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Thread: Proving irrationality

  1. #1
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    Proving irrationality

    Hi,

    Can anyone help with this one:
    Establish the following facts:
    √p is irrational for any prime p.
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  2. #2
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    Quote Originally Posted by lyla View Post
    Hi,

    Can anyone help with this one:
    Establish the following facts:
    √p is irrational for any prime p.
    Euclid's proof.
    --------------
    Assume,
    $\displaystyle \sqrt{p}=a$ a positive integer.
    By definition it is equivalent to say,
    $\displaystyle p=a^2$.
    Now the prime decomposition of $\displaystyle a^2$ has an even amount of prime factors because of the square. While $\displaystyle p$ does not. By uniquness this is impossible.

    Pythagorus' Proof.
    -------------------
    Assume that $\displaystyle \sqrt{p}=\frac{n}{m}$ where $\displaystyle n,m$ is a reduced fraction, meaning no common factors.
    Then,
    $\displaystyle p=\frac{n^2}{m^2}$
    $\displaystyle m^2p=n^2$.
    Note, the right hand side is divisible by $\displaystyle p$ because the left hand side. Meaning $\displaystyle p$ divides $\displaystyle n^2$. But then $\displaystyle n$ itself is divisible by $\displaystyle p$ by properties of prime numbers. That is $\displaystyle n=pk$.
    Subsitute,
    $\displaystyle m^2p=k^2p^2$
    $\displaystyle m^2=k^2p$
    But then the left hand side is divisble by $\displaystyle p$, that is $\displaystyle o$ divides $\displaystyle m^2$. But then $\displaystyle p$ divides $\displaystyle m$. Hence $\displaystyle n$ and $\displaystyle n$ have common factors, contrary to assumption.
    (This is also a similar approach to Fermat's principle of infinite descent.)

    Eisenstein Proof.
    -----------------
    The polynomial $\displaystyle x^2-p\in \mathbb{Z}[x]$ fits the conditions of Eisenstein irreducibility criterion for $\displaystyle p$. Thus, there are no solution in $\displaystyle \mathbb{Z}$ and hence none in $\displaystyle \mathbb{Q}$.

    Rational Roots Proof.
    ----------------
    The polynomial $\displaystyle x^2-p$ can only have zeros for $\displaystyle \pm 1,\pm p$. None of which work. Thus, it is irrational.
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  3. #3
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    I have to thank you very much for the help.
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