Fill in the blanks in the following proof: Theorem 1. The set of prime numbers is infinite. Proof. Suppose not. Then there is a complete, finite, which con- tains all of the prime numbers. P1 = 2, P2 = 3, P3 = 5, P4 = 7, ..., Pn Consider the integer formed by taking the product of all of these primes and then adding 1. N = (piP2P3P4. .- Pn) +1 This number is . than any of the primes in our list, in particular it cannot be prime itself and must therefore be divisible by at least one of the primes (say p;). On the other hand, N is one greater than a multiple of p;, so p; cannot divide N. This is a since we have shown that N is both divisible by p; and not divisible by pj. Hence the supposition is and the theorem is true.

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O Fill in the blanks in the following proof:
Theorem 1. The set of prime numbers is infinite.
Proof. Suppose not. Then there is a complete, finite,
which con-
tains all of the prime numbers.
P1 = 2, p2 = 3, P3 = 5, p1 = 7, ..., Pn
Consider the integer formed by taking the product of all of these primes and then
adding 1.
N = (PıP2P3P4... Pn) +1
This number is
than any of the primes in our list, in particular
it cannot be prime itself and must therefore be divisible by at least one of the primes
(say P;). On the other hand, N is one greater than a multiple of p;, so P; cannot
divide N. This is a
since we have shown that N is both divisible
by p; and not divisible by pj. Hence the supposition is
and the
theorem is true.
Transcribed Image Text:O Fill in the blanks in the following proof: Theorem 1. The set of prime numbers is infinite. Proof. Suppose not. Then there is a complete, finite, which con- tains all of the prime numbers. P1 = 2, p2 = 3, P3 = 5, p1 = 7, ..., Pn Consider the integer formed by taking the product of all of these primes and then adding 1. N = (PıP2P3P4... Pn) +1 This number is than any of the primes in our list, in particular it cannot be prime itself and must therefore be divisible by at least one of the primes (say P;). On the other hand, N is one greater than a multiple of p;, so P; cannot divide N. This is a since we have shown that N is both divisible by p; and not divisible by pj. Hence the supposition is and the theorem is true.
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