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

[QUOTE=kalikidoom;412554]...(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[/QUOTE]
That is false. For n=4, all are composite. Also for n=11, all are composite. :razz:

[QUOTE=kalikidoom;412554]...(2^(2^(2^n-1)-1)-1)...

I mean... isn't that at least a proof that there's an infinate amout of them?
[/QUOTE]

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[/QUOTE]

"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.

[QUOTE=kalikidoom;412554]<blah>... it always works[/QUOTE]For various values of "works".

[QUOTE=kalikidoom;412554]...(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[/QUOTE]

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.

[QUOTE=LaurV;412564]That is false. For n=4, all are composite. Also for n=11, all are composite. :razz:[/QUOTE]

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?

Well, the original poster said:
[QUOTE=kalikidoom;412554]...(2^(2^(2^n-1)-1)-1)...

I mean... isn't that at least a proof that there's an infinate amout of them?[/QUOTE]
He didn't say that he has a proof.

[QUOTE=alpertron;412571]Well, the original poster said:

He didn't say that he has a proof.[/QUOTE]

He suggested that a sequence of numbers [b]constituted[/b] 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 [I]might[/i] be a proof.

[QUOTE=R.D. Silverman;412573]He suggested that a sequence of numbers [b]constituted[/b] 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 [I]might[/i] be a proof.[/QUOTE]

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.

[QUOTE=science_man_88;412574]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.[/QUOTE]

Gibberish from someone who hasn't yet passed first year algebra.

[QUOTE=R.D. Silverman;412575]Gibberish from someone who hasn't yet passed first year algebra.[/QUOTE]

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.