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Math Help - Prove that 'a' is not element of Q (racional numbers)

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

    M = set
     M = \{x \in \mathbb{Q}   \mid (x \geq 0) \wedge (x^{2}< 2\}
    M \neq \emptyset

    Statement: This set does not have a supremum in \mathbb{Q}

    Prove with contradiction:
    Let's say there is an  a = sup M \in \mathbb{Q}

    Then we can assume each of the following 3 options:
    (1)  a^2 = 2
    (2)  a^2 < 2
    (3)  a^2 > 2

    a^2 = 2, can't be true because we know \sqrt{2} is not in \mathbb{Q} (won't prove that now..)

    Here's where I get stuck..

    a^2 < 2
    Let's say there is a h > 0  \ \ h \in \mathbb{Q} so that  (a + h)^2 < 2 \Rightarrow a + h \in M

    So we want that (a + h)^2 = a^2 + 2ah + h^2 < 2

    Then we say  0 < h < 1

    (a + h)^2 = a^2 + 2ah + h^2  < a^2 + h(2a +1) < 2

    This applies only if  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?
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  2. #2
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    Quote Originally Posted by metlx View Post
    M = set
     M = \{x \in \mathbb{Q}   \mid (x \geq 0) \wedge (x^{2}< 2\}
    M \neq \emptyset

    Statement: This set does not have a supremum in \mathbb{Q}

    Prove with contradiction:
    Let's say there is an  a = sup M \in \mathbb{Q}

    Then we can assume each of the following 3 options:
    (1)  a^2 = 2
    (2)  a^2 < 2
    (3)  a^2 > 2

    a^2 = 2, can't be true because we know \sqrt{2} is not in \mathbb{Q} (won't prove that now..)

    Here's where I get stuck..

    a^2 < 2
    Let's say there is a h > 0  \ \ h \in \mathbb{Q} so that  (a + h)^2 < 2 \Rightarrow a + h \in M

    So we want that (a + h)^2 = a^2 + 2ah + h^2 < 2

    Then we say  0 < h < 1

    (a + h)^2 = a^2 + 2ah + h^2  < a^2 + h(2a +1) < 2

    This applies only if  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 h<\frac{2-a^2}{2a+1} then (a+h)^2<2
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  3. #3
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    but why is 'a' less than 'h' then
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  4. #4
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    Quote Originally Posted by metlx View Post
    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. (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.
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  5. #5
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    so basically all i have to know is that such rational 'h' exists and it's smaller than  \frac{2 - a^2}{2a + 1}

    does this fraction tell me anything else?

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

    does this fraction tell me anything else?

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

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