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Math Help - Proof. Real number line.

  1. #1
    Member Greengoblin's Avatar
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    Proof. Real number line.

    Is there some theorem that shows that rational and irrational numbers together can be used to completely occupy a continuous infinite line? i.e. that the real numbers are the union of the two sets.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Greengoblin View Post
    Is there some theorem that shows that rational and irrational numbers together can be used to completely occupy a continuous infinite line? i.e. that the real numbers are the union of the two sets.
    Since the irrationals are defined to be those real numbers that are not rational, we are forced to the conclusion that the deals are the union of the rationals and irrationals.

    CB
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  3. #3
    Member Greengoblin's Avatar
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    Ok, but how can we show that the real numbers completely occupy a continuous infinite line? For example my book says:

    "It can be shown that these numbers can be used to 'label' every point on a continuous infinite line - the real line"
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Greengoblin View Post
    Ok, but how can we show that the real numbers completely occupy a continuous infinite line? For example my book says:

    "It can be shown that these numbers can be used to 'label' every point on a continuous infinite line - the real line"
    I am misunderstanding? It is as it is because that is how we define it. Let \mathbb{R} be a universal set and \mathbb{Z}\subset\mathbb{R}. Then define \mathbb{Q}\equiv\left\{x:\exists{a,b}\in\mathbb{Z}  \backepsilon~x=\frac{a}{b}\right\}, and then define \mathbb{I}\equiv\left\{x:\not\exists{a,b}\in\mathb  b{Z}\backepsilon~x=\frac{a}{b}\right\}\implies\mat  hbb{I}=\mathbb{Q}^c

    So by defintion of complement and universal set

    \mathbb{Q}\cup\mathbb{I}=\mathbb{Q}\cup\mathbb{Q}^  c=\mathbb{R}
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  5. #5
    Member Greengoblin's Avatar
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    OK, I understand that, but the book threw me off a bit when it said "it can be shown", as if it required an actual proof rather than just a definition. I think they should have made it clear.
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  6. #6
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    Slight correction:


    I am misunderstanding? It is as it is because that is how we define it. Let \mathbb{R} be a universal set and \mathbb{Z}\subset\mathbb{R}. Then define \color{red}\mathbb{Q}\equiv\left\{x:\exists{a,b}\b  ackepsilon~b\neq0\in\mathbb{Z}\backepsilon~x=\frac  {a}{b}\right\}, and then define \color{red}\mathbb{I}\equiv\left\{x:\not\exists{a,  b}\backepsilon~b\neq0\in\mathbb{Z}\backepsilon~x=\  frac{a}{b}\right\}\implies\mathbb{I}=\mathbb{Q}^c

    So by defintion of complement and universal set

    \mathbb{Q}\cup\mathbb{I}=\mathbb{Q}\cup\mathbb{Q}^  c=\mathbb{R}
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