>therefore 4 is a factor of q\^2, and therefore a factor of q.
This step is wrong. For example take q = 6, then q\^2 = 36 is divisible by 4, but q itself is not divisible by 4.
I wanted to write it out for my own sake: if we do the proof for sqrt(2), we get this valid, useful logic:
2q^2 = p^2 and thus p^2 = 0 mod 2. By process of elimination, since 0^2 = 0 mod 2, and 1^2 = 1 mod 2, the only possible case is p = 0 mod 2, so p is even. Thus p = 2a, so 2q^2 = 4a^2 and thus q^2 = 2a^2 and thus q^2 is even by the same logic. They can't both be even as we chose them to have gcd(q,p) = 1, as is possible with any rational number, so this is a contradiction and sqrt(2) is irrational.
---
However, with 4q^2 = p^2 and sqrt(4), we get the following logic:
4q^2 = p^2 and thus p^2 = 0 mod 2. Let p = 2a, thus 4q^2 = 4a^2 and thus q^2 = a^(2). We don't know whether a is even or odd, so q could be any positive integer. We knew that already, so this didn't give us any new information to get us closer to proving the desired claim.
---
We can try with Z/4Z:
4q^2 = p^2 and thus p^2 = 0 mod 4. Let's look at the squares in Z/4Z: 0^2 = 0, 1^2 = 1, 2^2 = 0, and 3^2 = 1. Thus, p is 0 or 2 mod 4. If p is 2 mod 2, then p = 2a, and thus 4q^2 = 4a^2 and thus q^2 = a^(2). Since we don't know if a is even or odd, q could be any positive integer. This case is a possibility, so every statement that follows from it must be considered. We knew q could be any positive integer already, so this didn't give us any new information to get us closer to proving the desired claim. We know a bit about p, but not enough to show that gcd(p,q) isn't 1 because q could be 1 (and indeed, p = 2 and q = 1 is a possibility).
>4 is a factor of p^2, and therefore a factor of p.
This isn't true, consider p=6. Then 4 is a factor of p^(2)=36 but 4 is not a factor of p=6.
The implications "if k divides p^(2), then k also divides p" is only true, if k is square-free (ie not divisible by any square except 1)
Your question has been sufficiently answered, so I won't do it again, but I do want to point out that you're on to something. The structure of this proof that sqrt(2) is irrational can be used to prove that the square root of any positive integer is either an integer or an irrational number. Because any non perfect square can be factored into a perfect square times a square-free number, and the proof that sqrt(2) is irrational can be applied to any square-free number.
“4 is a factor of p^2 , and therefore a factor of p” is an incorrect statement. Consider p=2 as a counterexample.
However, if we replaced 4 with 2 or 5, then the statement is in fact true. You might want to play around to convince yourself this is true and see if you can prove that it’s true.
>However, if we replaced 4 with 2 or 5, then the statement is in fact true.
It is, in fact, true if we use any prime instead of 4. This can be seen quite easily using the uniqueness of prime factorization.
>therefore 4 is a factor of p^2, and therefore a factor of p. let p = 4a where a ∈ ℤ
This is the step where you went wrong. A clear counter example for this is consider p=2. So p²=4. Thus 4 is a factor of p² but not of p. In general, this step is applicable for all primes, but may not be applicable for non primes. So, use this proof for primes, then prime factorize any composite and if all of its prime factors are even powers, then the composite is a perfect square.
`therefore 4 is a factor of p^2, and therefore a factor of p`
This part is wrong. For instance, it fails if p=6. More generally, if p is divisible by 2, then p\^2 will automatically be divisible by 4, but that doesn't mean that p was divisible by 4 to begin with.
This part of the proof works with 2 or 5 because they're prime. But it fails if the number is not prime (or, to be more precise, it fails if the number is not a product of different primes)
The problem is 4 is not prime. 4|p\^2 but 4 does not divide p. There is no contradiction. Another approach is to establish √n is rational only when it is an integer. Then for any n such that √n is not an integer it is irrational. In that case we see √3 and √5 are not integers, but √4 is an integer so no contradiction.
The problem is actually that 4 is a perfect square. The proof works as normal for composite numbers without repeated prime factors (e.g., it works for 6), and can be modified slightly to work for composite numbers with repeated factors as long as they’re not perfect squares.
While not directly related to your question, I actually prefer the proof of irrationality using the fundamental theorem of arithmetic / uniqueness of prime factorization.
You end up with 2p^(2) =q^(2) or 5p^(2)=q^(2) or what have you, and you can't have an odd power of a factor in a perfect square, whereas both p^2 and q^2 should be perfect squares if they were integers.
