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<math>1_x</math>
 
  
 
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Assume that there are only a finite number of prime number <math>p_1,p_2,p_3,......p_n</math>
Assume that there are only a finite number of prime number p_1,p_2,p_3,......p_n.  Then by using the fact from exercise 18 (Let p_1,p_2,p_3,....,p_n be primes.  Then p_1p_2.....p_n +1 is divisible by none of these primes), p_1p_2p_3....p_n +1 is not divisible by any prime.)  This means p_1p_2...p_n +1 (which is larger than our initial conditions) is itself prime.  This contradicts the assumption that p_1,p_2,...p_n is the list of all primes.
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.  Then by using the fact from exercise 18 (Let <math>p_1,p_2,p_3,....,p_n </math> be primes.  Then <math>p_1p_2.....p_n +1 </math> is divisible by none of these primes), <math>p_1p_2p_3....p_n +1</math>  is not divisible by any prime.)  This means <math>p_1p_2...p_n +1 </math> (which is larger than our initial conditions) is itself prime.  This contradicts the assumption that <math>p_1,p_2,...p_n </math> is the list of all primes.
  
 
~Angela
 
~Angela

Revision as of 15:06, 6 September 2008

Assume that there are only a finite number of prime number $ p_1,p_2,p_3,......p_n $ . Then by using the fact from exercise 18 (Let $ p_1,p_2,p_3,....,p_n $ be primes. Then $ p_1p_2.....p_n +1 $ is divisible by none of these primes), $ p_1p_2p_3....p_n +1 $ is not divisible by any prime.) This means $ p_1p_2...p_n +1 $ (which is larger than our initial conditions) is itself prime. This contradicts the assumption that $ p_1,p_2,...p_n $ is the list of all primes.

~Angela

Alumni Liaison

EISL lab graduate

Mu Qiao