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**metlx** M = set

$\displaystyle M = \{x \in \mathbb{Q} \mid (x \geq 0) \wedge (x^{2}< 2\}$

$\displaystyle M \neq \emptyset$

Statement: This set does not have a supremum in $\displaystyle \mathbb{Q}$

Prove with contradiction:

Let's say there is an $\displaystyle a = sup M \in \mathbb{Q}$

Then we can assume each of the following 3 options:

(1) $\displaystyle a^2 = 2 $

(2) $\displaystyle a^2 < 2 $

(3) $\displaystyle a^2 > 2 $

$\displaystyle a^2 = 2$, can't be true because we know $\displaystyle \sqrt{2} $ is not in $\displaystyle \mathbb{Q}$ (won't prove that now..)

Here's where I get stuck..

$\displaystyle a^2 < 2$

Let's say there is a $\displaystyle h > 0 \ \ h \in \mathbb{Q}$ so that $\displaystyle (a + h)^2 < 2 \Rightarrow a + h \in M$

So we want that $\displaystyle (a + h)^2 = a^2 + 2ah + h^2 < 2$

Then we say $\displaystyle 0 < h < 1$

$\displaystyle (a + h)^2 = a^2 + 2ah + h^2 < a^2 + h(2a +1) < 2$

This applies only if $\displaystyle a < h < \frac{2 - a^2}{2a + 1}$

That's what we had done at uni but I don't understand the last 2 paragraphs..

I know if we square a number that is less than 1 that we get a number that is less than the number we squared so I get it why the 'h' in brackets is replaced by 1.. but how do you know for sure that is still less than 2?