Question

(-11/3;47/40) intersectat cu Z cât da?
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Answer 1

$\left(-\dfrac{11}{3};\,\dfrac{47}{40}\right) = \left(\dfrac{-9-2}{3};\,\dfrac{40+7}{40}\right) =\\ \\ = \left(-3-\dfrac{2}{3};\,\,1+\dfrac{7}{40}\right) = \left(-3\dfrac{2}{3};\,\,1\dfrac{7}{40}\right)\\ \\ \\ \Rightarrow \left(-\dfrac{11}{3};\,\dfrac{47}{40}\right)\cap \mathbb{Z} = \{-3,-2,-1,0,1\}$