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The square roots of the primes are linearly independent over the field of rationals

I am reading a research article in which there is a theorem regarding square roots of square free numbers. A number is square free if it is not divisible by square of any prime number. The theorem states the following:

$$\text{The square roots of all positive square-free integers are linearly independent over }\mathbb{Q}.$$

Unfortunately, it provides a reference to an article to which I don't have an access at the moment. The article is titled, "Linear Algebra Methods in Combinatorics."

I have tried few things, but I didn't go far with any of that. Can somebody provides me a starting point from where I can work things out myself? I don't want a full solution. Just a starting pointer is appreciated.


marked as duplicate by Gerry Myerson, anon, Arturo Magidin, Qiaochu Yuan Jan 30 '12 at 2:59

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  • $\begingroup$ Thanks! I looked for if it is asked here, but couldn't find it. Do you know if this theorem has a name to it? $\endgroup$ – Jalaj Jan 30 '12 at 2:19
  • $\begingroup$ I don't think so. I'd call it folklore. $\endgroup$ – Qiaochu Yuan Jan 30 '12 at 2:20
  • 2
    $\begingroup$ I wouldn't call it a theorem, so much as a difficult exercise. I first came across (a version of) it in Stewart's Galois Theory text, where it was in an early chapter, way before the discussion of conjugates and automorphisms and Galois groups, and I wondered whether Stewart knew how hard it was to do from scratch. $\endgroup$ – Gerry Myerson Jan 30 '12 at 2:31

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