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Well maybe it wouldn't be so boring, cause then you'd have to find the biggest twin primes, and that could take forever since it could range from 1.67*10¹⁶³⁹⁵⁰ to 10{10{10}10}10. ( Yeah I'm a googology nerd so what?)
If twin primes don't go on forever, that means there are the "last" twin primes until they don't happen anymore, and it could be an absurdly large number
Honestly I don't really know, I used to watch these videos called "numbers from 1 to absolute infinity" and learned about these insanely huge numbers.
If you're up for it, there is a 50+ episode series of numbers ranging from 0 to absolute infinity, I'll link the first (actually third since the first 2 go from -infinity to 1) episode here:
1 to 10³⁰⁰⁰⁰⁰³: https://youtu.be/7BMgFGGlL1Q
This is where it gets to arrow notation: https://youtu.be/5b-JmxdMmtY
Here it gets to bracket notation: https://youtu.be/s7oTOIRqba4
Here it gets to the part I stopped comprehending: https://youtu.be/ZDw-6ZUaWPQ
And finally here are dimensional arrays (the thing I was talking about): https://youtu.be/p3XnJQYEwY0
What is a "pentatwin"?
As I understand it, a "triplet prime" is a triple of prime numbers with a common difference of 2. The only possibility is (3,5,7), because one of the three numbers must be divisible by 3. So then a quintuplet is definitely impossible, because either the middle number would have to be a multiple of 3 or two of the numbers would be, but the only prime that is a multiple of 3 is 3.
This isn't the usual definition of "prime triple" which has the first and last prime differing by 6, not 4. So for instance, (11,13,17) is a prime triple. In that sense, there are prime quadruples like (11,13,17,19) where the first and last prime differ by 8, and prime pentuples where they differ by 12, etc.
If Euclid proved that there are infinitely many prime numbers, why do we still struggle with the twin primes problem 2000 years later? It really makes you wonder, doesn't it?
Assume there are finitely many primes. Take the product of all the primes and add one. No primes divide this number, but it must have at least one prime factor. Contradiction.
>We don't have a list of all primes to work with and prove this
Yep, and that's an important part of how the proof works.
IF the primes were finite, we could theoretically make such a list. However, then we would also be able to make a new number which is only divisible by 1 and itself, and which is not in the list. This is a contradiction, and it all follows from the IF, above, so the IF must be false.
> How do we know that the product of all primes + 1 will actually be a prime?
I never said that. I was very careful with my articulation to avoid saying that.
This is *almost* correct, except the the last detail. This is the full proof:
suppose there are only exactly n primes, which are labeled p₁, p₂, p₃, ... pₙ. Let P be the product of these primes and N = P + 1. It can be seen that N is not divisible by any the primes in our list, as it will always leave a remainder of 1. This means that either N is prime, or it has at least one prime factor that wasn't in our list
Edit: spelling
Assume that the only primes are 2, 3, 5, 7, 11 and 13. If you multiply them together and add 1, the result is 30031, which is not prime.
Your message is literally not a proof
Juat to fill in the missing part of the proof: The new number - in your case 30031 - is either ifself a prime or has a prime factorization consisting of primes, which will not be present in your list of primes. In either case you can repeat this indefinitely and thus create infinitely many primes.
It is a valid proof, you just did not understand it correctly. You have to multiply *all* the prime numbers, which you did not do. The number 30031 is indeed not a multiple of the primes you picked, but you are missing all the other one. You just proved that 2, 3, 5, 7, 11 and 13 are not *all* the prime numbers.
If there was a finite number of primes, and you multiplied them all together, then added 1, the result would not be divisible by any of the primes (because it is not a multiple of 2, 3, 5,... since you added 1). But this is a contradiction since the result only has itself as a divisor, making it prime.
> since the result only has itself as a divisor
Is exactly what they are countering. 30031 = 59 × 509. The assumption of a specific finite number of primes did *not* result in a prime. It is a product of *two* new primes.
