3
$\begingroup$

Possible Duplicate:
$\sqrt a$ 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.

$\endgroup$

marked as duplicate by J. M. is a poor mathematician, t.b., Gerry Myerson, Pete L. Clark, tomasz Aug 16 '12 at 17:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ 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. $\endgroup$ – Ragib Zaman Jul 29 '12 at 11:11
  • $\begingroup$ If you're allowed to use the rational root theorem, consider the polynomial $x^2-p$... $\endgroup$ – J. M. is a poor mathematician Jul 29 '12 at 11:22
  • $\begingroup$ This shouldn't be a duplicate. Proof is simpler in this case where $p$ is prime. $\endgroup$ – 6005 Nov 19 '16 at 21:17
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The proof is correct. The downvotes are quite puzzling. If you have questions then please feel welcome to ask. $\endgroup$ – Bill Dubuque Jul 14 '15 at 18:45

Not the answer you're looking for? Browse other questions tagged or ask your own question.