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?

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edited Aug 5 "15 at 14:54 Bart Michels
inquiry Jul 25 "13 at 7:15

ankush981ankush981
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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\$?

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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\$.

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edited Jul 16 "15 at 18:08
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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.