When $ a^5 < 5 $ show that there exists b such that $ a<b, b^5<5 $

Question

Here'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 $a^5<5$.

Answer 1

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 $a<b$.