Elements of Algebra

106 ELEMENTS SECT. II. and for the denominator: $\begin{array}{r} 1+\sqrt{2} \\ 1-\sqrt{2} \\ \hline 1+\sqrt{2} \\ -\sqrt{2}-2 \\ \hline 1-2=-1 \end{array}$. Our new fraction therefore is $\frac{-\sqrt{2}-1}{-1}$; and if we again multiply the two terms by $-1$, we shall have for the numerator $\sqrt{2}+1$, and for the denominator $+1$. Now, it is easy to shew that $\sqrt{2}+1$ is equal to the proposed fraction $\frac{3+2\sqrt{2}}{1+\sqrt{2}}$; for $\sqrt{2}+1$ being multiplied by the divisor $1+\sqrt{2}$, thus, $\begin{array}{r} 1+\sqrt{2} \\ 1+\sqrt{2} \\ \hline 1+\sqrt{2} \\ \sqrt{2}+2 \end{array}$ we have $1+2\sqrt{2}+2=3+2\sqrt{2}$. Another example. Let $8-5\sqrt{2}$ be divided by $3-2\sqrt{2}$. This, in the first instance, is $\frac{8-5\sqrt{2}}{3-2\sqrt{2}}$; and multiplying the two terms of this fraction by $3+2\sqrt{2}$, we have for the numerator, $\begin{array}{r} 8-5\sqrt{2} \\ 3+2\sqrt{2} \\ \hline 24-15\sqrt{2} \\ 16\sqrt{2}-20 \\ \hline 24+\sqrt{2}-20=4+\sqrt{2} \end{array}$; and for the denominator, $\begin{array}{r} 3-2\sqrt{2} \\ 3+2\sqrt{2} \\ \hline 9-6\sqrt{2} \\ 6\sqrt{2}-8 \\ \hline 9-8=1 \end{array}$.