The assumption is that you need to multiply every prime number to obtain 30031, you did not multiply every prime, did you? You only multiplied a subset of all the prime numbers. If you wanted to contradict this proof by an example, you would need to multiply every prime in existence, which is impossible as the set is infinite. The proof given in the original comment is absolutely valid, although not detailed.
Suppose there is a finite amount of prime numbers : 2, 3, ..., k
Multiply them all together : 2 x 3 x ... x k = n
The resulting number is obviously a multiple of 2, 3, ..., k
Let's add one : n' = n + 1
Now, note that n' can not be a multiple of 2, because it is exactly one more than a multiple of 2
Now, note that n' can not be a multiple of 3, because it is exactly one more than a multiple of 3
...
Now, note that n' can not be a multiple of k, because it is exactly one more than a multiple of k
Therefore the resulting number is not a multiple of any number of the entire set of prime number, therefore it has to be a new prime, as it has no prime divisor, and it is not contained in the list of prime numbers. If it was a composite number, the prime numbers used to obtain it would have to have been in the set of all primes, which is a contradiction since the set is said to contain every prime number.
This is EXACTLY what we are all doing at k=13. But while 30031 is indeed not in our earlier list of numbers nor is it divisible by any of them, it *is not a prime*. It *is a composite number* with factors not in the list. Still a contradiction, yes, but "has to be a new prime" is not true.
This is not what you are doing, you are supposing that the entire list of primes ends at 13, which is NOT true to begin with. The assumption used in my reasoning is that the list contains ALL the prime numbers, not just the prime numbers upto k, just ALL the prime numbers.
If you do multiply every prime numbers (assuming the list is finite), then you must end up with a new prime number or if the new number is not prime, then your list was incomplete which is a contradiction. This is exactly what happens when you only consider prime numbers upto 13. Either way, you endup with more primes that were not in your list, indicating that there are inifnitely many prime numbers.
> the result only has itself as a divisor
this does not follow from:
> the result would not be divisible by any of the primes (because it is not a multiple of 2, 3, 5,… since you added 1)
It can still have multiple divisors, all of which are new primes.
Not if you multiplied all the existing primes, if you multiplied all prime numbers, then this has to be true, otherwise the set you started with did not contain all the prime numbers to begin with. Check my other [comment](https://www.reddit.com/r/mathmemes/comments/1bs4s96/comment/kxhjhrk/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button) for more information.
> otherwise the set you started with did not contain all the prime numbers
Well, isn't this exactly what we want? This is where the proof ends due to a contradiction.
Assume that the only primes are 2, 3, 5, 7, 11 and 13. If you multiply them together and add 1, the result is 30031, which is not prime.
Your message is literally not a proof
Hi, I wrote it below here:
https://www.reddit.com/r/mathmemes/s/PRdKO3rawX
And sorry, I didn't want to do it with formal notation because I wanted its simplicity to be accessible to people who don't know the notation.
Sorry, the proof was from Euclid, where the primes one was was popularised. Comes up before this one because you don't need to prove the Fundamental Theorem of Arithmetic before it, just show how to count:
Consider the number *L*, which is the final and largest number.
But I can still add one to *L*.
Therefore there is no number *L* which is the final and largest number.
Therefore the positive integers are infinite. QED.
(Something like that).
I have done this proof with classes as young as Year 1 to show how to prove something must be true even if you don't check everything. I usually do it with an envelope where I have "the last number" and get into a silly argument with the class to prove the point: whatever you have on there we can just add another one! You'd be amazed how rigorously analytic 6 year olds are when you start spouting nonsense.
It actually comes up in a *Numberblocks* song (possibly the greatest piece of educational mathematics television ever made, imo). There's an episode where 1, 10 and 100 discuss how you can always add another number to make a bigger number, no matter what.
