Is any set of distinct "reduced" roots linearly independent over the rationals? Definition: let $n, a \in \mathbb{N}$. The root $\sqrt[n]{a}$ is reducible if it can be written as $\sqrt[m]{b}$ for some natural numbers  $b$ and $m<n$, or if it can be written as $M\sqrt[n]{c}$ for naturals $M$ and $c<a$. Otherwise we say the root is reduced.
The condition that the roots be reduced is necessary to prevent cases such as
$$\sqrt{4}-\sqrt[4]{16}$$
or as
$$\sqrt{2}-\sqrt[4]{4}$$

Conjecture: a set of distinct reduced roots is linearly independent over the rationals. In other words,
$$\sum_{k=1}^n M_k\sqrt[N_k]a_k\notin \mathbb{Q}$$
for any rationals $M_k≠0$ and naturals $N_k, a_k$ such that the roots $\sqrt[N_k]{a_k}$ are distinct and reduced.

Is the conjecture above true?

The condition is sufficient to show that a pair of roots is independent over the rationals:
Lemma: any root $\sqrt[n]{a}$ is either irrational or an integer.
Theorem: let $\sqrt[n]{a}$ and $\sqrt[m]{b}$ be different reduced roots. Then $\sqrt[n]{a}$ and $\sqrt[m]{b}$ are linearly independent over $\mathbb{Q}$.
Proof: suppose without loss of generality that $n\leq m$. It is sufficient to show that $\sqrt[m]{b}/\sqrt[n]{a}$ is irrational. Suppose not, then
$$\frac{\sqrt[m]{b}}{\sqrt[n]{a}}=\frac{p}{q}\in\mathbb{Q}$$
If the expression was equal to an integer $k$ then $\sqrt[m]{b}=k\sqrt[n]{a}$ and the root $\sqrt[m]{b}$ wouldn't be reduced, contrary to hypothesis. Now choose $p,q$ so that these share no common divisors larger than $1$ and $q>1$. Then
$$\sqrt[m]{b^n}=\frac{p^n}{q^n}a$$
where the expression belongs in $\mathbb{Z}$ by lemma 1. Any prime that divides $q$ cannot divide $p$, since we have chosen these to have no common divisors. Thus $q^n$ must divide $a$. We have $q^nz=a$ for some positive integer $z$ and $q\sqrt[n]{z}=\sqrt[n]{a}$ with $q>1$ so that the root $\sqrt[n]{a}$ is not reduced, contrary to hypothesis. $\blacksquare$

Perhaps this is worth including.
 A: I think this is true (or if not true as stated, some small correction to it is true). I don't have a proof, I think the general case probably relies on being a bit more clever with roots of unity / Lagrange resolvents / representation theory. But here is a proof for the case of square and cube (and therefore sixth) roots.
First, as noted in the comments:
Fact 1. The extension $\mathbb{Q}(\sqrt{p_1}, \ldots, \sqrt{p_n})$, for distinct primes $p_i$, has degree $2^d$ and basis given by all products of square roots of subsets of the $p_i$. Actually a slightly more detailed statement is: the lattice of subfields is the lattice of subgroups of $(\mathbb{Z}/2\mathbb{Z})^n$.
Proof: This is obvious for $n=1$. In general, suppose $\sqrt{p_n} \in \mathbb{Q}(\sqrt{p_1}, \ldots, \sqrt{p_{n-1}})$. Then $\mathbb{Q}(\sqrt{p_n}) = \mathbb{Q}(\sqrt{d_i})$ for one of the quadratic subfields (listed according to the lattice description above). So it's enough to show:
Fact 2. For all distinct squarefree integers $d, d'$. $\mathbb{Q}(\sqrt{d}) \ne \mathbb{Q}(\sqrt{d'})$.
Proof. Suppose $\sqrt{d'} = a + b \sqrt{d}$. Apply the non-identity Galois automorphism. This can't fix $\sqrt{d'}$ since it's not rational, so it takes it to the other root and gives $-\sqrt{d'} = a - b \sqrt{d}$. Subtracting gives $2\sqrt{d'} = 2b \sqrt{d}$. This is a contradiction since $d'$ is squarefree.
I wrote out those proofs even though they're pretty standard, because I'm going to generalize them to the cubic case. I think the general case is probably along the same lines.
First let $K_2 = \mathbb{Q}(\sqrt{p} : p \text{ prime or } -1)$. Every finite subfield has degree $2^d$ and a lattice of subfields isomorphic to the lattice of subgroups of $(\mathbb{Z}/2\mathbb{Z})^d$. (Adjoin $i$ last to go from a subfield of $\mathbb{R}$ to a complex field.) Also, observe that $K_2$ contains the cube roots of unity $\zeta_3 = \tfrac{1}{2}(1 + i\sqrt{3})$.
Below, $\sqrt[3]{d}$ always refers to the real cube root.
Fact 3. Let $d \in \mathbb{Z}$ and let $K$ be any field extension of $\mathbb{Q}$ containing the cube roots of unity. Then $x^3 - d$ is irreducible or splits completely over $K$. In particular, $[K(\sqrt[3]{d}) : K]$ is $1$ or $3$.
Proof: If the polynomial factors, it has a root since it's degree $3$. So, we get the other roots too using the cube roots of unity. (This is one place where my argument is easier for cube roots!).
Now we can prove:
Theorem. For any distinct primes $p_1, \ldots, p_n$, $K_2(\sqrt[3]{p_1}, \ldots, \sqrt[3]{p_n})$ has degree $3^d$ and a lattice of subfields isomorphic to the lattice of subgroups of $(\mathbb{Z}/3\mathbb{Z})^n$. In particular:

*

*It has a $K_2$-basis given by $\{\sqrt[3]{d}\}$ where $d = p_1^{k_1} \cdots p_n^{k_n}$ with $0 \leq k_n \leq 2$. (These are all "reduced roots" in the sense you were defining, involving at most cube roots anyway.)

