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How many prime numbers can we list? Lets go one by one
07-29-2024, 08:50 PM
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#1
07-29-2024, 08:51 PM
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#2
- Getter_done
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****** plz
07-29-2024, 08:51 PM
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#3
- GaryRidgway
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1 is not a prime. 2 is.
5...
5...
07-29-2024, 08:54 PM
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#4
07-29-2024, 08:55 PM
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#5
07-29-2024, 08:55 PM
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#6
- LuigiMiami631
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Originally Posted By rstlne1⏩
Ron would like a word with you.wasn't expecting you to know this
this is unsettling
this is unsettling
07-29-2024, 08:55 PM
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#7
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Originally Posted By rstlne1⏩
Before I became retarded I was once a scholar of sortswasn't expecting you to know this
this is unsettling
this is unsettling
07-29-2024, 09:03 PM
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#8
07-29-2024, 09:03 PM
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#9
07-29-2024, 09:18 PM
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#10
- MiscMathematician
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1 used to be prime. then it got disrespected just like pluto
07-29-2024, 09:21 PM
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#11
07-29-2024, 09:52 PM
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#12
07-29-2024, 09:53 PM
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#13
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19.
07-29-2024, 09:58 PM
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#14
07-29-2024, 10:07 PM
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#15
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Originally Posted By rstlne1⏩
nope. has never been done17.
In after mathbrah
Mathbrah...is there a proof to show that there is an infinite number of prime numbers?
In after mathbrah
Mathbrah...is there a proof to show that there is an infinite number of prime numbers?
29
07-29-2024, 10:09 PM
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#16
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31.
07-29-2024, 10:11 PM
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#17
07-29-2024, 10:17 PM
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#18
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Originally Posted By rstlne1⏩
u know i cant resist the urgeI see
37
37
if there were say only 5 primes, a<b<c<d<e
then N = a*b*c*d*e + 1
is either prime or not. If N (which is > e) is prime, then there must be at least 6 primes, a contradiction.
If it is not prime, then N is not divisible by any of a,b,c,d,e (division by these yields a remainder of 1) but must be divisible by some other prime not on the list. contradiction
now just replace 5 by any number of primes there might be, same proof.
41
edit: yes, i said 39 by accident mademojie
07-29-2024, 10:23 PM
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#19
Originally Posted By MiscMathematician⏩
Assuming we are doing primes greater than 3 (nvm there's at least 5 primes)u know i cant resist the urge
if there were say only 5 primes, a<b<c<d<e
then N = a*b*c*d*e + 1
is either prime or not. If N (which is > e) is prime, then there must be at least 6 primes, a contradiction.
If it is not prime, then N is not divisible by any of a,b,c,d,e (division by these yields a remainder of 1) but must be divisible by some other prime not on the list. contradiction
now just replace 5 by any number of primes there might be, same proof.
41
if there were say only 5 primes, a<b<c<d<e
then N = a*b*c*d*e + 1
is either prime or not. If N (which is > e) is prime, then there must be at least 6 primes, a contradiction.
If it is not prime, then N is not divisible by any of a,b,c,d,e (division by these yields a remainder of 1) but must be divisible by some other prime not on the list. contradiction
now just replace 5 by any number of primes there might be, same proof.
