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Math Help - Proof by Construction.

  1. #1
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    Proof by Construction.

    I need to prove using the method of proof by construction that: Between every two distinct rational numbers lies an irrational number.

    I started my proof assuming that x and y are 2 rational numbers such that x<y.

    Since x and y are rational

    Therefore \exists_a, _b, _c, _d\in Z with   b and  d not = 0 such that x = \frac{a}{b} and y = \frac{c}{d}

    I need to construct an element q that can be between x and y and that has the properties of an irrational number. I thought about using the geometric mean of the numbers x and y: \sqrt{xy} but the geometric mean can be a rational number or an irrational number. I don't know if I have to prove this element q to be irrational all the time.

    Any assistance would be greatly appreciated, thank you.
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  2. #2
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    Suppose that \left\{ {a,b} \right\} \subseteq \mathbb{Q}\,\& \,a < b it follows that \sqrt 2 a < \sqrt 2 b.
    From that it follows  \left( {\exists r \in \mathbb{Q}} \right)\left[ {\sqrt 2 a < r < \sqrt 2 b} \right].
    Can you finish?
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  3. #3
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    When you say that \left\{ {a,b} \right\} \subseteq \mathbb{Q}\,\& \,a < b you are referring to two different numbers then when i say that x = \frac{a}{b} right?

    Quote Originally Posted by Plato View Post
    Suppose that \left\{ {a,b} \right\} \subseteq \mathbb{Q}\,\& \,a < b it follows that \sqrt 2 a < \sqrt 2 b.
    From that it follows  \left( {\exists r \in \mathbb{Q}} \right)\left[ {\sqrt 2 a < r < \sqrt 2 b} \right].
    Can you finish?
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  4. #4
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    And if that was the case, would you just have to divide each number by \sqrt{2} so that you would have a<\frac{r}{\sqrt{2}}<b and since \frac{r}{\sqrt{2}} is an irrational number the construction is complete?
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  5. #5
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    Quote Originally Posted by xwanderingpoetx View Post
    since \frac{r}{\sqrt{2}} is an irrational number the construction is complete?
    Well of course!
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