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Is it possible to know if $4-\sqrt{2}-\sqrt[3]{3}-\sqrt[5]{5} \gt 0$ without using decimal numbers?

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  • $\begingroup$ The original title was correct; now it's wrong. $\endgroup$ – George Law Dec 30 '16 at 18:38
  • $\begingroup$ sorry, I do not understand. What is wrong? (the proof can result in "no") $\endgroup$ – pasaba por aqui Dec 30 '16 at 18:39
  • $\begingroup$ In principle, either way, it should be possible to obtain close-enough rational approximations of the roots. $\endgroup$ – Mark Bennet Dec 30 '16 at 18:39
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    $\begingroup$ @pasabaporaqui You originally wrote $5-\sqrt2-\sqrt[3]3-\sqrt[5]5>0$ which was correct. $\endgroup$ – George Law Dec 30 '16 at 18:41
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    $\begingroup$ For a harder one, you might try $7-20\,\sqrt {2}+2\,\sqrt [3]{3}-\sqrt [5]{5}-\sqrt [6]{6}+16\,\sqrt [7 ]{7} $. $\endgroup$ – Robert Israel Dec 30 '16 at 18:50
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It is not hard to verify following inequalities (just power both sides and it should result into simple inequalities in natural numbers only):

\begin{align} \frac{4}{3} &< \sqrt{2} < \frac{5}{3}\\ \frac{4}{3} &< \sqrt[3]{3} < \frac{5}{3}\\ \frac{4}{3} &< \sqrt[5]{5} < \frac{5}{3}\\ \end{align} Summing these up will give you $$ 4 < \sqrt{2}+\sqrt[3]{3}+\sqrt[5]{5} < 5\\ $$

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  • $\begingroup$ Thanks for the answer. Is it possible infer a method for the general case from it? See by example comment of @Robert Israel $\endgroup$ – pasaba por aqui Dec 30 '16 at 18:54
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    $\begingroup$ @pasabaporaqui Well generally for each summand you can find arbitrary close rational bounds, they just won't be so obvious as in this example. $\endgroup$ – Sil Dec 30 '16 at 19:00

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