Does anybody know if there's a link between Goldbach's conjecture and the Riemann hypothesis? I mean, imagine there's a connection that asserts that Goldbach's conjecture implies Riemann hypothesis :)))
Relating to Brady's question: Do the primes appear to be a minimal set for generating all even numbers > 4 by adding two elements? If not, what primes can we skip and still get sums for the even numbers calculated so far?
The numbers 3, 5, and 7 are necessary because 6 = 3 + 3 and 12 = 5 + 7 are the only possible sums for 6 and 12. A commenter on the 210 video claims that 12 is the largest known even number with only one sum. This implies that as far as we know the Goldbach conjecture is still true for the set of prime numbers \ {11}, since if every even number > 12 has at least two sums then they have at least one sum that doesn't use 11. So I'd like to know the next prime we can eliminate after eliminating 11.
Could there exist mathematical truths that cannot be proven? If so, could that statement itself be proven to be true? If not, could it be proven to be false? (Don't say "axioms".)
Lilla Kiss Ah! I thought that had to do with any set of axioms being unable to prove its own consistency, but that's actually just the second part of it. Thanks :-)
With the method of decomposition of each pair in pairs of addends; Enfer Diez : They prove that in todo 2n there is always 2n - p = p_{1}. See : video Conjectura de Goldbach (Irrefutable).
From which even number on can even numbers be expressed as two distinct primes? Let's see. 2 NO 4 = 2+2 but those are not distinct so NO 6=3+3 so NO 8=3+5 YES. Will this keep going let's see. 10=3+7 YES 12=5+7 YES 14=3+11 YES 16=5+11 YES 18=7+11 YES 20=3+17 YES 22=5+17 YES 24=5+19 YES It seems to go on forever can somebody prove to me that an even number above 6 can always be expressed as two distinct primes? TY
You can equate goldbach's conjecture to every number greater than some value, is equidistant from two primes. Furthermore, the twin prime conjecturecan be restated as 6m-1,6m+1 is a goldbach partition of 12m infinitely often. The statement about two primes being a certain even number apart can be turned to general statements about goldbach partitions.
I think the difference is that no individual arithmetic progression necessarily has an infinite number of primes, but there is no limit on how many there can be in one.
Yes, I think so. Infinitely many primes in an arithmetic progression would mean that every arithmetic progression has infinitely many primes, which is false (2,4,6,8,10,...). Arbitrarily many primes in an arithmetic progression means that for any number k you can find an arithmetic progression with more than k primes. (Edit: too late :/)
The equations and proof that defines the prime number sequence were proven five years ago and subsequently the proofs to the Strong and Weak Goldbach conjectures as well as the Twin Primes at the same time. I enjoy your channel.
in case you didn't see my twitter comment you can also restate it as every number after a certain point is equidistant from two primes ( technically if you count distance=0 that's from 2 on, for distance>0 that's 4 on.)
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Interviews with David Eisenbud are always great!
His voice is so soothing!!
Uncle Petros and Goldbach's Conjecture
Likewise. I was once given the book as a present because of my (and I stress, layman's) interest in the problem.
Longer videos plssssss :)
woohoo, a new math book recommendation. Maybe ask them all for suggestions. I'd love to grow my math library a little.
Brady, I love hearing the questions you ask that catch the mathematicians! You're much smarter than you let on ;)
Bravo librarians! Those shelves are very tidy. Not the way I remember UC Berkeley libraries at all...
Thanks again for the extra footage! :-)
yayyyy! extra content as always!
Does anybody know if there's a link between Goldbach's conjecture and the Riemann hypothesis? I mean, imagine there's a connection that asserts that Goldbach's conjecture implies Riemann hypothesis :)))
Relating to Brady's question: Do the primes appear to be a minimal set for generating all even numbers > 4 by adding two elements? If not, what primes can we skip and still get sums for the even numbers calculated so far?
The numbers 3, 5, and 7 are necessary because 6 = 3 + 3 and 12 = 5 + 7 are the only possible sums for 6 and 12. A commenter on the 210 video claims that 12 is the largest known even number with only one sum. This implies that as far as we know the Goldbach conjecture is still true for the set of prime numbers \ {11}, since if every even number > 12 has at least two sums then they have at least one sum that doesn't use 11. So I'd like to know the next prime we can eliminate after eliminating 11.
@@martinepstein9826drg
It sounds like he's always in the middle of a yawn
"Number theory has exploded in recent years." says Eisenbud. "I know!" says Brady, giggling and rubbing his hands together.
