Question

两个不同的正数 $a$ 和 $b$ 各自与其倒数相差 $1$ 。求 $a + b$ ？

Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1.$ What is $a + b?$

$1$

$2$

$\sqrt{5}$

$\sqrt{6}$

$3$

正数 $x$ 与其倒数相差 $1$ 时， $x - \dfrac1x = 1$ 或 $x - \dfrac1x = -1$ ，即 $x^2 - x - 1 = 0$ 或 $x^2 + x - 1 = 0$ 。

正根为 $\dfrac{1+\sqrt5}{2}$ 和 $\dfrac{-1+\sqrt5}{2}$ ，它们互为倒数。它们的和是 $a + b = \sqrt5$ 。

因此，正确答案是 C。

A positive number $x$ differs from its reciprocal by $1$ when $x - \dfrac1x = 1$ or $x - \dfrac1x = -1,$ i.e. $x^2 - x - 1 = 0$ or $x^2 + x - 1 = 0.$

The positive roots are $\dfrac{1+\sqrt5}{2}$ and $\dfrac{-1+\sqrt5}{2},$ which are reciprocals of each other. Their sum is $a + b = \sqrt5.$

Thus, the correct answer is C.