Assume that $a,b,c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q$ are rational,prove $\sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q$,are rational.

I know that can be proved, would like to know that there is no easier way

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$\sqrt a + \sqrt b + \sqrt c = p \in \mathbb Q$,
$\sqrt a + \sqrt b = p- \sqrt c$,
$a+b+2\sqrt a \sqrt b = p^2+c-2p\sqrt c$,
$2\sqrt a\sqrt b=p^2+c-a-b-2p\sqrt c$,
$4ab=(p^2+c-a-b)+4p^2c-4p(p^2+c-a-b)\sqrt c$,
$\sqrt c=\frac{(p^2+c-a-b)+4p^c-4ab}{4p(p^2+c-a-b)}\in\mathbb Q$.

  • $\begingroup$ Looks good, the only thing that needs to be done is to mqke sure we are not dividing by $0$. $\endgroup$ – André Nicolas Apr 25 '12 at 1:29
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    $\begingroup$ Exactly; how do you know $p^2+c-a-b\ne0$? Also, the line that starts $4ab=$ should have $(p^2+c-a-b)^2$ in it. $\endgroup$ – Gerry Myerson Apr 25 '12 at 1:42

[See here and here for an introduction to the proof. They are explicitly worked special cases]

As you surmised, induction works, employing our prior Lemma (case $\rm\:n = 2\:\!).\:$ Put $\rm\:K = \mathbb Q\:$ in

Theorem $\rm\ \sqrt{c_1}+\cdots+\!\sqrt{c_{n}} = k\in K\ \Rightarrow \sqrt{c_i}\in K\:$ for all $\rm i,\:$ if $\rm\: 0 < c_i\in K\:$ an ordered field.

Proof $\: $ By induction on $\rm n.$ Clear if $\rm\:n=1.$ It is true for $\rm\:n=2\:$ by said Lemma. Suppose that $\rm\: n>2.$ It suffices to show one of the square-roots is in $\rm K,\:$ since then the sum of all of the others is in $\rm K,\:$ so, by induction, all of the others are in $\rm K$.

Note that $\rm\:\sqrt{c_1}+\cdots+\sqrt{c_{n-1}}\: =\: k\! -\! \sqrt{c_n}\in K(\sqrt{c_n})\:$ so all $\,\rm\sqrt{c_i}\in K(\sqrt{c_n})\:$ by induction.

Therefore $\rm\ \sqrt{c_i} =\: a_i + b_i\sqrt{c_n}\:$ for some $\rm\:a_i,\:\!b_i\in K,\:$ for $\rm\:i=1,\ldots,n\!-\!1$.

Some $\rm\: b_i < 0\:$ $\Rightarrow$ $\rm\: a_i = \sqrt{c_i}-b_i\sqrt{c_n} = \sqrt{c_i}+\!\sqrt{b_i^2 c_n}\in K\:\Rightarrow \sqrt{c_i}\in K\:$ by Lemma $\rm(n=2).$

Else all $\rm b_i \ge 0.\:$ Let $\rm\: b = b_1\!+\cdots+b_{n-1} \ge 0,\:$ and let $\rm\: a = a_1\!+\cdots+a_{n-1}.\:$ Then
$$\rm \sqrt{c_1}+\cdots+\!\sqrt{c_{n}}\: =\: a+(b\!+\!1)\:\sqrt{c_n} = k\in K\:\Rightarrow\:\!\sqrt{c_n}= (k\!-\!a)/(b\!+\!1)\in K$$

Note $\rm\:b\ge0\:\Rightarrow b\!+\!1\ne 0.\:$ Hence, in either case, one of the square-roots is in $\rm K.\ \ $ QED

Remark $ $ Note that the proof depends crucially on the positivity of the square-root summands. Without such the proof fails, e.g. $\:\sqrt{2} + (-\sqrt{2})\in \mathbb Q\:$ but $\rm\:\sqrt{2}\not\in\mathbb Q.\:$ It is instructive to examine all of the spots where positivity is used in the proof (above and Lemma), e.g. to avoid dividing by $\,0$.

See also this post on linear independence of square-roots (Besicovic's theorem).

  • $\begingroup$ The current solution, I do not quite understand. I would like to see your simple proof of, thank you $\endgroup$ – tianzhidaosunyouyu Apr 26 '12 at 11:52
  • $\begingroup$ @pxc417 I added further details - see above. $\endgroup$ – Bill Dubuque Apr 26 '12 at 23:31

Maybe not easier, but quite elegant :

Suppose that $a,b,c$ are all non zero. Let $K=\mathbb{Q}(\sqrt{a},\sqrt{b},\sqrt{c})$ and $n = [K: \mathbb{Q}]$. Then since $Tr_{K/\mathbb{Q}}(\sqrt{a}) = Tr_{\mathbb{Q}(\sqrt{a})/\mathbb{Q}} \circ Tr_{K/\mathbb{Q}(\sqrt{a})} (\sqrt{a})$, we have $$ Tr_{K/\mathbb{Q}}(\sqrt{a}) = \begin{cases} 0,& \text{if } \sqrt{a} \notin \mathbb{Q} \\ n\sqrt{a}, &\text{if } \sqrt{a} \in \mathbb{Q}, \end{cases}$$ and same for $\sqrt{b}$ and $\sqrt{c}$.

By hypothesis $\sqrt{a} + \sqrt{b} +\sqrt{c} \in \mathbb{Q}$, so $$ Tr_{K/\mathbb{Q}}(\sqrt{a}) + Tr_{K/\mathbb{Q}}(\sqrt{b}) + Tr_{K/\mathbb{Q}}(\sqrt{c}) = n\sqrt{a} + n \sqrt{b} + n \sqrt{c}.$$ It is easy to conclude that $\sqrt{a},\sqrt{b},\sqrt{c} \in \mathbb{Q}$.


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