prove that the supremum is $\sqrt{2}$

Question

show that $$\sup\{x\in \Bbb Q :x>0,x^2<2\}=\sqrt{2}$$

I have the solution but it is too complicated can someone explain it to me in a simple way? I would really appreciate it Thank you!

Answer 1

Firstly, $\sqrt{2}$ is an upperbound of this set. This is because for all $x$ in the set, $x^2\lt (\sqrt2)^2$ by definition.

Now, it needs to be shown that this is the least upper bound, or the supremum.

Suppose this is not the case. Let some $\alpha\in\mathbb Q$ be the supremum, where $\alpha$ must be less than $\sqrt2$, as $\sqrt2$ is an upper bound.

So, $$\alpha\lt \sqrt2$$ By the denseness of rationals, there exists some $y\in\mathbb Q$ such that $$\alpha\lt y\lt \sqrt2$$

But now, we have $y\in\mathbb Q$ and $$\alpha^2\lt y^2\lt 2$$ i.e, $y^2\lt 2$.

We have managed to find an element of the set, $y$, that is greater than the supremum, $\alpha$. This goes against the definition of a supremum, and is a contradiction.

Hence, the supremum of the set is $\sqrt2$.