Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

When a prime number $p$ divides $ ab $ then we have either $p$ divides $a$ or $p$ divides $b$. Prove that $ \sqrt p $ is not rational for any prime number $p$.

share|improve this question
    
I think OP meant to use Euclide's Lemma, not prove it. –  Ittay Weiss Mar 4 '13 at 7:52
    
@gnometorule only if my interpretation is correct ;) –  Ittay Weiss Mar 4 '13 at 7:55
add comment

3 Answers

Assume that $\sqrt p=\frac{s}{t}$, with $gcd(s,t)=1$. Then $pt^2=s^2$ and thus $p$ divides the right hand side (since it clearly divides the left hand side), so $p$ divides $s^2$. Using Euclid's lemma it follows that $p$ divides $s$. Write $s=pk$, and thus $pt^2=p^2k^2$. Divide by $p$ to obtain $t^2=pk^2$. Now $p$ divides the right hand side, so it also divides $t^2$. Euclid's lemma again shows that $p$ divides $t$, but this is a contradiction since we assumed $gcd(s,t)=1$. QED.

share|improve this answer
add comment

By induction, the given Prime Divisor Property (PDP): $ $ prime $\rm\,p\,$ divides $\rm\,ab\,\Rightarrow\, p\,$ divides $\rm\,a\,$ or $\rm\,b\,$ generalizes to $\,\rm\color{blue}{PDP^*}$: $ $ if a prime divides a product then it divides some factor.

If $\rm\, \sqrt{p}\, =\, r\,$ is rational then the polynomial $\rm\:x^2\! - p\:$ has a rational root $\rm\,x = r.\,$ Thus, by the Lemma below, the root is an integer $\rm\, x = n,\,$ hence $\rm\: p = x^2 = n^2,\:$ contra $\rm\,p\,$ is prime. $\ \ $ QED

Lemma $\ $ A rational root of $\rm\,f(x) = x^n\! + c_{n-1}x^{n-1}\!+\cdots+c_0\,$ is an integer, if all $\rm\,c_i$ are integers.

Proof $\ $ Suppose $\rm\:f(a/b) = 0.\:$ Wlog $\rm\:b>0\:$ and $\rm\:a/b\:$ is in least terms, so $\rm\:gcd(a,b)=1.\,$ Then $$\rm\: 0\, =\, b^n f(a/b) =\, a^n + c_{n-1} a^{n-1}\color{#C00}b+c_{n-2} a^{n-2} \color{#C00}{b^2} +\cdots+c_1 a\, \color{#C00}{b^{n-1}}\! + c_0 \color{#C00}{b^n}\:$$ If $\rm\:b>1\:$ then it has a prime factor $\rm\:p.\:$ Above, $\rm\,p\,$ divides the LHS $= 0,$ so it also divides the RHS. Since $\rm\,p\,$ divides $\rm\,\color{#C00}b\,$ it divides $\rm\,\color{#C00}{b^2,b^3,\ldots}\,$ so $\rm\,p\,$ divides all $\rm\color{#C00}{terms}$ after the first on the RHS, therefore $\rm\,p\,$ must also divide the first term $\rm\:a^n.\,$ Therefore, by $\rm\color{blue}{PDP^*}$, $\rm\,p\,$ divides $\rm\,a,\,$ contra $\rm\:gcd(a,b)=1.\:$ Thus $\rm\,b>1\,$ is impossible, so, since $\rm\,b>0,\,$ we infer $\rm\,b = 1,\:$ hence $\rm\, a/b\,$ is an integer. $\ \ $ QED

Remark $\ $ The Lemma is a special case of the well-known Rational Root Test, namely, the case where the polynomial is monic, i.e. it has leading coefficient $= 1.$ The proof easily extends to the general result: the denominator $\rm\,b\,$ divides the leading coefficient $\rm\,c_n,\,$ if $\rm\,a/b\,$ is in least terms.

share|improve this answer
add comment

Assume that $\sqrt p$ is a rational number $r/s$. We may assume that $r$ and $s$ are relatively prime integers.

Then we have $p=r^2/s^2$. Since $p$ is an integer and $r$ and $s$ are relatively prime, $s=1$ or $s=-1$. Then $p=r^2$.

But this contradicts that $p$ is a prime number.

share|improve this answer
1  
This proof is incorrect. –  Ittay Weiss Mar 4 '13 at 7:40
    
@IttayWeiss What is wrong with it? –  Ragib Zaman Mar 4 '13 at 7:45
    
@IttayWeiss Why?? Maybe I should have said that $s^2p=r^2$ and $p$ divides $r$. Then we see $p$ divides $s$. This contradicts the assumption that $r$ and $s$ are relatively prime. –  Primo Mar 4 '13 at 7:46
1  
@Primo yes, since OP was looking for a solution using Euclid's lemma, I think these things need to be spelled out (or at least pointed out). And when I remarked the proof is wrong, I should have said 'incomplete', I'm sorry. –  Ittay Weiss Mar 4 '13 at 7:52
    
Your argument seems to implicity use: $\, r,s\,$ coprime $\,\Rightarrow\ r^2,s^2\,$ coprime. That should be explicitly mentioned, and, more importantly, justified. –  Math Gems Mar 4 '13 at 17:41
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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