Proceeding as in the proof of $\sqrt 2$, let united state assume the $\sqrt 5$ is rational. This way for some distinct integers $p$ and also $q$ having no usual factor other than 1,

$$\fracpq = \sqrt5$$

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

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

This method that 5 divides $p^2$. This means that 5 divides $p$ (because every aspect must show up twice for the square to exist). So us have, $p = 5 r$ for part integer $r$. Prolonging the debate to $q$, we find that they have a common factor that 5, i beg your pardon is a contradiction.

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Is this proof correct?

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edited Aug 5 "15 in ~ 14:54

Bart Michels
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It is, however I think you must be a little bit more careful when explaining why $5$ divides $p^2$ implies $5$ divides $p$. If $4$ divides $p^2$ walk $4$ necessarily division $p$?

answer Jul 25 "13 in ~ 7:18

Michael AlbaneseMichael Albanese
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Yes, the evidence is correct. Utilizing this method you can display that $\sqrtp$ for any prime $p$ is irrational. During the evidence you essentially use the fact that when $p|u^2$ whereby $p$ is a prime, climate it means that $p|u$. This is true because that primes, yet is no true in general. You have the right to prove this as below Let $n|u^2,\ \gcd(n,u)=d$. Then, allow $n=rd,\ u=sd $. So, $$u^2=kn \Rightarrow s^2d^2=k r d\Rightarrow s^2d=kr$$ if we have actually $n\not\ u$, due to the fact that $\gcd(s,r)=1$, we have actually $$r|d$$ Then, through $d>1$, $n\not\ u$, however $\ n|u^2$. If $n$ is prime, climate $d=1\Rightarrow r=1$ unless $\ n|u$.

edited Jul 16 "15 in ~ 18:08
answer Jul 25 "13 at 7:20

Samrat MukhopadhyaySamrat Mukhopadhyay
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The variety of prime divisors of $p^2$ is even. Is the true for $5q^2$?

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reply Nov 1 "13 in ~ 10:19

Michael HoppeMichael Hoppe
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Yes, the proof is correct. However I think friend still need a lemma come reinforce her proof

Lemma: $$\textIf P|Q^2,\text wherein P is a prime, climate P|Q$$

Proof: through the unique factorization theorem,$Q$ is able to rewrited together a product of distinct prime numbers:$$Q = P_1^e_1P_2^e_2P_3^e_3\ldots P_k^e_k\tag1$$where $P_1,P_2,\ldots P_k$ are distinct prime numbers and also $e_1,e_2,\ldots e_k$ are positive integers. Then:$$Q^2 = P_1^2e_1P_2^2e_2P_3^2e_3\ldots P_k^2e_k\tag2$$By (1),(2), we know both $Q$ and $Q^2$ are a product of distinct prime numbers that belong come the set $\P_1,P_2,\ldots,P_k\$. Since $P|Q^2$ 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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edited Oct 31 "13 at 23:07
answered Oct 30 "13 at 8:29
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Let united state assume $√5$ is rational.

$$√5=\fracxy$$Square both sides of the equation above

$$5 =\fracx^2y^2$$

Multiply both sides by $y^2$

$$5 y^2 =\fracx^2 y^2$$

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

Another important concept prior to we end up our proof: prime factorization.

Key question: is the number of prime determinants for a number elevated to the second power an even or weird number?

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

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

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

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

There is a solid pattern here to finish that any type of number squared will have actually an even variety of prime factors

In stimulate words, $x^2$ has actually an even variety of prime factors.

Let"s end up the proof then!

$5 y^2 = x^2$

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

We just showed that

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

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

The number $5$ counts together $1$ prime factor, so $1$ + one even variety of prime components is one odd number of prime factors.

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

This is a contradiction due to the fact that a number cannot have an odd number of prime factors and also an even variety of prime components at the very same time.

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