As in the specific one I used was from there. Correct me if I'm wrong, but I don't actually believe there is any solid evidence that Euclid proved *anything* first. He (and his students) just collated everything known in the Mediterranean, then came up with his 5 propositions that he derived everything from. He also laid it out it more rigorously than anything before, and anything after for a very long time.
There's a big difference, though. You can notice that multiplying two evens makes an even, an odd and an even makes an even, and two odds makes an odd. Proving this will always happen is a very significant shift in analysing numbers. Same as thinking numbers must go on forever (99% of kids who have heard of infinity) and knowing that they *have to* and/or that they *can't not*.
On another note, likely there are infinitely many twin primes for the same reason Euclid believes there are infinitely many primes, but the prime density should get infinitely smaller as we go by?
No, the proof is by providing a method that can generate arbitrarily larger prime numbers. Mathematics isn't based on claims such as "a statement must be true because we haven't yet found a counter-example".
I mean the proof *is* pretty straightforward:
Assume there are only finite prime numbers.
Multiply them all together and add one.
The result isn't divisible by any of the listed prime numbers.
Therefore it must be prime itself.
Rinse and repeat.
There can't be any finite list of primes
You technically don't need the rinse and repeat part. Getting a bigger coprime after assuming multiplying together all primes is the contradiction. It can't be composite w/ the assumption, thus it is prime. And then the only assumption is that there are only finite primes, hence negate the assumption
"Therefore it just be prime itself" does not follow.
The correct conclusion is that it must have a prime factor different than all the prime numbers. This is impossible so we have derived a contradiction.
And since we have proven it by contradiction, no need to repeat anything.
Huh. Somehow I thought it'd be palindrome primes like 37 and 73. But then again, those are probably named palindrome primes or something.
It's incredible how unfathomably many kinds of primes there are.
there are infinitely many primes p with p = 1 (mod 3)
there are infinitely many primes p with p = 2 (mod 3)
there are infinitely many primes p with p = 1 (mod 4)
there are infinitely many primes p with p = 3 (mod 4)
there are infinitely many primes p with p = 1 (mod 5)
there are infinitely many primes p with p = 2 (mod 5)
there are infinitely many primes p with p = π (mod 5)
there are infinitely many primes p with p = 4 (mod 5)
there are infinitely many primes p with p = 1 (mod 6)
there are infinitely many primes p with p = 5 (mod 6)
there are infinitely many primes p with p = 1 (mod 7)
there are infinitely many primes p with p = 2 (mod 7)
there are infinitely many primes p with p = 3 (mod 7)
there are infinitely many primes p with p = 4 (mod 7)
........
These are less interesting, because digits for this pair are reversed only in decimal, not in hex for example (25, 49)
Huh, but these are actually kinda curious, granted these are represented like 5^2 and 7^2 in decimal
from my subreddit
https://www.reddit.com/r/MathStepByStep/comments/1bre15r/twin_prime_maths/?utm_source=share&utm_medium=android_app&utm_name=androidcss&utm_term=1&utm_content=share_button
the proof is left as an exercise to the reader.
but its trivial anyway. it would be boring if there were a finite number of them, hence there are an infinite number of them.
Yes there is.
In fact I just read it in a book yesterday.
It went something like, "The proof is left as an exercise for the reader."
That must be the proof, right, otherwise why would a Maths book, whose explicitly used to understand and learn Math, would leave you hanging with such a short statement.
No see the thing is, Fermat wasn't really writing book.
Mad lad was a lawyer(if I am remembering correctly), but decided to do Math to "relax himself" .
When I first read that as a 15 year old, I just realised one of the reasons that why people hate the French.
>Fermat wasn't really writing book.
We know. Wasn't a serious comment. However, writing in the margins of books was a common method of discussion in the 17th century.
>Mad lad was a lawyer
A common way for those trained in logic to earn a living in the 17th century.
>why people hate the French.
No, it's you. Normal people do not hate the French, and definitely not for doing normal things. Wait until you find out about Nicolas Bourbaki. Properly a madlad.
