Solution 3.1:4a

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The decimal number <math>0\textrm{.}16</math> can also be written as <math>16\cdot 10^{-2}</math> and then it is easier to see that, since <math>16 = 4\cdot 4 = 4^2</math> and <math>10^{-2} = (10^{-1})^2 = 0\textrm{.}1^2</math>,
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{{Displayed math||<math>\begin{align}
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\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]
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&= 4\cdot 0\textrm{.}1 = 0\textrm{.}4\,\textrm{.}
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\end{align}</math>}}
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Another alternative is, of course, to see directly that <math>0\textrm{.}16 = 0\textrm{.}4\cdot 0\textrm{.}4 = 0\textrm{.}4^2</math>, and then that <math>\sqrt{0\textrm{.}16} = \sqrt{0\textrm{.}4^2} = 0\textrm{.}4\,\textrm{.}</math>

Current revision

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{.}

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