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I've been studying a specific super-set of the rationals that I've been trying to prove is not closed under addition. The set is defined as all numbers of the form

$$ \prod_{i=1}^{\infty}{p_i^{\alpha_i}} $$

where $p_i$ is the ith prime number and $\alpha_i\in\mathbb{Q}$ (the exponents are rational) and the number of terms with non-zero alpha is finite. I've been calling this set $\mathbb{S}$ because of their semi-rational behavior, and the set is a super-set of the rationals. It is easy to show that $\mathbb{S}$ is closed under multiplication and division. What has been alluding me is a clear way to show that $\mathbb{S}$ is not closed under addition. In fact, what I've been trying to prove is this:

If $s$ and $t$ are numbers in $\mathbb{S}$, then $s+t\in\mathbb{S}$ if and only if $\frac{s}{t}\in\mathbb{Q}$.

I've tried going about it in similar ways of showing that radicals of products of distinct primes are irrational, but that's not quite enough to get the result. I've also fiddled with some polynomial properties to no avail, and the efforts are dismal enough to not show so I don't confuse anyone... If anyone could please give some advice/proof/counter-example, it would greatly appreciated.

Samuel
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  • How do you define addition at all for these forms of 'numbers'? – Berci Feb 21 '18 at 16:56
  • Perhaps I should state that $\mathbb{S}$ also includes $0$? Sorry for my lack of understanding, but are there multiple definitions of "addition"? – Samuel Feb 21 '18 at 17:05
  • Is your products finite, (ie $\alpha_i$ is zero for large $i$ ) ? – Kelenner Feb 21 '18 at 17:29
  • @kelenner: yes, I will update the question. Thanks! – Samuel Feb 21 '18 at 17:31
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    Note then that your set is ${x>0; \exists d\in \mathbb{N}, d\geq 1, x^d\in \mathbb{Q}}$ – Kelenner Feb 21 '18 at 17:38
  • That is true and one of the reasons I tried going down polynomial routes, but I'm afraid my polynomial knowledge isn't very deep beyond the basics. Just seems like $x^d$ hints to polynomial. – Samuel Feb 21 '18 at 17:42
  • I thought about the case when $\alhpa_n$ is not quasiconstant $0$. – Berci Feb 21 '18 at 23:58

1 Answers1

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You can just display a case where closure fails. We have $1,\sqrt 2\in \Bbb S$ but $1+\sqrt 2 \not \in \Bbb S$ because all the numbers in $\Bbb S$ have an integer power that is rational.

Ross Millikan
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  • But can you prove that $1+\sqrt{2}$ doesn't have an integer power that is rational? – Samuel Feb 21 '18 at 17:44
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    Yes. Just expand $(1+\sqrt 2)^n=\sum_{k=0}^n {n \choose k}1^k(\sqrt 2)^{n-k}$ by the binomial theorem. All the terms with $n-k$ odd are integers times $\sqrt 2$ and are positive so they can't cancel. – Ross Millikan Feb 21 '18 at 17:48
  • And why can't $a+b\sqrt{2}$ have an integer power that is rational, where $a$ and $b$ are integers? This would need to be true for your binomial theorem statement to hold. I think what you are getting at is that it can't have a integer power that is equal to 2. – Samuel Feb 21 '18 at 17:55
  • The binomial is already of the form $a+b\sqrt 2$. We just need that it is not rational, but $\sqrt 2$ is irrational and the rationals are closed under addition and multiplication. – Ross Millikan Feb 21 '18 at 17:58
  • Oh, alright, so $(1+\sqrt{2})^n=a+b\sqrt{2}$ which cannot be an integer. I hear yeah, great going. Thanks! – Samuel Feb 21 '18 at 17:59