Question
Is $\frac{8}{9 }$ the multiplicative inverse or $-1\frac{1}{8}$? Why or why not?
✓
Answer
Since multiplicative inverse of a rational number a is $\Big(\frac{1}{\text{a}}\Big),\text{ if }\text{a}\times\frac{1}{\text{a}}=1$
Therefore, $\frac{8}{9}\times\Big(-1\frac{1}{8}\Big)=\frac{8}{9}\times\frac{-9}{8}=-1$
But its product must be positive $1$.
Therefore, $\frac{8}{9}$ is not the multiplicative inverse of $\Big(-1\frac{1}{8}\Big)$
Therefore, $\frac{8}{9}\times\Big(-1\frac{1}{8}\Big)=\frac{8}{9}\times\frac{-9}{8}=-1$
But its product must be positive $1$.
Therefore, $\frac{8}{9}$ is not the multiplicative inverse of $\Big(-1\frac{1}{8}\Big)$
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