3
$\begingroup$

Let $E=\left\{ r\in\mathbb{Q}:r^{2}<3\right\}$. Prove that $\sup E=\sqrt{3}$.

Since $E$ is bounded from above by $\sqrt{3}$ and is nonempty, $\alpha:=\sup E$ must exist by the Least Upper Bound Principle. Now I am stuck. Suppose $\alpha<\sqrt{3}$. Then what?

Edit: It just hit me. If $\alpha < \sqrt{3}$ then since the rationals are dense in $\mathbb{R}$, there exists $q\in\mathbb{Q}$ such that $\alpha < q <\sqrt{3}$ and hence $\alpha^2 which contradicts the fact that $\alpha$ is the sup of $E$

  • 0
    What's your definition for real numbers? Different difinitions result in different proofs.2012-06-28

1 Answers 1

5

Let $r\in \bf E$ , $r = \dfrac m n $.

Then we need to prove that there always exists a $\dfrac{p}{q}$ such that

$ \frac{m}{n}<\frac p q < \sqrt 3$

By exploting the special properties of $\sqrt 3 $,we can do it:

We start with

$\eqalign{ & r < \sqrt 3 \cr & r + 1 < \sqrt 3 + 1 \cr & \frac{1}{{r + 1}} > \frac{1}{{\sqrt 3 + 1}} \cr & \frac{1}{{r + 1}} > \frac{{\sqrt 3 - 1}}{2} \cr & \frac{{r + 3}}{{r + 1}} > \sqrt 3 \cr} $

But now we've gotten a number greater than $\sqrt 3 $. So what we'll do is apply the process again, and the relation will be reversed:

$\frac{{Q + 3}}{{Q + 1}} < \sqrt 3 $

Letting ${Q = \frac{{r + 3}}{{r + 1}}}$ will give

$\frac{{2r + 3}}{{r + 2}} < \sqrt 3 $

All we need to do now is prove that

$r < \frac{{2r + 3}}{{r + 2}}$

But since $r+2>0$ we get that

$\eqalign{ & {r^2} + 2r < 2r + 3 \cr & {r^2} < 3 \cr} $

which is true by hypothesis. Thus, given any rational $r$, there exists another rational $q$ such that

$r

where $q = \frac{{2r + 3}}{{r + 2}}$

Just to make things clear, I'll add some extra information.

The recursion defined as $r_0=1$ $r_{n+1}=\frac{2 r_n+3}{r_n +2} $ converges monotonically (increasing) to $\sqrt 3 $. This in particular means that given any $\epsilon>0$ there is an $r \in \rm E$ such that $\sqrt 3 -\epsilon < r$ which means $\sqrt 3 $ is the supremum of the set.

This also plays the role of showing that $3$ is irrational.

  • 0
    @Peter Tamaroff: I take supremum of a subset of $\mathbb R$ is a least upper bound of that subset.2012-06-29