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

 
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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!
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'''Euclid's proof of the infinitude of primes''' is a classic and well-known [[proof]] by the Greek mathematician [[Euclid]] that there are infinitely many [[prime number]]s.
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==Proof==
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We proceed by [[proof by contradiction | contradiction]]. Suppose there are in fact only finitely many prime numbers, <math>p_1, p_2, p_3, \ldots, p_n</math>. Let <math>N = p_1 \cdot p_2 \cdot p_3 \cdots p_n + 1</math>. Since <math>N</math> leaves a remainder of 1 when divided by any of our prime numbers <math>p_k</math>, it is not divisible by any of them. However, the [[Fundamental Theorem of Arithmetic]] states that all positive integers have a unique prime factorization. Therefore, <math>N</math> must have a prime factor (possibly itself) that is ''not'' among our set of primes, <math>\{p_1, p_2, p_3, \ldots, p_n\}</math>. This means that <math>\{p_1, p_2, p_3, \ldots, p_n \}</math> does not contain all prime numbers, which contradicts our original assumption. Therefore, there must be infinitely many primes.
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==See Also==
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* [[Proof by contradiction]]
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* [[Number theory]]
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* [[Prime]]
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* [[Euclid]]
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[[Category:Proofs]]
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[[Category:Number theory]]
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[[category:Mathematics]]
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{{stub}}

Latest revision as of 10:45, 28 September 2024

Euclid's proof of the infinitude of primes is a classic and well-known proof by the Greek mathematician Euclid that there are infinitely many prime numbers.

Proof

We proceed by contradiction. Suppose there are in fact only finitely many prime numbers, $p_1, p_2, p_3, \ldots, p_n$. Let $N = p_1 \cdot p_2 \cdot p_3 \cdots p_n + 1$. Since $N$ leaves a remainder of 1 when divided by any of our prime numbers $p_k$, it is not divisible by any of them. However, the Fundamental Theorem of Arithmetic states that all positive integers have a unique prime factorization. Therefore, $N$ must have a prime factor (possibly itself) that is not among our set of primes, $\{p_1, p_2, p_3, \ldots, p_n\}$. This means that $\{p_1, p_2, p_3, \ldots, p_n \}$ does not contain all prime numbers, which contradicts our original assumption. Therefore, there must be infinitely many primes.

See Also

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