# Proof of fundamental theorem of arithmetic

1. Sep 29, 2014

### PcumP_Ravenclaw

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!!

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2. Sep 29, 2014

### Erland

A is the set of all natural numbers, not containing 0 and 1, which have unique prime factorizations. The goal is to prove with induction that A is all natural numbers except 0 and 1. Thus in the induction proof, the induction hypothesis is that all natural numbers (not equal to 0 or 1) less than or equal to n belong to A.

Both can occur.
Yes and yes.

This is assumed. It is the induction hypothesis.

If r=s then some pi must be greater qi, yes, but that is of no importance for the proof.

Hint: expand the parentheses at both sides, and use p1...pr = q1....qs.

Otherwise p1 must be a factor of some of q2, q3, ... ,qs, by the property that if a prime is a factor of a product of integers, it must be a factor of (at least) one of these integers.

Last edited: Sep 29, 2014
3. Oct 1, 2014

### PcumP_Ravenclaw

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!!

This is assumed. It is the induction hypothesis.

What is the induction hypothesis here?? Is it, if it can be proved that for n there is only one way to write the prime factors for A ={2,3,4,5,6,7.....n} then for n+1 also there is only one way to write the prime factors...

They have not shown the proof of the inductive step which is 'if n is true then n+1 is also true' so that it is true for all elements in A. Can you please show the anchor step and inductive step for this proof???

Expanding the parentheses

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

Hint: expand the parentheses at both sides, and use p1...pr = q1....qs.

p1p2...pr - p1q2q3....qs = q1q2....qs - p1q2.....qs
so now???

but how can you explicitly show that p1 has decreased by q1 - p1 or q1 increased by q1 - p1. can you please derive??

4. Oct 1, 2014

### Erland

An apparently (but not really) stronger version of induction is used here. The induction hypothesis can be expressed just as you wrote above

there is only one way to write the prime factors for A ={2,3,4,5,6,7.....n}

and the objective is to prove that this implies that

for n+1 also there is only one way to write the prime factors

This is done:

If p1p2...pr=q1q2...qs=n+1 (this being prime factorizations) with p1=q1, then p2p3...pr=q2q3...qs=m, say. Then, m is an integer <= n with two prime factorizations, so by the induction hypothesis, these prime factorizations are the same, so p2, p3, ..., pr is the same sequence of numbers as q2, q3, ..., qs, and since also p1=q1, the two prime factorizations of n+1 are the same.

We assumed that p1p2...pr=q1q2...qs, so this equality holds. Hence so does the original equality, with unexpanded parentheses.

It is not clear to me what you mean, but anyway:

p1 is a factor of the left side of p1(p2......pr - q2.......qs) = q2....qs(q1 - p1), hence also of the right side. Since p1 is a prime, it must be a factor of either some of q2, q3, ... qs, or of q1-p1. But q2, q3, ..., are primes, so none of them can be divisible with the smaller prime p1. So p1 divides q1-p1, hence it divides (q1-p1)+p1=q1, which is impossible by the same reason as just before. This is a contradiction. Hence the assummption that p1<q1 is false, so p1=q1.

5. Oct 2, 2014

### Erland

On reflection, this property needs not be used, and the author of the attached files did not intend this.

Instead, notice that m = p1(p2p3...pr - q2q3...qs) = q2q3...qs(q1-p1) <= n, so the induction hypothesis can be applied to m, which means that m has a unique prime factorization. From the left side, we see that p1 is one its prime factors, and from the right side, we see that q2, q3 ..., qs all are, and that all other prime factors must be factors in q1-p1. Since p1 < qj for all j, this means that p1 is none of the qj, so p1, being a prime factor of m, must be a factor in q1-p1, that is q1-p1=k*p1 for some k. Thus, q1 = (k+1)*p1, sp p1 is a factor in q1, which is impossible.