# Difference between revisions of "Euclid's proof of the infinitude of primes"

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− | + | We proceed by contradiction. Suppose there are only [[finite]]ly many [[prime number]]s; let's call them <math>p_1, p_2, p_3, \ldots, p_n</math>. Let <math>x=p_1\cdot p_2\cdot p_3 \cdots p_n + 1</math>. When <math>x</math> is divided by any of our primes <math>p_1, p_2, p_3, \ldots, p_n</math> it leaves a [[remainder]] of 1, so none of these primes divide <math>x</math>. Since every [[positive integer]] has at least one prime factor, <math>x</math> has some prime factor (possibly itself) not in the set <math>\{ p_1,p_2,p_3,\ldots,p_n\}</math>. Thus <math>\{ p_1,p_2,p_3,\ldots, p_n\}</math> does not contain all prime numbers. Contradiction! Our original assumption must have been false, so there are in fact infinitely many primes. |

## Revision as of 17:26, 15 January 2007

We proceed by contradiction. Suppose there are only finitely many prime numbers; let's call them . Let . When is divided by any of our primes it leaves a remainder of 1, so none of these primes divide . Since every positive integer has at least one prime factor, has some prime factor (possibly itself) not in the set . Thus does not contain all prime numbers. Contradiction! Our original assumption must have been false, so there are in fact infinitely many primes.