Goldbach conjecture: Every integer $n>3$ is halfway between $2$ primes.
$begingroup$
Prove that the following conjecture is equivalent to the strong Goldbach conjecture:
Every integer $n>3$ is halfway between $2$ primes.
I'm able to prove it, but i don't have much experience in writing proofs, witch is why i need help to find a proper way to explain it. I'd like to have a proof that is as "short and sweet" as the conjecture itself. The shorter the better!
What i have so far:
If $p$ and $q$ are a Goldbach's partition of an even integer $2n$, then:
$$ 2n=p+q $$
The midpoint between $p$ and $q$ is:
$$frac{p+q}{2}=frac{2n}{2}=n$$
Therefore, if an even integer $2n$ can be written as the sum of $2$ primes, $n$ is halfway between those $2$ primes.
number-theory prime-numbers integers prime-gaps goldbachs-conjecture
edited Jan 2 at 15:46
mrtaurho
6,09271641
asked Jan 2 at 15:22
François HuppéFrançois Huppé
365111
$endgroup$
add a comment |
$begingroup$
Prove that the following conjecture is equivalent to the strong Goldbach conjecture:
Every integer $n>3$ is halfway between $2$ primes.
I'm able to prove it, but i don't have much experience in writing proofs, witch is why i need help to find a proper way to explain it. I'd like to have a proof that is as "short and sweet" as the conjecture itself. The shorter the better!
What i have so far:
If $p$ and $q$ are a Goldbach's partition of an even integer $2n$, then:
$$ 2n=p+q $$
The midpoint between $p$ and $q$ is:
$$frac{p+q}{2}=frac{2n}{2}=n$$
Therefore, if an even integer $2n$ can be written as the sum of $2$ primes, $n$ is halfway between those $2$ primes.
number-theory prime-numbers integers prime-gaps goldbachs-conjecture
edited Jan 2 at 15:46
mrtaurho
6,09271641
asked Jan 2 at 15:22
François HuppéFrançois Huppé
365111
$endgroup$
$begingroup$
If an even number $n$ is the sum of two primes $a$ and $b$, where is $n/2$? And is $n/2$ integer?
$endgroup$
– user334732
Jan 2 at 15:28
$begingroup$
Hard to do this without seeing the proof you already have.
$endgroup$
– Randall
Jan 2 at 15:28
2
$begingroup$
If you give us the proof that you have, we will see if we can (and need to) improve it. Until then there isn't much we can do to help you.
$endgroup$
– Arthur
Jan 2 at 15:30
$begingroup$
yes i'm adding what i have so far! should not be long
$endgroup$
– François Huppé
Jan 2 at 15:31
add a comment |
$begingroup$
Prove that the following conjecture is equivalent to the strong Goldbach conjecture:
Every integer $n>3$ is halfway between $2$ primes.
I'm able to prove it, but i don't have much experience in writing proofs, witch is why i need help to find a proper way to explain it. I'd like to have a proof that is as "short and sweet" as the conjecture itself. The shorter the better!
What i have so far:
If $p$ and $q$ are a Goldbach's partition of an even integer $2n$, then:
$$ 2n=p+q $$
The midpoint between $p$ and $q$ is:
$$frac{p+q}{2}=frac{2n}{2}=n$$
Therefore, if an even integer $2n$ can be written as the sum of $2$ primes, $n$ is halfway between those $2$ primes.
number-theory prime-numbers integers prime-gaps goldbachs-conjecture
edited Jan 2 at 15:46
mrtaurho
6,09271641
asked Jan 2 at 15:22
François HuppéFrançois Huppé
365111
$endgroup$
Prove that the following conjecture is equivalent to the strong Goldbach conjecture:
Every integer $n>3$ is halfway between $2$ primes.
I'm able to prove it, but i don't have much experience in writing proofs, witch is why i need help to find a proper way to explain it. I'd like to have a proof that is as "short and sweet" as the conjecture itself. The shorter the better!
