Extending the Fundamental Theorem of Arithmetic to the rationals

[Warp]

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]

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.

 

[Warp]

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]

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]

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]

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 p_i 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]

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]

$2$ 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]

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.)

 

[Warp]

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]

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 $\mathbb{Q}$.

 

[Vanadium 50]

@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]

@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.

 

[Warp]

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]

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.

 

[Vanadium 50]

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]

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.

 

[FactChecker]

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.

 

[Warp]

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]

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]

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

 

[FactChecker]

@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]

@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]

The rest is an attempt to teach something which is not true. Sorry, that I do not think this is a good idea.

Theorems that establish a unique representation can have value in their own right and are very common in mathematics. Rational numbers have a unique simplest reduced form and that form has a unique representation derived from applying the Fundamental Theorem of Arithmetic separately to the numerator and the denominator. That can be rewritten as the product of powers of distinct primes, allowing negative exponents. And this product is unique up to the order of the factors. Your complaints may have merit regarding other issues, but they do not disprove the unique representation.

 

[fresh_42]

Theorems that establish a unique representation can have value in their own right and are very common in mathematics.

It is misleading. It has nothing to do with algebra and leads to the anywhere false impression of what prime means. It is as if I said: Every complex polynomial of odd degree has a zero. It is correct, but completely irrelevant and misses the target. As long as the OP doesn't plan to study abstract algebra at all, go ahead teaching misconceptions. But if he sometimes will study algebra, then you haven't done him a favor by using prime and field in the same context.

 

[FactChecker]

It is misleading. It has nothing to do with algebra and leads to the anywhere false impression of what prime means.

He asked a simple question and the answer is a simple "Yes, there is a similar statement that can be made about the rational numbers." The integers are a well-defined subset of the rationals that have additional properties and the prime numbers are a well-defined subset of the integers that have additional properties. To deny that is misleading.

 

[Infrared]

There are many contexts where it makes sense to talk about (integer) primes in the context of rational numbers. For example, every norm on $\mathbb{Q}$ is equivalent to either the trivial norm, the usual absolute value, or a $p$-adic norm. This is an important fact about $\mathbb{Q}$ that references primes in $\mathbb{Z}.$ Just because the $p_i$ are not prime elements of $\mathbb{Q}$ doesn't mean that they shouldn't be referenced in any theorem about $\mathbb{Q}!$

Anyway, the statement in the OP is of course true, even though the $p_i$ are not prime elements of $\mathbb{Q}.$ Uniqueness follows by clearing denominators and using unique factorization for the integers.

 

[mfb]

Only n, the product, was allowed to be rational. The primes stay untouched.

 

[Warp]

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.

I have been thinking that it's probably quite easy to show that the factorization is unique.

The only way that two rational numbers can be equal is if the numerator and denominator of one are multiples of the numerator and denominator of the other, using the same factor. In other words, $$\frac {a}{b} = \frac {c}{d} \Leftrightarrow \frac {a}{b} = \frac {ka}{kb}; a, b, c, d, k \in \mathbb{N}$$
However, if $k$ is not 1, then the fraction is not most-simplified. In other words, no two most-simplified fractions can be the same unless their numerator and denominator are identical.

 

[cormsby]

the statement is true for any positive rational number

If this is amended to include the condition "irreducible", then it seems clear that it is true.

 

[Svein]

I am not sure about the uniqueness, however. Could two different combinations of integers produce the same rational number?

Of course they can $\frac{1}{2}$ and $\frac{17}{34}$ represent the same rational number.

Rationals do not have a unique representation. Every rational number is a member of an equivalence class of combinations of integers.

 

[FactChecker]

Of course they can $\frac{1}{2}$ and $\frac{17}{34}$ represent the same rational number.

Rationals do not have a unique representation.

But they have a unique reduced form to lowest terms, p/q where GCD(p,q)=1. Then both p and q can be factored uniquely into primes where p and q have no common primes.

 

[fresh_42]

But they have a unique reduced form to lowest terms, p/q where GCD(p,q)=1. Then both p and q can be factored uniquely into primes where p and q have no common primes.

The crucial point is the title. The word extension suggests a form of generalization, which it is not. You can't extend from ring to field. No way.

 

[FactChecker]

The crucial point is the title. The word extension suggests a form of generalization, which it is not. You can't extend from ring to field. No way.

There are some formally defined uses of the word "extension", but extending a theorem is not one of those.
In any case, the original question has probably been answered.

 

[Infrared]

The crucial point is the title. The word extension suggests a form of generalization, which it is not. You can't extend from ring to field. No way.

It is a generalization in the sense that it is a correct statement which includes unique factorization in the integers as a special case, not in the sense that it deals with prime elements in $\mathbb{Q}.$