mersenneforum.org Official 'exchange of inanities' thread [Was: mm127 is prime, cuz I say so]
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 2015-10-13, 08:27 #1 kalikidoom   Oct 2015 18 Posts Official 'exchange of inanities' thread [Was: mm127 is prime, cuz I say so] ...(2^(2^(2^n-1)-1)-1)... I mean... isn't that at least a proof that there's an infinate amout of them? Try it with how meny primes you like... and with how meny 2^....-1 you like... it always works
2015-10-13, 13:22   #2
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

9,973 Posts

Quote:
 Originally Posted by kalikidoom ...(2^(2^(2^n-1)-1)-1)... I mean... isn't that at least a proof that there's an infinate amout of them? Try it with how meny primes you like... and with how meny 2^....-1 you like... it always works
That is false. For n=4, all are composite. Also for n=11, all are composite.

2015-10-13, 13:25   #3
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

11101001001002 Posts

Quote:
 Originally Posted by kalikidoom ...(2^(2^(2^n-1)-1)-1)... I mean... isn't that at least a proof that there's an infinate amout of them?
I see a sequence of numbers. I see no mathematical argument at all that suggests
every number in the sequence is prime. You have a strange notion about what
constitutes a proof.

Quote:
 Try it with how meny primes you like... and with how meny 2^....-1 you like... it always works
"It always works". Is this your proof? A simple assertion?

I suggest that you think about what you wrote.

For example, please prove to us that 2^(2^127-1) - 1 is prime.

Last fiddled with by R.D. Silverman on 2015-10-13 at 13:28

2015-10-13, 13:26   #4
retina
Undefined

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Jun 2006
My evil lair

22·1,627 Posts

Quote:
 Originally Posted by kalikidoom ... it always works
For various values of "works".

2015-10-13, 13:32   #5
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

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Quote:
 Originally Posted by kalikidoom ...(2^(2^(2^n-1)-1)-1)... I mean... isn't that at least a proof that there's an infinate amout of them? Try it with how meny primes you like... and with how meny 2^....-1 you like... it always works
1) n must be prime for 2^n-1 to be prime
2) by that same logic m=2^n-1 must be prime for 2^m-1= 2^(2^n-1)-1 to have a chance at being prime.
3) as LaurV pointed out not all n that are prime will allow 2^n-1 to be prime.

2015-10-13, 13:59   #6
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

22×5×373 Posts

Quote:
 Originally Posted by LaurV That is false. For n=4, all are composite. Also for n=11, all are composite.
I expect that the OP intended for n = 2.

However, I'd still like an answer to my question as to why the OP thinks that presenting a
sequence of numbers is in any way a "proof". There were no mathematical statements
asserting that some (set of) condition(s) is true, nor were there any logical statements.
It was just a list of numbers. How can this be a proof?

2015-10-13, 14:03   #7
alpertron

Aug 2002
Buenos Aires, Argentina

22×3×112 Posts

Well, the original poster said:
Quote:
 Originally Posted by kalikidoom ...(2^(2^(2^n-1)-1)-1)... I mean... isn't that at least a proof that there's an infinate amout of them?
He didn't say that he has a proof.

2015-10-13, 14:14   #8
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

22×5×373 Posts

Quote:
 Originally Posted by alpertron Well, the original poster said: He didn't say that he has a proof.
He suggested that a sequence of numbers constituted a proof.
It is hard to correct someone's misconceptions without knowing how and
why they believe what they believe. A simple statement that what was
posted is not a proof says very little./// We need to know why the OP thought
that simply presenting a sequence of numbers might be a proof.

Last fiddled with by R.D. Silverman on 2015-10-13 at 14:14 Reason: typo

2015-10-13, 14:24   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by R.D. Silverman He suggested that a sequence of numbers constituted a proof. It is hard to correct someone's misconceptions without knowing how and why they believe what they believe. A simple statement that what was posted is not a proof says very little./// We need to know why the OP thought that simply presenting a sequence of numbers might be a proof.
one thought I had is that they might think that because you can infinitely add to the representation that it represents and infinite amount of numbers and might intersect the primes infinitely often.

2015-10-13, 14:25   #10
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

22×5×373 Posts

Quote:
 Originally Posted by science_man_88 one thought I had is that they might think that because you can infinitely add to the representation that it represents and infinite amount of numbers and might intersect the primes infinitely often.
Gibberish from someone who hasn't yet passed first year algebra.

Last fiddled with by R.D. Silverman on 2015-10-13 at 14:26

2015-10-13, 14:29   #11
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by R.D. Silverman Gibberish from someone who hasn't yet passed first year algebra.
I didn't say it was mathematically sound, I was trying to think like the OP might be thinking. Yet assumes I plan on going back to school again.

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