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?
elementary-number-theory proof-verification radicals rationality-testing
edited Aug 5 "15 at 14:54
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asked Jul 25 "13 at 7:15
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5 Answers 5
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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 $p$?
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$.
edited Jul 16 "15 at 18:08
answered Jul 25 "13 at 7:20
Samrat MukhopadhyaySamrat Mukhopadhyay
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The number of prime divisors of $p^2$ is even. Is that true for $5q^2$?
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.
edited Oct 31 "13 at 23:07
answered Oct 30 "13 at 8:29
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Let us assume $√5$ is rational.
$$√5=fracxy$$Square both sides of the equation above
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.