3:40 The Prime Number Theorem actually says that the probability that a uniformly random integer in the interval [1, 2N] is prime is close to 1/log(2N) (and we know we're choosing an odd number, so we can double that). The size of a and b themselves don't need to factor into this at all. Of course, what you use in your heuristic argument is a refinement, and a more precise one at that, but it does mean we get into trouble concerning the sizes of a and b, which the regular PNT doesn't need to care about at all.
As 9 is the smallest odd composite, no number smaller than 18 can be the sum of two odd composites. Similarly, as the next odd composite is 15, every numbers smaller than 24 (except 18) cannot be the sum of two composites. Next odd composite is 21, which means from the numbers smaller than 40, only 9+9, 9+15, and 9+21 can be written as the sum of two composites. Less exhaustion that way.
It would be interesting to go into some of what's made Goldbach so intractable. There are lots of reasons we should *expect* it to be true, with the density of primes being only one. (Consequences of its falsehood would also be interesting.)
40=15+25, 42=21+21, 44=9+35. Any even number beyond 44 is of 6m + 40 or 42, or 44. So they can be sum of 2 odd composites (because 3 divides15, 21, and 9).
At around 3.5 min, you begin to estimate the # of ways of choosing a pair of (+ve) odd integers, a and b, that sum to 2N. You get N, the # of odd ints from 1 to 2N-1. But you've neglected that this counts those pairs twice, because they can be chosen in either order. So the actual count is ½N. This doesn't, however, make any real dent in your argument. Cool problem; cool video! I like! Fred
In each case, the m's are chosen to ensure the stuff in the parentheses is greater than 1, but in each of the cases, the composite constant we add (9, 35, 25) is divisible by m. Is this a coincidence? It's kind of a cool one, and I can't find any rhyme or reason to it needing to be. For instance, if m>=7 didn't work to prove 6m+2, m>=12 would've worked with (6m-63)+65=3(2m-21)+65, and 65 would not be divisible by 12. But it seems like a coincidence that, in all the minimum cases, the constant is divisible by m. Or is my math off and I'm missing something obvious?
In every case 6m+k, we need "a multiple of 3, that's k below a composite", so that you can subtract the multiple of 3 (giving you the possibility of factoring a 3 out) and leaving you with a composite. In other words, we need some number 3a and then some 3a+k that's composite, and if we use this then 6m+k = 6m-3a+3a+k = 3(2m-a)+(3a+k). So for the given a, m needs to be at least (a+3)/2 in order for 3(2m-a) to be composite. In other words, the minimum m is always (a+3)/2. We may also note that since 3a+k is an odd composite, and k is even, a must be odd. So your question is, rephrased: If a is the smallest number such that 3a+k is composite, is one of the factors of 3a+k always going to be (a+3)/2? If that is the case, then (3a+k)/((a+3)/2) is an integer, which means 2(3a+k)/(a+3) = 2(3a+9)/(a+3) + 2(k-9)/(a+3) is also an integer, so 2(k-9) is divisible by a+3. Remember that a+3 is even, and obviously greater than 3. For k=0: a+3 divides 2(-9) = -18, so smallest a+3 is 6. That gives us m = (a+3)/2 = 3. For k=2: a+3 divides 2(2-9) = -14, so smallest a+3 is 14. That gives us m = (a+3)/2 = 7. For k=4: a+3 divides 2(4-9) = -10, so smallest a+3 is 10. That gives us m = (a+3)/2 = 5. There is some structure here, though I'm not well enough versed in number theory to properly find it.
This is not required. For example, we can say that if m is greater than 7, then 6m+2=3(2m-6)+20, and we note that 20 is not divisible by 7. The idea is that you convert the left side into two composite terms, and in our example we adopted one of the two terms from Multiples of 3, whatever number we want m to be greater than or equal to, we can make the second term on the right side divisible or not divisible by that number because the condition is to add to the right side a multiple of 3, and since it is a common multiple of 3, it Any multiplier can be chosen that meets the need. I can translate my words with a mathematical proof. However, do not think that my words are incorrect.
Who here knows about the complex Goldbach Conjecture? All natural nⲷ2 mod 4 are the sums of 2 *natural gaussian primes* (primes that are 1 below a multiple of 4) Meanwhile all natural nⲷ0 mod 4 are the *differences* of 2 natural gaussian primes
Dear alreasy proved goldbach conjecture in 2021 in more than 4 ways since 2021 but confirming takes time . I dont know how to show. So submitted first for copyright . After copyright approval i will upload on different platforms and journal submission.please pray for my work to be accepted by all
Interesting video, but wildly distracting to see "likelyhood" misspelled on the board and to hear "composite" pronounced in such a peculiar way (why not the same stress pattern as "competent"?).😉 It would be just a bit too picky to expect any English-speaking academic to pronounce Goldbach as anything but Goldbark, but that said, don't a lot of people manage to pronounce the composer Bach's name with a German "ch" sound and not as Bark?
you know your predecessors 60 yrs ago have proved such adding-compositing could be strickly two primes based or shortly 2+2, which make your conjecture a joke.
