To find irrational numbers between two irrational numbers $a$ and $b$,we can use the form $\sqrt{a \cdot b}$,$\sqrt{a \cdot \sqrt{a \cdot b}}$,etc.,or simply find numbers whose decimal expansions are non-terminating and non-repeating.
$1$. First number: $\sqrt{\sqrt{2} \cdot \sqrt{5}} = \sqrt{\sqrt{10}} = 10^{\frac{1}{4}} \approx 1.778$.
$2$. Second number: $\sqrt{\sqrt{2} \cdot 10^{\frac{1}{4}}} = (2^{\frac{1}{2}} \cdot 10^{\frac{1}{4}})^{\frac{1}{2}} = 2^{\frac{1}{4}} \cdot 10^{\frac{1}{8}} = 2^{\frac{1}{4}} \cdot 2^{\frac{1}{8}} \cdot 5^{\frac{1}{8}} = 2^{\frac{3}{8}} \cdot 5^{\frac{1}{8}} \approx 1.682$.
$3$. Third number: $\sqrt{10^{\frac{1}{4}} \cdot \sqrt{5}} = (10^{\frac{1}{4}} \cdot 5^{\frac{1}{2}})^{\frac{1}{2}} = 10^{\frac{1}{8}} \cdot 5^{\frac{1}{4}} = 2^{\frac{1}{8}} \cdot 5^{\frac{1}{8}} \cdot 5^{\frac{1}{4}} = 2^{\frac{1}{8}} \cdot 5^{\frac{3}{8}} \approx 1.880$.
These values lie between $\sqrt{2} \approx 1.414$ and $\sqrt{5} \approx 2.236$.