What i have so far:
If $p$ and $q$ are a Goldbach's partition of an even integer $2n$, then:
$$ 2n=p+q $$
The midpoint between $p$ and $q$ is:
$$frac{p+q}{2}=frac{2n}{2}=n$$
Therefore, if an even integer $2n$ can be written as the sum of $2$ primes, $n$ is halfway between those $2$ primes.
number-theory prime-numbers integers prime-gaps goldbachs-conjecture
number-theory prime-numbers integers prime-gaps goldbachs-conjecture
edited Jan 2 at 15:46
mrtaurho
6,09271641
asked Jan 2 at 15:22
François HuppéFrançois Huppé
365111
edited Jan 2 at 15:46
mrtaurho
6,09271641
asked Jan 2 at 15:22
François HuppéFrançois Huppé
365111
edited Jan 2 at 15:46
mrtaurho
6,09271641
edited Jan 2 at 15:46
mrtaurho
6,09271641
edited Jan 2 at 15:46
mrtaurho
6,09271641
6,09271641
asked Jan 2 at 15:22
François HuppéFrançois Huppé
365111
asked Jan 2 at 15:22
François HuppéFrançois Huppé
365111
asked Jan 2 at 15:22
François HuppéFrançois Huppé
365111
365111
$begingroup$
If an even number $n$ is the sum of two primes $a$ and $b$, where is $n/2$? And is $n/2$ integer?
$endgroup$
– user334732
Jan 2 at 15:28
$begingroup$
Hard to do this without seeing the proof you already have.
$endgroup$
– Randall
Jan 2 at 15:28
2
$begingroup$
If you give us the proof that you have, we will see if we can (and need to) improve it. Until then there isn't much we can do to help you.
$endgroup$
– Arthur
Jan 2 at 15:30
$begingroup$
yes i'm adding what i have so far! should not be long
$endgroup$
– François Huppé
Jan 2 at 15:31
add a comment |
$begingroup$
If an even number $n$ is the sum of two primes $a$ and $b$, where is $n/2$? And is $n/2$ integer?
$endgroup$
– user334732
Jan 2 at 15:28
$begingroup$
Hard to do this without seeing the proof you already have.
$endgroup$
– Randall
Jan 2 at 15:28
2
$begingroup$
If you give us the proof that you have, we will see if we can (and need to) improve it. Until then there isn't much we can do to help you.
$endgroup$
– Arthur
Jan 2 at 15:30
$begingroup$
yes i'm adding what i have so far! should not be long
$endgroup$
– François Huppé
Jan 2 at 15:31
$begingroup$
If an even number $n$ is the sum of two primes $a$ and $b$, where is $n/2$? And is $n/2$ integer?
$endgroup$
– user334732
Jan 2 at 15:28
$begingroup$
If an even number $n$ is the sum of two primes $a$ and $b$, where is $n/2$? And is $n/2$ integer?
$endgroup$
– user334732
Jan 2 at 15:28
$begingroup$
Hard to do this without seeing the proof you already have.
$endgroup$
– Randall
Jan 2 at 15:28
$begingroup$
Hard to do this without seeing the proof you already have.
$endgroup$
– Randall
Jan 2 at 15:28
2
2
$begingroup$
If you give us the proof that you have, we will see if we can (and need to) improve it. Until then there isn't much we can do to help you.
$endgroup$
– Arthur
Jan 2 at 15:30
$begingroup$
If you give us the proof that you have, we will see if we can (and need to) improve it. Until then there isn't much we can do to help you.
$endgroup$
– Arthur
Jan 2 at 15:30
$begingroup$
yes i'm adding what i have so far! should not be long
$endgroup$
– François Huppé
Jan 2 at 15:31
$begingroup$
yes i'm adding what i have so far! should not be long
$endgroup$
– François Huppé
Jan 2 at 15:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So, let's do the equivalence.
Say Goldbach's conjecture is true, and take an integer $n>3$. Then there are primes $p, q$ such that $p+q = 2n$, and therefore $n = frac{p+q}2$ is the midpoint between $p$ and $q$.
On the other hand, let's say your conjecture is true, and let $2n>6$ be an even number. Then there are primes $p, q$ such that $n$ is the midpoint between $p$ and $q$. In other words, $frac{p+q}2 = n$, which transforms into $p+q = 2n$, and we have shown that the arbitrary even number $2n$ is the sum of two primes.
