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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.

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    $\begingroup$ 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 $a_i=e^i$. $\endgroup$
    – lulu
    Oct 23, 2018 at 14:45
  • $\begingroup$ @lulu That would do it. So generally, taking an irrational number to a power should do it too, right? $\endgroup$
    – The Dude
    Oct 23, 2018 at 14:49
  • $\begingroup$ No, square root of 2 doesnt work, for instance. $\endgroup$
    – alexp9
    Oct 23, 2018 at 14:49
  • $\begingroup$ No! you need a transcendental number. Won't work if you took $\sqrt 2$, say. $\endgroup$
    – lulu
    Oct 23, 2018 at 14:49
  • $\begingroup$ @Rhcpy99 exactly. $\endgroup$
    – lulu
    Oct 23, 2018 at 14:50

3 Answers 3

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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.

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  • $\begingroup$ Excellent! This is exactly what I am looking for. $\endgroup$
    – The Dude
    Oct 23, 2018 at 14:49
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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?

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  • $\begingroup$ Hmmm! Interesting! $\endgroup$
    – The Dude
    Oct 24, 2018 at 14:48
  • $\begingroup$ 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"). $\endgroup$
    – Piquito
    Oct 25, 2018 at 15:15
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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.

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  • $\begingroup$ Combined with the comment above, this is a good answer. $\endgroup$
    – The Dude
    Oct 24, 2018 at 14:47

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