Proving that there is no supremum/infimum

Question

I just saw a proof by contradiction for showing that there exists no supremum for the set

A = {x ∈ ℚ | 0 ≤ x² < 2} :

Assume b = sup(A) so b ∈ ℚ that's why b² ≠ 2 which means b² < 2 or b² > 2

if b² > 2 it follows that b > 2/b and (because b < 2) there exists an x = 0.5(b+2/b) and x < b

and now comes the step I don't understand...

ALSO x²-2 = (x-b)² > 0

the rest

x² > 2 => b cannot be the supremum

I do understand, so now my question is..

why is x²-2 = (x-b)² ?? and is the proof also valid if A = {x ∈ ℚ | 0 ≤ x² ≤ 2}?

Answer 1

Well you can just calculate: $$x^2-2=\left(\frac{1}{2}\left(b+\frac{2}{b}\right)\right)^2-2= \frac{1}{4}\left(b^2+4+\frac{4}{b^2}\right)-2 = \frac{1}{4}b^2-1+\frac{1}{b^2}$$ and $$(x-b)^2=\left(\frac{1}{2}\left(b+\frac{2}{b}\right)-b\right)^2= \left(\frac{1}{b}-\frac{1}{2}b\right)^2= \frac{1}{b^2}-1+\frac{1}{4}b^2.$$ So these are equal.

Yes, the proof is still valid if $A=\{x \in \mathbb{Q} : 0 \leq x^2 \leq 2\}$ because this is the same as the set $\{x \in \mathbb{Q} : 0 \leq x^2 <2\}$ (since there is no $x \in \mathbb{Q}$ with $x^2=2$).

Answer 2

We have

$(\forall x\in A)\;\;\; x<\sqrt{2}$

$\implies \sup A \leq \sqrt{2}$

suppose that $\sup A=M<\sqrt{2}.$

if we take $N=\lfloor \frac{1}{\sqrt{2}-M}\rfloor+1$

then

$\frac{1}{N}+M<\sqrt{2}$ and

$\frac{1}{N}+M >M$ which is in contradiction with $M=\sup A$.

finally $\sup A$ doesn't exist in $\mathbb Q$.