Lösung 3.1:4a
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Version vom 14:21, 22. Okt. 2008
The decimal number \displaystyle 0\textrm{.}16 can also be written as \displaystyle 16\cdot 10^{-2} and then it is easier to see that, since \displaystyle 16 = 4\cdot 4 = 4^2 and \displaystyle 10^{-2} = (10^{-1})^2 = 0\textrm{.}1^2,
| \displaystyle \begin{align} \sqrt{0\textrm{.}16} &= \sqrt{16\cdot 10^{-2}} = \sqrt{16}\cdot \sqrt{10^{-2}} = \sqrt{4^2}\cdot \sqrt{0\textrm{.}1^2}\\[5pt] &= 4\cdot 0\textrm{.}1 = 0\textrm{.}4\,\textrm{.} \end{align} | 
Another alternative is, of course, to see directly that \displaystyle 0\textrm{.}16 = 0\textrm{.}4\cdot 0\textrm{.}4 = 0\textrm{.}4^2, and then that \displaystyle \sqrt{0\textrm{.}16} = \sqrt{0\textrm{.}4^2} = 0\textrm{.}4\,\textrm{.}
 
		  