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Math Help - 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 \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 \sqrt{p} \in \mathbb{Q}. Thus \sqrt{p}=\frac{m}{n}, where m,n \in \mathbb{N} and this fraction is in lowest terms.

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

    Now we look back at our second statement, p=\frac{m^2}{n^2}. Substituting in a new expression for n^2 yields 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 \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 \sqrt{p} is irrational.

    The proof in outline form goes like this.
    Suppose otherwise. Then there is a smallest positive integer, K, such K\sqrt{p}\in \mathbb{Z}^+.
    0 < \sqrt p - \left\lfloor {\sqrt p } \right\rfloor < 1\, \Rightarrow \,0 < K\sqrt p - K\left\lfloor {\sqrt p } \right\rfloor   < K
    But \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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