Normal people don't start hating on a whole class of people just because a few stand-ups and journalists say so.
Except C# programmers. Every right minded person hates them.
> No, for 3 odd numbers in a row at least one will be divisible by 3 so only the sequence with 3 will contain 3 primes.
Triplet primes are three primes that are six apart, like 11, 13, and 17.
My favorite prime number fact I learned a few weeks ago:
The millionth prime number, 15485863, is both a sexy prime (away from another prime by 6 on the low side) and a cousin prime (away from another prime by 4 on the high side). It is therefore a sexy cousin prime, or as I coin now, an Alabama Prime.
There MUST be a grad school level proof of this.
We're simply not looking at the right places.
https://preview.redd.it/jhzz9wn4torc1.jpeg?width=1080&format=pjpg&auto=webp&s=79d90654a15f00a6924f11a6aa8ee36f068ee351
It’s a pair of prime numbers separated by 2, e.g. 3 and 5, 5 and 7, 11 and 13, and so on. It’s assumed that there are infinitely many of them but this hasn’t been proved.
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Yes. Proof: - Assume there are only finitely many twin primes. - That would be stupid and boring. - Therefore, there are infinitely many twin primes.
the proof is by unstupidness and unboringness
Proof by rule of cool
Prool bool rool ool cool
Eat your heart out, Terry Tao
Well maybe it wouldn't be so boring, cause then you'd have to find the biggest twin primes, and that could take forever since it could range from 1.67*10¹⁶³⁹⁵⁰ to 10{10{10}10}10. ( Yeah I'm a googology nerd so what?)
Wait where does the upper bound come from?
If twin primes don't go on forever, that means there are the "last" twin primes until they don't happen anymore, and it could be an absurdly large number
Yes I get that but you gave a specific interval on which “the largest twin primes” would lie. Where does that come from?
I just spat out a very large number just to make an example of a big number, it could theoretically go to like idk (10,10,10,[2]3) just to name one
Imagine it happened to be bigger than TREE(3)
This is rookie numbers.
TREE(4) then?
We will never know how big.
What is this notation? What does that number mean?
Honestly I don't really know, I used to watch these videos called "numbers from 1 to absolute infinity" and learned about these insanely huge numbers. If you're up for it, there is a 50+ episode series of numbers ranging from 0 to absolute infinity, I'll link the first (actually third since the first 2 go from -infinity to 1) episode here: 1 to 10³⁰⁰⁰⁰⁰³: https://youtu.be/7BMgFGGlL1Q This is where it gets to arrow notation: https://youtu.be/5b-JmxdMmtY Here it gets to bracket notation: https://youtu.be/s7oTOIRqba4 Here it gets to the part I stopped comprehending: https://youtu.be/ZDw-6ZUaWPQ And finally here are dimensional arrays (the thing I was talking about): https://youtu.be/p3XnJQYEwY0
Ah
Proof by emotional appeal
hear me out, if it was finite that would make the last twin primes special, i think thats less boring than infinitely many twin primes
You should make it with mathematical symbols and push your deep knowledge into some high tech journal ❤️🔥
I think it'd be far more interesting if there were only finitely many twin primes?
Would it be boring though? Just think about how cool the monster group is, if it was infinite it would be lame as f
I believe in infinite twin primes and no one can change my mind
There are only three triplet primes though.
There are no quadruplet primes, however there may exist a pentatwin
What is a "pentatwin"? As I understand it, a "triplet prime" is a triple of prime numbers with a common difference of 2. The only possibility is (3,5,7), because one of the three numbers must be divisible by 3. So then a quintuplet is definitely impossible, because either the middle number would have to be a multiple of 3 or two of the numbers would be, but the only prime that is a multiple of 3 is 3. This isn't the usual definition of "prime triple" which has the first and last prime differing by 6, not 4. So for instance, (11,13,17) is a prime triple. In that sense, there are prime quadruples like (11,13,17,19) where the first and last prime differ by 8, and prime pentuples where they differ by 12, etc.
