Official 'exchange of inanities' thread [Was: mm127 is prime, cuz I say so]
...(2^(2^(2^n1)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^n1)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^n1)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^(2^1271)  1 is prime. 
[QUOTE=kalikidoom;412554]<blah>... it always works[/QUOTE]For various values of "works".

[QUOTE=kalikidoom;412554]...(2^(2^(2^n1)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^n1 to be prime 2) by that same logic m=2^n1 must be prime for 2^m1= 2^(2^n1)1 to have a chance at being prime. 3) as LaurV pointed out not all n that are prime will allow 2^n1 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^n1)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. 
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