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Math Help - prove Q" =R

  1. #1
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    prove Q" =R

    Prove that all the accumulation points of the rationals are the Real Nos.

    We know from a theorem in Analysis that:

    Given a real No x and ε>0 there exists a rational No y such that :

    |x-y|<\epsilon and y\in Q

    or y\in (x-\epsilon,x+\epsilon) and y\in Q

    or (x-\epsilon,x+\epsilon)\cap Q\neq\emptyset

    Showing that y is a point of closure

    To show now that y is a point of accumulation we must additionaly show that

    y\neq x

    How do we show that??
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  2. #2
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    Quote Originally Posted by alexandros View Post
    Prove that all the accumulation points of the rationals are the Real Nos. How do we show that??
    Between any two numbers there is a rational number.

    \left( {\exists r} \right)\left[ {r \in \mathbb{Q} \cap \left( {x - \varepsilon ,x} \right)} \right] so r\not= x.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Between any two numbers there is a rational number.

    \left( {\exists r} \right)\left[ {r \in \mathbb{Q} \cap \left( {x - \varepsilon ,x} \right)} \right] so r\not= x.
    But given r and x if x-ε<r for all ε>0 ,then x\leq r .

    Suppose the opposite and x>r ,then put ε = x-r ,and x-(x-r)<r => 0<0 :contradiction

    Hence x can be equal to r
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  4. #4
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    Quote Originally Posted by alexandros View Post
    But given r and x if x-ε<r for all ε>0 ,then x\leq r .
    Suppose the opposite and x>r ,then put ε = x-r ,and x-(x-r)<r => 0<0 :contradiction
    Hence x can be equal to r
    You are seriously confusing two different concepts.
    If a~\&~b are any two real numbers then there is a rational number between them.

    The proper translation of 'between' implies that it is distinct from both.
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  5. #5
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    Quote Originally Posted by Plato View Post
    You are seriously confusing two different concepts.
    If a~\&~b are any two real numbers then there is a rational number between them.

    The proper translation of 'between' implies that it is distinct from both.
    .

    You are not listening to what i am saying and writing.

    PLease read the whole thread again ,if you are really want to help and not just force your ideas whether wrong or right.

    Start with any theorem you like and write a complete proof showing Q' = R,IF YOU wish to do so of course.

    Writing parts of the proof make me more confuse.

    The versions of the theorem concerning reals and rationals or irrationals that i know are:

    1)Given any two girls a<b ,there exist a rational or irrational y, such that: a<y<b

    2) Given any real x ,then for any ε>0 there exist a rational or irrational ,y such that

    |x-y|<ε.

    None of them proves that : Q' =R
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  6. #6
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    Quote Originally Posted by alexandros View Post
    .You are not listening to what i am saying and writing.
    PLease read the whole thread again ,if you are really want to help and not just force your ideas whether wrong or right.

    Start with any theorem you like and write a complete proof showing Q' = R,IF YOU wish to do so of course.
    Writing parts of the proof make me more confuse.
    The versions of the theorem concerning reals and rationals or irrationals that i know are:
    1)Given any two girls a<b ,there exist a rational or irrational y, such that: a<y<b
    2) Given any real x ,then for any ε>0 there exist a rational or irrational ,y such that
    |x-y|<ε.
    None of them proves that : Q' =R
    I beg your pardon.
    I taught this material for forty years.
    I know exactly what you are saying and how really confused you are.

    Every real number is a limit point of the set of rational numbers.
    Here is the proof. Suppose that \gamma\in\mathbb{R}.
    The is a rational number r_1\in (\gamma -1, \gamma ) & r_1>\gamma -1~\&~r_1\not=\gamma
    The is a rational number r_2\in (\max\{\gamma-\frac{1}{2},r_1\}, \gamma ) & r_2>r_1~\&~r_2\not=\gamma

    If n\ge 3 the is a rational number  r_n\in (\max\{\gamma-\frac{1}{n},r_{n-1}\}, \gamma ) & r_n> r_{n-1}~\&~r_n\not=\gamma

    Now the sequence \left( {r_n } \right) \to \gamma
    So \gamma\in\mathbb{Q}^{\prime} &  \mathbb{Q}^{\prime} =\mathbb{R}
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  7. #7
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    Quote Originally Posted by Plato View Post
    I beg your pardon.
    I taught this material for forty years.
    I know exactly what you are saying and how really confused you are.

    Every real number is a limit point of the set of rational numbers.
    Here is the proof. Suppose that \gamma\in\mathbb{R}.
    The is a rational number r_1\in (\gamma -1, \gamma ) & r_1>\gamma -1~\&~r_1\not=\gamma
    The is a rational number r_2\in (\max\{\gamma-\frac{1}{2},r_1\}, \gamma ) & r_2>r_1~\&~r_2\not=\gamma

    If n\ge 3 the is a rational number  r_n\in (\max\{\gamma-\frac{1}{n},r_{n-1}\}, \gamma ) & r_n> r_{n-1}~\&~r_n\not=\gamma

    Now the sequence \left( {r_n } \right) \to \gamma
    So \gamma\in\mathbb{Q}^{\prime} &  \mathbb{Q}^{\prime} =\mathbb{R}
    I wonder what axiom or theorem supports such sequence construction.
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  8. #8
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    Quote Originally Posted by alexandros View Post
    I wonder what axiom or theorem supports such sequence construction.
    Do you understand that between any two numbers there is a rational number?
    That is a theorem.

    If R\in\mathbb{R} then there is a rational number r_1 such that R-1<r_1<R.

    Thus there is a rational number r_2 such that r_1<r_2<R.

    Keep the up.
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  9. #9
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    Quote Originally Posted by Plato View Post

    Keep the up.
    .

    There is no theorem or axiom to justify the " Keep the up" process uppon which the constraction of your sequence is based.
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  10. #10
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    Alexandros: Please don't badger the contributors; we are fortunate to have their input and rely on their help. If you don't like what someone has to say, just move on. In my opinion, Plato seems very experienced and provided thoughtful input to you before the thread went personal.
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