# Rational numbers, supremum (Is my proof correct?)

## Homework Statement

Show that the set $\{x \in \mathbf Q; x^2< 2 \}$ has no least upper bound in $\mathbf Q$; using that if $r$ were one then $r^2=2$. Do this assuming that the real field haven't been constructed.

N/A

## The Attempt at a Solution

Attempt at proof:
$r\in Q$ is an upper bound of the set if and only if $r^2 \ge 2$.
Choose any $r_1 \in \mathbf Q$ that satisfies $r_1^2 >2$. Then $r_1$ is an upper bound of the set.
Set $r_2 = r_1-\frac{r_1^2-2}{r_1+2}<r_1 \Longrightarrow r_2^2=2+\frac{2(r_1^2-2)}{(r_1+2)^2}>2$.
Note that $r_2 \in \mathbf Q$ and $r_2 < r_1$. $r_2$ is also an upper bound since $r_2^2>2$.

This shows that for every upper bound $r_1$ with $r_1^2 > 2$ there's always possible to find an upper bound $r_2$ where $r_2<r_1$. Hence the only possible least upper bound satisfy $r^2=2$ but there is no $r\in \mathbf Q$ that satisfy that. Hence the set have no least upper bound in $\mathbf Q$.

Is the above correct? Anything I can do to improve it? I'm also wondering about the choice of $r_2$. I remembered that from an earlier example in our book but I'm not sure I could've come up with a choice like that by myself without a lot of trial and error. How would I go about finding a choice like that in the first place?

## Answers and Replies

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andrewkirk
Homework Helper
Gold Member
How did you get the following?
$r_2 = r_1-\frac{r_1^2-2}{r_1+2}<r_1 \Longrightarrow r_2^2=2+\frac{2(r_1^2-2)}{(r_1+2)^2}$
When I square $r_2$ I get 2 plus a fraction with the same denominator as yours but a messy quartic in $r_1$ in the numerator.

andrewkirk
Homework Helper
Gold Member
Given an upper bound $r$ such that $r{}^2=2+h$ for $h>0$, you can write an equation for a lower upper bound $s=r-k$ such that $s^2$ is 2 plus some positive fraction of $h$. Then, using the relationships you've established, solve for $k$.

How did you get the following?
When I square $r_2$ I get 2 plus a fraction with the same denominator as yours but a messy quartic in $r_1$ in the numerator.
Sorry, I should have clarified that step
$r_2 = r_1-\frac{r_1^2-2}{r_1+2} = \frac{r_1^2+2r_1-r_1^2+2}{r_1+2}=\frac{2r_1+2}{r_1+2}$
square
$r_2^2 = \frac{4(r_1^2+2r_1+1)}{(r_1+2)^2} = \frac{2(r_1^2+4r_1+4)+2(r_1^2-2)}{(r_1+2)^2}= 2+\frac{2(r_1^2-2)}{(r_1+2)^2}$.

Given an upper bound $r$ such that $r{}^2=2+h$ for $h>0$, you can write an equation for a lower upper bound $s=r-k$ such that $s^2$ is 2 plus some positive fraction of $h$. Then, using the relationships you've established, solve for $k$.
Thanks for responding! Is there something missing in my proof where I need this to complete it? When you say the relationship do you mean my value of $r_2$ or something else? I'm not entirely sure I understand.

andrewkirk