The specific question:

The product of three positive numbers is equal to 12. Of these three positive numbers, $n$ are not integers. What are all of the possible values of $n$?

The permissible values of $n$ are clearly 0, 1, 2, and 3.

My question is how would I mathematically determine the permissible values of $n$?

Also, I have an inclination that any integer could be composed of the products of any number of non-integers(fractions) from $-\infty$ to $\infty$, when constraints are put on the permissible range of values. Is this accurate, for a general case?

  • $\begingroup$ Well, you could try to find values of $x,y,z>0$ with $xyz=12$ and such that $n=0,1,2,3$. $\endgroup$ – Fimpellizieri May 11 '18 at 22:15
  • $\begingroup$ A good contrast to this question is the same with $n$ not being rational. Now $1$ is not permissible because two rationals times an irrational is always irrational. You can have an integer times a non-integer rational equal an integer, which is what makes $1$ be permissible. $\endgroup$ – Ross Millikan May 12 '18 at 3:12

The answer to the specific question itself has been answered by Rhys Hughes.

However, there is something that stands out to me in your question that I would like to go over. It seems you have an intuitive feeling for the answer (that's something that should not be underestimated) since you've stated "The permissible values of n are clearly 0, 1, 2, and 3." I would go so far as to say you know the answer to the problem. Yet you don't seem convinced that you've solved the problem since you have not "mathematically determined" the permissible values of n. I'm curious as to what you mean by "mathematically determine".

If you were dealing with the question:

"What are the values of $x$ for which $x^2 + x - 5 = 0$ has a solution?"

Then we can "mathematically determine" the values for $x$ that make the equation true using the quadratic formula. If you're wondering if there is something similar to that, a clean cut way we can show (an explicit formula for example) the permissible values of n, then you're out of luck. Many problems cannot be verified using an explicit formula. Instead you'll have to use properties that you know in order to verify a solution.

In writing this I assumed the least mathematical background so I apologize if I gave you less credit than you deserve but I wanted to cover the fundamentals.

Think about what your expectations to the solution are when you mention to "mathematically determine" something. Sometimes to "mathematically determine" something means to "logically determine" something which does not always imply there will be an explicit formula you can use.

This actually opens up a door into the philosophy of mathematics. When do we know we've proved something? I recommend you check out the following: Heuristic Arguments, and Paul Erdos use of his famous phrase 'THE BOOK'. Good luck with further problems.

Ok, I see what you mean now. Here is a proof to show we can express any integer as the product of any number of rational numbers.

Suppose you had an integer $n$, then $n$ is either prime or composite. (I'll treat the case if $n$ is prime since the composite case will follow from it.) Now let $p_i$ denote the $i$th prime number and define

$H_m = (p_1 * p_2 * ... * p_m)/(p_1 * p_2 * ... * p_m)$

Clearly, $H_m = 1$. We can express $H_m$ as the product of independent irreducible fractions such as

$H_m = (p_2 / p_1) * (p_3 / p_2) *...*(p_m / p_{m-1})*(p_1 / p_m)$

Now that we have expressed $H_m$ as the product of independent irreducible fractions we choose one of those fractions whose denominator does not divide our initial integer $n$ and multiply it by $n$. For example if $n$ is not an even prime then

$((n*p_2) / p_1) * (p_3 / p_2) *...*(p_m / p_{m-1})*(p_1 / p_m) = n*H_m = n*1 = n$

Notice then that we have expressed $n$ as the product of $m$ rational numbers. These are strictly rational numbers since they are irreducible fractions because no two different primes divide one another. Also since prime numbers are infinite then $m$ does not have an upper bound. Therefore you can express $n$ as the product of infinitely many rational numbers.

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  • 1
    $\begingroup$ I'm not looking for an equation, or anything like that. At the same time, I would prefer to have a method for finding permissible values. Rhys solution for finding $n = 3$ works for that. I guess I'm checking if there's a proof that could prove for any integer can be the product of any amount of $n$ non-integers. $\endgroup$ – ZeroPhase May 12 '18 at 0:08
  • $\begingroup$ Thanks. Just wanted to make sure my thinking was correct on the subject. If I see a question like this on the GRE I won't have to waste time trying to find fractions that make each possible size of $n$ work. I'll just select all of the above. $\endgroup$ – ZeroPhase May 12 '18 at 5:09
  • $\begingroup$ @ZeroPhase foe $n>3$ would it not be possible to simply extend my equation into $n$ variables: e.g. for $n=4$ take $\frac{w}{x}*\frac{x}{y}*\frac{y}{z}*\frac{z}{w}=1$, and multiply one of those by 12 where it won't become whole. $\endgroup$ – Rhys Hughes May 12 '18 at 10:05

$n=0,1,2$ are very easy to find

$n=0 \to 1*3*4$

$n=1\to 0.5*1*24$

$n=2\to 0.5*0.5*48$

For $n=3$, note: $\frac{x}{y}*\frac{y}{z}*\frac{z}{x}=1$

Hence let's say $x=3, y=4, z=5$: we can then insert:


$$\frac{3}{4}*\frac{48}{5}*\frac{5}{3}=12$$ for $n=3$

I'd say there's probably infinitely many ways for each of these.

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