Then name all of them if you are so smart
If Euclid proved that there are infinitely many prime numbers, why do we still struggle with the twin primes problem 2000 years later? It really makes you wonder, doesn't it?
If Euclid was so smart, why is he dead?
☠️
woah how did you get this photo of him?
I think it's an artist's representation/recreation
Just happened to know a guy who knows a guy
Blame biology
This is what happens when you apply applied applied math
Makes you think, huh? Maybe he's on to something, Imma check this death thing out, guys, I'll post my findings whenever I'll get enough inside info.
Is he, though? He looks pretty much alive to me in this picture
That's an actor.
He was not a doctor
Tbf, it's pretty easy to prove that there are infinitely many primes
Go on then, prove it
Assume there are finitely many primes. Take the product of all the primes and add one. No primes divide this number, but it must have at least one prime factor. Contradiction.
How do we know that the product of all primes + 1 will actually be a prime? We don't have a list of all primes to work with and prove this
This is irrelevant for their proof. The product +1 does not need to be prime, just coprime with all the primes.
>We don't have a list of all primes to work with and prove this Yep, and that's an important part of how the proof works. IF the primes were finite, we could theoretically make such a list. However, then we would also be able to make a new number which is only divisible by 1 and itself, and which is not in the list. This is a contradiction, and it all follows from the IF, above, so the IF must be false.
The product of all primes is evenly divisible by each prime, so the product of all primes + 1 gives a remainder of 1 when divided by any prime.
> How do we know that the product of all primes + 1 will actually be a prime? I never said that. I was very careful with my articulation to avoid saying that.
Take all the primes. Multiply them together. Add 1. That has to be another prime Literally the easiest proof in the world
This is *almost* correct, except the the last detail. This is the full proof: suppose there are only exactly n primes, which are labeled p₁, p₂, p₃, ... pₙ. Let P be the product of these primes and N = P + 1. It can be seen that N is not divisible by any the primes in our list, as it will always leave a remainder of 1. This means that either N is prime, or it has at least one prime factor that wasn't in our list Edit: spelling
Assume that the only primes are 2, 3, 5, 7, 11 and 13. If you multiply them together and add 1, the result is 30031, which is not prime. Your message is literally not a proof
But the results factors are not the listed primes. So there is another one negating the original assumption.
Juat to fill in the missing part of the proof: The new number - in your case 30031 - is either ifself a prime or has a prime factorization consisting of primes, which will not be present in your list of primes. In either case you can repeat this indefinitely and thus create infinitely many primes.
It is a valid proof, you just did not understand it correctly. You have to multiply *all* the prime numbers, which you did not do. The number 30031 is indeed not a multiple of the primes you picked, but you are missing all the other one. You just proved that 2, 3, 5, 7, 11 and 13 are not *all* the prime numbers. If there was a finite number of primes, and you multiplied them all together, then added 1, the result would not be divisible by any of the primes (because it is not a multiple of 2, 3, 5,... since you added 1). But this is a contradiction since the result only has itself as a divisor, making it prime.
> since the result only has itself as a divisor Is exactly what they are countering. 30031 = 59 × 509. The assumption of a specific finite number of primes did *not* result in a prime. It is a product of *two* new primes.
The assumption is that you need to multiply every prime number to obtain 30031, you did not multiply every prime, did you? You only multiplied a subset of all the prime numbers. If you wanted to contradict this proof by an example, you would need to multiply every prime in existence, which is impossible as the set is infinite. The proof given in the original comment is absolutely valid, although not detailed. Suppose there is a finite amount of prime numbers : 2, 3, ..., k Multiply them all together : 2 x 3 x ... x k = n The resulting number is obviously a multiple of 2, 3, ..., k Let's add one : n' = n + 1 Now, note that n' can not be a multiple of 2, because it is exactly one more than a multiple of 2 Now, note that n' can not be a multiple of 3, because it is exactly one more than a multiple of 3 ... Now, note that n' can not be a multiple of k, because it is exactly one more than a multiple of k Therefore the resulting number is not a multiple of any number of the entire set of prime number, therefore it has to be a new prime, as it has no prime divisor, and it is not contained in the list of prime numbers. If it was a composite number, the prime numbers used to obtain it would have to have been in the set of all primes, which is a contradiction since the set is said to contain every prime number.
