# Proving that $\sqrt {2} ,\sqrt {4},1$ are linearly independent over rationals

I was trying to prove that $\sqrt {2} ,\sqrt {4}$ and $1$ are linearly independent using elementary knowledge of rational numbers. I also saw this which was in a way close to the question I was thinking about. But I could not come up with any proof using simple arguments. So if someone could give a simple proof, it would be great.

My try:

$a \sqrt {2}+b\sqrt {4}+c=0$ Then taking $c$ to the other side cubing on both sides we get $2a^3+4b^3+6ab(a+b)=-c^3$. I could not proceed further from here.

Apart from the above question i was also wondering how one would prove that $\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\sqrt{11},\sqrt{13}$ are linearly independent. Here assuming $a\sqrt{2}+b\sqrt{3}+c\sqrt{5}+...=0$ and solving seems to get complicated. So how does one solve problems of this type?

• I didn't read the accept answer in the linked question, but it seems exactly what you want applied in a different case. – Git Gud Jun 10 '14 at 10:16
• @GitGud yes i want a proof for a different case but I tries to use the same method like cubing but could not conclude anything – happymath Jun 10 '14 at 10:18
• Your edit just changed the question entirely. If you have new questions, please ask them separately. – Shaun Jun 10 '14 at 10:29
• I accept your apology. Please don't do it again :) – Shaun Jun 10 '14 at 11:11
• You may want to check math.stackexchange.com/questions/30687/… – Macavity Jun 10 '14 at 11:12

Let $\alpha=\sqrt {2}$ and suppose $a+b\alpha+c\alpha^2=0$ with $a,b,c\in \mathbb Z$, which we may assume coprime.

Then $a\alpha+b\alpha^2+2c=0$ and $a\alpha^2+2b+2c\alpha=0$.

This means that the matrix below is singular $$\pmatrix{ a & b & c \\ 2c & a & b \\ 2b & 2c & a}$$ Its determinant must be zero: $$a^3-6 a b c+2 b^3+4 c^3=0$$ This implies that $a$ is even: $a=2A$. So $$4A^3-6 A b c+ b^3+2 c^3=0$$

This implies that $b$ is even: $b=2B$. So $$2A^3-6 A B c+ 4B^3+ c^3=0$$ This implies that $c$ is even. This contradicts they being coprime.

• thanks for the awesome proof – happymath Jun 10 '14 at 13:06
• Using $a$ and $\alpha$ makes this a bit confusing at first. – robjohn Dec 17 '15 at 7:51

Suppose $1,\sqrt2,\sqrt4$ are linearly dependant. This means that there is a nonzero polynomial $P \in \Bbb Q[X]$ of degree at most $2$ such that $P(\sqrt2)=0$.

However, we know that $Q = X^3-2$ also satisfies $Q(\sqrt2) = 0$. By taking the greatest common divisor of $P$ and $Q$, we obtain a strict divisor $R$ of $Q$ (because the degree of $R$ is less than the degree of $Q$).

By Eisenstein's criterion, $Q$ is irreducible, which contradicts the existence of $R$.

If you know about the quadratic reciprocity law and Dirichlet's theorem about primes in arithmetic progression, you can show that the family of $\sqrt p$ are linearly independant :

If a relation existed between the square roots of a family of primes $p_0, \ldots, p_n$, then you can express $\sqrt{p_0}$ in terms of all the others. Such a formula means that $X^2 - p$ has a root in $\Bbb Q[T_1,\ldots,T_n]/(T_1^2-p_i)\ldots(T_n^2-p_n)$. If we have a prime $q$ such that (1) $q$ doesn't divide any denominator in the coefficients of the root, and (2) $p_i$ has a square root mod $q$ for $1 \le i \le n$ but not for $i=0$, you get a contradiction by looking at the root modulo $q$.
But such a prime exists because by the quadratic reciprocity law, (2) is equivalent to a modular condition, and then Dirichlet's theorem shows that there are infinitely many primes satisfying it, so there is one that satisfy (1).

In fact this shows even more : the family of $\sqrt n$ with $n \in \Bbb Z$ and $n$ squarefree, is linearly independant.

• Hmm.. let me remind that the question was, how to prove all this using only elementary knowledge about rationals. – Vladimir Jun 10 '14 at 11:46
• @Vladimir I don't think the second question can be answered with only elementary knowledge about rationals. Of course, feel free to contradict me on this. – mercio Jun 10 '14 at 11:57
• I absolutely agree. That is why I have answered only the first:) – Vladimir Jun 10 '14 at 11:58
• @mercio thanks a lot for your answer – happymath Jun 10 '14 at 13:03

Here's a simpler proof.

Let $\alpha=\sqrt {2}$ and suppose $a+b\alpha+c\alpha^2=0$ with $a,b,c\in \mathbb Z$.

Then $a\alpha+b\alpha^2+2c=0$.

Now $0=b(a+b\alpha+c\alpha^2)-c(a\alpha+b\alpha^2+2c)=(ab-2c^2)+(b^2-ac)\alpha$.

Since $\alpha$ is irrational, we must have $ab=2c^2$ and $ac=b^2$. This implies $ab^3=2ac^3$. Since $\alpha$ is irrational, we must have $a=0$ and so $b=0$ and $c=0$.