41
If you multiply two odd numbers, you'll always get an odd number (too lazy to write proof)
So you'll end up getting an odd number above
Adding 1 will give you an even number for N
Not prime in their example
Tbh I made this all up but seems legit
07-29-2024, 10:25 PM
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#20
07-29-2024, 10:27 PM
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#21
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47 smack ass piss it
07-29-2024, 10:35 PM
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#22
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Originally Posted By rstlne1⏩
since we know 2 is prime, we may as well assume it is equal to a, but the proof works generally without call to parity.Assuming we are doing primes greater than 3
If you multiply two odd numbers, you'll always get an odd number (too lazy to write proof)
So you'll end up getting an odd number above
Adding 1 will give you an even number for N
Not prime in their example
Tbh I made this all up but seems legit
If you multiply two odd numbers, you'll always get an odd number (too lazy to write proof)
So you'll end up getting an odd number above
Adding 1 will give you an even number for N
Not prime in their example
Tbh I made this all up but seems legit
this proof can apply to any unique factorization domain with characteristic zero (which basically means 1+1+1+... never equals zero)
for example the polynomial x^2+1 is prime with coefficients considered over the real numbers (can't be factored into smaller polynomials). there are an infinite number of such unfactorable polynomials. same exact proof
07-29-2024, 10:44 PM
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#23
- 4skinfermenter
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Originally Posted By rstlne1⏩
It's an ez proof.Assuming we are doing primes greater than 3 (nvm there's at least 5 primes)
If you multiply two odd numbers, you'll always get an odd number (too lazy to write proof)
So you'll end up getting an odd number above
Adding 1 will give you an even number for N
Not prime in their example
Tbh I made this all up but seems legit
If you multiply two odd numbers, you'll always get an odd number (too lazy to write proof)
So you'll end up getting an odd number above
Adding 1 will give you an even number for N
Not prime in their example
Tbh I made this all up but seems legit
(2k+1)(2m+1)=4mk+2(k+m)+1 is odd.
07-29-2024, 10:46 PM
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#24
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Originally Posted By MiscMathematician⏩
Technically, that is irreducibility, not primality. The two are the same in a GCD domain, which that polynomial ring is.since we know 2 is prime, we may as well assume it is equal to a, but the proof works generally without call to parity.
this proof can apply to any unique factorization domain with characteristic zero (which basically means 1+1+1+... never equals zero)
for example the polynomial x^2+1 is prime with coefficients considered over the real numbers (can't be factored into smaller polynomials). there are an infinite number of such unfactorable polynomials. same exact proof
this proof can apply to any unique factorization domain with characteristic zero (which basically means 1+1+1+... never equals zero)
for example the polynomial x^2+1 is prime with coefficients considered over the real numbers (can't be factored into smaller polynomials). there are an infinite number of such unfactorable polynomials. same exact proof
07-29-2024, 10:51 PM
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#25
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[QUOTE=4skinfermenter post_id=1704330081]Technically, that is irreducibility, not primality. The two are the same in a GCD domain, which that polynomial ring is.[/QUOTE]true, R[x] is both though. but yeah technicality
07-29-2024, 11:00 PM
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#26
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Originally Posted By SmackIllegals⏩
53.47 smack ass piss it
Prove by induction that for every integer n > 0, 1 + 2 + 3 +...+ n = n(n+1)/2.
07-29-2024, 11:05 PM
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#27
- 4skinfermenter
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Originally Posted By SpeakethTruth⏩
Base case. 1=1(2)/2.53.
Prove by induction that for every integer n > 0, 1 + 2 + 3 +...+ n = n(n+1)/2.
Prove by induction that for every integer n > 0, 1 + 2 + 3 +...+ n = n(n+1)/2.
Assume the statement S(n) holds for n. Then S(n)+n+1=n(n+1)/2 + n+1=n(n+1)/2+2(n+1)/2=(n+1)(n+2)/2.
By induction, the statement is proved.
07-29-2024, 11:08 PM
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#28
- elterrible987
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4096
07-29-2024, 11:09 PM
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#29
- elterrible987
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Originally Posted By 4skinfermenter⏩
thats amazing but youre still a virginBase case. 1=1(2)/2.
Assume the statement S(n) holds for n. Then S(n)+n+1=n(n+1)/2 + n+1=n(n+1)/2+2(n+1)/2=(n+1)(n+2)/2.
By induction, the statement is proved.
Assume the statement S(n) holds for n. Then S(n)+n+1=n(n+1)/2 + n+1=n(n+1)/2+2(n+1)/2=(n+1)(n+2)/2.
By induction, the statement is proved.
07-29-2024, 11:12 PM
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#30
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Originally Posted By elterrible987⏩
It's not amazing, and not the most informative way to show the proof.thats amazing but youre still a virgin
There are several ways. I like this one.
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