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Prove that $\mathbb{Q}^n$is dense subset of $\mathbb{R^n}$ using the fact that $\mathbb{Q}$ is dense subset of $\mathbb{R}$

I know that $\mathbb{Q}$ is dense subset of $\mathbb{R}$ That is, between any two real numbers, there exists a rational number

then how to prove that $\mathbb{Q}^n$is dense subset of $\mathbb{R^n}$

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    $\begingroup$ Q dense in R means that for any real number r you can find a sequence of rational numbers converging to q. Well for any element of $R^n$, $r=(r_1,\cdots, r_n)$ you could find n sequences of rational numbers converging to each component of r $\endgroup$ – Max Mar 26 '18 at 14:16
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Let $(x_1,\ldots,x_n)\in\mathbb{R}^n$ and let $r>0$. For each $k\in\{1,2,\ldots,n\}$, let $q_k\in\mathbb Q$ be such that$$x_n<q_n<x_n+\frac r{\sqrt n}.$$Then\begin{align}\bigl\|(x_1,\ldots,x_n)-(q_1,\ldots,q_n)\bigr\|&=\sqrt{\sum_{k=1}^n(x_k-q_k)^2}\\&<\sqrt{\sum_{k=1}^n\left(\frac r{\sqrt n}\right)^2}\\&=r.\end{align}

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  • $\begingroup$ The fact that such a $q_n$ exists uses the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$, am I correct? That would imply any open interval in $\mathbb{R}$ includes an element of $\mathbb{Q}$ and hence, the existence of $q_n$. $\endgroup$ – zaira Oct 25 '20 at 5:25
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    $\begingroup$ Yes, I used the fact that $\Bbb Q$ is dense in $\Bbb R$. $\endgroup$ – José Carlos Santos Oct 25 '20 at 6:34
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Let $$x=(x_1,x_2,...,x_n)\in \mathbb R^n$$

for each $1\le i \le n$ the interval $$(x_i-\epsilon,x_i+\epsilon)$$ includes a rational point $r_i$

The point $$ r=(r_1,r_2,...,r_n)$$ is included in the open box of size $2\epsilon$ around $x$.

Thus $Q^n$ is dense in $\mathbb R^n.$

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