# Is number rational?

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

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

• Equivalently you only need to show that $\sqrt{2}+\sqrt{3}$ is not rational. May 2, 2014 at 18:02
• $\dfrac{a}{a+1}$ is rational iff $a$ is rational. May 2, 2014 at 18:04
• Why on earth are there votes to close this splendid question? May 2, 2014 at 19:31
• What does $(\mathbb Q(\sqrt{2}, \sqrt{3}):\mathbb Q)$ denote? May 3, 2014 at 12:21

I was curious as to how hard it would be to show that $\sqrt{2} + \sqrt{3}$ 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 $x = \sqrt{2} + \sqrt{3}$ as a root.

$$x \; = \; \sqrt{2} + \sqrt{3}$$ $$x - \sqrt{3} \; = \; \sqrt{2}$$ $$\left( x - \sqrt{3} \right)^4 \; = \; \left( \sqrt{2} \right)^4$$ $$x^4 - 4x^{3}\sqrt{3} + 6x^{2}\sqrt{9} - 4x\sqrt{27} + \sqrt{81} \; = \; 2$$ $$x^4 - 4x^{3}\sqrt{3} + 6x^{2}\sqrt{9} - 12x + 3\sqrt{3} \; = \; 2$$ $$\left(x^4 - 12x - 2\right) \; + \; \sqrt{3}\left( 3 - 4x^3\right) \; + \; \sqrt{9}\left(6x^2\right) \;\;\; = \;\;\; 0$$ 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 $\;b = \sqrt{3}\left( 3 - 4x^3\right)\;$ and $\;c = \sqrt{9}\left(6x^2\right).$

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 $$\left(x^4 - 12x - 2\right)^3 \; + \; \left(\sqrt{3}\right)^3\left( 3 - 4x^3\right)^3 \; + \; \left(\sqrt{9}\right)^3\left(6x^2\right)^3$$ $$- \;\;\; 3 \cdot \left(x^4 - 12x - 2\right) \cdot \left(\sqrt{3}\right)\left( 3 - 4x^3\right) \cdot \left(\sqrt{9}\right)\left(6x^2\right) \;\;\; = \;\;\; 0$$ 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 $\sqrt{2} + \sqrt{3}$ 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 $\sqrt{2} + \sqrt{3}.$ [Want proof? Since $\sqrt{2} + \sqrt{3}$ is a sum of two positive real numbers, it follows that $-1$ and $-73$ are eliminated. Also, $\sqrt{2} + \sqrt{3}$ is greater than $\sqrt{1} + \sqrt{1} = 1+1 = 2,$ so $1$ is eliminated. Finally, $\sqrt{2} + \sqrt{3}$ is less than $\sqrt{16} + \sqrt{27} = 2 + 3 = 5,$ so $73$ is eliminated.]

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

• This is quite interesting, and really illustrates how powerful Galois theory is for reasoning about roots of polynomials. May 2, 2014 at 20:25

As noted in the comments, it suffices to show that $\sqrt{2}+\sqrt{3}$ is irrational. We can indeed use the fact that $[\mathbb Q(\sqrt{2},\sqrt{3}):\mathbb Q]=12$ to show this. Note that \begin{align*} [\mathbb Q(\sqrt{2},\sqrt{3}):\mathbb Q] &=[\mathbb Q(\sqrt{2}+\sqrt{3},\sqrt{3}):\mathbb Q]\\ &=[\mathbb Q(\sqrt{2}+\sqrt{3},\sqrt{3}):\mathbb Q(\sqrt{2}+\sqrt{3})][\mathbb Q(\sqrt{2}+\sqrt{3}):\mathbb Q]\\ &\le 3[\mathbb Q(\sqrt{2}+\sqrt{3}):\mathbb Q] \end{align*} thus $[\mathbb Q(\sqrt{2}+\sqrt{3}):\mathbb Q]\ge 4$, so $\sqrt2+\sqrt3$ cannot be rational.

• What does $(\mathbb Q(\sqrt{2}, \sqrt{3}):\mathbb Q)$ denote? May 3, 2014 at 12:00
• @mathh I assumed the OP meant the degree of the extension. May 3, 2014 at 21:13