Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I prove that $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? $\mathbb{Q}$ is the rational field.

I want to know the detail about the proof. Thanks in advance.

Actually I know any two of them and three of them are linearly independent.

share|cite|improve this question
@t.b.: Surely not...the question here is "does there exist $a, b, c, d\in\mathbb{Q}$ not all zero such that $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}=0$", while the question you link to boils down to $\sqrt{2}\sqrt{3}=\sqrt{6}$, which is an illegal move here... – user1729 Jan 6 '12 at 15:13
@t.b.: Not quite (thought not unrelated). For example, $\sqrt{6}$ is certainly in $\mathbb{Q}(\sqrt{2},\sqrt{3})$, but isn't in the $\mathbb{Q}$-vector space generated by $1$, $\sqrt{2}$, and $\sqrt{3}$. – Cam McLeman Jan 6 '12 at 15:15
Actually, @user1729, Bill Dubuque's answer in the linked question does in fact answer this question (and some more). And the fact that $b\sqrt{2} + c\sqrt{3} \not\in Q(\sqrt{6})$ (as demonstrated below by Martin) is basically how one shows (in an elementary way) that $Q(\sqrt{2},\sqrt{3}) \neq Q(\sqrt{6})$ (Alvaro's proof in the linked question). That is to say I agree with t.b. that this question is an abstract duplicate of the other, though whether it should be closed as such is another matter. – Willie Wong Jan 6 '12 at 15:39
See this blog post by Qiaochu, and the linked file from there. – Arturo Magidin Jan 6 '12 at 15:39
@WillieWong: Yeah, I had a similar paragraph written before I decided to scrap it in favor of "though not unrelated." In any case, I more or less agree, though it's unclear to me the extent to which the existence of repeatable answers implies that the questions themselves are duplicate. – Cam McLeman Jan 6 '12 at 16:56
up vote 18 down vote accepted

I have no doubt that many slick solutions will be given here. I'll try to post an elementary one; which uses perhaps the most straightforward approach I can imagine. (The only things that are needed are some algebraic manipulation and basic properties of rational numbers; such as that the only solution of $x^2=6y^2$ in $\mathbb Q$ are $x=y=0$.)

This is equivalent to showing that if $$a+b\sqrt6=c\sqrt3+d\sqrt2$$ then $a=b=c=d=0$.

By squaring both sides of the above equation we get $$ \begin{align*} a^2+6b^2+2ab\sqrt6&=3c^2+2d^2+2cd\sqrt6\\ 2(ab-cd)\sqrt6=3c^2+2d^2-a^2-6b^2 \end{align*} $$ Since $a,b,c,d\in\mathbb Q$, this implies $$ \begin{align*} ab-cd&=0\\ 3c^2+2d^2-a^2-6b^2&=0 \end{align*} $$ which is the same as $$ \begin{align*} ab&=cd\\ 3c^2+2d^2&=a^2+6b^2 \end{align*} $$

Suppose that $b\ne0$, $c\ne0$. (I'll leave the solution of these cases to the reader.) Then we can rewrite the first equation as $\frac ac = \frac db = x$, where $x\in\mathbb Q$. Now the second equation becomes $$ \begin{align*} 3c^2+2x^2b^2&=x^2c^2+6b^2\\ x^2(2b^2-c^2)&=3(2b^2-c^2)\\ (x^2-3)(2b^2-c^2)&=0 \end{align*}$$ This implies that $x^2=3$ or $2b^2=c^2$. None of them has non-zero solutions in rational numbers.

share|cite|improve this answer
Is there any general method for solving such problems? – Norbert Jan 6 '12 at 15:33
@Norbert I think that having look at these questions could help… and… Perhaps someone who has better knowledge of field theory and Galois theory than me could give you a better answer. – Martin Sleziak Jan 6 '12 at 15:45
@Martin Sleziak:I think I have got your idea.Thanks.And I have read a version of proof that it first gives any two of them and any three of them are linearly independent.After having read your solution,I want to ask whether the version I read before is useful or not.Is it necessary to do like that?And one more question,is there any proof that based on any other different theory and can be presented in an elementary way like yours? – Andylang Jan 6 '12 at 16:21
@Andylang: The parts which I "left for the reader" are using the fact that three/two of them are linearly independent. (But you probably don't need to do all possible triples.) For different proofs just check the comments under your question. – Martin Sleziak Jan 6 '12 at 16:41
@Martin Sleziak:Got it. – Andylang Jan 9 '12 at 16:14

Observe that $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} = 0$ if and only if $(a + b\sqrt{2}) + (c + d\sqrt{2})\sqrt{3} = 0$. Hence it is enough to show that $\sqrt{3}$ and $1$ are independent over $\mathbb{Q}(\sqrt{2})$, but this is equivalent to showing that $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$, or that $\sqrt{3}$ cannot be written as $a + b\sqrt{2}$. This is obvious (square both sides and play with the result).

share|cite|improve this answer

HINT $\ $ Specialize the Lemma below to $\rm\ K = \mathbb Q,\ a,b\ =\ 2,3\:. $

LEMMA $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K]\ =\ 4\ $ if $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\ $ all are not in $\rm\:K\:$ and $\rm\: 2\: \ne\: 0\:$ in $\rm\:K\:.$

Proof $\ \ $ Let $\rm\ L = K(\sqrt{b})\:.\:$ Then $\rm\: [L:K] = 2\:$ via $\rm\:\sqrt{b} \not\in K\:,\:$ so it is sufficient to prove $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ It fails only if $\rm\:\sqrt{a} \in L = K(\sqrt{b})\ $ and then $\rm\ \sqrt{a}\ =\ r + s\ \sqrt{b}\ $ for $\rm\ r,s\in K\:.\:$ But that is impossible since squaring yields $\rm(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\ \sqrt{b}\:,\: $ which contradicts hypotheses as follows:

$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad rs \ne 0\ \ \Rightarrow\ \ \sqrt{b}\ \in\ K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\:,\:$ using $\rm\:2 \ne 0$

$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\ s = 0\ \ \Rightarrow\ \ \ \sqrt{a}\ \in\ K\ \ $ via $\rm\ \sqrt{a}\ =\ r \in K$

$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\ r = 0\ \ \Rightarrow\ \ \sqrt{a\:b}\in K\ \ $ via $\rm\ \sqrt{a}\ =\ s\ \sqrt{b}\:,\: \ $times $\rm\:\sqrt{b}\quad\quad$ QED

REMARK $\ $ By induction, the lemma easily generalizes to algebraic extensions generated by adjoining $\rm\:n\:$ square-roots, see my post here on Besicovic's Theorem, which includes references to generalizations by Mordell and Siegel. These results are elementary special cases of the Galois theory of Kummer extensions.

share|cite|improve this answer
A general explanation.It is useful.Thank you – Andylang Jan 9 '12 at 16:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.