Consider $c_1\sqrt{2}+c_2\sqrt{3}+c_3\sqrt{5}=0$. Then $c_1\sqrt{2}+c_2\sqrt{3}=-c_3\sqrt{5}$. Squaring both sides we will have $2c_1^2+3c_2^2+2\sqrt{6}c_1c_2=5c_3^2$. If either $c_1$ or $c_2$ turns out to be $0$ then we will either have $c_2\sqrt{3}+c_3\sqrt{5}=0$ implying $3c_2^2=5c_3^2$ which gives $\left(\frac{c_2}{c_3}\right)^2=\frac{5}{3}$ which is not possible . Similarly for the case when $c_2$ is $0$. (It is obvious when both $c_1$ and $c_2$ are $0$.) Hence if $c_1$ and $c_2$ are both non-zero then $$-\sqrt{6}=\frac{2c_1^2+3c_2^2-5c_3^2}{2c_1c_2}.$$

Now observe that the R.H.S is a rational no but the L.H.S is not.

• I apologise for having changed the question entirely. Please have a look at it. – happymath Jun 10 '14 at 10:38

First show that $1$ and $\sqrt2$ are linearly independent. (This should be relatively easy.)

Then in order for $1$, $\sqrt2$ and $\sqrt4$ to be linearly dependent we must have $$\sqrt{4}=a+b\sqrt2$$ for some $a,b\in\mathbb Q$. (Since $\sqrt{4}$ is a linear combination of $1$ and $\sqrt2$.)

If we multiply the above equation by $\sqrt2$, we get \begin{gather*} 4=a\sqrt2+b\sqrt4=a\sqrt2+b(a+b\sqrt2)=(a+b^2)\sqrt2+ab\\ 4-ab=(a+b^2)\sqrt2 \end{gather*} Since $1$ and $\sqrt2$ are linearly independent, we get $4-ab=a+b^2=0$.

So we get \begin{align*} ab&=4\\ b^2&=-a \end{align*} which yields $b^2=-\frac4b$ and $b^3=-4$. This means that there is no rational solution.

(Notice that $ab=4$ implies $b\ne0$, hence we could divide the first equation by $b$ to get $a=\frac4b$.)

We argue by contradiction. Obviously, $a\ne0$ and $b\ne0$. Then $x=\sqrt2$ is a root of the square equation with rational coefficients and hence can be represented in the form $$x=A+\sqrt{B} \qquad\text{or}\qquad x=A-\sqrt B$$ with rational $A$ and $B\ge0$. In fact, $B>0$, otherwise $x$ would be rational. We have $$2=x^3=A^3\pm3A^2\sqrt B+3AB\pm B\sqrt B=A^3+3AB\pm(B+3A^2)\sqrt B$$ Since $B>0$, it follows that $B+3A^2\ne0$ and hence $\sqrt B$ is rational. Thus, so is $x$. Contradiction.

• nice proof thanks a lot – happymath Jun 10 '14 at 13:07

The linear independence of the square roots, can be noted from these propositions.

1. Suppose that some (1:) $a^n = \sum^{n-1}_{x=0} z_x a^x$. If $a$ were rational, then the reduced form is some $a=p/q$ where $\gcd(p,q)=1$. If we multiply the defining equation by $q^n$, then all the right-hand terms are miltiples of q, where the LHS is not, unless $q=1$. This means that equation 1 defines a set that is integer or irrational. We can call these numbers 'irrational integers' $\mathbb{Y}$.

2. All of the roots of integers solve (1:), in the form that $a^n = z$ gives $a=\sqrt[n]{z}$.

3. Every irrational integer divides some integer. (Proof: multiply (1:) by $1/a$, this gives an expression in a = $z_0 / a$.

4. The linear independence of square roots $p$, $q$ derives directly from (2), in that the $\sqrt{pq}$ is itself an irrational integer, either in Z or Y. We let $e+f\sqrt{p}+g\sqrt{q}+h\sqrt{pq}=0$ and let $(e+h\sqrt{pq})^2=(f\sqrt{p}+g\sqrt{q})^2$ or $(e^2+h^2pq)+2eh\sqrt{pq} = pf^2+qg^2 + 2fg\sqrt{pq}$.

5. Similarly, we assert that $e^2f^2 = qg^2h^2$ and $e^2g^2=pg^2h^2$, but we have demonstrated that this can not happen in the integers, so $\sqrt{pq}$ is linearly independent from $1,\ \sqrt{p}, \sqrt{q}$.

1. If any two numbers $a,b,c,\dots$ in $a+b\sqrt{z_1}+c\sqrt{z_2}+\dots$ are non-zero, then the square contains a sqruare root term. Since some $n\sqrt{z_n}$, not in the above set, has a square with no square-root term, then $n\sqrt{z_n}=d\sqrt{z_d}$, exactly. But if $z_n$ is not co-square with any of the list, then its square root can not be in the named span.

This implies that all of the square roots of square-free numbers are linearly independent, viz 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, &c.

Note that this proof generalises to all powers. A root number is not a member of a span of numbers, unless it is an integer multiple of exactly one member.

• here in the answer are $z_x$ integers? and if i am not wrong you are talking about algebraic integers right? – happymath Dec 18 '15 at 11:47