Solution 1.1:4d

From Förberedande kurs i matematik 1

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Current revision (13:55, 13 September 2008) (edit) (undo)
 
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We have that
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:<math>\biggl(\dfrac{4}{\sqrt{2}}\biggr)^2=\dfrac{4^2}{(\sqrt{2}\,)^2}=\dfrac{16}{2}=8</math>
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is a natural number, integer and rational number.
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Current revision

We have that

\displaystyle \biggl(\dfrac{4}{\sqrt{2}}\biggr)^2=\dfrac{4^2}{(\sqrt{2}\,)^2}=\dfrac{16}{2}=8

is a natural number, integer and rational number.