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I arrived at an idea of considering rational linear combinations of square roots of non-square naturals and of the irrationality of those combinations.

So suppose that we have some $n$-tuple $(\sqrt{a_1},...,\sqrt{a_n})$ where $a_1,...,a_n$ are natural numbers which are not squares so that $\sqrt{a_1},...,\sqrt{a_n}$ are all irrational.

Now the functions $\text{rl}$ can be defined as $\text{rl}(\sqrt{a_1},...,\sqrt{a_n} ; r_1,...,r_n)=\displaystyle \sum_{k=1}^n r_k \cdot\sqrt{a_k}$ where $r_1,...,r_n$ are non-negative rational numbers and at least one of them is $\neq 0$.

I am of the opinion that $\text{rl}(\sqrt{a_1},...,\sqrt{a_n} ; r_1,...,r_n)=\displaystyle \sum_{k=1}^n r_k \cdot\sqrt{a_k}$ is always an irrational number no matter what input we choose for the functions $\text{rl}$.

Is that true?

To demistify this notation, this questions is really:

Choose $n$ numbers so that all of them are square roots of non-square natural numbers. Then form their linear combination with non-negative rational coefficients so that at least one of the coefficients is not equal to zero. Is that rational linear combination always an irrational number?

  • @MarkBennet The rational numbers must be non-negative and at least one of them not equal to zero, that is stated in the body of the question. –  Oct 22 '19 at 06:16
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    Possible duplicate of The square roots of different primes are linearly independent over the field of rationals – Greg Martin Oct 22 '19 at 06:16
  • @GregMartin Is it really a duplicate, my question is not only about square roots of primes? Can those answers be adapted to my question also? –  Oct 22 '19 at 06:18
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    @Donkey I noticed the non-negative condition late. With an additional condition you can allow negative rationals. Really the heart of the proof is about showing that each added square root creates a field extension of degree $2$ as in Bill Dubuque's anger to the linked question. – Mark Bennet Oct 22 '19 at 06:20
  • It's not quite true as stated, e.g. $\sqrt{8} - 2\sqrt{2} = 0$. You can rule out this silliness by assuming your naturals are square-free. In that case, you can apply the answer Greg linked to the field generated by the primes appearing in the factorizations of your naturals. – Joshua P. Swanson Oct 22 '19 at 06:34
  • @JSwanson: Donkey said "non-negative rational numbers". – user21820 Oct 22 '19 at 10:21
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    @user21820 I am pointing out that if you get the conditions right you can allow non-negative rationals - it is then harder to state, but in some respects more natural (because being rational or irrational is related to the whole field $\mathbb Q$ and not just the positive rationals). – Mark Bennet Oct 22 '19 at 13:55
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    @user21820 I haven't voted to close and deliberately made a comment rather than an answer. The suggested duplicate shows that certain sets of square roots are linearly independent over $\mathbb Q$. Here there can be multiple terms involving the same square root, but since (a) you can make all the $a_i$ square free, and (b) the coefficients of these surds are all positive, the linear combination of the same square roots cannot be zero and by the linked question the sum can't be in $\mathbb Q$ – Mark Bennet Oct 22 '19 at 15:38
  • @MarkBennet: You are right; I somehow missed that trivial solution. Thanks, and because of that it seems reasonable to consider this a duplicate after all, contrary to my earlier comment. I suppose I shall delete my now useless comments. – user21820 Oct 22 '19 at 15:46
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    @user21820 FWIW I agree with Mark Bennet. Furthermore, it suffices to adjoin square roots of all the primes up to a certain point to the rationals. I used that technique in this old answer. So IMO the claim of this question does follow from that of the linked thread. I guess it is a matter of opinion whether that is sufficient to make this a duplicate? – Jyrki Lahtonen Oct 22 '19 at 20:35

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