a. Use mathematical induction and Euclid’s lemma to prove that for every positive integer s. if p and are prime numbers and , then for some i with .
b. The uniqueness part of the unique factorization theorem for the integers says that given any integer n if for some positive integers r and s and prime numers and then and for every integer i with .
Use the result of part (a) to fill in the details of the following sketch of a proof: Suppose than n is an integer with two different prime faclorizations: . All the prime factors that appear on both sides can be cancelled (as many times as they appear on both sides) to arrive at the situation where , and for any integers i and j. Then use part (a) to deduce a contradiction. and conclude that the prime factorization of n is unique except, possibly, for the order in which the prime factors are written.
For all positive integers are prime numbers and for some with using mathematical induction and with Euclid’s lemma
Mathematical induction and Euclid’s lemma.
Mathematical Induction: This method has two steps to be followed:
Prove the statement for the first possible value .
The statement becomes, “if are prime numbers and ”
By definition of a prime number, it has only two factors itself and .
Since, is a prime number so it is divisible by only.
But it is divisible by . So, has to be either .
It is given that is a prime number so, it can’t be . So, .
Therefore, the statement holds true for .
Assume that the statement holds true for .
Therefore, it is assumed that if are prime numbers and for some is true .
The next case in the series is .
The following statement has to be proved as true.
If are prime numbers and for some
The uniqueness part of the unique factorization theorem for integers.
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