Let $p, q, r, s$ be rational and $p\sqrt{2}+q\sqrt{5}+r\sqrt{10}+s=0$. What does $2p+5q+10r+s$ equal?
I tried messing with both statements. But I usually just end up stuck or hit a dead end.
(I'm new to the site. I'm very sorry if this post is mal-written. please correct me on anything you can notice)
On
Suppose $(a+b\sqrt{2})^2=5$, with $a,b\in\mathbb{Q}$. Then $$ a^2+2ab\sqrt{2}+2b^2=5 $$ If $ab\ne0$, we get $$ \sqrt{2}=\frac{5-a^2-2b^2}{2ab}\in\mathbb{Q} $$ which is a contradiction. Thus either $a=0$ or $b=0$. If $a=0$, then $2b^2=5$, a contradiction; if $b=0$, then $a^2=5$, a contradiction.
Since the degree of $\sqrt{5}$ over $\mathbb{Q}(\sqrt{2})=2$, because the minimal polynomial is $x^2-5$, we conclude that every element of $\mathbb{Q}(\sqrt{2},\sqrt{5})$ can be written in a unique way as $$ x+y\sqrt{5} $$ with $x,y\in\mathbb{Q}(\sqrt{2})$. Now $$ 0=p\sqrt{2}+q\sqrt{5}+r\sqrt{10}+s=(s+p\sqrt{2})+(q+r\sqrt{2})\sqrt{5} $$ implies $s+p\sqrt{2}=0$ and $q+r\sqrt{2}=0$. Therefore $s=p=q=r=0$, because an element in $\mathbb{Q}(\sqrt{2})$ can be written in a unique way as $a+b\sqrt{2}$ with $a,b\in\mathbb{Q}$.
The key result is that if $t$ is algebraic over the field $F$ and its degree is $2$, then every element of $F(t)$ can be written in a unique way as $a+bt$, with $a,b\in F$.
For rational $p,q,r,s$ equality $$p\sqrt2+q\sqrt5+r\sqrt{10}+s = (s+p\sqrt2)+\sqrt5(q+r\sqrt2) =0$$ means $p=q=r=s=0,$ so $$\boxed{2p+5q+10r+s=0.}$$