Find a rational number that differs from the number
a) $\alpha=\sqrt{2}$;
b) $\alpha=2+\sqrt{5}$;
c) $\alpha=3+\sqrt{7}$ by no more than 0.0001.
a) The continued fraction expansion of the number $\sqrt{2}$ is known from problem $\underline{60613}$, and the convergents are calculated in the remark to problem $\underline{60614}$. Since $\frac{P_{5}}{Q_{5}}-\sqrt{2}10000$, the fraction $P_{3 /} Q_{3}=305 / 72$ fits.

b) Let's expand the number $\beta=3+\sqrt{7}$ into a continued fraction.

$b_{0}=[\beta]=5 ; \quad \beta_{1}=\frac{1}{\sqrt{7}-2}=\frac{\sqrt{7}+2}{3}, \quad b_{1}=\left[\beta_{1}\right]=1 ; \beta_{2}=\frac{3}{\sqrt{7}-1}=\frac{\sqrt{7}+1}{2}, \quad b_{2}=\left[\beta_{2}\right]=1$;

$\beta_{3}=\frac{2}{\sqrt{7}-1}=\frac{\sqrt{7}+1}{3}, \quad b_{3}=\left[\beta_{3}\right]=1 ; \quad \beta_{4}=\frac{3}{\sqrt{7}-2}=\sqrt{7}+2=\beta-1$.

Therefore, $\beta-1=[4,(1,1,1,4)]$, and $\beta=[5,(1,1,1,4)]$. Calculating the convergents, we get $P_{7}=271$, $Q_{7}=48, Q_{8}=223$. Since

$Q_{7} Q_{8}>10000$, the fraction $P^{7} / Q_{7}=271 / 48$ fits.

## Answer

a) ${ }^{99} / 70$;
b) ${ }^{305} / 72 ;$
c) $271 / 48$.