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$\sqrt a$ is either an integer or an irrational number.

I'm a total beginner and any help with this proof would be much appreciated. Not even sure where to begin.

Prove that for each prime number $p$, the square root of p is irrational.

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Welcome to math.SE. Have you seen any proofs of the fact that the square root of 2 is irrational? If so, try to generalize its ideas, and edit your post when you've got more questions. –  Ragib Zaman Jul 29 '12 at 11:11
    
If you're allowed to use the rational root theorem, consider the polynomial $x^2-p$... –  J. M. Jul 29 '12 at 11:22
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marked as duplicate by J. M., t.b., Gerry Myerson, Pete L. Clark, tomasz Aug 16 '12 at 17:22

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3 Answers

Suppose $\sqrt{p}=\frac{m}{n}$, where $m,n$ are relatively prime integers and $n\neq 0$. Then by squaring you get $n^2\cdot p=m^2$, so $p$ divides $m^2$. As $p$ is prime we must have $p^2$ divides $m^2$. On the other hand then $p^2$ divides $pn^2$ so $p$ divides $n$. Hence $m$ and $n$ are not relatively prime so we reached a contradiction.

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If $\sqrt p=\frac{a}{b}$ where (a,b)=1 and a>1 as p>1,

then $p=\frac{a^2}{b^2}$.

Now as p is integer, $b^2$ must divide $a^2$, which is impossible unless b=1 as (a,b)=1.

If b=1, p=$a^2$ which can not be prime as a>1.

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Hint $\rm\ gcd(m,n)\!=\!1,\: \frac{m}n =\! \sqrt{p}\,\Rightarrow m^2\!=\!pn^2\!.\,$ Euclid's Lemma $\rm\Rightarrow m^2|\:p\,\Rightarrow m^2\! = 1= pn^2\Rightarrow\Leftarrow$

Remark $\ $ This is a key step employed in one proof of uniqueness of squarefree decompositions, of which $\rm\:m^2 \ne pn^2\:$ is a special case.

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