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Thread: Check my proof that sqrt{prime} is always irrational

  1. #1
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    Check my proof that sqrt{prime} is always irrational

    The proof that $\displaystyle \sqrt{2}$ is irrational is well-known and I never did the proof that extended this to all primes, so I will attempt this now.

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    Assume towards a contradiction that $\displaystyle \sqrt{p} \in \mathbb{Q}$. Thus $\displaystyle \sqrt{p}=\frac{m}{n}$, where $\displaystyle m,n \in \mathbb{N}$ and this fraction is in lowest terms.

    By algebra, $\displaystyle p=\frac{m^2}{n^2} \Rightarrow n^2=\frac{m^2}{p}$. If $\displaystyle n^2$ is not an integer, then clearly n is not an integer. This isn't possible by definition of n, so $\displaystyle n^2$ must be an integer.

    Given this conclusion, p divided m^2 evenly. This could lead to two possibilities. (1) m=p and (2)m contains a factor of p. Since p is prime the only factor other than 1 is p itself. So m can be written as $\displaystyle kp$, where $\displaystyle k \in \mathbb{N}$. If (1) is the case, then m/n is not in lowest terms by out initial assumption, thus cannot be true. If (2) is the case then $\displaystyle n^2=\frac{k^2p^2}{p}=k^2p$.

    Now we look back at our second statement, $\displaystyle p=\frac{m^2}{n^2}$. Substituting in a new expression for n^2 yields $\displaystyle p=\frac{m^2}{k^2p}$. This implies that m and n have a common factor and thus possibility (2) cannot work. Since (1) and (2) are not possible, the initial assumption is false and $\displaystyle \sqrt{p} $ is not in Q.

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    I just read through it and I think it's ok. I'm sure there are ways to shorten it and I'm sure I didn't explain some things well enough. The big thing I think might get criticized is justifying when I say two things have a common factor. Anyway, let me know how I did and how I can improve.

    Thanks
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  2. #2
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    This is an even more general result.
    If p is a non-square positive integer then $\displaystyle \sqrt{p}$ is irrational.

    The proof in outline form goes like this.
    Suppose otherwise. Then there is a smallest positive integer, K, such $\displaystyle K\sqrt{p}\in \mathbb{Z}^+$.
    $\displaystyle 0 < \sqrt p - \left\lfloor {\sqrt p } \right\rfloor < 1\, \Rightarrow \,0 < K\sqrt p - K\left\lfloor {\sqrt p } \right\rfloor < K$
    But $\displaystyle \left( {K\sqrt p - K\left\lfloor {\sqrt p } \right\rfloor } \right) \in \mathbb{Z}^ + \;\& \,\left(K\sqrt p - {K\left\lfloor {\sqrt p } \right\rfloor } \right)\sqrt p \in \mathbb{Z}^ +$
    That violates the minimal nature of K.
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  3. #3
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    Interesting, Plato. Thanks.

    Did my proof look ok in general though?
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