# Thread: Prove that 'a' is not element of Q (racional numbers)

1. ## Prove that 'a' is not element of Q (racional numbers)

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 $\displaystyle \mathbb{Q}$

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?

2. Originally Posted by 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 $\displaystyle \mathbb{Q}$

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?
Well, as your last line says -- if $\displaystyle h<\frac{2-a^2}{2a+1}$ then $\displaystyle (a+h)^2<2$

3. but why is 'a' less than 'h' then

4. Originally Posted by metlx
but why is 'a' less than 'h' then
h is not greater that a, a+h is. The idea is to find a rational h, s.t. $\displaystyle (a+h)^2<2$, and as you know the sum of rationals is a rational, so you found an element of M greater than its supremum. You can always find such h because the rationals are dense.

5. so basically all i have to know is that such rational 'h' exists and it's smaller than $\displaystyle \frac{2 - a^2}{2a + 1}$

does this fraction tell me anything else?

6. Originally Posted by metlx
so basically all i have to know is that such rational 'h' exists and it's smaller than $\displaystyle \frac{2 - a^2}{2a + 1}$

does this fraction tell me anything else?

The idea is if there exists a h s.t. $\displaystyle 0< h < \frac{2 - a^2}{2a + 1}$ then that h has the property that $\displaystyle (a+h)^2<2$.

Now the reason why I can find a h s.t. $\displaystyle 0< h < \frac{2 - a^2}{2a + 1}$ is because the rationals are dense.