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 LawCommented Dec 30, 2016 at 18:38
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$\begingroup$ sorry, I do not understand. What is wrong? (the proof can result in "no") $\endgroup$– pasaba por aquiCommented Dec 30, 2016 at 18:39
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$\begingroup$ In principle, either way, it should be possible to obtain close-enough rational approximations of the roots. $\endgroup$– Mark BennetCommented Dec 30, 2016 at 18:39
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2$\begingroup$ @pasabaporaqui You originally wrote $5-\sqrt2-\sqrt[3]3-\sqrt[5]5>0$ which was correct. $\endgroup$– George LawCommented Dec 30, 2016 at 18:41
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1$\begingroup$ 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 IsraelCommented Dec 30, 2016 at 18:50
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1 Answer
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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 $\endgroup$ Commented Dec 30, 2016 at 18:54
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3$\begingroup$ @pasabaporaqui Well generally for each summand you can find arbitrary close rational bounds, they just won't be so obvious as in this example. $\endgroup$– SilCommented Dec 30, 2016 at 19:00