- #1

PcumP_Ravenclaw

- 106

- 4

Dear all,

Please help me understand the proof by induction for only one way of expressing the product of primes up to the order of the factors.

Please see the two attachments from the textbook "alan F beardon, algebra and geometry"

A is a set of all natural numbers excluding 1 and 0??

r and s in p1...pr = q1...qs can be same or two different numbers?

2 <= p1 <= ... <= pr means that the magnitude of p is progessively increasing from p1 to pr?

same condition 2 <= q1 <= ... <= qs for q1 also?

Two cases to consider (1) p1 = q1 (2) p1 < q1

in case (1) how is it possible to say that when p2...pr = q2...qs = m and 2 <= m <= n. "This means that the two factorizations of m are the same(up to order)" I thought this is what we are trying to prove! but here it is assumed even though there are a few prime factors involved from p2 to pr and q2 to qs and although the final product is m the prime factors may be different right which is our hypothesis??

in case (2) if p1 < q1 all other p terms should be greater than or equal to their corresponding q terms (i.e. p2 >= q2, etc.) only then will the equation

p1...pr = q1...qs

will hold and we can say that there is only one way to write a number as a product of primes disregarding the order of the factors

so in trying to make the two sides equal for the case p1 < q1 the following equation is obtained?

p1(p2...pr - q2...qs) = q2...qs(q1 - p1)

can you please derive this equation for the case p1 < q1 from the previous equation??

i understand that the q terms are less than n+1 because q1 is less because q1 = ( q1 - p1)

why should p1 be a prime factor of q1 - p1??

I cannot understand beyond this point!

Thanks in advance...

Please help me understand the proof by induction for only one way of expressing the product of primes up to the order of the factors.

Please see the two attachments from the textbook "alan F beardon, algebra and geometry"

A is a set of all natural numbers excluding 1 and 0??

r and s in p1...pr = q1...qs can be same or two different numbers?

2 <= p1 <= ... <= pr means that the magnitude of p is progessively increasing from p1 to pr?

same condition 2 <= q1 <= ... <= qs for q1 also?

Two cases to consider (1) p1 = q1 (2) p1 < q1

in case (1) how is it possible to say that when p2...pr = q2...qs = m and 2 <= m <= n. "This means that the two factorizations of m are the same(up to order)" I thought this is what we are trying to prove! but here it is assumed even though there are a few prime factors involved from p2 to pr and q2 to qs and although the final product is m the prime factors may be different right which is our hypothesis??

in case (2) if p1 < q1 all other p terms should be greater than or equal to their corresponding q terms (i.e. p2 >= q2, etc.) only then will the equation

p1...pr = q1...qs

will hold and we can say that there is only one way to write a number as a product of primes disregarding the order of the factors

so in trying to make the two sides equal for the case p1 < q1 the following equation is obtained?

p1(p2...pr - q2...qs) = q2...qs(q1 - p1)

can you please derive this equation for the case p1 < q1 from the previous equation??

i understand that the q terms are less than n+1 because q1 is less because q1 = ( q1 - p1)

why should p1 be a prime factor of q1 - p1??

I cannot understand beyond this point!

Thanks in advance...