Thread: The existence of root 2

1. The existence of root 2

Hi guys the analysis continues.

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

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

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

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

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

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

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

SOLUTION 2: This is where I am having trouble. I have let $\displaystyle z = y - \delta$ and again assumed that $\displaystyle \delta < 1$ and $\displaystyle y > 1$. 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 $\displaystyle x^2 = 2$. [Hint: one way of doing this is to show that both $\displaystyle x^2 < 2$ and $\displaystyle x^2 > 2$ 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.

2. Originally Posted by slevvio
QUESTION 2: Show too that if $\displaystyle y \in \mathbb{Q}$ has $\displaystyle y^2 > 2$ and $\displaystyle y > 0$, then there is $\displaystyle 0 < z < y$ with $\displaystyle z^2 > 2$.
SOLUTION 2: Let $\displaystyle z=\frac{2y+2}{y+2}$.
Notice that$\displaystyle z=\frac{2y+2}{y+2}=y-\frac{y^2-2}{y+2}$.
Now all you have to do is show $\displaystyle z^2>2$.

3. How did you know to choose that particular z ?

4. its guess and check.

5. 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.

6. 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

7. 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 $\displaystyle z^2\le2$ then do the algebra see a contradiction.

8. Thanks a lot plato

9. 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?

10. 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 $\displaystyle x^2 = 2$. [Hint: one way of doing this is to show that both $\displaystyle x^2 < 2$ and $\displaystyle x^2 > 2$ lead to a contradiction using the previous step.
Well we have shown $\displaystyle \left\{ {x\in \mathbb{Q}:1 < x^2 < 2} \right\}$ does not have a maximal term but is bounded above. Therefore, it has a supremum.
Also $\displaystyle \left\{ {x\in \mathbb{Q}:1 x^2 > 2} \right\}$ does not have a minimal term but is bounded below. Therefore, it has an infimum.
Clearly those are equal cannot be rational.