Solution 1.1:4d
From Förberedande kurs i matematik 1
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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. | |
<!--<center> [[Image:1_1_4d.gif]] </center>--> | <!--<center> [[Image:1_1_4d.gif]] </center>--> |
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.