Thus either conjecture may be used to prove the other, and they are equivalent.
(I'm assuming that the specifics of whether Goldbach's conjecture starts at $4$ or $6$ or $8$ isn't the important part of the conjecture. If you include those cases, then no, the two aren't entirely equivalent.)
answered Jan 2 at 15:28
ArthurArthur
121k7121207
$endgroup$
$begingroup$
The Goldbach's conjecture starts is not really important for the main objective, but still it is interesting.. who knows, maybe 1 is prime after all, this would bring us to n>1 :)
$endgroup$
– François Huppé
Jan 2 at 15:55
$begingroup$
What about a case where n is prime, with only 1 Goldbach partition (n,n) ? Wouldn't it respects Goldbach's conjecture but not mine ?
$endgroup$
– François Huppé
Jan 5 at 1:22
add a comment |
Your Answer
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1 Answer
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$begingroup$
So, let's do the equivalence.
Say Goldbach's conjecture is true, and take an integer $n>3$. Then there are primes $p, q$ such that $p+q = 2n$, and therefore $n = frac{p+q}2$ is the midpoint between $p$ and $q$.
On the other hand, let's say your conjecture is true, and let $2n>6$ be an even number. Then there are primes $p, q$ such that $n$ is the midpoint between $p$ and $q$. In other words, $frac{p+q}2 = n$, which transforms into $p+q = 2n$, and we have shown that the arbitrary even number $2n$ is the sum of two primes.
Thus either conjecture may be used to prove the other, and they are equivalent.
(I'm assuming that the specifics of whether Goldbach's conjecture starts at $4$ or $6$ or $8$ isn't the important part of the conjecture. If you include those cases, then no, the two aren't entirely equivalent.)
answered Jan 2 at 15:28
ArthurArthur
121k7121207
$endgroup$
$begingroup$
The Goldbach's conjecture starts is not really important for the main objective, but still it is interesting.. who knows, maybe 1 is prime after all, this would bring us to n>1 :)
$endgroup$
– François Huppé
Jan 2 at 15:55
$begingroup$
What about a case where n is prime, with only 1 Goldbach partition (n,n) ? Wouldn't it respects Goldbach's conjecture but not mine ?
$endgroup$
– François Huppé
Jan 5 at 1:22
add a comment |
$begingroup$
So, let's do the equivalence.
Say Goldbach's conjecture is true, and take an integer $n>3$. Then there are primes $p, q$ such that $p+q = 2n$, and therefore $n = frac{p+q}2$ is the midpoint between $p$ and $q$.
On the other hand, let's say your conjecture is true, and let $2n>6$ be an even number. Then there are primes $p, q$ such that $n$ is the midpoint between $p$ and $q$. In other words, $frac{p+q}2 = n$, which transforms into $p+q = 2n$, and we have shown that the arbitrary even number $2n$ is the sum of two primes.
Thus either conjecture may be used to prove the other, and they are equivalent.
(I'm assuming that the specifics of whether Goldbach's conjecture starts at $4$ or $6$ or $8$ isn't the important part of the conjecture. If you include those cases, then no, the two aren't entirely equivalent.)
answered Jan 2 at 15:28
ArthurArthur
121k7121207
$endgroup$
$begingroup$
The Goldbach's conjecture starts is not really important for the main objective, but still it is interesting.. who knows, maybe 1 is prime after all, this would bring us to n>1 :)
$endgroup$
– François Huppé
Jan 2 at 15:55
$begingroup$
What about a case where n is prime, with only 1 Goldbach partition (n,n) ? Wouldn't it respects Goldbach's conjecture but not mine ?
$endgroup$
– François Huppé
Jan 5 at 1:22
add a comment |
$begingroup$
So, let's do the equivalence.
Say Goldbach's conjecture is true, and take an integer $n>3$. Then there are primes $p, q$ such that $p+q = 2n$, and therefore $n = frac{p+q}2$ is the midpoint between $p$ and $q$.
On the other hand, let's say your conjecture is true, and let $2n>6$ be an even number. Then there are primes $p, q$ such that $n$ is the midpoint between $p$ and $q$. In other words, $frac{p+q}2 = n$, which transforms into $p+q = 2n$, and we have shown that the arbitrary even number $2n$ is the sum of two primes.