Yes, nice conjecture 👍
Ahh, the Numberphile2 people. The real math(s) nuts.
Lemme get my tea real quick.
Alright, now I can enjoy this with class.
My grandfather solved the Goldbach problem. We are from Kazakhstan. We just dont know how to publish it.
Wow it's so lonely here lol
Ahmad Max ikr?
Hahaha ... you can tell it's a math library by all of the yellow Springer-Verlag textbooks.
I have waited for this for many days
Is it just my machine or is the audio really fuzzy?
do an episode on five twenty
Will anybody who solves the Golbach win the Fields Medal or Abel Prize?? What other prizes would be on offer??
Could there exist mathematical truths that cannot be proven? If so, could that statement itself be proven to be true? If not, could it be proven to be false? (Don't say "axioms".)
Lookup the continuum hypothesis. It was proven that starting from the ZFC axioms, you can neither prove it or disprove it.
Joren Heit, *yes*, and *yes*.
I think Gödel's incompleteness theorems are also about this topic
Lilla Kiss Ah! I thought that had to do with any set of axioms being unable to prove its own consistency, but that's actually just the second part of it. Thanks :-)
Hexameron So that would be a great example! Thanks, I didn't know about this :-)
With the method of decomposition of each pair in pairs of addends; Enfer Diez : They prove that in todo 2n there is always 2n - p = p_{1}. See : video Conjectura de Goldbach (Irrefutable).
From which even number on can even numbers be expressed as two distinct primes? Let's see.
2 NO
4 = 2+2 but those are not distinct so NO
6=3+3 so NO
8=3+5 YES.
Will this keep going let's see.
10=3+7 YES
12=5+7 YES
14=3+11 YES
16=5+11 YES
18=7+11 YES
20=3+17 YES
22=5+17 YES
24=5+19 YES
It seems to go on forever can somebody prove to me that an even number above 6 can always be expressed as two distinct primes? TY
Lol probably no one will be able to proof this or otherwise goldbach conjecture should be proven
If this can be proven, goldbach's conjecture is also proven
26=7+19
28=5+23
30=11+19
32=13+19
34=11+23
36=13+23
38=9+29
40=11+29
You can equate goldbach's conjecture to every number greater than some value, is equidistant from two primes. Furthermore, the twin prime conjecturecan be restated as 6m-1,6m+1 is a goldbach partition of 12m infinitely often. The statement about two primes being a certain even number apart can be turned to general statements about goldbach partitions.
@@htmlguy88 what does 12m mean?
Why did he correct himself with 'arbitrarily many primes' instead of just saying 'infinitely many primes' at 3:09? Is there a difference?
I think the difference is that no individual arithmetic progression necessarily has an infinite number of primes, but there is no limit on how many there can be in one.
Yes, I think so. Infinitely many primes in an arithmetic progression would mean that every arithmetic progression has infinitely many primes, which is false (2,4,6,8,10,...). Arbitrarily many primes in an arithmetic progression means that for any number k you can find an arithmetic progression with more than k primes. (Edit: too late :/)
infinite and arbitrarily large are not the same thing. there are arbitrarily large natural numbers, but there are no infinite natural numbers
There are infinitely many natural numbers just as there are arbitrarily many natural numbers
Oh right, that makes a lot more sense. I guess it's just a consequence of communicating through language instead of maths. Thanks!
The equations and proof that defines the prime number sequence were proven five years ago and subsequently the proofs to the Strong and Weak Goldbach conjectures as well as the Twin Primes at the same time. I enjoy your channel.
“Here at Berkeley“
Oh god...
I have discovered a truly marvelous proof of the Goldbach conjecture, which this comment section is too narrow to contain...
33rd! So few comments, it's beautiful!
I proved this conjecture, really
in case you didn't see my twitter comment you can also restate it as every number after a certain point is equidistant from two primes ( technically if you count distance=0 that's from 2 on, for distance>0 that's 4 on.)
Why should I use -1/12 to solved Goldbach conjeture?
This proof is so elementary.
My comment can be written in the form of n^3
Guess the value of n :) will comment on the first person who gets it correct
Ashley2 Khoo your mama
n = cuberoot (My comment can be written in the form of n^3)
My comment # cannot be expressed as the sum of two primes... Poor number 11
Cyanide Cloud it can be expressed as the sum of one prime.
hsha
The (2 + 5)th comment! Woohoo