Ta có:
$2 = \sqrt4$
$= \sqrt{2 + 2}$
$= \sqrt{2 + \sqrt4}$
$=\sqrt{2 + \sqrt{2 + 2}}$
$= \sqrt{2 + \sqrt{2 + \sqrt4}}$
$= \sqrt{2 + \sqrt{2 + \sqrt{2 + 2}}}$
$=\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt4}}}$
$ =\underbrace{\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+\dots+\sqrt4}}}}}_{\text{2020 dấu căn}}$
Do $\sqrt2 < \sqrt4$
nên $\underbrace{\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+\dots+\sqrt2}}}}}_{\text{2020 dấu căn}} < \underbrace{\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+\dots+\sqrt4}}}}}_{\text{2020 dấu căn}}$
hay $\underbrace{\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+\dots+\sqrt2}}}}}_{\text{2020 dấu căn}} < 2$
