I have been thinking about how to show that the set $ S = \{ \text{x is rational} , x^2 < 2 \} $ has no least upper bound in the rationals space, in a way that is intuitive for me. In that sense, I have been trying to formalize what I believe is a good idea:
I have to show that given any rational number whose square is more than $2$, there is still a smaller one whose square is also greater than $2$, that is, it is not the least upper bound.
Since the “input” rational is arbitrary, that is what is making my formalization harder to be made.
My idea is the following:
Take that input number, which I will denote by $q$ and using $1$ as a sort of pivot, add to it (almost) the entire difference $q-1$, that is , a fraction $\alpha = (n-1)/n$ of it, where $n$ is large enough (it will depend on $q$) to have that the “output” number $q’=1+\alpha*[q-1]$ (which is rational since rational plus rational is rational) satisfies: $q’<q$ and also $(q’)^2 > 2$
Could you please tell me what you think and how I can further formalize the idea to prove it is a valid proof, in the sense that it shows a way to construct a lower rational upper bound?
Ps : So far, I think that for the mentioned form of $\alpha$, taking $n > 1/(q-2)$ is big enough to ensure that $q' = 1+\alpha*[q-1]$ has a square bigger than two (the fact that is smaller than q is true for any alpha).
NOTE : I am not interested in proving the fact since I have read and understood other proofs already; my question is related to my approach and how to completely turn my "semi-proof" into a proof. Thank you in advance!