Thus either conjecture may be used to prove the other, and they are equivalent.
(I'm assuming that the specifics of whether Goldbach's conjecture starts at $4$ or $6$ or $8$ isn't the important part of the conjecture. If you include those cases, then no, the two aren't entirely equivalent.)
answered Jan 2 at 15:28
ArthurArthur
121k7121207
$endgroup$
So, let's do the equivalence.
Say Goldbach's conjecture is true, and take an integer $n>3$. Then there are primes $p, q$ such that $p+q = 2n$, and therefore $n = frac{p+q}2$ is the midpoint between $p$ and $q$.
On the other hand, let's say your conjecture is true, and let $2n>6$ be an even number. Then there are primes $p, q$ such that $n$ is the midpoint between $p$ and $q$. In other words, $frac{p+q}2 = n$, which transforms into $p+q = 2n$, and we have shown that the arbitrary even number $2n$ is the sum of two primes.
Thus either conjecture may be used to prove the other, and they are equivalent.
(I'm assuming that the specifics of whether Goldbach's conjecture starts at $4$ or $6$ or $8$ isn't the important part of the conjecture. If you include those cases, then no, the two aren't entirely equivalent.)
answered Jan 2 at 15:28
ArthurArthur
121k7121207
edited Jan 2 at 15:48
answered Jan 2 at 15:28
ArthurArthur
121k7121207
answered Jan 2 at 15:28
ArthurArthur
121k7121207
answered Jan 2 at 15:28
ArthurArthur
121k7121207
121k7121207
$begingroup$
The Goldbach's conjecture starts is not really important for the main objective, but still it is interesting.. who knows, maybe 1 is prime after all, this would bring us to n>1 :)
$endgroup$
– François Huppé
Jan 2 at 15:55
$begingroup$
What about a case where n is prime, with only 1 Goldbach partition (n,n) ? Wouldn't it respects Goldbach's conjecture but not mine ?
$endgroup$
– François Huppé
Jan 5 at 1:22
add a comment |
$begingroup$
The Goldbach's conjecture starts is not really important for the main objective, but still it is interesting.. who knows, maybe 1 is prime after all, this would bring us to n>1 :)
$endgroup$
– François Huppé
Jan 2 at 15:55
$begingroup$
What about a case where n is prime, with only 1 Goldbach partition (n,n) ? Wouldn't it respects Goldbach's conjecture but not mine ?
$endgroup$
– François Huppé
Jan 5 at 1:22
$begingroup$
The Goldbach's conjecture starts is not really important for the main objective, but still it is interesting.. who knows, maybe 1 is prime after all, this would bring us to n>1 :)
$endgroup$
– François Huppé
Jan 2 at 15:55
$begingroup$
The Goldbach's conjecture starts is not really important for the main objective, but still it is interesting.. who knows, maybe 1 is prime after all, this would bring us to n>1 :)
$endgroup$
– François Huppé
Jan 2 at 15:55
$begingroup$
What about a case where n is prime, with only 1 Goldbach partition (n,n) ? Wouldn't it respects Goldbach's conjecture but not mine ?
$endgroup$
– François Huppé
Jan 5 at 1:22
$begingroup$
What about a case where n is prime, with only 1 Goldbach partition (n,n) ? Wouldn't it respects Goldbach's conjecture but not mine ?
$endgroup$
– François Huppé
Jan 5 at 1:22
add a comment |
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$begingroup$
If an even number $n$ is the sum of two primes $a$ and $b$, where is $n/2$? And is $n/2$ integer?
$endgroup$
– user334732
Jan 2 at 15:28
$begingroup$
Hard to do this without seeing the proof you already have.
$endgroup$
– Randall
Jan 2 at 15:28
2
$begingroup$
If you give us the proof that you have, we will see if we can (and need to) improve it. Until then there isn't much we can do to help you.
$endgroup$
– Arthur
Jan 2 at 15:30
$begingroup$
yes i'm adding what i have so far! should not be long
$endgroup$
– François Huppé
Jan 2 at 15:31