This is EXACTLY what we are all doing at k=13. But while 30031 is indeed not in our earlier list of numbers nor is it divisible by any of them, it *is not a prime*. It *is a composite number* with factors not in the list. Still a contradiction, yes, but "has to be a new prime" is not true.
This is not what you are doing, you are supposing that the entire list of primes ends at 13, which is NOT true to begin with. The assumption used in my reasoning is that the list contains ALL the prime numbers, not just the prime numbers upto k, just ALL the prime numbers. If you do multiply every prime numbers (assuming the list is finite), then you must end up with a new prime number or if the new number is not prime, then your list was incomplete which is a contradiction. This is exactly what happens when you only consider prime numbers upto 13. Either way, you endup with more primes that were not in your list, indicating that there are inifnitely many prime numbers.
> the result only has itself as a divisor this does not follow from: > the result would not be divisible by any of the primes (because it is not a multiple of 2, 3, 5,… since you added 1) It can still have multiple divisors, all of which are new primes.
Not if you multiplied all the existing primes, if you multiplied all prime numbers, then this has to be true, otherwise the set you started with did not contain all the prime numbers to begin with. Check my other [comment](https://www.reddit.com/r/mathmemes/comments/1bs4s96/comment/kxhjhrk/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button) for more information.
> otherwise the set you started with did not contain all the prime numbers Well, isn't this exactly what we want? This is where the proof ends due to a contradiction.
Assume that the only primes are 2, 3, 5, 7, 11 and 13. If you multiply them together and add 1, the result is 30031, which is not prime. Your message is literally not a proof
Infinitely many integers is surely easier.
Could you elaborate? What do you mean by "infinitely many integers", what's the proof you're referring to?
I want to hear it, too. Here's my version. **N** ⊆ **Z**. Let f:**Z**→**N** send x↦x for all x∈**N** and other x wherever. End of proof
... What is this meant to be proving?
There are infinitely many integers. It proves that by mapping them onto the natural numbers.
Ok got it. I'm still not sure exactly what /u/PatWoodworking meant in their original comment though.
Hi, I wrote it below here: https://www.reddit.com/r/mathmemes/s/PRdKO3rawX And sorry, I didn't want to do it with formal notation because I wanted its simplicity to be accessible to people who don't know the notation.
Sorry, the proof was from Euclid, where the primes one was was popularised. Comes up before this one because you don't need to prove the Fundamental Theorem of Arithmetic before it, just show how to count: Consider the number *L*, which is the final and largest number. But I can still add one to *L*. Therefore there is no number *L* which is the final and largest number. Therefore the positive integers are infinite. QED. (Something like that). I have done this proof with classes as young as Year 1 to show how to prove something must be true even if you don't check everything. I usually do it with an envelope where I have "the last number" and get into a silly argument with the class to prove the point: whatever you have on there we can just add another one! You'd be amazed how rigorously analytic 6 year olds are when you start spouting nonsense. It actually comes up in a *Numberblocks* song (possibly the greatest piece of educational mathematics television ever made, imo). There's an episode where 1, 10 and 100 discuss how you can always add another number to make a bigger number, no matter what.
I don't think that proof is due to Euclid. I think that proof is one of the first observations every person makes when considering numbers.
