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

• Sep 17th 2009, 12:44 PM
Jameson
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
• Sep 17th 2009, 01:05 PM
Plato
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.
• Sep 17th 2009, 01:08 PM
Jameson
Interesting, Plato. Thanks.

Did my proof look ok in general though?