I solved Goldbach conjecture. The key is that all primes, be they even or odd, is and must be, the difference of two squares. Thus we may substitute odd with prime, i.e., If (2n=odd +odd') we can substitute odd with prime, so Goldbach conjecture is true.
@@stephenaustin3026 1. Consider a Pythagorean spiral . 2. Square all sides creating a QECian spiral. 3. Note that every non-unit side is SIMULTANEOUSLY both even (a height) and odd (a hypotenuse) thus every non-unit side is composite. 4. Every non unit side is and must be a difference of two squares (hypotenuse^2-base^2) 5. Since every odd is a difference of 2 squares and every prime, including 2, is odd (all even integers can be proven to be the difference of two squares) then all non-unit integers are composite.
Suppose that p and q are two prime numbers and that n, n1, n2 are even numbers, it is obvious that when multiplying a prime number (which is an odd number) by 2, the result is an even number, that is: 2xp=n1…………. (1) 2xq=n2…………. (2) Adding equations (1) and (2), we get: n1+n2=2p+2q…… (3) The sum of the even numbers n1 and n2 is a new even number n: n=n1+n2 So equation (3) becomes: n=2p+2q…….… (4) Let's divide equation (4) by 2: n/2=p+q…….… (5) Noting that dividing the even number n here will not give an odd number (because it results from multiplying an odd number by 2) because it is the sum of two even numbers, and therefore the result of the division will be even. Assuming that N = n/2, where N is a new even number, equation (5) becomes: N=p+q The result is that an even number N is the sum of two prime numbers, which is a text of Goldbach's strong conjecture. Is this acceptable? Kindly explain, please
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3:40 The Prime Number Theorem actually says that the probability that a uniformly random integer in the interval [1, 2N] is prime is close to 1/log(2N) (and we know we're choosing an odd number, so we can double that). The size of a and b themselves don't need to factor into this at all.
Of course, what you use in your heuristic argument is a refinement, and a more precise one at that, but it does mean we get into trouble concerning the sizes of a and b, which the regular PNT doesn't need to care about at all.
As 9 is the smallest odd composite, no number smaller than 18 can be the sum of two odd composites. Similarly, as the next odd composite is 15, every numbers smaller than 24 (except 18) cannot be the sum of two composites. Next odd composite is 21, which means from the numbers smaller than 40, only 9+9, 9+15, and 9+21 can be written as the sum of two composites. Less exhaustion that way.
You missed 15+21, but yes, easier argument
@@silver6054 well, to my shame I have to admit that I estimated it to be 46 🙄
@@OliverJennrich 🤣
You have to consider all odd composites less than 40-9, therefore also 25+9 and 27+9.
You missed 15+15 (though its already covered by 9+21), 21+15, 25+9, 27+9 (again double covered)
It would be interesting to go into some of what's made Goldbach so intractable. There are lots of reasons we should *expect* it to be true, with the density of primes being only one. (Consequences of its falsehood would also be interesting.)
0:50 nice way of making concepts more clear 👍
40=15+25, 42=21+21, 44=9+35. Any even number beyond 44 is of 6m + 40 or 42, or 44. So they can be sum of 2 odd composites (because 3 divides15, 21, and 9).
I can't see how that follows. Can you give a worked example for each case ofthe constants 40, 42 and 44?
At around 3.5 min, you begin to estimate the # of ways of choosing a pair of (+ve) odd integers, a and b, that sum to 2N. You get N, the # of odd ints from 1 to 2N-1.
But you've neglected that this counts those pairs twice, because they can be chosen in either order. So the actual count is ½N.
This doesn't, however, make any real dent in your argument.
Cool problem; cool video! I like!
Fred
I think it's really cool how, while the conjecture itself is intractable, an extremely similar statement is! Good video
The difference is that we can't provide a constructive proof for Goldbach .
14:58
1:09 two plus two is four minus one that three quick mafs.
for goldbach conjecture heuristic, multiplicative property on probability equation does not hold as a and b are dependent variables, right?
This is a very educational video, thank you.
In each case, the m's are chosen to ensure the stuff in the parentheses is greater than 1, but in each of the cases, the composite constant we add (9, 35, 25) is divisible by m. Is this a coincidence? It's kind of a cool one, and I can't find any rhyme or reason to it needing to be.
For instance, if m>=7 didn't work to prove 6m+2, m>=12 would've worked with (6m-63)+65=3(2m-21)+65, and 65 would not be divisible by 12. But it seems like a coincidence that, in all the minimum cases, the constant is divisible by m.
Or is my math off and I'm missing something obvious?
In every case 6m+k, we need "a multiple of 3, that's k below a composite", so that you can subtract the multiple of 3 (giving you the possibility of factoring a 3 out) and leaving you with a composite.
In other words, we need some number 3a and then some 3a+k that's composite, and if we use this then 6m+k = 6m-3a+3a+k = 3(2m-a)+(3a+k).
So for the given a, m needs to be at least (a+3)/2 in order for 3(2m-a) to be composite. In other words, the minimum m is always (a+3)/2.
