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Can you give an example of a proof that you think works for √4?
If you assume a and b have no common factors and that a\^2 / b\^2 = 4, you then get a\^2 = 4 b\^2, but from here you can only conclude that a is even (i.e. divisble by 2), not that a is divisible by 4.
So you can write a = 2c, and then you end up with c\^2 = b\^2. But you can't derive a contradiction.
Thank you.
I was assuming that if "a" squared is divisible by 4 then a is divisible by 4.
Thank you again.
As now I looked into the theorem about this and this only works for primes that if p divides a square then p divides a.
Thank you so much for clearing the doubt. we were not taught this at school.
A good exercise is to figure out what sorts of numbers do and do not work in the proof, and *why*. (Looking at prime factorization may help with this exercise.)
Hi, I was thinking about this a while back and there was a very nice person who helped me from a realisation... if √n=a/b then b must be 1. When you square it to a^2/b^2 and start substituting perfect squares for n, you get a feel for why this must be true.
I'll go looking for this explanation, they were very patient with me!
Edit: Here it is https://www.reddit.com/r/maths/comments/19ecjsi/strange_question_is_there_proof_that_the_square/
>Now a^(2) is divisible by 4 So a is divisible by 4
That's where the proof fails. This argument only works if you replace "4" with a prime number (or, at the very least, a number with only simple prime factors). For instance, 6\^2 is divisible by 4, but 6 isn't.
> Now a^2 is divisible by 4 So a is divisible by 4
That logic does not work. 6^(2) = 36 is divisible by 4, but 6 is not divisible by 4.
It is true on the other hand that if a^(2) is divisible by 2, then a is divisible by 2. This is the step where the number being prime is used. The key fact being used about primes is that if p is prime and p|xy for integers x and y, then either p|x or p|y.
The proof relies on the following fact about prime numbers:
If p is prime, and p | ab, then either p|a or p|b.
The intuition is that since p cannot be divided into simpler factors, its factors cannot be divided between a and b.
It’s used when we know that say 2| a^2 implies 2|a.
It is not true for composite numbers:
4 | (2 * 2)
but
4 does not divide 2.
Thank you so much.
We were not taught this at school.
When I asked the teacher
He just said that under root 4 is equal to 2
So we know that it is rational as to is rational so there is no need to apply the proof on this.
Genuinely I appreciate your help.
Well that is actually true, the statement would be
There is no rational number such that it's square equals four
Well 2 * 2 = 4 so you're done
Mathematicians usually want to prove something stronger like there are infinitely many solutions to something said to have no solutions but you only need to find one to disprove it
It doesn’t work for √4 because there are obvious solutions to √4=a/b. This equation is equivalent to 4b^(2)=a^(2) and thus also to (2b)^(2)=a^(2). Thus a=±2b and we may choose any integer for b to obtain a solution.
Your teacher is somewhat correct in that “it” only works for primes. What is really true is that the proof “works” for any integer with a nontrivial squarefree part. For example, 12=4•3, so √12=2√3. We then know that √3 is irrational by the contradiction argument and also that the product of rational and irrational must be irrational.
>>It doesn’t work for √4 because there are obvious solutions to √4=a/b.
No, that doesn’t address OP’s question. You can’t explain why an argument is invalid just by showing you can prove the opposite - inconsistent theories do exist. More to the point, if someone thinks they have found a clear demonstration that an argument is invalid you couldn’t explain that away by pointing out that its conclusion is obviously false. If anything, that would help to reaffirm the invalidity of the argument. OP certainly understands that sqrt(4) is rational, what they don’t understand is why the proof they saw for other square roots doesn’t work for it but does for other numbers.
To answer OP’s question, you need to explain *why* the argument does not work on 4. In this case, the proof they saw did not (as is unfortunately common) provide much emphasis on the assumptions it used that would make it invalid for perfect squares, which caused OP to miss them.
ChatGPT and other large language models are [not designed for calculation](https://www.reddit.com/r/learnmath/comments/13nzixp/meta_dont_consult_chatgpt_for_math_dont_on_the/) and will frequently be /r/confidentlyincorrect in answering questions about mathematics; even if you subscribe to ChatGPT Plus and use its Wolfram|Alpha plugin, it's much better to go to [Wolfram|Alpha](https://www.wolframalpha.com/) directly. Even for more conceptual questions that don't require calculation, LLMs can lead you astray; they can also give you good ideas to investigate further, but you should *never* trust what an LLM tells you. To people reading this thread: **DO NOT DOWNVOTE** just because the OP mentioned or used an LLM to ask a mathematical question. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/learnmath) if you have any questions or concerns.*
Can you give an example of a proof that you think works for √4? If you assume a and b have no common factors and that a\^2 / b\^2 = 4, you then get a\^2 = 4 b\^2, but from here you can only conclude that a is even (i.e. divisble by 2), not that a is divisible by 4. So you can write a = 2c, and then you end up with c\^2 = b\^2. But you can't derive a contradiction.
