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I come along this problem while messing around with some other problems. So the question is:

Is $\left\{ a \sqrt{3} - b \sqrt{2} \ | \ a, b \in \mathbb{N} \right\}$ dense in $\mathbb{R}$?

I know for sure that:

  • $A = \left\{ a \sqrt{3} + b \sqrt{2} \ | \ a, b \in \color{red}{\mathbb{Z}} \right\}$ is dense in $\mathbb{R}$.
  • $B = \left\{ a \sqrt{3} + b \sqrt{2} \ | \ a, b \in \mathbb{N} \right\}$ is not dense in $\mathbb{R}$.

But what about $\left\{ a \sqrt{3} - b \sqrt{2} \ | \ a, b \in \mathbb{N} \right\}$? I suspect that it may be dense; however, I cannot prove it. Can someone please give me a push?

Thanks very much in advance, :*

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    $\begingroup$ We may reduce to these two cases: 1) $\left\{ a \sqrt{3} - b \sqrt{2} \mid a, b \in \mathbb{N} \right\}$ contains positive numbers arbitrarily close to $0$, and 2) $\left\{ a \sqrt{3} - b \sqrt{2} \mid a, b \in \mathbb{N} \right\}$ contains negative numbers arbitrarily close to $0$. Can you see why this is necessary and sufficient? $\endgroup$
    – Arthur
    Jul 18, 2021 at 15:34
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    $\begingroup$ You are correct. It can be seen to be equivalent to $$\{a\sqrt{\frac{3}{2}}-b\mid a,b\in\mathbb N\}$$ is dense. Then you can show the general theorem that, for $\alpha\in R,$ $$\{a\alpha-b\mid a,b\in \mathbb N\}$$ is dense in $\mathbb R$ if and only if $\alpha$ is irrational. $\endgroup$ Jul 18, 2021 at 15:35
  • $\begingroup$ @Arthur Ah, I see it now. Thank you so much, Arthur :* $\endgroup$
    – user49685
    Jul 18, 2021 at 15:51
  • $\begingroup$ @ThomasAndrews. Thank you so much for your reply, so this is basically the same as the proof for $\{n \alpha - \lfloor n \alpha \rfloor : n \in \mathbb{N} \}$ dense in $[0; 1]$ where $\alpha$ is an irrational number, right? $\endgroup$
    – user49685
    Jul 18, 2021 at 15:53
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    $\begingroup$ That’s one approach, yes. @user49685 $\endgroup$ Jul 18, 2021 at 16:17

1 Answer 1

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Note that if $\sqrt{2}a+\sqrt{3}b$ is dense in $\mathbb R$ then it is also dense if we disallow $(a,b)=(0,0)$.

Take $\varepsilon > 0$ and $c\in \mathbb R$. We must find an $a,b \in \mathbb Z$ so that $a\sqrt{3} + b\sqrt{2} \in (c-\varepsilon,c+\varepsilon)$ with $a$ positive and $b$ negative.

Take $a_1,b_1 \in \mathbb Z$ so that $a_1\sqrt{3} + b_1\sqrt{2} \in (c-\varepsilon/2, c+\varepsilon/2)$.

For each integer value of $a$ there is a minimum for $|a\sqrt{3}+b\sqrt{2}|$ where $b$ can take values on $\mathbb Z$, call it $f(a)$.

For each integer value of $b$ there is a minimum for $|a\sqrt{3}+b\sqrt{2}|$ where $a$ can take values on $\mathbb Z$, call it $g(b)$.

Let $\varepsilon'$ be a positive real smaller than $\varepsilon/2$ and smaller than $f(a)$ for all $|a| \leq |a_1|$ and smaller than $g(b)$ for all $|b| \leq |b_1|$, and smaller than $\sqrt{2}$.

Take $a_2,b_2\in \mathbb Z$ so that $a_2\sqrt{3} + b_2\sqrt{2} \in (-\epsilon',\epsilon')$, we can assume $a_2>a_1$ and $b_2< -b_1$ (otherwise multiply both by $-1$). Note that we must have $|a_2| > |a_1|$ because $\epsilon'$ is less than the $f(a)$ and we must have $|b_2| > |b_1|$ because $\epsilon'$ is less than the $g(b)$ and both values cant have the same sign because $\varepsilon' < \sqrt{2}$.

Notice $(a_1+a_2)\sqrt{3} + (b_1+b_2)\sqrt{2} \in (c-\varepsilon,c + \varepsilon)$ with $a_1+a_2$ positive and $b_1+b_2$ negative.

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    $\begingroup$ Thank you very much for your help, Yorch. I think we can improve the answer just a little bit by removing the condition $\epsilon' < \sqrt{2}$; because, we have $f(0) = \sqrt{2}$; and $\epsilon'$ is already set to be less than every instance of $f(a)$ for $|a| < |a_1|$. Thank you so much for your help. I really appreciate it. :* $\endgroup$
    – user49685
    Jul 18, 2021 at 17:58

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