# Fundamental Theorem of Arithmetic Problem

1. Aug 20, 2009

### Fisicks

Problem number 4 on the image has me stumped. I understand the problem (obviously not enough) and what its saying, I'm just having trouble putting it into a proof. Can i get a hint to get me started? Thanks

http://img40.imageshack.us/i/asdasdjql.jpg/" [Broken]

Last edited by a moderator: May 4, 2017
2. Aug 20, 2009

### srijithju

I think you should pay attention to the fact t1 , t2, etc . are greater than OR EQUAL TO zero.

3. Aug 20, 2009

### CRGreathouse

I think you mean g_i and h_i?

Fisicks, I don't know what problem you have. The basic idea is simple: sort the q_i and s_i in order and put them into one big list. When a number comes from both lists, add the exponents; otherwise take the exponent from where it came (e_i or f_i). Example:

(3^2 * 2^1) * (7^2 * 2^2 * 3^3) =
(2^1 * 3^2) * (2^2 * 3^3 * 7^2) =
2^(1+2) * 3^(2+3) * 7^2 =
2^3 * 3^5 * 7^2

4. Aug 20, 2009

### Fisicks

greathouse why are you multiplying exponents, its just representing a and b to have the same bases of primes, and what i need help with is the proof part, not the what is the problem saying part.

Last edited: Aug 20, 2009
5. Aug 21, 2009

### CRGreathouse

Sorry, misread the problem. Your problem is easier -- just ignore the part about adding, and take missing primes' exponents to be 0.

6. Aug 21, 2009

### Fisicks

yeah i get that part!!! i need to know how you would write a formal proof to show that, cause thats what the problem is asking.

7. Aug 21, 2009

### CRGreathouse

Without knowing the requirements for your "formal" proof I couldn't possibly start. I don't imagine you mean what I mean when I say 'formal'.

8. Aug 21, 2009

### Fisicks

i don't know what more you want, the question is posted right up there in that link lol. all i want to know is how to answer that question.

9. Aug 21, 2009

### ramsey2879

I think you are having a problem of not being able to see a forest when a thousand trees are right in front of you. It is formal to state that a*b is a number that according to problem 3 has a unique prime fractorization, the rest is a fairly simple exercise.

10. Aug 21, 2009

### Dragonfall

Yes, for a truly formal proof, see "Principia Mathematica". Not sure where the theorem in question is; probably in volume 2 or 3.

11. Aug 21, 2009

### ramsey2879

Dragonfall isn't it true that the only theorem needed is that provided by problem 3. I believe that problem 3 should be easy to give a formal proof. All the trees in problem 3 are those provided by "a" and "b" and how to obtain the forest that comprises them is simply to multiply "a" and "b". So what does problem three say about the product of "a" and "b"?

12. Aug 22, 2009

### Dragonfall

Problem 3 isn't enough, it just "collects" equal primes into single powers. In problem 4 you have to "split" the product by introducing powers of 0. A proof along the lines suggested by CRGreathouse is what the problem's asking for, I think.

You can use problem 3 to say that a*b has a unique factorization, but you won't be able to say that a can be expressed with the prime factors of b directly, which is required for problem 3 to apply.

13. Aug 22, 2009

### ramsey2879

I think all problem 4 was asking was to apply the fact that p^0 is equal to 1 to expand on the proof of problem 3. Call it an axiom or a proven fact or what ever, but I dont think problem 4 requires one to prove that p^0 is 1.

Last edited: Aug 22, 2009
14. Aug 23, 2009

### Dragonfall

Not having seen the rest of the book, I don't know how problem 3 really relates to problem 4. But the fact is that to use the result of 3, you need to have the prime factorization. However, 1 isn't a prime, so you can't just use the fact that p^0=1.

15. Aug 23, 2009

### ramsey2879

You don't understand because problem 4 specifically allows zero as the exponent of some of the primes in the prime factorization of a*b to be applied in the expression of "a" and "b" as an expression of common primes to a non-negative power multiplied together respectively. if p^6 is part of the prime factorization of a*b then the p part of "a" is p^i and that of "b" must be p^(6-i) where the i can go from 0 to 6.