Question

Find rational numbers a and b such that

(i) $\frac{\sqrt{2}-1}{\sqrt{2}+1}=a+b\sqrt{2}$

(ii) $\frac{{\displaystyle 2-\sqrt{5}}}{{\displaystyle 2+\sqrt{5}}}=a\sqrt{5}+b$

(iii) $\frac{{\displaystyle \sqrt{3}+\sqrt{2}}}{{\displaystyle \sqrt{3}-\sqrt{2}}}=a+b\sqrt{6}$

(iv) $\frac{{\displaystyle 5+2\sqrt{3}}}{{\displaystyle 7+4\sqrt{3}}}=a+b\sqrt{3}$

Answer 1

(i)
$$\begin{gathered} \frac{\sqrt{2}-1}{\sqrt{2}+1} \\ =\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1} \\ =\frac{{\left(\sqrt{2}-1\right)}^2}{{\left(\sqrt{2}\right)}^2-1^2} \end{gathered}$$
$$\begin{gathered} =\frac{2+1-2\sqrt{2}}{2-1} \\ =3-2\sqrt{2} \end{gathered}$$
$$\begin{gathered} \therefore \frac{\sqrt{2}-1}{\sqrt{2}+1}=3+\left(-2\right)\sqrt{2}=a+b\sqrt{2} \\ \Rightarrow a=3,b=-2 \end{gathered}$$

(ii)
$$\begin{gathered} \frac{{\displaystyle 2-\sqrt{5}}}{{\displaystyle 2+\sqrt{5}}} \\ =\frac{{\displaystyle 2-\sqrt{5}}}{{\displaystyle 2+\sqrt{5}}}\times \frac{{\displaystyle 2-\sqrt{5}}}{{\displaystyle 2-\sqrt{5}}} \\ =\frac{{\left(2-\sqrt{5}\right)}^2}{{\left(2\right)}^2-{\left(\sqrt{5}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{4+5-4\sqrt{5}}{4-5} \\ =\frac{9-4\sqrt{5}}{-1} \\ =-9+4\sqrt{5} \end{gathered}$$
$$\begin{gathered} \therefore \frac{{\displaystyle 2-\sqrt{5}}}{{\displaystyle 2+\sqrt{5}}}=4\sqrt{5}+\left(-9\right)=a\sqrt{5}+b \\ \Rightarrow a=4,b=-9 \end{gathered}$$

(iii)
$$\begin{gathered} \frac{{\displaystyle \sqrt{3}+\sqrt{2}}}{{\displaystyle \sqrt{3}-\sqrt{2}}} \\ =\frac{{\displaystyle \sqrt{3}+\sqrt{2}}}{{\displaystyle \sqrt{3}-\sqrt{2}}}\times \frac{{\displaystyle \sqrt{3}+\sqrt{2}}}{{\displaystyle \sqrt{3}+\sqrt{2}}} \\ =\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{3+2+2\times \sqrt{3}\times \sqrt{2}}{3-2} \\ =5+2\sqrt{6} \end{gathered}$$
$$\begin{gathered} \therefore \frac{{\displaystyle \sqrt{3}+\sqrt{2}}}{{\displaystyle \sqrt{3}-\sqrt{2}}}=5+2\sqrt{6}=a+b\sqrt{6} \\ \Rightarrow a=5,b=2 \end{gathered}$$

(iv)
$$\begin{gathered} \frac{{\displaystyle 5+2\sqrt{3}}}{{\displaystyle 7+4\sqrt{3}}} \\ =\frac{{\displaystyle 5+2\sqrt{3}}}{{\displaystyle 7+4\sqrt{3}}}\times \frac{{\displaystyle 7-4\sqrt{3}}}{{\displaystyle 7-4\sqrt{3}}} \\ =\frac{35-20\sqrt{3}+14\sqrt{3}-24}{{\left(7\right)}^2-{\left(4\sqrt{3}\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{11-6\sqrt{3}}{49-48} \\ =11-6\sqrt{3} \end{gathered}$$
$$\begin{gathered} \therefore \frac{{\displaystyle 5+2\sqrt{3}}}{{\displaystyle 7+4\sqrt{3}}}=11+\left(-6\right)\sqrt{3}=a+b\sqrt{3} \\ \Rightarrow a=11,b=-6 \end{gathered}$$