Number theory: Can all rational numbers $>1$ be expressed as a product of rational numbers $>1$?

Question

Let G denote the set of rational numbers that are greater than or equal to 1. Call an element $x \in G$ a G-prime if it cannot be factored as $x = yz$, where $y, z \in G$, unless $y = 1$ or $z = 1$. Find all G-primes. Is it the case that every element of G can be factored as a product of G-primes?

My attempt: I first started with the set G denoting all rational numbers $\geq1$. I know a G-prime to be 1, as 1 can only be expressed as $1=1\times1$ or $1= ab$, where one of $a\neq b$, and one of $a,b < 1$ and thus not in the set G. However, finding another rational that meets the prime criteria eludes me. At the same time, I cannot find a way to generalize my finding of 1 into a proof. In that case, I find the answer to the second question to be no, as not every element of G is 1, and thus cannot be expressed as products of 1.

Answer 1

If $x$ is in $G$, take the average of $x$ with $1$, and use that as a factor of $x$. Like $$x=\frac{x+1}2\cdot\frac{2x}{x+1}$$

Answer 2

Let your rational number be $\dfrac{p}{q}$, where $p,q\in\mathbb{Z^+}$ and $p>q$.

$$\frac{p}{q}=\left(\frac{p}{q+\frac{1}{2}}\right)\left(\frac{q+\frac{1}{2}}{q}\right)$$

Answer 3

Here's the thing: Whenever you have "factors of a rational number" you should immediately think "but any rational number (except $0$) can be a factor of any other rational number because we can divide any two rational numbers to get a rational number".

If we want to find the factors of $r$ and take any (non-zero) $q$ then $q$ is a factor of $r$ because $r = q\cdot \frac rq$.

So to have $1 < r = qt$ and $q,t$ are both rationals should always be possible.

If we let $q$ be any rational, $q> 1$, then we can have $r = qt$ if $t=\frac rq$. In order to have $t =\frac rq > 1$ we need $\frac rq > 1$ or in other words $r > q$ (Note, we can do this because we are assuming $q$ is positive.)

And that's that. We are done. If $r > 1$, let $q$ be any rational so that $1 < q < r$ and let $t =\frac rq$. $t=\frac rq > \frac rr =1$ and $qt = q\cdot \frac rq = r$. $r$ is not G-prime.

OLD ANSWER KEPT FOR COMPARISON

If $r > 1$ then $r = q\cdot \frac rq$ for any rational $q$ (and $\frac rq$ is rational).

We just need to show we can always find a $q > 1$ so that $\frac rq > 1$.

But if $q> 0$ then $\frac rq > 1\iff r> q$. So we just need to find any $q$ so that $1 < q < r$.

There's an infinite number of such $q$ but if we have to find one we can just take the average of $r$ and $1$.

....

So for any rational $r > 1$ there is rational $q$ so that $1 < q < r$. Then $w=\frac rq > 1$ and is rational, and $r = wq$.

So G-primes (other than one) do not exist.

A bunch of editorial comment about why we can always find rationals between two rationals follow:

===

I am taking it for granted that it is known that between any two rationals there are an infinite number of rationals. People often argue this by pointing out if $r < q$ (and $r$ and $q$ is rational) then the average of $r$ and $q$ ($\frac {r+q}2$) is rational and $r < \frac {r+q}2 < q$.

I always feel a little weary as the act of even trying to calculate a rational number between $r$ and $q$ implies such numbers might be hard to find and require calculations, whereas I feel one must develop the intuition that OF COURSE there are a rational numbers between $r$ and $q$ and the interval $(r,q)$ is just jammed packed with them and you just need to step in the interval and roll around and you will be washing them out of your underwear for weeks.

If we need a naive argument take $r = \frac mn$ and $q = \frac st$ where $m,n,s,t$ are integers and $n,t$ are positive and $r < q$. Put them over a common denominator $r = \frac {mt}{nt}$ and $q=\frac{sn}{nt}$ and $mt < sn$. Now to find a gazillion rational numbers between $r$ and $q$ just multiply the numerator and denominators by $K$ where $K$ is a huge a goose-honking number as you like. Say $K =$ one hundred gazillion.

