# Extending the Fundamental Theorem of Arithmetic to the rationals

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The Fundamental Theorem of Arithmetic essentially states that any positive whole number n can be written as:

##n = p_1^{a_1} \cdot p_2^{a_2} \cdot p_3^{a_3} \cdot \dots##

where ##p_1##, ##p_2##, ##p_3##, etc. are all the primes, and ##a_1##, ##a_2##, ##a_3##, etc. are non-negative integers. (The theorem also essentially states that no two distinct combinations of the latter will produce the same natural number.)

I have been thinking that it's relatively easy to prove that if we lift the restriction of the exponents being non-negative, ie. we allow any integer as exponent, then the statement is true for any positive rational number n. In other words, any positive rational number n can be written as the above type of product (where the exponents are integers).

(The proof is easy to sketch by realizing that in the most-simplified forms of rational numbers the numerator and denominator are relatively prime, ie. they don't share any prime factors, which means the same prime number is never needed for both the numerator and the denominator.)

I am not sure about the uniqueness, however. Could two different combinations of integers produce the same rational number? I think that completeness (ie. any rational number can be written as a finite product like the above) is relatively easy to prove, but I'm not so sure about uniqueness.

Also, I'm not sure how this "extended" version of the theorem could be succinctly expressed. The original theorem can be expressed as "any positive whole number can be written as a unique product of primes", but this extended version is more difficult to express in such a simple manner.

fresh_42
Mentor
An extension to the rationals does not make sense. The theorem is about primes, and there are no primes in a field like ##\mathbb{Q}##. So if you ask for a decomposition of a rational number ##n\in \mathbb{Q},## then ##n=n## is already a solution.

The idea is the question: How does a decomposition of an integer into irreducible components look like? Now irreducible and prime are the same in that case, so the question becomes: How does a decomposition of an integer into prime components look like? However, there are no irreducible elements in a field, because all numbers are either ##0## or units.

I don't see why not. Every positive rational number can be expressed as the ratio between two unique products of primes which do not share any primes. When the Fundamental Theorem of Arithmetic is expressed in the form I used in my post, it naturally extends to the (positive) rationals by simply lifting the restriction of the exponents being non-negative. The exact same equation applies in both cases.

fresh_42
Mentor
Sure, you can write a rational number as a primitive quotient, and then apply the FTA to numerator and denominator. But you cannot generalize the theorem itself because the conditions become empty statements. You will always have to distinguish the two sides of the quotient, for otherwise ##10=2\cdot 5 = \dfrac{1}{\dfrac{1}{2}}\cdot\dfrac{1}{\dfrac{1}{5}}## would be allowed. But if you distinguish numerator and denominator, then it is no more then FTA applied twice.

Try to formulate the theorem and you will see that you cannot do it.

Office_Shredder
Staff Emeritus
Gold Member
Every non-zero rational number can uniquely be written as a product ##\pm \prod p_i^{\alpha_i}## where the ##p##s are distinct prime integers and the ##\alpha##s are non zero integers (unique up to reordering)

mfb
Mentor
2 and 5 are primes, 1/2 and 1/5 are not.
Every non-zero rational number can uniquely be written as a product ##\pm \prod p_i^{\alpha_i}## where the ##p##s are distinct prime integers and the ##\alpha##s are non zero integers (unique up to reordering)
Alternatively, ##q = \pm \prod p_i^{\alpha_i}## where pi is the sequence of (all) prime numbers and the integers ##\alpha_i## are non-zero for a finite number of elements (i.e. there is an N such that ##\alpha_i=0 \,\forall i>N##). This is a unique representation as well.

fresh_42
Mentor
2 and 5 are primes, 1/2 and 1/5 are not.
##2## and ##5## aren't either! You cannot formulate a theorem which doesn't have to apply FTA twice in order to get the expression. There is no way to remain in the field without referring to its ring of integers.

Office_Shredder
Staff Emeritus
Gold Member
##2## and ##5## aren't either! You cannot formulate a theorem which doesn't have to apply FTA twice in order to get the expression. There is no way to remain in the field without referring to its ring of integers.

2 and 5 are prime numbers in the ring of integers.

Do you think what I wrote in #5 is wrong? If your only claim is it's not a very deep theory that tells you interesting new things to try to extend to more abstract objects, that might be true.

fresh_42
Mentor
2 and 5 are prime numbers in the ring of integers.

Do you think what I wrote in #5 is wrong?
Of course not.
If your only claim is it's not a very deep theory that tells you interesting new things to try to extend to more abstract objects, that might be true.
My claim is that the theorem makes only sense for rings, not for fields. If you want to apply it to a quotient field, then it is the version of the integer ring applied twice. A formulation which only refers to the rational numbers is not possible, because there are no primes. Hence the OP's suggestion is simply FTA in its ordinary version applied to numerator and denominator. It is no theorem of rational numbers, i.e. nothing new under the sun. And the reason is, that there are no irreducible or prime rational numbers.

FTA over the rationals is a contradiction in itself. (For nitpickers: a tautology.)

2 and 5 are primes, 1/2 and 1/5 are not.
The FTA is not a statement about primes, but a statement about the natural numbers. The FTA doesn't say that 2 and 5 are primes, it just says that 2 and 5 can be expressed as a product of primes. The FTA doesn't define the prime numbers, it simply states that all positive whole numbers can be composed of prime numbers.

