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Math Help - x,y\in\mathbb{R} x<y, there exist an irrational between x and y

  1. #1
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    x,y\in\mathbb{R} x<y, there exist an irrational between x and y

    If x,y\in\mathbb{R}, x < y. Show that if, x and y are rational, then \exists an irrational number u, x < u < y.

    m,n,r,s\in\mathbb{Z}, \ n\neq 0, \ s\neq 0

    x=\frac{m}{n}, \ y=\frac{r}{s}

    \frac{m}{n}<u<\frac{r}{s}

    I am not sure what to do now.

    Contradiction, contrapositive??
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  2. #2
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    Here's a simple solution using basic set theory:

    The interval (x,y) is uncountable.
    Since there are only countably many rational numbers, there must be an irrational number in the interval.

    Note: this proof doesn't require x and y to be rational, only that x<y.
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    If x,y\in\mathbb{R}, x < y. Show that if, x and y are rational, then \exists an irrational number u, x < u < y.
    This is standard proof in any basic analysis course.
    But it depends upon this theorem: Between any two real numbers there is a rational number.

    In this case \left( {\exists r \in \mathbb{Q}} \right)\left[ {x\sqrt 2  < r < y\sqrt 2 } \right].

    So \frac{r}{\sqrt{2}} is between x~\&~y.
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  4. #4
    Senior Member Tinyboss's Avatar
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    While DrSteve's answer is very elegant and is how I'd answer it if asked, you may not have those concepts available to you yet. Since the sum of a rational and an irrational is irrational, it suffices to find an irrational number between 0 and y-x. Look at the "Archimedean property" of the reals. Is there an n such that \sqrt2/n<y-x?
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