We may also note that since 3a+k is an odd composite, and k is even, a must be odd.
So your question is, rephrased:
If a is the smallest number such that 3a+k is composite, is one of the factors of 3a+k always going to be (a+3)/2?
If that is the case, then (3a+k)/((a+3)/2) is an integer, which means
2(3a+k)/(a+3) = 2(3a+9)/(a+3) + 2(k-9)/(a+3)
is also an integer, so 2(k-9) is divisible by a+3.
Remember that a+3 is even, and obviously greater than 3.
For k=0: a+3 divides 2(-9) = -18, so smallest a+3 is 6. That gives us m = (a+3)/2 = 3.
For k=2: a+3 divides 2(2-9) = -14, so smallest a+3 is 14. That gives us m = (a+3)/2 = 7.
For k=4: a+3 divides 2(4-9) = -10, so smallest a+3 is 10. That gives us m = (a+3)/2 = 5.
There is some structure here, though I'm not well enough versed in number theory to properly find it.
This is not required. For example, we can say that if m is greater than 7, then 6m+2=3(2m-6)+20, and we note that 20 is not divisible by 7. The idea is that you convert the left side into two composite terms, and in our example we adopted one of the two terms from Multiples of 3, whatever number we want m to be greater than or equal to, we can make the second term on the right side divisible or not divisible by that number because the condition is to add to the right side a multiple of 3, and since it is a common multiple of 3, it Any multiplier can be chosen that meets the need. I can translate my words with a mathematical proof. However, do not think that my words are incorrect.
Can you make a video on uncountably infinite dimensional vector spaces and whatever Interesting properties you can find about them.
"You don't mean this very gold Asian box?"
Who here knows about the complex Goldbach Conjecture?
All natural nⲷ2 mod 4 are the sums of 2 *natural gaussian primes* (primes that are 1 below a multiple of 4)
Meanwhile all natural nⲷ0 mod 4 are the *differences* of 2 natural gaussian primes
Does anyone know of the pattern to which all numbers are constructed ??
The product pattern is quite simple. Would this make a good video?
The proof of the list by exhaustion works. I started grinding it out by hand and boy am I tired.
I ran out of space in my margin.
"Likelihood"
We should exclude a = 1
Everyone know in our university this statement was came from statement word replacement
Yes, your are taking about etymology, evolution of lexicon?
The two even is not independent
Very interesting stuff 👌
Every even integer is the sum of at least two prime integers (meaning it includes negative whole numbers) 🙂
Dear alreasy proved goldbach conjecture in 2021 in more than 4 ways since 2021 but confirming takes time . I dont know how to show. So submitted first for copyright . After copyright approval i will upload on different platforms and journal submission.please pray for my work to be accepted by all
Interesting video, but wildly distracting to see "likelyhood" misspelled on the board and to hear "composite" pronounced in such a peculiar way (why not the same stress pattern as "competent"?).😉 It would be just a bit too picky to expect any English-speaking academic to pronounce Goldbach as anything but Goldbark, but that said, don't a lot of people manage to pronounce the composer Bach's name with a German "ch" sound and not as Bark?
COMposite 😊
Goldbach wonders fill your eyes, smiles await you when you rise... ;)
you know your predecessors 60 yrs ago have proved such adding-compositing could be strickly two primes based or shortly 2+2, which make your conjecture a joke.
I solved Goldbach conjecture. The key is that all primes, be they even or odd, is and must be, the difference of two squares. Thus we may substitute odd with prime, i.e.,
If (2n=odd +odd') we can substitute odd with prime, so Goldbach conjecture is true.
The difference of two squares is always composite.
@@stephenaustin3026 1. Consider a Pythagorean spiral .
2. Square all sides creating a QECian spiral.
3. Note that every non-unit side is SIMULTANEOUSLY both even (a height) and odd (a hypotenuse) thus every non-unit side is composite.
4. Every non unit side is and must be a difference of two squares
(hypotenuse^2-base^2)
5. Since every odd is a difference of 2 squares and every prime, including 2, is odd (all even integers can be proven to be the difference of two squares) then all non-unit integers are composite.
Hello,
I have been able to demonstrate that Goldbach's conjecture is true.
Suppose that p and q are two prime numbers and that n, n1, n2 are even numbers, it is obvious that when multiplying a prime number (which is an odd number) by 2, the result is an even number, that is:
2xp=n1…………. (1)
2xq=n2…………. (2)
Adding equations (1) and (2), we get:
n1+n2=2p+2q…… (3)
The sum of the even numbers n1 and n2 is a new even number n:
n=n1+n2
So equation (3) becomes:
n=2p+2q…….… (4)
Let's divide equation (4) by 2:
n/2=p+q…….… (5)
Noting that dividing the even number n here will not give an odd number (because it results from multiplying an odd number by 2) because it is the sum of two even numbers, and therefore the result of the division will be even.
Assuming that N = n/2, where N is a new even number, equation (5) becomes:
N=p+q
The result is that an even number N is the sum of two prime numbers, which is a text of Goldbach's strong conjecture.
Is this acceptable?
Kindly explain, please
lol