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Thread: There is a rational and an irrational number in any interval

  1. #1
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    There is a rational and an irrational number in any interval

    Prove that in any interval there is a rational and an irrational number.
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  2. #2
    Super Member Aryth's Avatar
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    Dr. Math: Go here

    This is a proof of just what you are asking.
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  3. #3
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    Lemma: If $\displaystyle x-y>1$ then $\displaystyle \left( {\exists k \in Z} \right)\left[ {y < k < x} \right]$.
    Proof:
    By the floor function, $\displaystyle \left\lfloor y \right\rfloor \leqslant y < \left\lfloor y \right\rfloor + 1\,\mbox{or}\,\left\lfloor y \right\rfloor + 1 \leqslant y + 1 < x \Rightarrow \quad y < \left\lfloor y \right\rfloor + 1 < x$.


    Between any two numbers there is a rational number.
    Proof:
    If $\displaystyle a<b$ then $\displaystyle \left( {\exists K \in Z^ + } \right)\left[ {\frac{1}
    {{b - a}} < K} \right]$.
    But that means that $\displaystyle 1 < Kb - Ka$ so by the lemma $\displaystyle \left( {\exists J \in Z} \right)\left[ {Ka < J < Kb} \right] \Rightarrow \quad a < \frac{K}{J} < b$.

    Between any two numbers there is a irrational number.
    Proof:
    If $\displaystyle a<b$ then $\displaystyle a\sqrt 2 < b\sqrt 2 $ so by the above $\displaystyle \left( {\exists r \in Q} \right)\left[ {a\sqrt 2 < r < b\sqrt 2 } \right]$.
    But $\displaystyle a < \frac{r}{{\sqrt 2 }} < b$ and $\displaystyle \frac{r}{{\sqrt 2 }}$ is irrational .
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  4. #4
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Plato View Post
    But $\displaystyle a < \frac{r}{{\sqrt 2 }} < b$ and $\displaystyle \frac{r}{{\sqrt 2 }}$ is irrational.
    $\displaystyle \frac{r}{\sqrt{2}}$ is rational if $\displaystyle r=0$.
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  5. #5
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    Quote Originally Posted by JaneBennet View Post
    $\displaystyle \frac{r}{\sqrt{2}}$ is rational if $\displaystyle r=0$.
    True.
    But if a < 0 < b, then we can apply density of rationals to $\displaystyle a\sqrt{2}<0$ to obtain a 'r' such that $\displaystyle a\sqrt{2} < r < 0$......
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  6. #6
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Plato View Post
    Between any two numbers there is a irrational number.

    Proof:

    If $\displaystyle a<b$ then $\displaystyle a\sqrt 2 < b\sqrt 2 $ so by the above $\displaystyle \left( {\exists r \in Q} \right)\left[ {a\sqrt 2 < r < b\sqrt 2 } \right]$.

    But $\displaystyle a < \frac{r}{{\sqrt 2 }} < b$ and $\displaystyle \frac{r}{{\sqrt 2 }}$ is irrational .
    Quote Originally Posted by JaneBennet View Post
    $\displaystyle \frac{r}{\sqrt{2}}$ is rational if $\displaystyle r=0$.
    Quote Originally Posted by Isomorphism View Post
    True.
    But if a < 0 < b, then we can apply density of rationals to $\displaystyle a\sqrt{2}<0$ to obtain a 'r' such that $\displaystyle a\sqrt{2} < r < 0$......
    Bah! The last part should be done this way.

    If $\displaystyle a<b$ then $\displaystyle a+\sqrt 2 < b+\sqrt 2 $ so by the above $\displaystyle \left( {\exists r \in \mathbb{Q}} \right)\left[ {a+\sqrt 2 < r < b+\sqrt 2 } \right]$.

    But $\displaystyle a < r-\sqrt 2 < b$ and $\displaystyle r-\sqrt 2$ is irrational.
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