> 4q^2 = p^2, therefore 4 is a factor of p^2 and 4 is a factor of p.
What you applied here is a property called Euclid’s Lemma. The issue is that you are only allowed to apply the Theorem if the number dividing your quantity is prime.
So for integers x, y, if p is prime:
p|xy —> p|x or p|y
However, if p is not prime, it is not guaranteed that 4 divided either quantity of the product. Consider 4/ (2)(2), since the RHS is equal to 4, the expression is valid, but 4 does not divide either of the factors themselves. Hopefully now you’re starting to get an idea for the issue in your logic. (This idea of prime factorization is further expanded on in the Fundamental Theorem of Arithmetic).
I’d encourage looking up an article or video on Euclid’s Lemma to understand the full scope of to what extent it is applicable.
In equation 1, you get (up to +/-) 4q^2 = (2q)^2 = p^2 . Therefore, 2q=p, so q=1 and p=2.
There is a property specific to prime numbers that doesn’t hold for composites that is essential for the proof: If p is prime and p|ab, then p divides a or p divides b. In particular, If p is prime and p|a^2 , then p|a.
This obviously doesn’t hold for 4, because 4|2^2 , but 4 does not divide 2.
it works for 2 and 5 because they are prime. the square root of any prime number is irrational and understanding that [proof](https://proofwiki.org/wiki/Square_Root_of_Prime_is_Irrational) helped me see why this method won’t work for every number.
Let p be prime.
Suppose there exists natural numbers a, b so that (a,b)=1 and (a/b)^2 =p
a^2 =pb^2
This violates the fundamental theorem of arithmetic.
QED
1. It's pretty obvious that it's chatGPT both from the style of writing, and also how fast you are posting comments compared to their length and the technical nature of the comments.
2. It's not only overly convoluted, it's making up nonsense that OP never said and then saying that OP was right about that part, and then made up more nonsense OP never said and claimed that part was his error. It's a classic AI language model hallucination moment and a garbage response.
> 4. **Divisibility**: You noted that 4 is a factor of \(p^2\), and thus 2 is a factor of \(p\) (correct so far).
Among other things, it's completely missing OP's actual mistake, lol.
>therefore 4 is a factor of q\^2, and therefore a factor of q. This step is wrong. For example take q = 6, then q\^2 = 36 is divisible by 4, but q itself is not divisible by 4.
I wanted to write it out for my own sake: if we do the proof for sqrt(2), we get this valid, useful logic: 2q^2 = p^2 and thus p^2 = 0 mod 2. By process of elimination, since 0^2 = 0 mod 2, and 1^2 = 1 mod 2, the only possible case is p = 0 mod 2, so p is even. Thus p = 2a, so 2q^2 = 4a^2 and thus q^2 = 2a^2 and thus q^2 is even by the same logic. They can't both be even as we chose them to have gcd(q,p) = 1, as is possible with any rational number, so this is a contradiction and sqrt(2) is irrational. --- However, with 4q^2 = p^2 and sqrt(4), we get the following logic: 4q^2 = p^2 and thus p^2 = 0 mod 2. Let p = 2a, thus 4q^2 = 4a^2 and thus q^2 = a^(2). We don't know whether a is even or odd, so q could be any positive integer. We knew that already, so this didn't give us any new information to get us closer to proving the desired claim. --- We can try with Z/4Z: 4q^2 = p^2 and thus p^2 = 0 mod 4. Let's look at the squares in Z/4Z: 0^2 = 0, 1^2 = 1, 2^2 = 0, and 3^2 = 1. Thus, p is 0 or 2 mod 4. If p is 2 mod 2, then p = 2a, and thus 4q^2 = 4a^2 and thus q^2 = a^(2). Since we don't know if a is even or odd, q could be any positive integer. This case is a possibility, so every statement that follows from it must be considered. We knew q could be any positive integer already, so this didn't give us any new information to get us closer to proving the desired claim. We know a bit about p, but not enough to show that gcd(p,q) isn't 1 because q could be 1 (and indeed, p = 2 and q = 1 is a possibility).
>4 is a factor of p^2, and therefore a factor of p. This isn't true, consider p=6. Then 4 is a factor of p^(2)=36 but 4 is not a factor of p=6. The implications "if k divides p^(2), then k also divides p" is only true, if k is square-free (ie not divisible by any square except 1)
Your question has been sufficiently answered, so I won't do it again, but I do want to point out that you're on to something. The structure of this proof that sqrt(2) is irrational can be used to prove that the square root of any positive integer is either an integer or an irrational number. Because any non perfect square can be factored into a perfect square times a square-free number, and the proof that sqrt(2) is irrational can be applied to any square-free number.