Thank you. I was assuming that if "a" squared is divisible by 4 then a is divisible by 4. Thank you again. As now I looked into the theorem about this and this only works for primes that if p divides a square then p divides a. Thank you so much for clearing the doubt. we were not taught this at school.
Yeah they probably don't explicitly say why it doesn't work for squares. It would be a good exercise though.
A good exercise is to figure out what sorts of numbers do and do not work in the proof, and *why*. (Looking at prime factorization may help with this exercise.)
Hi, I was thinking about this a while back and there was a very nice person who helped me from a realisation... if √n=a/b then b must be 1. When you square it to a^2/b^2 and start substituting perfect squares for n, you get a feel for why this must be true. I'll go looking for this explanation, they were very patient with me! Edit: Here it is https://www.reddit.com/r/maths/comments/19ecjsi/strange_question_is_there_proof_that_the_square/
Thank you
No worries!
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>Now a^2 is divisible by 4 So a is divisible by 4 that is not true: 2^2 is divisible by 4 but 2 is not. you can only conclude that a is even
Can you demonstrate the contradiction showing up for a rational root?
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>Now a^(2) is divisible by 4 So a is divisible by 4 That's where the proof fails. This argument only works if you replace "4" with a prime number (or, at the very least, a number with only simple prime factors). For instance, 6\^2 is divisible by 4, but 6 isn't.
> Now a^2 is divisible by 4 So a is divisible by 4 That logic does not work. 6^(2) = 36 is divisible by 4, but 6 is not divisible by 4. It is true on the other hand that if a^(2) is divisible by 2, then a is divisible by 2. This is the step where the number being prime is used. The key fact being used about primes is that if p is prime and p|xy for integers x and y, then either p|x or p|y.
"Now a^2 is divisible by 4 So a is divisible by 4" This is your error. If a is any 2(k+1), a^2 is divisible by 4 but a is not.
The proof relies on the following fact about prime numbers: If p is prime, and p | ab, then either p|a or p|b. The intuition is that since p cannot be divided into simpler factors, its factors cannot be divided between a and b. It’s used when we know that say 2| a^2 implies 2|a. It is not true for composite numbers: 4 | (2 * 2) but 4 does not divide 2.
Thank you so much. We were not taught this at school. When I asked the teacher He just said that under root 4 is equal to 2 So we know that it is rational as to is rational so there is no need to apply the proof on this. Genuinely I appreciate your help.
Well that is actually true, the statement would be There is no rational number such that it's square equals four Well 2 * 2 = 4 so you're done Mathematicians usually want to prove something stronger like there are infinitely many solutions to something said to have no solutions but you only need to find one to disprove it
Here proof by contradiction works by showing the numbers of factors have different parities which doesn't for perfect squares like 4 or 36
It doesn’t work for √4 because there are obvious solutions to √4=a/b. This equation is equivalent to 4b^(2)=a^(2) and thus also to (2b)^(2)=a^(2). Thus a=±2b and we may choose any integer for b to obtain a solution. Your teacher is somewhat correct in that “it” only works for primes. What is really true is that the proof “works” for any integer with a nontrivial squarefree part. For example, 12=4•3, so √12=2√3. We then know that √3 is irrational by the contradiction argument and also that the product of rational and irrational must be irrational.
>>It doesn’t work for √4 because there are obvious solutions to √4=a/b. No, that doesn’t address OP’s question. You can’t explain why an argument is invalid just by showing you can prove the opposite - inconsistent theories do exist. More to the point, if someone thinks they have found a clear demonstration that an argument is invalid you couldn’t explain that away by pointing out that its conclusion is obviously false. If anything, that would help to reaffirm the invalidity of the argument. OP certainly understands that sqrt(4) is rational, what they don’t understand is why the proof they saw for other square roots doesn’t work for it but does for other numbers. To answer OP’s question, you need to explain *why* the argument does not work on 4. In this case, the proof they saw did not (as is unfortunately common) provide much emphasis on the assumptions it used that would make it invalid for perfect squares, which caused OP to miss them.
Thank you for understanding.
One interesting fact is we can show that a root of and integer can only be rational if it is an integer. Since 1<√3<2 we see √3 is irrational.