Sum of two irrational radicals is irrational?

Question

If $a,b,m$ and $n$ are positive integers such that $\sqrt[m]{a}$ and $\sqrt[n]{b}$ are irrational numbers, how can we prove that the sum $\sqrt[m]{a}+\sqrt[n]{b}$ is also irrational?

Answer 1

Let $a'$, $b'$, $m'$ and $n'$ be positive integers such that $\sqrt[m']{a'}$ and $\sqrt[n']{b'}$ are irrational. Let $a$, $b$, $m$ and $n$ be the minimal positive integers such that $\sqrt[m]{a}=\sqrt[m']{a'}$ and $\sqrt[n]{b}=\sqrt[n']{b'}$, so that their minimal polynomials are $f_a=X^m-a$ and $f_b=X^n-b$, respectively. Note that $m,n>1$ as $\sqrt[m]{a}$ and $\sqrt[n]{b}$ are irrational.

Suppose $\sqrt[m]{a}+\sqrt[n]{b}=q\in\Bbb{Q}$. Then $\sqrt[m]{a}$ and $\sqrt[n]{b}$ are roots of $f_b(q-X)$ and $f_a(q-X)$, respectively, which shows that $f_a$ divides $f_b(q-X)$ and $f_b$ divides $f_a(q-X)$, respectively. In particular we see that $m\leq n$ and $n\leq m$, so $n=m$, and hence $f_a=cf_b(q-X)$ for some nonzero constant $c\in\Bbb{Q}$. Then $$X^m-a=f_a=cf_b(q-X)=c(q-X)^n-cb=c(q-X)^m-cb,$$ which immediately shows that $q=0$ because $m,n>1$. It follows that $c=\pm1$ and $a=cb$. Because $a$ and $b$ is positive it follows that $c=1$ and so $$\sqrt[m]{a}+\sqrt[n]{b}=2\sqrt[m]{a}=2\sqrt[n]{b},$$ which is irrational because $\sqrt[n]{b}$ is irrational.

Answer 2

Actually, Pierre Wantzel proved a more general result already in 1843, in his important paper Classification des nombres incommensurables d'origine algébrique (section 6). He showed that if we have a sum: $$ \alpha = \sum_i b_i \sqrt[n_i]{a_i}, $$ where the $a_i$ and $b_i$ are rational, the $n_i$ are integers greater than $1$, and the radicals $\sqrt[n_i]{a_i}$ are all irrational and moreover pairwise linearly independent over the integers, then $\alpha$ must be irrational.

In terms of modern algebra, his proof can be understood as follows. First assume $\alpha \neq 0$. Then the result follows because the trace of $\alpha$ is zero (it being a linear combination of the $\sqrt[n_i]{a_i}$, each of which is a zero of a polynomial of the form $X^{n}-a$, whose irreducibility Wantzel easily establishes in sec. 4), and the trace of a non-zero rational number is non-zero. (Of course Wantzel did not use the term "trace" but simply talked about the sum of all the conjugates of $\alpha$.)

In the remaining case of $\alpha = 0$, we simply divide the whole equation by one of the terms on the right-hand side, move the non-zero rational term over to the left, and apply the argument for the case $\alpha \neq 0$.

Answer 3

A minor modification of this answer and this answer leads up to the following general statement (whose proof does not need the existence of minimal polynomial):

Statement for the general case

Let $a,b$ be two rational numbers and $m,n$ be two positive integers such that $\sqrt[m]{a}+\sqrt[n]{b}$ is a nonzero rational number. Then both $\sqrt[m]{a}$, $\sqrt[n]{b}$ are rational numbers.

Remark: Here all radicals are real numbers (i.e. $\sqrt[m]{a}\in\mathbb{R}\Longrightarrow a\geq 0$ when $2\mid m$ and $b\geq 0$ when $2\mid n$).

We first give out a lemma:

Lemma (it is also proved here)

For $a\in \mathbb{N},n\in\mathbb{Z}^{+}$, if there exists $m\in\mathbb{Z^{+}}$ such that $(\sqrt[n]{a})^m\in\mathbb{Q}$, Then we have $\exists b\in\mathbb{Z}$ satisfying $b^{\dfrac{n}{(m,n)}}=a$.

