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

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, :*

• 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? Jul 18, 2021 at 15:34
• 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. Jul 18, 2021 at 15:35
• @Arthur Ah, I see it now. Thank you so much, Arthur :* Jul 18, 2021 at 15:51
• @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? Jul 18, 2021 at 15:53
• That’s one approach, yes. @user49685 Jul 18, 2021 at 16:17

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.

• 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. :* Jul 18, 2021 at 17:58