Lösung 3.1:8c

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Since \displaystyle 2\textrm{.}5^{2} = 2\textrm{.}5\cdot 2\textrm{.}5 = 6\textrm{.}25\,, then \displaystyle 2\textrm{.}5 = \sqrt{6\textrm{.}25}\,, and then we see that \displaystyle \sqrt{7} is greater than 2.5 since \displaystyle 7^{1/2} > 6\textrm{.}25^{1/2}\,.