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Math Help - Proof that in any open interval there is infinitley many irrationals

  1. #1
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    Proof that in any open interval there is infinitley many irrationals

    Is this proof ok?


    If we take the open interval (a,b) and shift it (ax,bx). Where x=\sqrt(p) where p is a prime.

    Then we have that as the rationals are dense in the reals ( we can have a rational arbitrarly close to any real) we have:

    ax<q<bx\ \ s.t. q\in\mathbb{Q}\rightarrow a<\frac{q}{x}<b.

    Therefore as \frac{q}{x} and we have \frac{q}{x}\in(a,b)
    we have an irrational in any open interval.

    Now as there are infinitely many primes we have infinitley many irrationals of the form \sqrt(p), so we have infinitley many irrationals in the interval.

    thanks for any help.
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  2. #2
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    Re: Proof that in any open interval there is infinitley many irrationals

    Quote Originally Posted by hmmmm View Post
    If we take the open interval (a,b) and shift it (ax,bx). Where x=\sqrt(p) where p is a prime.
    Then we have that as the rationals are dense in the reals ( we can have a rational arbitrarly close to any real) we have:
    ax<q<bx\ \ s.t. q\in\mathbb{Q}\rightarrow a<\frac{q}{x}<b.
    Therefore as \frac{q}{x} and we have \frac{q}{x}\in(a,b)
    we have an irrational in any open interval.
    Now as there are infinitely many primes we have infinitley many irrationals of the form \sqrt(p), so we have infinitley many irrationals in the interval.
    Here is one comment. Let \mathcal{I}=\mathbb{R}\setminus\mathbb{Q}
    You have proven that any open interval contains an irrational.
    Once you find i_1\in (a,b)\cap\mathcal{I} then find i_2\in (a,i_1)\cap\mathcal{I}.
    If n\ge 3 find i_n\in (a,i_{n-1})\cap\mathcal{I}.
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  3. #3
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    Re: Proof that in any open interval there is infinitley many irrationals

    Yeah thanks for the help there, thought that it was a bit poor at the end there thanks.
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    Re: Proof that in any open interval there is infinitley many irrationals

    Quote Originally Posted by hmmmm View Post
    Is this proof ok?


    If we take the open interval (a,b) and shift it (ax,bx). Where x=\sqrt(p) where p is a prime.

    Then we have that as the rationals are dense in the reals ( we can have a rational arbitrarly close to any real) we have:

    ax<q<bx\ \ s.t. q\in\mathbb{Q}\rightarrow a<\frac{q}{x}<b.

    Therefore as \frac{q}{x} and we have \frac{q}{x}\in(a,b)
    we have an irrational in any open interval.

    Now as there are infinitely many primes we have infinitley many irrationals of the form \sqrt(p), so we have infinitley many irrationals in the interval.

    thanks for any help.
    Another possibility would be to note that if there were finitely many irrationals in (a,b) then (a,b) would be countable...but I'm sure you this isn't true.
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