Supremum Proof Concerning Sqrt[2]

[MI6]

The following is my book's proof that $\sup\left\{x\in\mathbb{Q}:x>0, ~ x^2<2\right\} = \sqrt{2}.$

I don't follow the bit where it says "if [tex]s[/tex] were irrational, then [tex]w = \frac{\lfloor(n+1)s\rfloor}{n+1}+\frac{1}{n+1}.[/tex]"

Could someone please elaborate on that bit? How does s being irrational imply that?

 

[Deveno]

all we are trying to do is find SOME rational number between s and s + 1/n.

the claim is that w is one such rational number.

note that since $\lfloor (n+1)s \rfloor < (n+1)s$ (s is NOT an integer, since s is irrational)

$\dfrac{\lfloor (n+1)s \rfloor}{n+1} < s$

so:

$w = \dfrac{\lfloor (n+1)s \rfloor}{n+1} + \dfrac{1}{n+1} < s + \dfrac{1}{n+1}$

on the other hand, we also have:

$(n+1)s - 1 < \lfloor (n+1)s \rfloor$

so:

$s = \dfrac{ns + s - 1}{n+1} + \dfrac{1}{n+1} < \dfrac{\lfloor (n+1)s \rfloor}{n+1} + \dfrac{1}{n+1} = w$

(all we are doing is explicitly finding a rational number between s and s + 1/(n+1)...we need a rational number to use because the set A only has rational members).

 

[Opalg]

The following is my book's proof that $\sup\left\{x\in\mathbb{Q}:x>0, ~ x<2\right\} = 2.$

It is true that $\sup\left\{x\in\mathbb{Q}:x>0, ~ x<2\right\} = 2$, but that is not what the book proves. The result is that if $s = \sup\left\{x\in\mathbb{Q}:x>0, ~ x^2<2\right\}$ then $s^2=2$. Or, to put it another way, $s=\sqrt2$.