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Why Mathematicians Prefer the Alternative Definition
:You misunderstand the argument. Since you multiplied only the first 7 primes to get 510510, your hypothesis (the thing to be contradicted in this proof by contradiction) is that those 7 numbers constitute the complete finite set of primes. And in fact, 510511 is not divisible by any of them. Therefore they are not the complete set of primes. And the same argument can be made for any hypothesized finite set. Therefore, no finite set can be the complete set of primes. [[User:Ga ohoyt|Ga ohoyt]] 20:15, 20 February 2008 (EST)
:You misunderstand the argument. Since you multiplied only the first 7 primes to get 510510, your hypothesis (the thing to be contradicted in this proof by contradiction) is that those 7 numbers constitute the complete finite set of primes. And in fact, 510511 is not divisible by any of them. Therefore they are not the complete set of primes. And the same argument can be made for any hypothesized finite set. Therefore, no finite set can be the complete set of primes. [[User:Ga ohoyt|Ga ohoyt]] 20:15, 20 February 2008 (EST)
==Why Mathematicians Prefer the Alternative Definition==
Because mathematician make a distinction between irreducible elements (p irred. <=> (a|p => a unit or a=p)) and prime elements (p prime <=> (for all a,b: p|ab => p|a or p|b). While these two properties coincide on <math>\mathbf{Z}</math>, they differ e.g., on <math>\mathbf{Z}[\sqrt{-5}]</math>: here 3 is irreducible, but not a prime, as <math>21 = 3 \cdot 7 = (1 + 2 \sqrt{-5})(1 - 2 \sqrt{-5})</math> --[[User:DiEb|DiEb]] 13:03, 29 August 2008 (EDT)