<s>[INST] When $ a^5 < 5 $ show that there exists b such that $ aHere's my approach.
Since $ a^5 < 5, a<\sqrt[5]{5} $
By density of rational number, there must be integer $m$ and natural number $n$ such that
$ a< \frac{m}{n} < \sqrt[5]{5}$
If I let $b= \frac{m}{n} $,   then $ a<b$ and $ b<\sqrt[5]{5} $
Since, $ b< \sqrt[5]{5}$, $ b^5 < 5 $,
therefore there exists $b$ such that $a<b$ and $b^5<5$
Is there anything that I should fix or add?
I wonder if I have to show that there exists real number $a$ such that $a^5<5$.
 [/INST] Let $b=\frac{a+\sqrt[5]5}{2}.$
Thus, since $a<\sqrt[5]5,$ we see that $$b<\frac{\sqrt[5]5+\sqrt[5]5}{2}=\sqrt[5]5,$$ which gives $$b^5<5.$$
Also, $$b-a=\frac{a+\sqrt[5]5}{2}-a=\frac{\sqrt[5]5-a}{2}>0,$$ which says $a<b$.
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