Proceeding together in the proof of $sqrt 2$, let united state assume that $sqrt 5$ is rational. This way for some unique integers $p$ and also $q$ having actually no common factor various other than 1,

$$fracpq = sqrt5$$

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

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

This way that 5 divides $p^2$. This method that 5 divides $p$ (because every element must appear twice because that the square come exist). So we have, $p = 5 r$ for some integer $r$. Extending the dispute to $q$, we uncover that they have actually a common factor of 5, i m sorry is a contradiction.

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Is this proof correct?

elementary-number-theory proof-verification radicals rationality-testing
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edited Aug 5 "15 at 14:54

Bart Michels
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inquiry Jul 25 "13 at 7:15

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It is, however I think you have to be a little bit much more careful once explaining why $5$ divides $p^2$ means $5$ divides $p$. If $4$ divides $p^2$ go $4$ necessarily division $p$?

answer Jul 25 "13 in ~ 7:18

Michael AlbaneseMichael Albanese
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Yes, the evidence is correct. Utilizing this technique you can display that $sqrtp$ for any type of prime $p$ is irrational. During the proof you basically use the reality that once $p|u^2$ where $p$ is a prime, then it implies that $p|u$. This is true because that primes, however is not true in general. You deserve to prove this as below Let $n|u^2, gcd(n,u)=d$. Then, allow $n=rd, u=sd $. So, $$u^2=kn Rightarrow s^2d^2=k r dRightarrow s^2d=kr$$ if we have actually $n ot u$, since $gcd(s,r)=1$, we have $$r|d$$ Then, through $d>1$, $n ot u$, but $ n|u^2$. If $n$ is prime, then $d=1Rightarrow r=1$ uneven $ n|u$.

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

Samrat MukhopadhyaySamrat Mukhopadhyay
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The variety of prime divisors that $p^2$ 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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Let us assume $√5$ is rational.

$$√5=fracxy$$Square both political parties of the equation above

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

Multiply both sides by $y^2$

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

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

Another essential concept before we end up our proof: element factorization.

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

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

$6^2 = 6 × 6 = 2 × 3 × 2 × 3$ ($4$ element 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 right here to break up that any type of number squared will have an even variety of prime factors

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

Let"s complete 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 actually the same variety of prime factors.

We simply showed that

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

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

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

$5 y^2$ is the very same number together $x^2$. However, $5 y^2$ offers an odd variety of prime factor while $x^2$ offers an even variety of prime factors.

This is a contradiction because a number cannot have actually an odd number of prime factors and an even variety of prime determinants at the exact same time.

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