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.

I have to prove that $\sqrt 5$ is irrational.

Proceeding as in the proof of $\sqrt 2$, let us assume that $\sqrt 5$ is rational. This means for some distinct integers $p$ and $q$ having no common factor other than 1,

$$\frac{p}{q} = \sqrt5$$

$$\Rightarrow \frac{p^2}{q^2} = 5$$

$$\Rightarrow p^2 = 5 q^2$$

This means that 5 divides $p^2$. This means that 5 divides $p$ (because every factor must appear twice for the square to exist). So we have, $p = 5 r$ for some integer $r$. Extending the argument to $q$, we discover that they have a common factor of 5, which is a contradiction.

Is this proof correct?

share|improve this question
    
See this –  Mahdi Khosravi Jul 25 '13 at 7:18
    
Yes, it is correct. –  DonAntonio Jul 25 '13 at 7:28
    
Forgive my ignorance, but isn't it a little simpler than this? Here's my argument. Since $2^2 < (\sqrt{5})^2 < 3^2,$ and since the positive square root function is strictly increasing, thus $2 < \sqrt{5} < 3.$ Since there are not natural numbers between $2$ and $3$, this means that $\sqrt{5}$ is non-natural. But, I think that, if the square root of a natural number is rational, then its square root is natural. The contrapositive is that if its square root is non-natural, then its square root is non-rational. So $\sqrt{5}$ is irrational. –  goblin Jul 25 '13 at 10:05
1  
@user18921: I believe the issue with your argument is that it relies on the implicit assumption that $\sqrt 5$ "exists" and can be meaningfully compared to 2 and 3. (That is, in what set is $\sqrt 5$ supposed to exist?) –  Jean Hominal Jul 25 '13 at 10:48
1  
@user18921: "if the square root of a natural number is rational, then its square root is natural. The contrapositive is that if its square root is non-natural, then its square root is non-rational" -- yes, if you prove this, then the irrationality of $\sqrt{5}$ follows. But it takes a bit more work to prove your statement than to prove the special case, that $\sqrt{5}$ is irrational. –  ShreevatsaR Jul 25 '13 at 10:51

4 Answers 4

up vote 9 down vote accepted

It is, but I think you need to be a little bit more careful when explaining why $5$ divides $p^2$ implies $5$ divides $p$. If $4$ divides $p^2$ does $4$ necessarily divide $p$?

share|improve this answer
    
In that case I'd say that 2 divides $p$, because 2 divides $4$. Funnily, the statement $4 \, | \, p^2 \Rightarrow 4 \, | \, p$ also seems to hold except for $p=2$. What is the finer detail I'm missing? –  dotslash Jul 25 '13 at 7:28
1  
@dotslash The implication ($p$ divides $ab$ $\implies$ $p$ divides $a$ or $p$ divides $b$) holds if $p$ is prime. In particular, the implication ($p$ divides $a^2$ $\implies$ $p$ divides $a$) holds if $p$ is prime. In fact, this property (the one described in the first sentence of this comment) is sometimes taken to be the definition of a prime number. Can you prove this property using the definition of "prime number" with which, I assume, you're familiar (i.e., a positive integer $p$ is prime if and only if it has two distinct positive divisors: $1$ and $p$ itself)? –  Amitesh Datta Jul 25 '13 at 7:31
    
Your conclusion that $2$ divides $p$ is the correct one. The implication $a\mid p^2 \Rightarrow a\mid p$ for all integers $p$ is only true if $a$ is a product of distinct primes, which $4$ is not. There are cases with $p \neq 2$ where $4\mid p^2$ but $4\nmid p$, they occur precisely when $p = 2l$ for some odd number $l$. –  Michael Albanese Jul 25 '13 at 7:45
    
The easiest solution may be via the Fundamental theorem of arithmetic. Both p and q have a unique prime factorization. From that it follows that p² and q² have unique prime factorizations too, and each power is even. That includes the power of 5. It also means that 5q² has an odd power of 5, which contradicts the assumption that 5q²=p² has an even power of 5. –  MSalters Jul 25 '13 at 13:44

Yes, the proof is correct. Using this method you can show that $\sqrt{p}$ for any prime $p$ is irrational. During the proof you essentially use the fact that when $p|u^2$ where $p$ is a prime, then it implies that $p|u$. This is true for primes, but is not true in general. You can prove this as below Let $n|u^2,\ \gcd(n,u)=d$. Then, let $n=rd,\ u=sd $. So, $$u^2=kn \Rightarrow s^2d^2=k r d\Rightarrow s^2d=kr$$ if we have $n\not{|}\ u$ then $\gcd(s,r)=1$, and hence $$r|d$$ Then, with $d>1$, $n\not{|}\ u$, but $\ n|u^2$. If $n$ is prime, then $d=1\Rightarrow r=1$ unless $\ n|u$.

share|improve this answer
    
I just cannot wrap my head around this argument. One has $\gcd(s,r)=1$ regardless of whether $n$ divides $u$, just by factoring $d$ from $d=\gcd(rd,sd)$. But it beats me how this is instrumental in proving $r\mid d$, unless it is by: $r$ divides $kr=s^2d$ and since $r,s$ relatively prime $r$ must divide $d$. But this uses "if $\gcd(r,s)=1$ and $r\mid sa$ then $r\mid a$", which already generalises Euclid's lemma you are trying to prove. Using it, you could say "for $p$ prime if $p\mid u^2$ but $p\nmid u$ then $\gcd(p,u)=1$ so by the above $p\mid u$ (the other factor of $u^2$), a contradiction". –  Marc van Leeuwen Oct 30 '13 at 11:19

Yes, the proof is correct. But I think you still need a lemma to reinforce your proof

Lemma: $$\text{If }P|Q^2,\text{ where P is a prime, then } P|Q$$

Proof: By the unique factorization theorem,$Q$ is able to rewrited as a product of distinct prime numbers: $$Q = P_1^{e_1}P_2^{e_2}P_3^{e_3}\ldots P_k^{e_k}\tag{1}$$ where $P_1,P_2,\ldots P_k$ are distinct prime numbers and $e_1,e_2,\ldots e_k$ are positive integers. Then: $$Q^2 = P_1^{2e_1}P_2^{2e_2}P_3^{2e_3}\ldots P_k^{2e_k}\tag{2}$$ By (1),(2), we know both $Q$ and $Q^2$ are a product of distinct prime numbers that belong to the set $\{P_1,P_2,\ldots,P_k\}$. Because $P|Q^2$ and $P$ is also a prime, It implies $P\in\{P_1,P_2,\ldots,P_k\}$. Hence, P|Q, which complete the proof.

share|improve this answer
    
Using the Unique Factorization Theorem to prove this lemma is putting the cart before the horse: a somewhat more general statement then the lemma (namely Euclid's lemma: if a prime divides a product, it divides at least one of the factors) is a virtually inevitable preliminary result used in proving the UFT. –  Marc van Leeuwen Oct 30 '13 at 10:21
    
@MarcvanLeeuwen Thanks for reminding, I have never thought about that :). I just wanted to provide another way to prove this lemma when I was writing this answer. –  sundaycat Oct 30 '13 at 17:33

The number of prime divisors of $p^2$ is even. Is that true for $5q^2$?

share|improve this answer

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.