# Prove $4-\sqrt{2}-\sqrt[3]{3}-\sqrt[5]{5} \gt 0$

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

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

\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 $$4 < \sqrt{2}+\sqrt[3]{3}+\sqrt[5]{5} < 5\\$$