Does $\sqrt{2}+\sqrt{3}+\sqrt{5}\in\mathbb{Q}$ imply that $\sqrt{2}+\sqrt{3}-\sqrt{5}\in\mathbb{Q}$?

Question

Suppose $\sqrt{2}+\sqrt{3}+\sqrt{5}\in\mathbb{Q}$. Does this imply that imply that $\sqrt{2}+\sqrt{3}-\sqrt{5}\in\mathbb{Q}$?

If you only have two of these radicals, it follows easily as if $\sqrt{2}+\sqrt{3}\in\mathbb{Q}$ then $\sqrt{2}-\sqrt{3}=\frac{-1}{\sqrt{2}+\sqrt{3}}$ so $\sqrt{2}-\sqrt{3}\in\mathbb{Q}$, but I can't seem to use this method for sums of three radicals.

If this is true for sums of three radicals, then it leads to an alternative proof of $\sqrt{2}+\sqrt{3}+\sqrt{5}\in\mathbb{R-Q}$

Suppose $\sqrt{2}+\sqrt{3}+\sqrt{5}\in\mathbb{Q}$.Then $\sqrt{2}+\sqrt{3}-\sqrt{5}\in\mathbb{Q}$, so $\sqrt{2}+\sqrt{3}+\sqrt{5}-(\sqrt{2}+\sqrt{3}-\sqrt{5})=2\sqrt{5}\in\mathbb{Q}$ which clearly false. Hence, $\sqrt{2}+\sqrt{3}+\sqrt{5}\in\mathbb{R-Q}$

I am not looking for proofs of $\sqrt{2}+\sqrt{3}+\sqrt{5}\in\mathbb{R-Q}$ ; I am just wondering if there is a way to make this method work.

Answer 1

If $\sqrt{2}+\sqrt{3}+\sqrt{5} \in \Bbb{Q}$ and $\sqrt{2}+\sqrt{3}-\sqrt{5} \in \Bbb{Q}$, then the difference is in $\Bbb{Q}$, and that would imply that $\sqrt{5} \in \Bbb{Q}$

Answer 2

One has that

$$a-\sqrt{b}=\frac{1}{a+\sqrt{b}}\cdot(a^2-b)$$

Take $a=\sqrt{2}+\sqrt{3}$ and $b=5$. Then

$$\sqrt{2}+\sqrt{3}-\sqrt{5}=\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\cdot(5+2\sqrt{6}-5)=\frac{2}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\cdot\sqrt{6}$$

So, under the assumption that $\sqrt{2}+\sqrt{3}+\sqrt{5} \in \mathbb{Q}$, it is true that $\sqrt{2}+\sqrt{3}-\sqrt{5}\in \mathbb{Q}$ if and only if $\sqrt{6}\in\mathbb{Q}$.

Answer 3

$\sqrt{2} + \sqrt{3} + \sqrt{5}$ is not in $\Bbb{Q}$, so $\sqrt{2} + \sqrt{3} + \sqrt{5} \in \Bbb{Q}$ implies any proposition you like.

Answer 4

I suspect that your question is mis-phrased, in that you appear to be using the symbol $\mathbb{Q}$ to represent any field containing the given element, rather than its usual usage for 'the rationals'. If this is true, then the answer to the question you are really asking would be 'yes' - the extension of a field by the root of any quadratic [such as '$x^2$-5'] is automatically a splitting field, i.e it contains both roots [because their sum is the linear coefficient of the quadratic]. So $\mathbb{Q}(\sqrt{2})$ contains $\sqrt{2}$ and -$\sqrt{2}$, $\mathbb{Q}(\sqrt{2})(\sqrt{3})$ contains $\sqrt{2}$, -$\sqrt{2}$, $\sqrt{3}$ and -$\sqrt{3}$, and finally $\mathbb{Q}(\sqrt{2})(\sqrt{3})(\sqrt{5})$ contains $\sqrt{2}$, -$\sqrt{2}$, $\sqrt{3}$,-$\sqrt{3}$, $\sqrt{5}$ and -$\sqrt{5}$.