Question

Quantas frações da forma n/n + 1 são menores do que 7/9 , sabendo que n é um número inteiro positivo

Answer 1

$\mathrm{\dfrac{n}{n+1}\ \textless \ \dfrac{7}{9}\ \to\ \dfrac{n}{n+1}-\dfrac{7}{9}\ \textless \ \dfrac{7}{9}-\dfrac{7}{9}\ \to}\\\\ \mathrm{\to\ \dfrac{n}{n+1}-\dfrac{7}{9}\ \textless \ 0\ \to\ \dfrac{9n-7(n+1)}{9(n+1)}\ \textless \ 0\ \to}\\\\ \mathrm{\to\ \dfrac{9n-7n-7}{9n+9}\ \textless \ 0\ \to\ \dfrac{2n-7}{9n+9}\ \textless \ 0}\\\\\\ \textrm{Para que a inequa\c{c}\~ao seja verdadeira, temos}\\ \textrm{duas possibilidades:}\ \mathrm{2n-7>0\ e\ 9n+9<0}\\ \mathrm{ou\ 2n-7<0\ e\ 9n+9>0.\ Logo:}$

$\mathrm{2n-7\ \textgreater \ 0\ \to\ 2n\ \textgreater \ 7\ \to\ n\ \textgreater \ \dfrac{7}{2}}\\\\ \mathrm{9n+9\ \textless \ 0\ \to\ 9n\ \textless \ -9\ \to\ n\ \textless \ -1}\\\\ \mathrm{n\ \textgreater \ \dfrac{7}{2}\ \ e\ \ n\ \textless \ -1\ \to\ imposs\'ivel}}\\\\ \mathrm{2n-7\ \textless \ 0\ \to\ 2n\ \textless \ 7\ \to\ n\ \textless \ \dfrac{7}{2}}\\\\ \mathrm{9n+9\ \textgreater \ 0\ \to\ 9n\ \textgreater \ -9\ \to\ n\ \textgreater \ -1}\\\\ \mathrm{n\ \textless \ \dfrac{7}{2}\ \ e\ \ n\ \textgreater \ -1\ \to\ -1\ \textless \ n\ \textless \ \dfrac{7}{2}}\\\\ \mathrm{Por\'em,\ como\ n\in\mathbb{Z^+}:}\ \mathbf{0 \leq n\leq 3}$

$\textrm{Portanto, teremos 4 possibilidades:}\\\\ \mathrm{\mathbf{I.}\ \dfrac{0}{0+1}=\dfrac{0}{1}}\ \ \| \ \ \mathrm{\mathbf{II.}\ \dfrac{1}{1+1}=\dfrac{1}{2}}\\\\ \mathrm{\mathbf{III.}\ \dfrac{2}{2+1}=\dfrac{2}{3}}\ \ \| \ \ \mathrm{\mathbf{IV.}\ \dfrac{3}{3+1}=\dfrac{3}{4}}\\\\ \mathrm{Resposta\ \to\ \textbf{4\ fra\c{c}\~oes}}$