Example 3. Compare the number $a$ with one, if

$$
a=0.99999^{1.00001} \cdot 1.00001^{0.99999} .
$$
S o l u t i o n. Let's represent the numbers in the given expression as follows:

$$
0.99999=1-\alpha, \quad 1.00001=1+\alpha
$$

where $\alpha=0.00001$. Then

$$
a=(1-\alpha)^{1+\alpha}(1+\alpha)^{1-\alpha}=\left(1-\alpha^{2}\right)\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}
$$

Since $1-\alpha^{2}<1$, then $a<1$

b) $\frac{1}{2}<5$
$$

156. Prove that

$$
\begin{gathered}
\log _{\sqrt{2}+\sqrt{3}}(4 \sqrt{2}+3 \sqrt{\overline{3}}) \cdot \log _{\sqrt{6}+1}(\sqrt{3}-\sqrt{2})+ \\
+\log ^{2} \sqrt{6}+7 \\
(2 \sqrt{6}+5)+1=0
\end{gathered}
$$

## § 2. Systematic Fractions and Irrationalities