im done with math in school, but i still watch your videos for entertainment. your enthusiasm is so infectious. thanks for making the world a better place, especially for those lucky enough to have you teach them in person!
Ik it might be trivial but if you let n=ab with a,b≠1 you exclude the case for n=1 which is also part of your contra positive, which I can only assume means you'd need to address it separately later.
I think the easiest way to prove this is to think in base 2: 2ⁿ - 1 = 111...111 - there will be exactly n 1s in the base 2 representation of the number. So if n is composite, i.e. n = xy for integers x and y greater than 1, then 2ⁿ - 1 must have 2ˣ - 1, i.e. the number with x 1s in its base 2 representation, as a factor.
The MERSENNE PRIMES formula is part of the following formula 1. Let a, b, n be natural numbers with a>b - {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is always a natural number. If {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is prime then a-b is prime. If a-b is composite, then {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is composite. - If [(a^n)-(b^n)]/(a-b) is prime, then n is prime. If n is composite then [(a^n)-(b^n)]/(a-b) is composite. 2. Let a, b, n be natural numbers where n is odd - With a+b being odd, {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is always a natural number. If {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is prime then a+b is prime. If a+b is composite then {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is composite. - If [(a^n)+(b^n)]/(a+b) is prime, then n is prime. If n is composite, then [(a^n)+(b^n)]/(a+b) is composite.
Both the statement of the question and the proof given are a little sloppy. First up, it doesn't state what set n lives in (e.g. n = log_2(12) satisfies 2^n - 1 prime, but n is not prime). It should say that n is in N (or Z). Then in the proof, n=ab is written with a,b in Z, so I guess the assumption is that n is in Z. In that case, we should really say a and b are not 1 OR -1. [7 = (-7)*(-1), for example] But with a, b possibly negative, you can't then use the factorisation rule given below. So a better thing to do would be to assume n is positive, pick a, b from N, then the proof is alright. Deal with the case of n negative separately. C-
Hum. So n not prime -> 2^n -1 not prime If n is an even number bigger than 2 we have n = 2p -> 2^(2p) - 1 = (2^p-1)*(2^p+1) , since n > 2 --> p > 1 and 2^p - 1 > 1. So we have a decomposition into prime factors and 2^n-1 is not prime If we try for n multiple of 3 bigger than 3 itself, we have n = 3p -> 2^(3p) - 1 = (2^p - 1 ) * ( 2^(2p) + 2^p + 1) which means as above that 2^n-1 with n a multiple of 3 > 3 is not prime Now, I "just" need to proof that 2^(kp) - 1^(kp) can be factored as (2^p - 1) ( some positive integer) for any number k or any prime number k. 2^(kp)-1^(kp) = ( 2^p - 1 ) ( 2^((k-1)p) + 2^((k-2)p) + ... + 2^p + 1 ) with both side > 1 as long as p > 1 If you distribute the product above, you'll see that you obtain 2^(kp) - 1^(kp) This means that as long as n is not prime, it can be expressed as n=k*p with k>1 and p > 1, and 2^n-1 is not prime Eheh, seems quite similar to his proof :)
Cool! Thank you=) I was searching for the explanation of the exercise 25 in chapter 5 in the Book Of Proof (3rd edition) by Richard Hammack - just in case someone is doing the same =)
13:11 you said a, b belong to Z not N. Thus there could be negative powers. If there are negative odd powers, then we are adding a bunch of stuff but some of that stuff is negative. So it's not really proven here that the right expression cannot be 1.
This is so genius but logical and easy at the same time... wow that's just why I love maths
im done with math in school, but i still watch your videos for entertainment. your enthusiasm is so infectious.
thanks for making the world a better place, especially for those lucky enough to have you teach them in person!
Mesmerising - both in logic and use of technology.
This question came up on my final exam last night (I wasn't able to solve it)--now this video came up on my feed.
Funny how that works!
Wow that was so elegant, thanks for posting.
One of your best works
Elegant proof, love your content, Jesus bless.
That was awesome, I was never good at math in school but this was a pretty straight forward logical deduction.
Thanks for the video! I find these super interesting to watch
Wow thanks so much for your content, Mr Woo
Ik it might be trivial but if you let n=ab with a,b≠1 you exclude the case for n=1 which is also part of your contra positive, which I can only assume means you'd need to address it separately later.
1 isnt a prime number therefore you can ignore the case where n=1
@@adrianpilot159 but we're using the contra positive where n is *not* prime
@@ARKGAMING my bad, u would need to check the case n=1 later
True, 1 is non-prime so has to be verified separately.
