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

Is it possible to know if $4-\sqrt{2}-\sqrt{3}-\sqrt{5} \gt 0$ 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 $5-\sqrt2-\sqrt3-\sqrt5>0$ which was correct. – George Law Dec 30 '16 at 18:41
• For a harder one, you might try $7-20\,\sqrt {2}+2\,\sqrt {3}-\sqrt {5}-\sqrt {6}+16\,\sqrt [7 ]{7}$. – Robert Israel Dec 30 '16 at 18:50

## 1 Answer

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

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