*The only degree-3 sub-extensions are of the form $K_2(\sqrt[3]{d})$ for some $d$ as above.

Proof: For $n=1$, by Fact 3 it's enough to show that $\sqrt[3]{d} \notin K_2$. But if this were true, then $\sqrt[3]{d}$ would be in a finite subfield of $K_2$, and those all have degrees given by powers of $2$, a contradiction.
For $n \geq 2$, suppose $\sqrt[3]{p_n} \in K_2(\sqrt[3]{p_1}, \ldots, \sqrt[3]{p_{n-1}})$. Then $K_2(\sqrt[3]{p_n}) = K_2(\sqrt[3]{d})$ for $d$ one of the integers described above. So it's enough to show that this doesn't happen -- in other words we're reducing to the case $n=2$.
Fact 4. If $d, d'$ are distinct integers of the form $p_1^{k_1} \cdots p_n^{k_n}$ with $0 \leq k_i \leq 2$, then $K_2(\sqrt[3]{d}) \ne K_2(\sqrt[3]{d'})$ unless the vectors $\vec{k}$ and $\vec{k}'$, viewed as elements of $(\mathbb{Z}/3\mathbb{Z})^n$, generate the same subgroup. (Note that this exception doesn't occur above since $p_n$ is the new prime and $d$ isn't divisible by it at all.)
Proof: Similar to the Galois trick above. The extension is Galois since $K_2$ contains the cube roots of unity. Suppose
$$\sqrt[3]{d'} = a + b \sqrt[3]{d} + c (\sqrt[3]{d})^2$$
for some $a,b,c \in K_2$. Apply the Galois automorphism sending $\sqrt[3]{d} \mapsto \zeta_3 \sqrt[3]{d}$. This can't fix $\sqrt[3]{d'}$ since, as noted above, $\sqrt[3]{d'} \notin K_2$. So we get
$$\begin{align*} \zeta_3^\epsilon \sqrt[3]{d'} &= a + b \zeta_3 \sqrt[3]{d} + c \zeta_3^2 (\sqrt[3]{d})^2 \\ \zeta_3^{2\epsilon}\sqrt[3]{d'} &= a + b \zeta_3^2 \sqrt[3]{d} + c \zeta_3 (\sqrt[3]{d})^2, \end{align*}$$
where $\epsilon$ is either $1$ or $2$. Divide out the $\zeta_3^\epsilon$ and $\zeta_3^{2\epsilon}$ factors and add up all three equations to get
$$3\sqrt[3]{d'} = (1 + \zeta_3^{2\epsilon} + \zeta_3^\epsilon)a + (1 + \zeta_3^{1+2\epsilon} + \zeta_3^{2+\epsilon})b\sqrt[3]{d} + (1 + \zeta_3^{2+2\epsilon} + \zeta_3^{1+\epsilon})c(\sqrt[3]{d})^2.$$
If $\epsilon = 1$, this says $$3\sqrt[3]{d'} = 3b \sqrt[3]{d}.$$
If $\epsilon = -1$, it instead says $$3\sqrt[3]{d'} = 3c (\sqrt[3]{d})^2.$$
So $d' = b^3 d$ or $d' = c^3 d^2$. So $b$ or $c$ is the cube root of a rational number, and since $b, c \in K_2$, the only possibility here is for them $b$ or $c$ to be a rational number times a cube root of unity. Since we used positive real cube roots, we can assume $b, c$ are actually rational and positive. Then by examining prime factorizations, we see that the exponent vectors of $d$ and $d'$ span the same subgroup of $(\mathbb{Z}/3\mathbb{Z})^n$.
A: Your claim is indeed true. Suppose that we have an identity
$$
\sum_{i=1}^m M_i\sqrt[N_i]{a_i} = 0 \tag{1}
$$
where the $M_i (1\leq i \leq m)$ are rational numbers, the $a_i$ are positive integers and the roots $r_i=\sqrt[N_i]{a_i}$ are distinct and reduced (note that $1=\sqrt[1]{1}$ is itself a reduced root). Denote by $p_1 \lt p_2 \lt \ldots \lt p_k$ the prime appearing in the factorization of some $a_i$. Then for each $i$, we have (nonnegative, rational) exponents $e_{i,1},\ldots,e_{i,n}$ such that $r_i=p_1^{e_{i,1}}\ldots p_1^{e_{i,n}}$ with $0\leq e_{i,n} \lt 1$ (because $r_i$ is reduced) and the uples $(e_{i,1},\ldots,e_{i,n})$ are pairwise distinct (because the $r_i$ are distinct).
Denote by $N$ the least common multiple of the $N_i$ ; then for each $i$ and $j$, the denominator of $e_{i,j}$ divides $N_i$ which in turn divides $N$. So $g_{i,j}=Ne_{i,j}$ is an integer, and we have $0 \leq g_{i,j} \lt N$. Then
(1) can be rewritten as
$$
\sum_{i=1}^m M_i\alpha_1^{g_{i,1}}\alpha_2^{g_{i,2}}\ldots\alpha_n^{g_{i,n}} = 0 \tag{2}
$$
where $\alpha_i=\sqrt[N]{p_i}$  for $1\leq i \leq m$.
But by Besicovich' result (which is also the subject of an older MSE question)
we have $[{\mathbb Q}(\alpha_1,\alpha_2,\ldots,\alpha_n):{\mathbb Q}]=N^n$, so all the $M_i$ must be zero in (2), which finishes the proof.
