Chapter 6.4, Problem 38WE

To determine

To Show:

The reciprocal of $\frac{\sqrt{5}+1}{2}$ is also the conjugate of $\frac{\sqrt{5}+1}{2}$ .

Explanation of Solution

Given:

Given expression:

  $\frac{\sqrt{5}+1}{2}$

Formula Used:

  $\left(a+b\right)\left(a-b\right)=a^2-b^2$

Calculation:

Reciprocal of $x\text{ is }\frac{1}{x}$ .

So, the reciprocal of $\frac{\sqrt{5}+1}{2}\text{ is }\frac{1}{\frac{\sqrt{5}+1}{2}}=\frac{2}{\sqrt{5}+1}.$

Since the denominator cannot have radicals , we need to rationalize the denominator by multiplying and dividing the number by the conjugate of the denominator.

  $\begin{array}{l}\frac{2}{\sqrt{5}+1}\\ =\frac{2}{\sqrt{5}+1}\cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}\\ =\frac{2\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\\ =\frac{2\left(\sqrt{5}-1\right)}{{\left(\sqrt{5}\right)}^2-{\left(1\right)}^2}\mathrm{.................}[\left(a+b\right)\left(a-b\right)=a^2-b^2]\\ =\frac{2\left(\sqrt{5}-1\right)}{5-1}\\ =\frac{2\left(\sqrt{5}-1\right)}{2\cdot 2}\\ =\frac{\sqrt{5}-1}{2}\end{array}$

So, the reciprocal of $\frac{\sqrt{5}+1}{2}\text{ is }\frac{\sqrt{5}-1}{2}\mathrm{................}(1)$

The expressions $a\sqrt{b}+c\sqrt{d}\text{ and }a\sqrt{b}-c\sqrt{d}$ are conjugates.

So, the conjugate of $\frac{\sqrt{5}+1}{2}\text{ is }\frac{\sqrt{5}-1}{2}\mathrm{..................}(2)$

Hence, from the equation (1) and (2), we have that the reciprocal and the conjugate of $\frac{\sqrt{5}+1}{2}$ are equal.