You just nailed it buddy 🤩
I think the easiest way to prove this is to think in base 2: 2ⁿ - 1 = 111...111 - there will be exactly n 1s in the base 2 representation of the number. So if n is composite, i.e. n = xy for integers x and y greater than 1, then 2ⁿ - 1 must have 2ˣ - 1, i.e. the number with x 1s in its base 2 representation, as a factor.
Thank You.
Amazing. Things like this make you remember the cube identity for ever.
Oh Boy.... geography is hard! 😂
"This is a really cute sort of question" in other words this is too easy you should challenge yourself
The MERSENNE PRIMES formula is part of the following formula
1. Let a, b, n be natural numbers with a>b
- {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is always a natural number. If {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is prime then a-b is prime. If a-b is composite, then {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is composite.
- If [(a^n)-(b^n)]/(a-b) is prime, then n is prime. If n is composite then [(a^n)-(b^n)]/(a-b) is composite.
2. Let a, b, n be natural numbers where n is odd
- With a+b being odd, {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is always a natural number. If {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is prime then a+b is prime. If a+b is composite then {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is composite.
- If [(a^n)+(b^n)]/(a+b) is prime, then n is prime. If n is composite, then [(a^n)+(b^n)]/(a+b) is composite.
Thank you so much, but can you share me what kind of camera do you use?
love your content prof
How do you make this video please with tablet and you explaining. Very nice! can you prove that there no perfect odd number??
Awesome!!!!!!
Thanks.
Forgot at the end?
QED!
There we go, now it's done.
Both the statement of the question and the proof given are a little sloppy. First up, it doesn't state what set n lives in (e.g. n = log_2(12) satisfies 2^n - 1 prime, but n is not prime). It should say that n is in N (or Z).
Then in the proof, n=ab is written with a,b in Z, so I guess the assumption is that n is in Z. In that case, we should really say a and b are not 1 OR -1. [7 = (-7)*(-1), for example] But with a, b possibly negative, you can't then use the factorisation rule given below. So a better thing to do would be to assume n is positive, pick a, b from N, then the proof is alright. Deal with the case of n negative separately.
C-
Awesome video! Could you do a video on why the Lucas-Lehmer test is correct also?
Wow! Amazing proof. ❤ from 🌏 😊
Si n es entero positivo y no es primo entonces puede ser la unidad o compuesto. Faltaría analizar el caso n=1. ¡Muy buen video!
Ese caso es trivial.
Which app is being used?
This was in 2022 HSC!
My teacher, why did the comments in Arabic distinguish his position? I see you from the Arab world, Egypt
Hum.
So n not prime -> 2^n -1 not prime
If n is an even number bigger than 2 we have n = 2p -> 2^(2p) - 1 = (2^p-1)*(2^p+1) , since n > 2 --> p > 1 and 2^p - 1 > 1. So we have a decomposition into prime factors and 2^n-1 is not prime
If we try for n multiple of 3 bigger than 3 itself, we have n = 3p -> 2^(3p) - 1 = (2^p - 1 ) * ( 2^(2p) + 2^p + 1) which means as above that 2^n-1 with n a multiple of 3 > 3 is not prime
Now, I "just" need to proof that 2^(kp) - 1^(kp) can be factored as (2^p - 1) ( some positive integer) for any number k or any prime number k.
2^(kp)-1^(kp) = ( 2^p - 1 ) ( 2^((k-1)p) + 2^((k-2)p) + ... + 2^p + 1 ) with both side > 1 as long as p > 1
If you distribute the product above, you'll see that you obtain 2^(kp) - 1^(kp)
This means that as long as n is not prime, it can be expressed as n=k*p with k>1 and p > 1, and 2^n-1 is not prime
Eheh, seems quite similar to his proof :)
Cool! Thank you=)
I was searching for the explanation of the exercise 25 in chapter 5 in the Book Of Proof (3rd edition) by Richard Hammack - just in case someone is doing the same =)
Nice argument. I don't think number theory was covered much when I was in school.
13:11 you said a, b belong to Z not N. Thus there could be negative powers. If there are negative odd powers, then we are adding a bunch of stuff but some of that stuff is negative. So it's not really proven here that the right expression cannot be 1.
They still aren’t negative. Example: 5^(-2)= 1/25
Sir love from india
This problem is nice about an arithmetic property:
If 2^n-1 is prime then n is prime.
I do not understand you because I am an Arab
Luckily, math is an universal language :)
@@beyza2004 yes
@arqam station But manages to post a comment in perfectly cromulent English?
@@Grizzly01 Probably using Google translate
👍👍