Certain step that $\{x\in \mathbb{Q}\mid x^2<2\}$ has no supremum in $\mathbb{Q}$

Question

Let $A=\{x\in \mathbb{Q}\mid x^2<2\}$. Then $A$ has no least upper bound in $\mathbb{Q}$.

The proof I read is: ''If it did, say $r$, then it can be shown that $r^2=2$. But we know that there is no such rational number.''

Question: How can it be shown that $r^2=2$ in the first sentence?

If $B$ is a set, and if $\sup B=k$, wouldn't it mean that $y\leq k$ for all $y\in B$ and there is no upper bound lesser than $k$? Or, is it using the fact that: ''There exists a real number $x$, such that $x^2=2$'', then replace $x$ by $r$, as supremum is unique?

Answer 1

Suppose there exists $r=\sup\ A$. In order to show $r^2=2$, let us show that none of $r^2<2$ and $r^2>2$ can happen.

Let us suppose $r^2<2$. In particular, $2-r^2>0$. By Archimedian property, there is $n\in\mathbb N\setminus\{0\}$, such that:

$$n\cdot (2-r^2)>2\cdot r+1.$$

This implies \begin{align*} \frac{1}{n}<\frac{2-r^2}{2\cdot r+1}. \end{align*} Hence, using that

$$\frac{1}{n^2}\leq \frac{1}{n},$$ it follows

\begin{align*} \left(r+\frac{1}{n}\right)^2&=r^2+\frac{2\cdot r}{n}+\frac{1}{n^2}\\ &\leq r^2+\frac{2\cdot r}{n}+\frac{1}{n}\\ &=r^2+\frac{2\cdot r+1}{n}\\ &<r^2+(2\cdot r+1)\cdot \frac{2-r^2}{2\cdot r+1}\\ &=r^2+2-r^2\\ &=2. \end{align*}

Since

$$r+\frac{1}{n}\in \mathbb Q\quad \textrm{and}\quad \left(r+\frac{1}{n}\right)^2<2$$ one has

$$r+\frac{1}{n}\in A.$$

This contradicts the fact that $r$ is an upper bound of $A$ for

$$r<r+\frac{1}{n}.$$

Now, it is left to show $r^2>2$ can not happen. If $r^2>2$, then $r^2-2>0$ and by Archimedian property, there exists $n\in \mathbb N\setminus\{0\}$ such that

$$ n\cdot (r^2-2)>2\cdot r. $$ This imples

$$ r^2-\frac{2\cdot r}{n}>2. $$ Notice also $$ r^2-\frac{2\cdot r}{n}>2\implies r-\frac{2}{r}>\frac{2}{n}>\frac{1}{n}\implies r-\frac{1}{n}>0. $$ Now, \begin{align*} \left(r-\frac{1}{n}\right)^2&=r^2-\frac{2\cdot r}{n}+\frac{1}{n^2}>r^2-\frac{2\cdot r}{n}>2 \end{align*}

Since $r=\sup\ A$, there exists $x_0\in A$ such that

$$r-\frac{1}{n}<x_0.$$ Since $r-\frac{1}{n}>0$, it follows

$$x_0^2>\left(r-\frac{1}{n}\right)^2>2.$$ This contradicts $x_0\in A$.

Answer 2

As already commented, in the answer in this post, if one assumes by contradiction that $r∈\Bbb Q$ is the supremum of $A ,$ here is how to prove by cases that $r^2<2$ and $r^2>2$ both imply contradiction, without using the Archimedean property:

• If $x^2<2$ (and $x>0$ ), $$y:=\frac2{{\frac1x}+\frac x2}$$ will be $>x$ and still $y^2<2$ (this is Heron's method). So, $r^2$ cannot be $<2.$

• Similarly, if $u^2>2$ (and $u>0$ ), $$v:=\frac{u+\frac2u}2$$ will be $<u$ and still $v^2>2.$ So, $r^2$ cannot be $>2.$