(i) 相对于固定原点 $O$，点 $A$、$B$ 和 $C$ 的位置向量分别为 $\mathbf{a},\mathbf{b},\mathbf{c}$。

点 $A$、$B$ 和 $C$ 共线，且 $B$ 在线段 $AC$ 之间。

已知 $AB:AC=1:3$，证明

(ii) 已知 $n\in\mathbb{N}$，用反证法证明：如果 $n^2$ 是 3 的倍数，那么 $n$ 也是 3 的倍数。

因为 $AB:AC=1:3$ 且 $B$ 位于 $A,C$ 之间，所以 $BC=2AB$。分别用位置向量写出 $\overrightarrow{AB}$ 与 $\overrightarrow{BC}$，再整理即可得到目标式。

Since $AB:AC=1:3$, we have

In vector form,

and

Therefore,

Hence,

as required.

由共线及点的顺序，向量 $\overrightarrow{AB}$ 与 $\overrightarrow{AC}$ 同向。直接把长度比写成 $\overrightarrow{AC}=3\overrightarrow{AB}$，代入位置向量并展开括号即可。

Since $B$ lies between $A$ and $C$ and $AB:AC=1:3$,

Substituting the position vectors gives

Adding $\mathbf{a}$ to both sides,

as required.

反设 $n^2$ 是 3 的倍数，但 $n$ 不是。任何不是 3 的倍数的自然数都可写成 $3p+1$ 或 $3p+2$；分别平方后，结果除以 3 都余 1，与假设矛盾。

Assume, for a contradiction, that $n^2$ is a multiple of $3$ but $n$ is not a multiple of $3$.

Then, for some integer $p$, either

or

In the first case,

Thus $n^2$ leaves remainder $1$ when divided by $3$.

In the second case,

Again, $n^2$ leaves remainder $1$ when divided by $3$.

In neither case is $n^2$ a multiple of $3$, contradicting the assumption. Therefore,

也可以用模 3 算术表达同一个反证。若 $n$ 不是 3 的倍数，则它模 3 只能为 1 或 2；两种余数平方后都模 3 余 1。

Assume, for a contradiction, that

but

Then $n\equiv1\pmod3$ or $n\equiv2\pmod3$. However,

and

Thus $n^2\equiv1\pmod3$, contradicting $n^2\equiv0\pmod3$. Hence,

不是 3 的倍数的整数也可统一写成 $3p+1$ 或 $3p-1$。分别展开平方，两式都等于 3 的倍数加 1；这与反设中 $n^2$ 是 3 的倍数矛盾。

Assume, for a contradiction, that $n^2$ is a multiple of $3$ but $n$ is not.

For some integer $p$, either $n=3p+1$ or $n=3p-1$.

The two cases give

In either case, $n^2$ leaves remainder $1$ when divided by $3$, contradicting the assumption that $n^2$ is a multiple of $3$.

Therefore, if $n^2$ is a multiple of $3$, then $n$ is a multiple of $3$.