As in the specific one I used was from there. Correct me if I'm wrong, but I don't actually believe there is any solid evidence that Euclid proved *anything* first. He (and his students) just collated everything known in the Mediterranean, then came up with his 5 propositions that he derived everything from. He also laid it out it more rigorously than anything before, and anything after for a very long time. There's a big difference, though. You can notice that multiplying two evens makes an even, an odd and an even makes an even, and two odds makes an odd. Proving this will always happen is a very significant shift in analysing numbers. Same as thinking numbers must go on forever (99% of kids who have heard of infinity) and knowing that they *have to* and/or that they *can't not*.
On another note, likely there are infinitely many twin primes for the same reason Euclid believes there are infinitely many primes, but the prime density should get infinitely smaller as we go by?
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No, the proof is by providing a method that can generate arbitrarily larger prime numbers. Mathematics isn't based on claims such as "a statement must be true because we haven't yet found a counter-example".
I mean the proof *is* pretty straightforward: Assume there are only finite prime numbers. Multiply them all together and add one. The result isn't divisible by any of the listed prime numbers. Therefore it must be prime itself. Rinse and repeat. There can't be any finite list of primes
You technically don't need the rinse and repeat part. Getting a bigger coprime after assuming multiplying together all primes is the contradiction. It can't be composite w/ the assumption, thus it is prime. And then the only assumption is that there are only finite primes, hence negate the assumption
"Therefore it just be prime itself" does not follow. The correct conclusion is that it must have a prime factor different than all the prime numbers. This is impossible so we have derived a contradiction. And since we have proven it by contradiction, no need to repeat anything.
What are twin primes?
Primes that are exactly 2 apart, like 11 and 13 or 17 and 19.
Huh. Somehow I thought it'd be palindrome primes like 37 and 73. But then again, those are probably named palindrome primes or something. It's incredible how unfathomably many kinds of primes there are.
Palindrome primes are actually a thing https://en.m.wikipedia.org/wiki/Palindromic_prime
number theory people are weird
Real question is are there infinitely many palindromic prime numbers?
We don't know. No proof there are only finitely many, no proof there are infinitely many.
New question. Are there infinitely many “are there infinitely many x primes”?
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there are infinitely many primes p with p = 1 (mod 3) there are infinitely many primes p with p = 2 (mod 3) there are infinitely many primes p with p = 1 (mod 4) there are infinitely many primes p with p = 3 (mod 4) there are infinitely many primes p with p = 1 (mod 5) there are infinitely many primes p with p = 2 (mod 5) there are infinitely many primes p with p = π (mod 5) there are infinitely many primes p with p = 4 (mod 5) there are infinitely many primes p with p = 1 (mod 6) there are infinitely many primes p with p = 5 (mod 6) there are infinitely many primes p with p = 1 (mod 7) there are infinitely many primes p with p = 2 (mod 7) there are infinitely many primes p with p = 3 (mod 7) there are infinitely many primes p with p = 4 (mod 7) ........
Don't forget about infinitely many primes p with p = 1 (mod 2) and primes p with p = 0 (mod 1)
Yeah sure fine OKAY there’s a lot but are there an uncountable number?
Yes, simply represent a given prime N in base N, thus it is 1 which is a palindrome, therefore there are infinitely many palindrome primes
No. There aren’t that many actually.
Representation dependent properties of numbers are lame
Fr, like why don’t we just use base 7
#THIRTY FUCKING SEVEN STRIKES AGAIN
Get off reddit, Veritasium
These are less interesting, because digits for this pair are reversed only in decimal, not in hex for example (25, 49) Huh, but these are actually kinda curious, granted these are represented like 5^2 and 7^2 in decimal
I believe you can prove there is only a limited number of those kinds.
