Alternative way of thinking about irrationality of $\sqrt{2}$

Question

A user on stack exchange suggested to think of the following problem as a good way to distinguish between algebraic mindset and an analysis mindset:

a) Prove $\sqrt{2}$ is irrational by expressing it is equal to $\frac{m}{n}$ and then come up with some irreducibility argument (lowest terms). (We did this in my introduction to maths course)

b) Prove that whenever some rational $a^2 <2$, there is some other rational $b>a$, such that $b^2 <2$

I found this second way of thinking about it extremely interesting and have been messing around with inequalities like $0< a^2 -1<1$, but I think I'm completely missing the point on the second one, to be quite frank, I'm kind of lost. I do get that the point is that you will never reach $\sqrt2$ since it's a limit point of the set of integers leading up to it, but it's not contained within $\mathbb{Q}$. This is the feature that makes $\mathbb{Q}$ not closed (does not contain all of its limit points).

Answer 1

If $a>0$ and $a^2<2$ then $b^2>2$ when $b=2/a$. If $a$ is close to $\sqrt2$ we'd expect $b$ to be roughly as close to $\sqrt2$ so $c=\frac12(a+b)$ should be closer. But $c^2>2$. So try $d=2/c$. Your task: prove $d^2<2$ and $d>a$.

Answer 2

You can take $$b=a+\frac {1}{n} $$ with $n $ such that

$$a <a+\frac {1}{n}<\sqrt {2}$$

or

$$n=\lfloor \frac {1}{\sqrt {2}-a} \rfloor +1.$$