Proof of lemma

Without loss of generality, we may suppose $(m,n)=1$, then $\exists\ x,y\in\mathbb{N}$ such that $xm-yn=1$. So $\sqrt[n]{a}=\dfrac{(\sqrt[n]{a})^{mx}}{a^y}\in\mathbb{Q}$, implying $\sqrt[n]{a}\in\mathbb{Z}$, which proves the proposition.

Proof of the general case

Some assumptions first:

(1) If $a<0,b<0$ (there must be $2\nmid mn$ in this case), just replace $(a,b)$ with $(-a,-b)$, so we might assume $a \geq 0$.

(2) If $ab=0$ or $\min(m,n)=1$, the proposition holds, so we might assume $ab \neq 0,\min(m,n)\geq 2$.

(3) Let $q=\sqrt[m]{a}+\sqrt[n]{b}\in\mathbb{Q}$. Since $a, b$ can be multiplied by a positive integer satisfying that $a, b, q$ all become integers, we only need to prove the proposition in this case, i.e. we may assume $a,b,q\in\mathbb{Z}$.

(4) Moreover, it may be assumed that there is no $r\geq 2$, satisfying $r\mid m$ and $\sqrt[ r]{a}\in\mathbb{Z}$ (as may replace $(a,m)$ with $(\sqrt[ r]{a},\dfrac{m}{r})$), and there is no $s\geq 2$, satisfying $s\mid n$ and $\sqrt[s]{b}\in\mathbb{Z}$.

Now, for $f(x)=x^m-a,g(x)=(q-x)^n+b$, let $h(x)$ be the greatest common factor of $f(x),g(x)$ in $\mathbb{Q}[x]$ and $\deg(h)=k$, and we have $h(\sqrt[m]{a })=0$. In addition, as each root of $h(x)$ (in $\mathbb{C}$) is a root of $f(x)$ and must have the form of $\sqrt[m]{a }\cdot \omega_1$ where $\omega_1$ is a $m$th root of unity.

As $|h(0)|=|\prod\limits_{z\in\mathbb{C}}z|$ and each $z$ has the form of $\sqrt[m]{a }\cdot \omega_1$, there exists $\omega$ such that $|h(0)|=|(\sqrt[m]{a })^k\cdot \omega|$ and $\omega$ is a $m$th root of unity. Thus $\omega\in\mathbb{R}$ and $\omega$ must be $\pm 1$, which derives $|h(0)|=|(\sqrt[m]{a })^k|\Longrightarrow (\sqrt[m]{a })^k\in\mathbb{Z}$.

From the lemma, we sees that, by setting $r=\dfrac{m}{(m,k)}$, $r|m$ and $\sqrt [r]{a}\in\mathbb{Z}$. Thus, according to assumption $(4)$, must have $r=1\Longrightarrow m\mid k$. As $k=\deg(h)\leq \deg(f)=m$, we derives $m=k$ and $f(x)\mid g(x)$.

Note that for $g(x)=(q-x)^n+b$, the coefficient of $x^{n-1}$ is $(-1)^{n-1} qn$. As $m\geq 2$, if $x^m-a|(q-x)^n+ b$ holds, must have $(-1)^{n-1}qn=0$. So $q=0$, a contradiction.

Remark

If we replace $\sqrt[m]{a}+\sqrt[n]{b}$ with $\sqrt[m]{a}-\sqrt[n]{b}$ (respectively, $g(x)=(x-q)^n-b$), the proposition and its proof holds in a similar way.

Thus, the statement can be rephrased as the following:

Let $a_1,a_2,c_1,c_2\in\mathbb{Q}$ and $n_1,n_2$ be two positive integers such that $c_1\sqrt[n_1]{a_1}+c_2\sqrt[n_2]{a_2}$ is a nonzero rational number. Then both $\sqrt[n_1]{a_1}$, $\sqrt[n_2]{a_2}$ are rational numbers.

Answer 4

One way to show that a sum of square roots is an irrational number(let me consider this case which includes the idea for tackling the general problem- although whether it can be put into practice is not clear) is to notice that you can write a sum of square roots as a single square root.

Consider $\sqrt2$ + $\sqrt5$. You want to find a number $z$ such that $z=(\sqrt2+\sqrt5)^2$ So $\sqrt z=\sqrt{(2+2\sqrt2\sqrt5+5)}$. That is $\sqrt{(7+2\sqrt{10 })}$ ,which is a square root of an irrational number. And a square root of an irrational number is always irrational.