# Create a set of N numbers with no common rational factor

Question:

So I want to create a set of real numbers $$\{a\}_{N} = \{a_{1}, a_{2}, \ldots, a_{N}\}$$ such that if there exists a common factor between all of the elements, it must be irrational.

In this way, the fraction $$\frac{a_{i}}{a_{j}} \notin \mathbb{Q} \; \forall i \neq j \; \; (1 \leq i,j \leq N)$$.

How can I create this set?

Solution Attempt:

We know that the square root of any prime number is irrational, therefore, just pick the set $$\{a\}_{N}$$ to be a set of square roots of distinct prime numbers.

What is tripping me up is that, the division of two irrational numbers can still be rational. An easy example is $$\frac{2 \sqrt{2}}{3\sqrt{2}} = \frac{2}{3} \in \mathbb{Q}$$

Disclaimer:

I am not a number theorist, and have never taken a course on number theory, but I converted another problem I am working on to this problem. If I can solve this problem, I can solve the other problem, however, I don't know if this problem has a solution, and if it does, how to find it.

• Speaking of "common factors" in this context is confusing, but the requirement that $i\neq j\implies \frac {a_i}{a_j}\notin \mathbb Q$ is clear. Just take $a_i=e^i$.
– lulu
Oct 23, 2018 at 14:45
• @lulu That would do it. So generally, taking an irrational number to a power should do it too, right? Oct 23, 2018 at 14:49
• No, square root of 2 doesnt work, for instance. Oct 23, 2018 at 14:49
• No! you need a transcendental number. Won't work if you took $\sqrt 2$, say.
– lulu
Oct 23, 2018 at 14:49
• @Rhcpy99 exactly.
– lulu
Oct 23, 2018 at 14:50

## 3 Answers

Let $$\alpha$$ be any transcendental number (such as $$e$$ or $$\pi$$). Then let $$a_i=\alpha^i$$.

To see that this works, suppose that $$\frac {a_i}{a_j}\in \mathbb Q$$ for some $$i\neq j$$. Then we'd have $$\alpha^i-c\times \alpha^j=0$$ for $$c\in \mathbb Q$$ which would be a polynomial with rational coefficients satisfied by $$\alpha$$, contradicting transcendence.

• Excellent! This is exactly what I am looking for. Oct 23, 2018 at 14:49

HINT.-Take the primes $$2,3,5,7,11,13,\cdots p_N$$

You do have an example with the set $$\{a\}_{N} = \{\sqrt2, \sqrt{2\cdot3},\sqrt{2\cdot3\cdot5}\cdots\sqrt{2\cdot3\cdot5\cdots.p_N}\}$$

NOTE.-For all non-zero real numbers $$x,k$$ you do have $$x=kx_1$$ for some real $$x_1$$. What is the "common factor" you want to say?

• Hmmm! Interesting! Oct 24, 2018 at 14:48
• The problem is, dear friend, that you can always have a RATIONAL "common factor" for any set of real numbers (unless you want to dispense with the basic principle that "any quantity can be replaced by its equal"). Oct 25, 2018 at 15:15

Well, your solution attempt also works. It's not that much different or more difficult to prove that $$\frac{\sqrt{p}}{\sqrt{q}}$$ is irrational for different primes $$p,q$$ than proving $$\sqrt{2}$$ is irrational.

Your 'tripping me up' fear is not without reason, of course, but you circumvented it by chosing roots of primes, not just any number that isn't a square.

• Combined with the comment above, this is a good answer. Oct 24, 2018 at 14:47