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How can I prove that

nonzero integer linear combination of two rational independent irrational numbers is still a irrational number?That is to say, given two irrational numbers a and b, if a/b is a irrational number too, then for any m,n is nonzero integer, we have that the number ma+nb is a irrational number, why?

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  • $\begingroup$ Did you try a proof by contradiction ? $\endgroup$ – lmsteffan Sep 5 '14 at 12:05
  • $\begingroup$ $(1-\sqrt{2})+\sqrt{2}\in \mathbb{Q}\setminus\{0\}$ $\endgroup$ – user72870 Sep 5 '14 at 12:05
  • $\begingroup$ Actually your second sentence ("That is to say...") is not a translation of the first one.See en.wikipedia.org/wiki/Rational_dependence $\endgroup$ – lmsteffan Sep 5 '14 at 12:25
  • $\begingroup$ I'd like to increase that what condition to make my conclusion right,maybe if a/b is not an algebaic number? $\endgroup$ – David Chan Sep 5 '14 at 12:46
  • $\begingroup$ I think your first sentence is correct, and your second could be "That is to say, if the only pair of integers $k_1$, $k_2$ such that $k_1 a + k_2 b = 0$ is the trivial solution in which $k_1 = k_2 = 0$, then for any non-zero integers $m$, $n$, $ma+nb$ is irrational. And maybe you can prove that. $\endgroup$ – lmsteffan Sep 5 '14 at 12:54
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That's not true: Take $a=\sqrt{2} -1$, $b=\sqrt{2}$. Then $\frac{a}{b} = \frac{1}{\sqrt{2}} - 1 $ isn't rational, but $a-b=1$

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    $\begingroup$ I believe you should have $a=1-\sqrt2$... $\endgroup$ – gebruiker Sep 5 '14 at 12:09
  • $\begingroup$ However $\sqrt{2}$ and $\sqrt{2} -1$ are not rationally independent. $\endgroup$ – lmsteffan Sep 5 '14 at 12:24
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Now that I know that “rationally (in)dependent” merely means linearly (in)dependent over $\mathbb Q$, I see that you have two distinct conditions: the first is for $\{a_1,a_2,\cdots,a_n\}$ to be rationally independent; the second is for $\{1,a_1,a_2,\cdots,a_n\}$ to be rationally independent. Of course the second implies the first. but not the converse. Your question was whether independence in the first sense implies independence in the second sense, and @PenasRaul’s perfect answer gives a counterexample.

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