Proceeding together in the proof of $\sqrt 2$, let us assume the $\sqrt 5$ is rational. This means for some distinct integers $p$ and also $q$ having actually 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 way that 5 divides $p$ (because every factor must appear twice because that the square to exist). So we have, $p = 5 r$ for some integer $r$. Extending the debate to $q$, we uncover that they have a usual factor the 5, which is a contradiction.

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Is this evidence correct?


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edited Aug 5 "15 at 14:54
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Bart Michels
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It is, yet I think you have to be a little bit much more careful when explaining why $5$ divides $p^2$ implies $5$ divides $p$. If $4$ divides $p^2$ does $4$ necessarily division $p$?


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answered Jul 25 "13 at 7:18
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Michael AlbaneseMichael Albanese
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Yes, the proof is correct. Utilizing this method you can present that $\sqrtp$ for any kind of prime $p$ is irrational. During the evidence you basically use the truth that when $p|u^2$ where $p$ is a prime, climate it indicates that $p|u$. This is true for primes, however is no true in general. You can prove this as listed 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 $n\not\ u$, because $\gcd(s,r)=1$, we have $$r|d$$ Then, with $d>1$, $n\not\ u$, however $\ n|u^2$. If $n$ is prime, climate $d=1\Rightarrow r=1$ uneven $\ n|u$.


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edited Jul 16 "15 in ~ 18:08
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Samrat MukhopadhyaySamrat Mukhopadhyay
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The number of prime divisors the $p^2$ is even. Is the true for $5q^2$?


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answer Nov 1 "13 in ~ 10:19
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Michael HoppeMichael Hoppe
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Yes, the proof is correct. Yet I think girlfriend still require a lemma to reinforce your proof

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

Proof: by the unique factorization theorem,$Q$ is able come rewrited together a product of distinctive 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 hopeful integers. Then:$$Q^2 = P_1^2e_1P_2^2e_2P_3^2e_3\ldots P_k^2e_k\tag2$$By (1),(2), we understand both $Q$ and also $Q^2$ are a product of distinctive prime numbers the belong to the collection $\P_1,P_2,\ldots,P_k\$. Because $P|Q^2$ and also $P$ is additionally a prime, It implies $P\in\P_1,P_2,\ldots,P_k\$. Hence, P|Q, which finish the proof.


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edited Oct 31 "13 at 23:07
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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 political parties by $y^2$

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

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

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

Key question: is the number of prime components for a number increased to the 2nd power an also or strange 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 also number)

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

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

Let"s end up the evidence 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 variety of prime factors.

We simply showed that

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

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

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

$5 y^2$ is the exact same number together $x^2$. However, $5 y^2$ gives an odd variety of prime aspect while $x^2$ gives an even variety of prime factors.

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

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