# Prove that $\sqrt{10} - \sqrt6 - \sqrt5 + \sqrt3$ is irrational

I tried the methods shown in Can $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m$ are perfect squares? but I cannot extend them well into 4 numbers.

• $\sqrt{10}-\sqrt{5} = \sqrt{5}\cdot (\sqrt{2}-1)$ and $\sqrt{6}-\sqrt{3} = \sqrt{3}\cdot (\sqrt{2}-1)$ so your expression can be rewritten as $(\sqrt{5}-\sqrt{3})\cdot (\sqrt{2}-1)$. Does this help? – postmortes Nov 23 '15 at 19:05
• If you can prove that $a=4\sqrt{30}-6\sqrt{15}-16\sqrt{2}$ is irrational, then you're done, since $\sqrt{10}-\sqrt{6}-\sqrt{5}+\sqrt{3} = \sqrt{a+24}$. – João Victor Bateli Romão Nov 23 '15 at 19:29
• Possible duplicate of $\sqrt{m_1}+\sqrt{m_2}+ \cdots + \sqrt{m_n}$ is Irrational – Watson Nov 26 '18 at 20:44

Let $a=\sqrt6+\sqrt5$ and $b=\sqrt{10}+\sqrt3$. You want to show that $b-a\notin\mathbb Q$.
Notice that \begin{align*} a^2&=11+2\sqrt{30}\\ b^2&=13+2\sqrt{30} \end{align*} hence $$b^2-a^2=(b-a)(b+a)=2$$ and $b+a=\frac2{b-a}$.
Now if $b-a\in\mathbb Q$, then also $b+a=\frac2{b-a}\in\mathbb Q$ and $$b=\frac{(b-a)+(b+a)}2\in\mathbb Q,$$ a contradiction. (To prove that $b$ is irrational you can use methods from the thread you linked.)
$$\alpha=\sqrt{10}-\sqrt{6}-\sqrt{5}+\sqrt{3} = \left(\sqrt{2}-1\right)\left(\sqrt{5}-\sqrt{3}\right)$$ Hence: $$\alpha^2 = \left(3-2\sqrt{2}\right)\left(8-2\sqrt{15}\right)$$ and since $\sqrt{15}$ does not belong to $\mathbb{Q}(\sqrt{2})$, $\alpha^2$ is not a rational number, hence $\alpha\not\in\mathbb{Q}$.