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Math Help - The existence of root 2

  1. #1
    Senior Member slevvio's Avatar
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    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.
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  2. #2
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    Quote Originally Posted by slevvio View Post
    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.
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  3. #3
    Senior Member slevvio's Avatar
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    How did you know to choose that particular z ?
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    Senior Member Sampras's Avatar
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    its guess and check.
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  5. #5
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    Quote Originally Posted by slevvio View Post
    How did you know to choose that particular z ?
    I picked some number that acually works.
    Of course its is not unique.
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  6. #6
    Senior Member slevvio's Avatar
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    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
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  7. #7
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    Quote Originally Posted by slevvio View Post
    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.
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  8. #8
    Senior Member slevvio's Avatar
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    Thanks a lot plato
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  9. #9
    Senior Member slevvio's Avatar
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    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?
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  10. #10
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    Quote Originally Posted by slevvio View Post
    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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