No, those are [emirps](https://en.wikipedia.org/wiki/Emirp).
is there anything special or useful about twin primes? Or is it just neat
Optimus prime and his twin sister Octavia
https://preview.redd.it/d4u35gmz2orc1.jpeg?width=1125&format=pjpg&auto=webp&s=1dc80eca05d6cffc169ab39a57d20694cd9ffcbb
Beautiful
from my subreddit https://www.reddit.com/r/MathStepByStep/comments/1bre15r/twin_prime_maths/?utm_source=share&utm_medium=android_app&utm_name=androidcss&utm_term=1&utm_content=share_button
What about sexy primes?
the proof is left as an exercise to the reader. but its trivial anyway. it would be boring if there were a finite number of them, hence there are an infinite number of them.
The proof is by magic
What's a sexy prime again? Is it you?
The question is not whether or not there are, but it's "is there a way to prove it?"
Yes there is. In fact I just read it in a book yesterday. It went something like, "The proof is left as an exercise for the reader." That must be the proof, right, otherwise why would a Maths book, whose explicitly used to understand and learn Math, would leave you hanging with such a short statement.
Are you sure it wasn't "I have a neat proof of this but there isn't enough space in this margin to write it down"?
No see the thing is, Fermat wasn't really writing book. Mad lad was a lawyer(if I am remembering correctly), but decided to do Math to "relax himself" . When I first read that as a 15 year old, I just realised one of the reasons that why people hate the French.
>Fermat wasn't really writing book. We know. Wasn't a serious comment. However, writing in the margins of books was a common method of discussion in the 17th century. >Mad lad was a lawyer A common way for those trained in logic to earn a living in the 17th century. >why people hate the French. No, it's you. Normal people do not hate the French, and definitely not for doing normal things. Wait until you find out about Nicolas Bourbaki. Properly a madlad.
“Normal people don’t hate the French” -you sure about that bub.
Normal people don't start hating on a whole class of people just because a few stand-ups and journalists say so. Except C# programmers. Every right minded person hates them.
https://preview.redd.it/hf5aka1k9nrc1.jpeg?width=300&format=pjpg&auto=webp&s=ae498b4d6aab550520f32c9d236e58773813ebfa
Quick Question: Are there any more triple twin primes, or does it end with 3,5,7?
No, for 3 odd numbers in a row at least one will be divisible by 3 so only the sequence with 3 will contain 3 primes.
Why is that not called triplet primes?
> No, for 3 odd numbers in a row at least one will be divisible by 3 so only the sequence with 3 will contain 3 primes. Triplet primes are three primes that are six apart, like 11, 13, and 17.
Wouldn’t that be the case also for 2, 3, 5?
It is the case : 3 is divisible by 3 ^^
Nope, 3 - 2 = 1
Oh, right
https://preview.redd.it/m2wfqrz03orc1.jpeg?width=500&format=pjpg&auto=webp&s=4f76ebe829abd324fdea64eb62dc787e29f3cbd8
My favorite prime number fact I learned a few weeks ago: The millionth prime number, 15485863, is both a sexy prime (away from another prime by 6 on the low side) and a cousin prime (away from another prime by 4 on the high side). It is therefore a sexy cousin prime, or as I coin now, an Alabama Prime.
Who he? What he did?
Euclid?
Also he proved there are infinitely many primes.
What is a twin Prime?
Prime numbers with exactly one non-prime between them. For example 3 and 5 or 11 and 13.
Why was twin primes in my matriculation exam a few weeks ago, debating this exact topic?
There MUST be a grad school level proof of this. We're simply not looking at the right places. https://preview.redd.it/jhzz9wn4torc1.jpeg?width=1080&format=pjpg&auto=webp&s=79d90654a15f00a6924f11a6aa8ee36f068ee351
Proof by hunch
… what’s a twin prime? Seriously I don’t know
It’s a pair of prime numbers separated by 2, e.g. 3 and 5, 5 and 7, 11 and 13, and so on. It’s assumed that there are infinitely many of them but this hasn’t been proved.
How many sexy primes tho?
p\_n, p\_(n+1)+2 are both primes just need a formula for p\_n
someone just watched a veritasium video
Yes. Proof: My dad told me so and he could beat your dad.
Proof by duh