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Math Help - 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 x-y>1 then \left( {\exists k \in Z} \right)\left[ {y < k < x} \right].
    Proof:
    By the floor function, \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 a<b then \left( {\exists K \in Z^ +  } \right)\left[ {\frac{1}<br />
{{b - a}} < K} \right].
    But that means that 1 < Kb - Ka so by the lemma \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 a<b then a\sqrt 2  < b\sqrt 2 so by the above \left( {\exists r \in Q} \right)\left[ {a\sqrt 2  < r < b\sqrt 2 } \right].
    But a < \frac{r}{{\sqrt 2 }} < b and \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 a < \frac{r}{{\sqrt 2 }} < b and \frac{r}{{\sqrt 2 }} is irrational.
    \frac{r}{\sqrt{2}} is rational if r=0.
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  5. #5
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    Quote Originally Posted by JaneBennet View Post
    \frac{r}{\sqrt{2}} is rational if r=0.
    True.
    But if a < 0 < b, then we can apply density of rationals to a\sqrt{2}<0 to obtain a 'r' such that 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 a<b then a\sqrt 2 < b\sqrt 2 so by the above \left( {\exists r \in Q} \right)\left[ {a\sqrt 2 < r < b\sqrt 2 } \right].

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

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

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