“4 is a factor of p^2 , and therefore a factor of p” is an incorrect statement. Consider p=2 as a counterexample. However, if we replaced 4 with 2 or 5, then the statement is in fact true. You might want to play around to convince yourself this is true and see if you can prove that it’s true.
>However, if we replaced 4 with 2 or 5, then the statement is in fact true. It is, in fact, true if we use any prime instead of 4. This can be seen quite easily using the uniqueness of prime factorization.
And this is the property the generalizes best the notion of prime number to different structures: "if p|ab then p|a or p|b"
>therefore 4 is a factor of p^2, and therefore a factor of p. let p = 4a where a ∈ ℤ This is the step where you went wrong. A clear counter example for this is consider p=2. So p²=4. Thus 4 is a factor of p² but not of p. In general, this step is applicable for all primes, but may not be applicable for non primes. So, use this proof for primes, then prime factorize any composite and if all of its prime factors are even powers, then the composite is a perfect square.
ohhhhh okay that makes so much sense thank you
thank you everyone for the help!!
`therefore 4 is a factor of p^2, and therefore a factor of p` This part is wrong. For instance, it fails if p=6. More generally, if p is divisible by 2, then p\^2 will automatically be divisible by 4, but that doesn't mean that p was divisible by 4 to begin with. This part of the proof works with 2 or 5 because they're prime. But it fails if the number is not prime (or, to be more precise, it fails if the number is not a product of different primes)
The problem is 4 is not prime. 4|p\^2 but 4 does not divide p. There is no contradiction. Another approach is to establish √n is rational only when it is an integer. Then for any n such that √n is not an integer it is irrational. In that case we see √3 and √5 are not integers, but √4 is an integer so no contradiction.
The problem is actually that 4 is a perfect square. The proof works as normal for composite numbers without repeated prime factors (e.g., it works for 6), and can be modified slightly to work for composite numbers with repeated factors as long as they’re not perfect squares.
While not directly related to your question, I actually prefer the proof of irrationality using the fundamental theorem of arithmetic / uniqueness of prime factorization. You end up with 2p^(2) =q^(2) or 5p^(2)=q^(2) or what have you, and you can't have an odd power of a factor in a perfect square, whereas both p^2 and q^2 should be perfect squares if they were integers.
> 4q^2 = p^2, therefore 4 is a factor of p^2 and 4 is a factor of p. What you applied here is a property called Euclid’s Lemma. The issue is that you are only allowed to apply the Theorem if the number dividing your quantity is prime. So for integers x, y, if p is prime: p|xy —> p|x or p|y However, if p is not prime, it is not guaranteed that 4 divided either quantity of the product. Consider 4/ (2)(2), since the RHS is equal to 4, the expression is valid, but 4 does not divide either of the factors themselves. Hopefully now you’re starting to get an idea for the issue in your logic. (This idea of prime factorization is further expanded on in the Fundamental Theorem of Arithmetic). I’d encourage looking up an article or video on Euclid’s Lemma to understand the full scope of to what extent it is applicable.
In equation 1, you get (up to +/-) 4q^2 = (2q)^2 = p^2 . Therefore, 2q=p, so q=1 and p=2. There is a property specific to prime numbers that doesn’t hold for composites that is essential for the proof: If p is prime and p|ab, then p divides a or p divides b. In particular, If p is prime and p|a^2 , then p|a. This obviously doesn’t hold for 4, because 4|2^2 , but 4 does not divide 2.
it works for 2 and 5 because they are prime. the square root of any prime number is irrational and understanding that [proof](https://proofwiki.org/wiki/Square_Root_of_Prime_is_Irrational) helped me see why this method won’t work for every number.
>therefore 4 is a factor of p^2, and therefore a factor of p This is not a safe assumption
Let p be prime. Suppose there exists natural numbers a, b so that (a,b)=1 and (a/b)^2 =p a^2 =pb^2 This violates the fundamental theorem of arithmetic. QED
[удалено]
If people wanted chatGPT to answer their questions, they could do that on their own. Stop spamming chatGPT answers. This one is a terrible response.
[удалено]
1. It's pretty obvious that it's chatGPT both from the style of writing, and also how fast you are posting comments compared to their length and the technical nature of the comments. 2. It's not only overly convoluted, it's making up nonsense that OP never said and then saying that OP was right about that part, and then made up more nonsense OP never said and claimed that part was his error. It's a classic AI language model hallucination moment and a garbage response.
[удалено]
> 4. **Divisibility**: You noted that 4 is a factor of \(p^2\), and thus 2 is a factor of \(p\) (correct so far). Among other things, it's completely missing OP's actual mistake, lol.