My extension proposal simply extends that idea to all positive rational numbers. What I noticed is that the only thing needed for this is lifting the restriction of the prime exponents being positive. The theorem still works, and now covers all positive rational numbers, not just the whole numbers.

fresh_42
Mentor
The FTA is not a statement about primes, but a statement about the natural numbers.
It is a statement about the ring of integers. It needs the concept of primes to be phrased.
The FTA doesn't say that 2 and 5 are primes, it just says that 2 and 5 can be expressed as a product of primes. The FTA doesn't define the prime numbers, it simply states that all positive whole numbers can be composed of prime numbers.
Right. You said it yourself: you first need to know what a prime number is.
My extension proposal simply extends that idea to all positive rational numbers.
Wrong. It uses FTA and applies it to numerator and denominator. It does not extend anything, it applies a known fact twice. ##\mathbb{Z}\times \mathbb{Z}## as you use it, is fundamentally something different than ##\mathbb{Q}## and is important that you see this! You cannot define a prime number in the field of rational numbers. O.k., you can define them, but there are none!
What I noticed is that the only thing needed for this is lifting the restriction of the prime exponents being positive.
Wrong. You needed the entire theorem before you could even phrase your statement. It is not a generalization, it is an application.
The theorem still works, and now covers all positive rational numbers, not just the whole numbers.
Wrong. The theorem does not exist over the rational numbers! You confuse ##\mathbb{Z}^2## with ##\mathbb{Q}##.

Staff Emeritus
@fresh_42 , I don't think @Warp understands your reasoning. It doesn't look like he has run across the concept of ring and field.

fresh_42
Mentor
@fresh_42 , I don't think @Warp understands your reasoning. It doesn't look like he has run across the concept of ring and field.
Yes, but the question is whether he wants to understand the concept of prime elements or not. It is important to understand that there are no prime rational numbers. And it is important to understand the difference between an application and a generalization. We don't need ring or field as terms for neither of those. But the terms could be a motivation to learn what's behind the theorem. Rings with that property even have a special name.

It is important to understand that there are no prime rational numbers.
For someone who is so nitpicky about exact details, you have been extraordinarily careless in that statement (which is technically false). You might want to rephrase.

FactChecker
Gold Member
The Fundamental Theorem of Arithmetic is simply a statement about a unique representation of integers in terms of the product of powers of primes, where each prime appears only once. It certainly can be applied to rational numbers, since they are represented by ratios of integers. There is a unique representation of any rational number as the ratio of products of powers of prime numbers, where each prime number appears only once.
The OP expresses some doubt about how to express a theorem for the rationals. I think it would be a good exercise to look up the exact statement of the Fundamental Theorem of Arithmetic and see how it could be stated to apply to the rationals.

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Staff Emeritus
I think it would be a good exercise to look up the exact statement of the Fundamental Theorem of Arithmetic and see how it could be stated to apply to the rationals.

I agree. I also think where the thread went south was here:

Also, I'm not sure how this "extended" version of the theorem could be succinctly expressed.

A crisp definition of exactly what we are talking about would have helped.

fresh_42
Mentor
For someone who is so nitpicky about exact details, you have been extraordinarily careless in that statement (which is technically false). You might want to rephrase.
Nope,
It is important to understand that there are no prime rational numbers.
is completely correct. You will not find a prime in ##\mathbb{Q}##.

Edit: It is not nitpicky. It is about understanding. That the others say you are right isn't helpful. You can only define primality for numbers that do not have an inverse. However, all nonzero rational numbers do have an inverse. The point is, that you cannot leave the integers without destroying the meaning of a prime number. Hence you should not phrase FTA over the rationals, simply because you will learn the wrong message. The fact that others refuse to see their misguiding advice doesn't make them right.

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FactChecker
Gold Member
Any rational number can be reduced to lowest terms and then both the numerator and denominator of the reduced fraction can be factored uniquely into their unique prime representations.

You will not find a prime in ##\mathbb{Q}##.
All integers are rational numbers (they belong to ##\mathbb{Q}##). All the prime numbers are integers. Therefore all prime numbers are rational.

By saying that prime numbers are not rational you are saying that they are irrational numbers.

Does your definition of ##\mathbb{Q}## not include the integers?

fresh_42
Mentor
All integers are rational numbers (they belong to ##\mathbb{Q}##). All the prime numbers are integers. Therefore all prime numbers are rational.
And this is exactly what you got wrong and refuse to learn. Repetition doesn't make it right. The condition to be prime does not make any sense within ##\mathbb{Q}##. It can only be phrased in ##\mathbb{Z}##.
By saying that prime numbers are not rational you are saying that they are irrational numbers.
No. I am saying that the word 'prime' becomes void over the rationals:
"A number ##p## is prime if ##p## is not invertible, and from ##p|(a\cdot b)## it follows that ##p|a## or ##p|b.##"
There is simply no rational number for which this is true. Only integers.
Does your definition of ##\mathbb{Q}## not include the integers?
It includes the integers, it does not include a meaningful way to define the term 'prime'.

mfb
Mentor
@fresh_42: It's clear to everyone that OP means "primes in the integers" throughout the thread.

FactChecker
Gold Member
@fresh_42: It's clear to everyone that OP means "primes in the integers" throughout the thread.
I'm not sure that it was originally clear, and the comments about rationals not having primes were reasonable. But I think that it was appropriate to steer the thread toward prime factorization of numerator and denominator since that has a reasonable answer.

fresh_42
Mentor
@fresh_42: It's clear to everyone that OP means "primes in the integers" throughout the thread.
No, it is not. Read the first post:
I have been thinking that it's relatively easy to prove that if we lift the restriction of the exponents being non-negative, ie. we allow any integer as exponent, then the statement is true for any positive rational number n.
The theorem reduces to a tautology over rational numbers. The distinction between ##\mathbb{Z}^2## and ##\mathbb{Q}## is crucial! Everything else gives the OP a false impression, which we can see throughout the thread. It makes no sense to confirm a false impression. The question has been answered in the first 3 posts. The rest is an attempt to teach something which is not true. Sorry, that I do not think this is a good idea.

FactChecker