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-   -   Official 'exchange of inanities' thread [Was: mm127 is prime, cuz I say so] (https://www.mersenneforum.org/showthread.php?t=20542)

 kalikidoom 2015-10-13 08:27

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

 LaurV 2015-10-13 13:22

[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:

 R.D. Silverman 2015-10-13 13:25

[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.

 retina 2015-10-13 13:26

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

 science_man_88 2015-10-13 13:32

[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.

 R.D. Silverman 2015-10-13 13:59

[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?

 alpertron 2015-10-13 14:03

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.

 R.D. Silverman 2015-10-13 14:14

[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.

 science_man_88 2015-10-13 14:24

[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.

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

[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.

 science_man_88 2015-10-13 14:29

[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.

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