so that $b^{2}c^{2} = 4a^{4}$ or $2ab^{3} = 4a^{4}$ or $b^{3} = 2a^{3}$. This is not possible unless $a = 0, b = 0$ because $\sqrt[3]{2}$ is irrational. We have reached a contradiction and hence $k$ is irrational.
Note: In terms of abstract algebra the above is a very elementary proof that polynomial $x^{3} - 2$ is irreducible over $\mathbb{Q}$ and $[\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}] = 3$. A proof more in line with techniques of abstract algebra is done via the concept of GCD. The GCD of polynomials in $(1)$ and $(2)$ also has $\sqrt[3]{2}$ as its root and since it is irrational it follows that the GCD has degree $2$ so that the polynomial in $(1)$ is the GCD. Hence $ax^{2} + bx + c$ is a factor of $x^{3} - 2$ and thus $x^{3} - 2$ has a linear factor so that $(2)$ has a rational root and thus the contradiction.
\begin{align*} 2\sqrt 3+3\sqrt[3] 2=\frac{p}{q} &\Rightarrow 3\sqrt[3] 2=\frac{p}{q}-2\sqrt 3\\ &\Rightarrow 54=\left(\frac{p}{q}-2\sqrt 3\right)^3\\ &\Rightarrow 54=\frac{p^3}{q^3}-6\frac{p^2}{q^2}\sqrt{3}+36\frac{p}{q}-24\sqrt{3}\\ &\Rightarrow \sqrt{3}=\frac{p^3+36pq^2-54q^3}{6q(p^2+4q^2)} \in \mathbb{Q}\\ \end{align*}