The existence of root 2

**slevvio** · Sep 30th 2009, 09:51 AM

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.

**Plato** · Sep 30th 2009, 12:46 PM

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 $z^2>2$ .

**slevvio** · Sep 30th 2009, 03:50 PM

How did you know to choose that particular z ?

**Sampras** · Sep 30th 2009, 04:19 PM

its guess and check.

**Plato** · Sep 30th 2009, 05:11 PM

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.

**slevvio** · Oct 1st 2009, 01:36 AM

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

**Plato** · Oct 1st 2009, 02:52 AM

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.

**slevvio** · Oct 1st 2009, 03:01 AM

Thanks a lot plato

**slevvio** · Oct 1st 2009, 03:54 AM

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?

**Plato** · Oct 1st 2009, 06:35 AM

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.

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