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

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'''Euclid's proof of the infinitude of primes''' is a [[proof]] by [[Euclid]] that the number of [[prime number]]s is [[infinite]].
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==Proof==
 
We proceed by [[proof by contradiction | 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.
 
We proceed by [[proof by contradiction | 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.
  
[[Euclid | Back to Euclid]]
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==See Also==
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*[[Number theory]]
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*[[Euclid]]
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[[Category:Proofs]]
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[[Category:Number theory]]

Revision as of 20:26, 11 January 2008

Euclid's proof of the infinitude of primes is a proof by Euclid that the number of prime numbers is infinite.

Proof

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

See Also