− | This is proved by contradiction. Suppose there is a finite number of primes and let them be <math>p_1,p_2,p_3,...,p_n</math>. Let <math>x=p_1p_2p_3\cdots p_n</math>. Then we have <math>x+1=p_1p_2p_3\cdots p_n+1</math>. When divided by any of the primes <math>p_1,p_2,p_3,...,p_n</math>, <math>x+1</math> leaves a remainder of 1 implying that either <math>x+1</math> is prime or that it has some other prime factors not in the set <math>\{ p_1,p_2,p_3,...,p_n\}</math>. In any case we have it so that <math>\{ p_1,p_2,p_3,...,p_n\}</math> does not contain all prime numbers. Contradiction!
| + | #REDIRECT [[Euclid's Theorem]] |