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,

\$\$fracpq = sqrt5\$\$

\$\$Rightarrow fracp^2q^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.

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Is this proof correct?

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

ankush981ankush981
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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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answered Jul 25 "13 at 7:18 Michael AlbaneseMichael Albanese
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Yes, the proof is correct. Using this method you can show that \$sqrtp\$ 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 dRightarrow s^2d=kr\$\$ if we have \$n ot u\$, since \$gcd(s,r)=1\$, we have \$\$r|d\$\$ Then, with \$d>1\$, \$n ot u\$, but \$ n|u^2\$. If \$n\$ is prime, then \$d=1Rightarrow r=1\$ unless \$ n|u\$.

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edited Jul 16 "15 at 18:08
answered Jul 25 "13 at 7:20 \$endgroup\$
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The number of prime divisors of \$p^2\$ is even. Is that true for \$5q^2\$?

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Folshort
answered Nov 1 "13 at 10:19 Michael HoppeMichael Hoppe
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Yes, the proof is correct. But I think you still need a lemma to reinforce your proof

Lemma: \$\$ extIf P|Q^2, ext 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_1P_2^e_2P_3^e_3ldots P_k^e_k ag1\$\$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_1P_2^2e_2P_3^2e_3ldots P_k^2e_k ag2\$\$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 \$PinP_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
SundayCatSundayCat
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Let us 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 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.

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