canceling double fractions how?

Question

I had this example: $$ \frac{\frac{11}{5}}{2} = \frac{11}{10} $$

then:

$$ \frac{2\frac{1}{5}}{2} = \frac{11}{10} $$

$$ \frac{1}{5} \not= \frac{11}{10} $$

is this right canceling of double equation?

Answer 1

I always regarded the triple (or quadruple) fraction notation to be confusing and wrong, so instead (even to this day) I always make myself read fractions like this:

The fraction $\frac{\frac{a}{b}}{c}$ is equivalent to $\frac{a}{b} \div c$ and therefore $\frac{a}{b} \times \frac{1}{c} = \frac{a}{bc}$. For quadruples it works the same: $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$.

So using this I would say $\frac{\frac{11}{5}}{2} = \frac{11}{5} \times \frac{1}{2} = \frac{11}{10}$.

Answer 2

$2\frac{1}{5}$ does not mean $2 \times \frac{1}{5}$. It means $2+\frac{1}{5}$.

Answer 3

No, it isn't. Multiplying $$\frac{\frac{11}{5}}{2}$$ by $$\frac{5}{5}\ (=1)$$ gives you $$\frac{\frac{11}{5}}{2}=\frac{\frac{11}{5}}{2}\cdot\frac{5}{5}=\frac{\frac{11}{5}\cdot 5}{2\cdot 5}=\frac{11}{10}.$$

Answer 4

You can think like this: $$\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\frac{a}{b}\times\dfrac{1}{\dfrac{c}{d}}=\dfrac{a}{b}\times\dfrac{d}{c}$$

Answer 5

we have $$\frac{11}{5}=2\frac{1}{5}$$ thus $$\frac{2\frac{1}{5}}{2}=1+\frac{1}{10}=\frac{11}{10}$$