Prove that there is no largest rational number $q$ such that $q^3 < 2$

Question

I know how to prove such things where $q < 2$, but $q^3 < 2$ seems a bit tricky for me. Could someone give me a lead please?

Answer 1

I expand on my comment. Let $q>0$ be rational such that $q^3<2$ so that $r=2-q^3>0$. Let $h>0$ be rational and $p=q+h$ and consider $$2-p^3=r-h(3q^2+3qh+h^2)$$ Let $h<1$ so that $$3q^2+3qh+h^2<3q^2+3q+1$$ and if we take $h<r/(3q^2+3q+1)$ ie $h<\min(1,r/(3q^2+3q+1))$ then we can see that $$h(3h^2+3qh+h^2)<r$$ and hence $2-p^3>0$. Thus the proof is complete. The argument works in exactly the same manner if the exponent $3$ is replaced by any positive integer, for example $100$.

The fundamental idea here is the fact that there is no end to supply of small positive rationals. Given any positive rational, there is a smaller one. This fact studied when learning fractions and their manipulation is the key aspect behind the proofs in calculus and strangely it is never emphasized in entire high school curriculum.