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anonymous

  • 5 years ago

There are infinitely many prime numbers. Proof. Following Euclid’s proof, we shall show that to every prime p there is a greater one. Assume that p is the greatest prime number, and let q = 1 · 2 · 3 · . . . · p. Then, q + 1 is not divisible by 2, 3, . . . , p. It follows that q is divisible only by 1 and itself, and thus, it is a prime number greater than p. This, however, contradicts the hypothesis that p is the greatest prime, and it follows that there is no greatest prime. In other words, the set of primes is infinite. Help clarify how 1*2*3..*p + 1 is not divisible by 1,2,3 etc..

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  1. KingGeorge
    • 5 years ago
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    Well, we know that 1*2*3*...*p is divisible by 1, 2, 3, ..., p because it's the product of those numbers. Thus, if we add 1, we know that it can't be divisible by any number 2, 3, 4, ..., p. Does this help clarify?

  2. anonymous
    • 5 years ago
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    So q would be odd? How can we verify that in any instance it is not divisible by 3 for example?

  3. KingGeorge
    • 5 years ago
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    Well, q would be even since it's divisible by two. But therefore q+1 is odd. Likewise, q is divisible by 3, but q+1 obviously can't. This argument can be continued up to p, where we know p divides q, but therefore, p can't divide q+1. However, this is a proof that there is a prime greater than p, not necessarily that q is prime. Take 5 as an example. 5! is 120, so 5!+1 is 121. This is 11*11, and 11 is both prime and greater than 5, but 121 is not prime.

  4. anonymous
    • 5 years ago
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    Ok. So if it's not prime then what does it prove? That there's a possibility that it's square root is prime?

  5. KingGeorge
    • 5 years ago
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    All it proves is that there exists a prime larger than p. It doesn't say what that prime is, just that it exists.

  6. KingGeorge
    • 5 years ago
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    You can try this with various values of p and using wolfram alpha you can find and factor q+1

  7. anonymous
    • 5 years ago
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    Ok. Thank you!

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