[Solved]Metic Recall Fundamental Theorem Arith Factorisation Natural Number Greater Order Factors Q37040795
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metic Recall that the Fundamental theorem of arith factorisation of a natural number greater than order of the factors. Here is a proof of this resul unhe proof by contradiction metic states that the p Suppose we could find a natural number, n, with factorisations. We can assume that the two p no prime factors in common, because if t common factors to obtain a smaller natural nu factorisations and no common prime factors. Let factorisations of n be piPPho and let q be a prime prime factorisation of n. n, with two differe nt Pri they did, we couldot one of the rime a prime number in the Now, since qln, we can write n Omodg, and therefo PIP2… Pk Omodg. We know that none of the any of the ps equal a multiple of q either, because they ar oda. We know that none of the ps are e equal to g, nor can all prime numbers. Therefore, using our results from Section2.ie 12.4, all of the be used as divisors modulo q pPhrt Divide both sides of the congruence piP2 …Pk Omodą in turn P2, Pk factorisation of n must be unique. This concludes the proof. Exercises 1 Prove the divisibility properties listed in Section 12.2 2 Let n be any natural number greater than 1. Explain why the numbers n! + 2, n! +3, n! +4, n! + n must all be composite (This exercise shows that it is possible to find arbitrarily long sequences of consecutive composite numbers.) A list of all the prime numbers less than or equal to n can be generated in the following way: Begin with an empty list of primes For each number from 2 to n in turn, test whether it is divisible by any of the primes in the list, and append it to the list if it isn’t. Write this procedure as an algorithm Another method for generating all the prime numbers less than or equal to n, called the Sieve of Eratosthenes, is carried out as follows: Begin with a list of all the natural numbers from 2 ton In the first step, remove from the list all the multiples of 2 (he first number on the list) except 2 itself. In the second step, remove from the list all the multiples of 3 (the next number o 3 4 Show transcribed image text metic Recall that the Fundamental theorem of arith factorisation of a natural number greater than order of the factors. Here is a proof of this resul unhe proof by contradiction metic states that the p Suppose we could find a natural number, n, with factorisations. We can assume that the two p no prime factors in common, because if t common factors to obtain a smaller natural nu factorisations and no common prime factors. Let factorisations of n be piPPho and let q be a prime prime factorisation of n. n, with two differe nt Pri they did, we couldot one of the rime a prime number in the Now, since qln, we can write n Omodg, and therefo PIP2… Pk Omodg. We know that none of the any of the ps equal a multiple of q either, because they ar oda. We know that none of the ps are e equal to g, nor can all prime numbers. Therefore, using our results from Section2.ie 12.4, all of the be used as divisors modulo q pPhrt Divide both sides of the congruence piP2 …Pk Omodą in turn P2, Pk factorisation of n must be unique. This concludes the proof. Exercises 1 Prove the divisibility properties listed in Section 12.2 2 Let n be any natural number greater than 1. Explain why the numbers n! + 2, n! +3, n! +4, n! + n must all be composite (This exercise shows that it is possible to find arbitrarily long sequences of consecutive composite numbers.) A list of all the prime numbers less than or equal to n can be generated in the following way: Begin with an empty list of primes For each number from 2 to n in turn, test whether it is divisible by any of the primes in the list, and append it to the list if it isn’t. Write this procedure as an algorithm Another method for generating all the prime numbers less than or equal to n, called the Sieve of Eratosthenes, is carried out as follows: Begin with a list of all the natural numbers from 2 ton In the first step, remove from the list all the multiples of 2 (he first number on the list) except 2 itself. In the second step, remove from the list all the multiples of 3 (the next number o 3 4
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Answer to metic Recall that the Fundamental theorem of arith factorisation of a natural number greater than order of the factors. … . . .
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