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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?

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    $\begingroup$ See this $\endgroup$ Jul 25, 2013 at 7:18
  • $\begingroup$ Yes, it is correct. $\endgroup$
    – DonAntonio
    Jul 25, 2013 at 7:28
  • $\begingroup$ 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. $\endgroup$ Jul 25, 2013 at 10:05
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    $\begingroup$ @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?) $\endgroup$ Jul 25, 2013 at 10:48
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    $\begingroup$ @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. $\endgroup$ Jul 25, 2013 at 10:51

5 Answers 5

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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 $p$?

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  • $\begingroup$ 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? $\endgroup$
    – ankush981
    Jul 25, 2013 at 7:28
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    $\begingroup$ @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)? $\endgroup$ Jul 25, 2013 at 7:31
  • $\begingroup$ 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 $l$. $\endgroup$ Jul 25, 2013 at 7:45
  • $\begingroup$ 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. $\endgroup$
    – MSalters
    Jul 25, 2013 at 13:44
  • $\begingroup$ @MichaelAlbanese Since 6 is a product of distinct primes, $ 6 \vert a^2 \implies 6 \vert a $ holds true? $\endgroup$ Feb 7, 2021 at 7:01
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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 $n\not{|}\ u$, since $\gcd(s,r)=1$, we have $$r|d$$ Then, with $d>1$, $n\not{|}\ u$, but $\ n|u^2$. If $n$ is prime, then $d=1\Rightarrow r=1$ unless $\ n|u$.

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  • $\begingroup$ 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". $\endgroup$ Oct 30, 2013 at 11:19
  • $\begingroup$ @MarcvanLeeuwen, sorry for my inappropriate use of the preposition "then", I meant to say "since", let me correct it. $\endgroup$ Jul 16, 2015 at 18:07
  • $\begingroup$ what is the meaning of this symbol: $\not{|}$ $\endgroup$
    – user25849
    Sep 21, 2020 at 10:10
  • $\begingroup$ @user25849, it means "does not divide". $\endgroup$ Sep 24, 2020 at 8:02
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Let us assume $√5$ is rational.

$$√5=\frac{x}{y}$$ Square both sides of the equation above

$$5 =\frac{x^2}{y^2}$$

Multiply both sides by $y^2$

$$5 y^2 =\frac{x^2 }{y^2}$$

We get $5 y^2 = x^2$

Another important concept before we finish our proof: Prime factorization.

Key question: is the number of prime factors for a number raised to the second power an even or odd number?

For example, $6^2$, $12^2$, and $15^2$

$6^2 = 6 × 6 = 2 × 3 × 2 × 3$ ($4$ prime factors, so even number)

$12^2 = 12 × 12 = 4 × 3 × 4 × 3 = 2 × 2 × 3 × 2 × 2 × 3$ ($6$ prime factors, so even number)

$15^2 = 15 × 15 = 3 × 5 × 3 × 5$ ($4$ prime factors, so even number)

There is a solid pattern here to conclude that any number squared will have an even number of prime factors

In order words, $x^2$ has an even number of prime factors.

Let's finish the proof then!

$5 y^2 = x^2$

Since $5 y^2$ is equal to $x^2$, $5 y^2$ and $x^2$ must have the same number of prime factors.

We just showed that

$x^2$ has an even number of prime factors, $y^2$ has also an even number of prime factors.

$5 y^2$ will then have an odd number of prime factors.

The number $5$ counts as $1$ prime factor, so $1$ + an even number of prime factors is an odd number of prime factors.

$5 y^2$ is the same number as $x^2$. However, $5 y^2$ gives an odd number of prime factor while $x^2$ gives an even number of prime factors.

This is a contradiction since a number cannot have an odd number of prime factors and an even number of prime factors at the same time.

The assumption that square root of $5$ is rational is wrong. Therefore, square of $5$ is irrational.

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

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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.

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    $\begingroup$ 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. $\endgroup$ Oct 30, 2013 at 10:21
  • $\begingroup$ @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. $\endgroup$
    – sundaycat
    Oct 30, 2013 at 17:33

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