The existence of root 2

• Sep 30th 2009, 09:51 AM
slevvio
The existence of root 2
Hi guys the analysis continues.

Let $A = \{y \in \mathbb{Q} : y \ge 0, y^2 < 2 \}$

QUESTION 1: Show, without assuming the existence of $\sqrt{2}$, that A does not have a maximal element, i.e. prove that if $y \in A$ then there exists $z > y$ with $z \in A$.

SOLUTION 1: Let $z = y + \delta$. We need to find delta so small that z > y but $z^2 < 2$. We require $y^2 < z^2 < 2 \iff y^2 < (y+\delta)^2 < 2 \iff y^2 < y^2 + 2 \delta y + {\delta}^2 <2$
$\iff 0 < 2 \delta y + {\delta}^2 < 2 - y^2$

Take $\delta < 1$, then ${\delta}^2 < \delta$ .

Also it is safe to assume that y < 2. So $2\delta y < 4 \delta$

So $2 \delta y + {\delta}^2 < \delta + 4 \delta = 5 \delta$.
We get a small enough delta therefore if we choose $5\delta < 2-y^2 \iff \delta < \frac{ 2-y^2}{5}$ i.e. choose $\delta = \frac{2-y^2}{10}$. Indeed $\delta > 0$ since $2 - y^2 > 0$.

QUESTION 2: Show too that if $y \in \mathbb{Q}$ has $y^2 > 2$ and $y > 0$, then there is $0 < z < y$ with $z^2 > 2$.

SOLUTION 2: This is where I am having trouble. I have let $z = y - \delta$ and again assumed that $\delta < 1$ and $y > 1$. Any help would be appreciated:)

QUESTION 3: A is non-empty and bounded above therefore by the completeness axiom the supremum of A exists. Write x = sup A and show that $x^2 = 2$. [Hint: one way of doing this is to show that both $x^2 < 2$ and $x^2 > 2$ lead to a contradiction using the previous step.]

SOLUTION 3: I don't really understand how to use the previous steps to solve this part. Again any help would be appreciated.
• Sep 30th 2009, 12:46 PM
Plato
Quote:

Originally Posted by slevvio
QUESTION 2: Show too that if $y \in \mathbb{Q}$ has $y^2 > 2$ and $y > 0$, then there is $0 < z < y$ with $z^2 > 2$.

SOLUTION 2: Let $z=\frac{2y+2}{y+2}$.
Notice that $z=\frac{2y+2}{y+2}=y-\frac{y^2-2}{y+2}$.
Now all you have to do is show $z^2>2$.
• Sep 30th 2009, 03:50 PM
slevvio
How did you know to choose that particular z ?
• Sep 30th 2009, 04:19 PM
Sampras
its guess and check.
• Sep 30th 2009, 05:11 PM
Plato
Quote:

Originally Posted by slevvio
How did you know to choose that particular z ?

I picked some number that acually works.
Of course its is not unique.
• Oct 1st 2009, 01:36 AM
slevvio
Why is z squared bigger than 2? Perhaps if I see this I will understand why that z is a good choice. I can see that z is less than y
• Oct 1st 2009, 02:52 AM
Plato
Quote:

Originally Posted by slevvio
Why is z squared bigger than 2? Perhaps if I see this I will understand why that z is a good choice. I can see that z is less than y

Suppose that $z^2\le2$ then do the algebra see a contradiction.
• Oct 1st 2009, 03:01 AM
slevvio
Thanks a lot plato
• Oct 1st 2009, 03:54 AM
slevvio
Would it be possible to get a hint for part 3? I can cope with that if A is a set in the reals but because x squared might not be rational how can I pick a z bigger than it still in the set?
• Oct 1st 2009, 06:35 AM
Plato
Quote:

Originally Posted by slevvio
QUESTION 3: A is non-empty and bounded above therefore by the completeness axiom the supremum of A exists. Write x = sup A and show that $x^2 = 2$. [Hint: one way of doing this is to show that both $x^2 < 2$ and $x^2 > 2$ lead to a contradiction using the previous step.

Well we have shown $\left\{ {x\in \mathbb{Q}:1 < x^2 < 2} \right\}$ does not have a maximal term but is bounded above. Therefore, it has a supremum.
Also $\left\{ {x\in \mathbb{Q}:1 x^2 > 2} \right\}$ does not have a minimal term but is bounded below. Therefore, it has an infimum.
Clearly those are equal cannot be rational.