problem canceling in multiplication of fractions

Question

now that i have a grasp on formatting mathematical fractions and whatnot i present you with my latest confusion

$$3\frac15 \times 1\frac23 \times 2\frac 34$$

converted into improper fractions

$$\frac{16}5 \times \frac 53 \times \frac{11}4$$

this much i understand. the following bit is whats driving me insane

$$\frac41 \times \frac13 \times \frac{11}1$$

i simply dont get how it goes from the improper fractions to the next set of fractions. apparently its called canceling but i cant see any sense in it

Answer 1

First, $\frac{16}5 \times \frac 53 \times \frac{11}4$ is the same as $\frac{16}1 \times \frac 13 \times \frac{11}4$, since we can cancel the $5$ from both numerator and denominator. And in turn, $\frac{16}1 \times \frac 13 \times \frac{11}4$ is the same as $\frac41 \times \frac13 \times \frac{11}1$, since we can cancel a factor of $4$ from both numerator and denominator.

It might be able to see the validity of these cancellations more easily if, instead of writing $\frac{16}5 \times \frac 53 \times \frac{11}4 = \frac{16}1 \times \frac 13 \times \frac{11}4 = \frac41 \times \frac13 \times \frac{11}1$, we instead write the equivalent $\frac{16\times5\times11}{5\times3\times4} = \frac{16\times1\times11}{1\times3\times4} = \frac{4\times1\times11}{1\times3\times1}$.

Answer 2

We know that $\dfrac{a}{b}\times\dfrac{c}{d}=\dfrac{a\times c}{b\times d}$ and that we can commute $\dfrac{a}{b}\times\dfrac{c}{d}=\dfrac{c}{b}\times\dfrac{a}{d}=\dfrac{c}{d}\times\dfrac{a}{b}$ and so on, of course in the nominator and denominatior. Then, in your exercise: $$\frac{16}5 \times \frac 53 \times \frac{11}4=\dfrac{16}{4}\times\dfrac{5}{5}\times\dfrac{11}{3}=\dfrac{4}{1}\times\dfrac{1}{1}\times\dfrac{11}{3}=\dfrac{4}{1}\times\dfrac{1}{3}\times\dfrac{11}{1}$$

Answer 3

The theory behind cancelling is: (according to my logic) EQUIVALENT FRACTIONS. If you know a bit of algebraic notation, then equivalent fractions are conceptualised as: Let a, b be two real numbers. Then, for any real c $≠$ 0, $$\frac{a}{b}=\frac{ca}{cb}.$$ For example, we can write the fraction $\frac 34$ as: $$\displaystyle\frac 34 = \frac {3\times 4}{4\times 4}= \frac {3\times 2.7534}{4\times 2.7534} , etc .$$ Now, during the “cancellation” procedure, we go reverse. That is, we take a fraction $\displaystyle \frac {p}{q}$; find some c such that p=ca and q= cb, and write $$\frac {p}{q}= \frac {ca}{cb}= \frac {a}{b}.$$ Also, $\displaystyle\frac {m}{n}\times \frac{r}{s}$ can be written as $\displaystyle\frac{m \times r}{n \times s}$. Thus, $$\frac {16}{5}\times \frac {5}{3}\times \frac {11}{4}= \frac {16\times 5\times 11}{5\times 3\times 4}= \frac {4 \times 4 \times 5 \times 11}{5 \times 3 \times 4}$$ Now you can remove the factors which appear in both the numerator and denominator, so removing 4 and 5 gives you $\displaystyle\frac{4 \times 11 \times 1 \times 1}{3 \times 1 \times 1}$.

Answer 4

Since you know about equivalent fractions:

$$ \frac{a \times b}{a \times c} = \frac{b}{c} $$

I will try to explain in those terms.

You also know that $$ \frac{r}{s} \times \frac{t}{u} = \frac{r \times t}{s \times u}. $$

First use that to write your product as a single fraction with the numerator and denominator as a product: $$ \frac{16}5 \times \frac 53 \times \frac{11}4 = \frac{16 \times 5 \times 11}{5 \times 3 \times 4}. $$

Now both the numerator and denominator have $5$ as a factor, so your knowledge of equivalent fractions says you can cancel it: $$ \frac{16 \times 5 \times 11}{5 \times 3 \times 4} = \frac{16 \times 11}{ 3 \times 4}. $$ The fact that the factor of $5$ comes from different fractions in the original problem does not matter.

Now use the fact that $16 = 4 \times 4$ to cancel a $4$.

What's left, $$ \frac{4 \times 11}{3}, $$ is in lowest terms. There is no more cancelling possible and you are done.

Once you understand these principles you can do what your book or teacher (the "it" in your question) does without having to go through those steps explicitly.