The as $mt<sn$ are integers so the are at least one unit apart then $r=\frac{mtK}{ntK} < \frac {snK}{ntK}$ and $mtK < snK$ and there are at least $K$ integers between them. So $r=\frac{mtK+1}{ntK}<\frac {mtK+2}{ntK}< ......< \frac {mtK + 4567}{ntK}< \frac {mtK+4568}{ntK}<....< \frac{snK-2}{ntK}< \frac {snK-1}{ntK} < \frac {snK}{ntK}=q$ are at least $K = $ one hundred gazillion rational numbers between $r$ and $q$.

(Also note this also shows that we can find a rational number as close to an original rational number as we like, so let's not hear any silliness about "What's the smallest rational number we can get that is bigger than $r$")

Answer 4

If $p>q>0$, then for any $k>0$ we have $\frac{p}{q}>\frac{p+k}{q+k}>1$.

So using $k=1$ or $k=pq$, an example is $$\left(\frac{p+1}{q+1}\right)\left(\frac{p+pq}{q+pq}\right) =\frac{p^2+p^2q+p+pq}{q^2+pq^2+q+pq}=\frac{p}{q}.$$

Answer 5

For every $x \in G$ there are many solutions $x = yz$ with $y,z \in G$.

Namely for every $1<y<x$ and $z = x/y$, we have that $y,z \in G$ and $zy=x$.

This leads to the different solutions of the current answers like taking

• $y = \frac{1+x}{2}$

• $y = \frac{p+k}{q+k}$

• $y = \frac{p}{q+1/2}$

where $x = p/q$ (obviously with $p>q$).

---

This idea was already mentioned in the comment by fleablood.

---

We can generalize the result when $G$ is the set of all rational numbers greater than $L$. In this case we look for numbers $y>L$ and $z>L$ such that $x = yz > L^2$, and the G-primes are the numbers in G such $x \leq L^2$. If $L<1$, then there are no G-primes.

Answer 6

Hint: $\,\ a_3 > \color{#c00}{a_2} > a_1 >0\,\Rightarrow\, \displaystyle \frac{a_3}{a_1} = \frac{a_3}{\color{#c00}{a_2}}\,\frac{\color{#c00}{a_2}}{a_1},\, $ where $\,\color{#0a0}{\text{each factor $> 1$}}$.

---

Generally we can use Multiplicative Telescopy $ \ \displaystyle f(n) = f(a\!-\!1) \prod_{\large k\,=\,a}^{\large n} \:\!\dfrac{f(k)}{f(k\!-\!1)} $

e.g. $\,(m\!+\!n)!\,\ge\, m!\,n!\,$ by $\,\dfrac{(m\!+\!n)!}{m!\,n!} = \dfrac{m\!+\!1}{1}\,\dfrac{m\!+\!2}{2}\cdots \dfrac{m\!+\!n}{n},\ $ $\,\color{#0a0}{\text{each factor $> 1$}}$ if $\,m>1$

e.g. simpler $\ n^{n-1}\ge n!\ \ $ by $\ \dfrac{n^{n-1}}{n!}=\, \dfrac{n}{n}\,\dfrac{n}{n\!-\!1}\cdots \dfrac{n}{3}\,\dfrac{n}{2}\, \ge\, 1,\,$ with $\,\color{#0a0}{\text{each factor $> 1$}}$

Answer 7

Let $z$ be the truncation of $\sqrt{x}$ after the first nonzero digit following the decimal point. Then $1<z<x$ and $z\in\mathbb Q$. Therefore $x= z \cdot \frac{x}{z}$ where both $z$ and $\frac{x}{z}$ are greater than $1$.

Answer 8

The answers showing how to "factor" with rational numbers are correct, but they don't fully answer the question. The full answer is: Every element of G other than 1 can be written as the product of two elements of G. Therefore, the only "G-prime" is the number 1 since it cannot be written in this way.

Simple proof without formulas: Take a rational number $r > 1$. There is always a rational number strictly between two rational numbers, so choose any $q$ such that $1 < q < r$. Then $r/q > 1$, and $r$ can be "factored" into $q$ and $r/q$ since their product is $r$.