Is number rational?

How can we check if number $a=\frac{ \sqrt[4]{2}+\sqrt[3]{3}}{\sqrt[4]{2}+\sqrt[3]{3} +1}$ is rational?

Is there any smart solution? Another assignment is to find $\left( \mathbb{Q}(\sqrt[4]{2},\sqrt[3]{3}):\mathbb{Q} \right)$ which is twelve, maybe we can somehow use it for checking whether $a$ is rational or not?

Solution 1:

I was curious as to how hard it would be to show that $\sqrt[4]{2} + \sqrt[3]{3}$ is irrational by using ordinary high school precalculus methods, where the rational root theorem for polynomials with integer coefficients is applied.

The first step is to get a polynomial with integer coefficients having $x = \sqrt[4]{2} + \sqrt[3]{3}$ as a root.

$$x \; = \; \sqrt[4]{2} + \sqrt[3]{3}$$ $$ x - \sqrt[3]{3} \; = \; \sqrt[4]{2}$$ $$ \left( x - \sqrt[3]{3} \right)^4 \; = \; \left( \sqrt[4]{2} \right)^4$$ $$ x^4 - 4x^{3}\sqrt[3]{3} + 6x^{2}\sqrt[3]{9} - 4x\sqrt[3]{27} + \sqrt[3]{81} \; = \; 2$$ $$ x^4 - 4x^{3}\sqrt[3]{3} + 6x^{2}\sqrt[3]{9} - 12x + 3\sqrt[3]{3} \; = \; 2$$ $$ \left(x^4 - 12x - 2\right) \; + \; \sqrt[3]{3}\left( 3 - 4x^3\right) \; + \; \sqrt[3]{9}\left(6x^2\right) \;\;\; = \;\;\; 0$$ Now I'll make use of the identity $$(a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \;\; = \;\; a^3 + b^3 + c^3 - 3abc$$ for $\;a = \left(x^4 - 12x - 2\right)\;$ and $\;b = \sqrt[3]{3}\left( 3 - 4x^3\right)\;$ and $\;c = \sqrt[3]{9}\left(6x^2\right).$

Specifically, I'll multiply both sides of the equation above (the equation having $0$ on its right hand side) by the "$x$-expression equivalent" of what $a^2 + b^2 + c^2 - ab - ac - bc$ equals. However, there is no need to actually write out this "$x$-expression equivalent", since the left hand side will be converted into $a^3 + b^3 + c^3 - 3abc$ (which I will write out) and the right hand side will still be $0.$ Thus, after multiplying both sides by the "$x$-expression equivalent" of $a^2 + b^2 + c^2 - ab - ac - bc,$ we get $$ \left(x^4 - 12x - 2\right)^3 \; + \; \left(\sqrt[3]{3}\right)^3\left( 3 - 4x^3\right)^3 \; + \; \left(\sqrt[3]{9}\right)^3\left(6x^2\right)^3$$ $$ - \;\;\; 3 \cdot \left(x^4 - 12x - 2\right) \cdot \left(\sqrt[3]{3}\right)\left( 3 - 4x^3\right) \cdot \left(\sqrt[3]{9}\right)\left(6x^2\right) \;\;\; = \;\;\; 0$$ The key to keeping this from getting really messy is to remember that for the rational root test we only need the leading coefficient and the constant coefficient. $$\left(x^{12} \; + \; \ldots \; - \; 8 \right) \;\; + \;\; 3\left(27 \; + \; \ldots \; - \; 64x^9\right) \;\; + \;\; 9\cdot6^3x^6$$ $$- \;\;\; 54x^2\left(x^4 - 12x - 2\right)\left(3 - 4x^3\right) \;\;\; = \;\;\; 0$$ Clearly, the leading coefficient is $1$ (coefficient of $x^{12},$ which only appears once --- in the left-most parenthesized expression) and the constant coefficient is $-8 + (3)(27) = 73$ (note that only the two parenthesized expressions containing an ellipsis contribute to the constant coefficient). Therefore, the only possible rational roots of the equation above, which has $\sqrt[4]{2} + \sqrt[3]{3}$ as a root, are factors of $73,$ and hence belong to the set $\{1, \, -1, \, 73, \, -73\}.$ Clearly, none of these four integers is equal to $\sqrt[4]{2} + \sqrt[3]{3}.$ [Want proof? Since $\sqrt[4]{2} + \sqrt[3]{3}$ is a sum of two positive real numbers, it follows that $-1$ and $-73$ are eliminated. Also, $\sqrt[4]{2} + \sqrt[3]{3}$ is greater than $\sqrt[4]{1} + \sqrt[3]{1} = 1+1 = 2,$ so $1$ is eliminated. Finally, $\sqrt[4]{2} + \sqrt[3]{3}$ is less than $\sqrt[4]{16} + \sqrt[3]{27} = 2 + 3 = 5,$ so $73$ is eliminated.]

Therefore, since $\sqrt[4]{2} + \sqrt[3]{3}$ is a solution to the equation above and $\sqrt[4]{2} + \sqrt[3]{3}$ differs from all of the possible rational roots of the equation, it follows that $\sqrt[4]{2} + \sqrt[3]{3}$ is not rational.

Solution 2:

As noted in the comments, it suffices to show that $\sqrt[4]{2}+\sqrt[3]{3}$ is irrational. We can indeed use the fact that $[\mathbb Q(\sqrt[4]{2},\sqrt[3]{3}):\mathbb Q]=12$ to show this. Note that $$\begin{align*} [\mathbb Q(\sqrt[4]{2},\sqrt[3]{3}):\mathbb Q] &=[\mathbb Q(\sqrt[4]{2}+\sqrt[3]{3},\sqrt[3]{3}):\mathbb Q]\\ &=[\mathbb Q(\sqrt[4]{2}+\sqrt[3]{3},\sqrt[3]{3}):\mathbb Q(\sqrt[4]{2}+\sqrt[3]{3})][\mathbb Q(\sqrt[4]{2}+\sqrt[3]{3}):\mathbb Q]\\ &\le 3[\mathbb Q(\sqrt[4]{2}+\sqrt[3]{3}):\mathbb Q] \end{align*}$$ thus $[\mathbb Q(\sqrt[4]{2}+\sqrt[3]{3}):\mathbb Q]\ge 4$, so $\sqrt[4]2+\sqrt[3]3$ cannot be rational.