7.4 $R^{n}$ is the $n$-dimensional Euclidean space. If a set of points in $R^{n}$ ensures that every point in $R^{n}$ is at an irrational distance from at least one point in this set, find the minimum number of points $g(n)$ in this set.

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［Solution］If $n=1$, it is obvious that $g(1)=2$.
If $n \geqslant 2$, we will prove that $g(n)=3$.
First, we prove that $g(n)>2$.
For any two points $A, B$ in the $R^{n}$ space, we can draw the perpendicular bisector (plane) $\pi$ of $A B$. Clearly, there always exists a rational number $r>\frac{1}{2} A B$, and with $A$ as the center and $r$ as the radius, we can draw a circular arc (sphere) that intersects $\pi$. Taking one intersection point $P$, we have
$$
P A=P B=r
$$

which is a rational number.
Take three points $A, B, C$ on the same line in $R^{n}$ space, such that $A B=1, B C=\sqrt[3]{2}$. Now we will prove that for any point $P$ in $R^{n}$ space, the lengths $P A, P B, P C$ must include at least one irrational number.
If $P B$ is irrational, the proposition is obvious.
If $P B$ is rational, let $P B=r$, and set $\angle P B C=\alpha$. By the cosine rule,
$$
\begin{array}{rlr}
P A^{2} & =P B^{2}+A B^{2}-2 P B \cdot A B \quad \cos (\pi-\alpha) \\
& =r^{2}+1+2 r \cos \alpha . \\
P C^{2} & =P B^{2}+B C^{2}-2 P B \cdot B C \quad \cos \alpha \\
& =r^{2}+\sqrt[3]{4}-2 \sqrt[3]{2} r \cos \alpha .
\end{array}
$$

If $\cos \alpha=\frac{\sqrt[3]{2}}{2 r}-\frac{k}{\sqrt[3]{2}}$ (where $k$ is a rational number), then $P A$ is irrational. If $\cos \alpha \neq \frac{\sqrt[3]{2}}{2 r}-\frac{k}{\sqrt[3]{2}}$ (where $k$ is a rational number), then $P C$ is irrational. In summary, we have proven that when $n \geqslant 2$, $g(n)=3$.