Lösung 1.1:4d
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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- | + | We have that | |
:<math>\biggl(\dfrac{4}{\sqrt{2}}\biggr)^2=\dfrac{4^2}{(\sqrt{2}\,)^2}=\dfrac{16}{2}=8</math> | :<math>\biggl(\dfrac{4}{\sqrt{2}}\biggr)^2=\dfrac{4^2}{(\sqrt{2}\,)^2}=\dfrac{16}{2}=8</math> | ||
- | + | is a natural number, integer and rational number. | |
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Version vom 13:55, 13. Sep. 2008
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.