problem canceling in multiplication of fractions

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

• Hey do you know about equivalent fractions? Commented Jun 19, 2022 at 19:07
• They are canceling in their heads. $\require{cancel}$$\frac {\color{red}{\cancel{16}}^4}{\color{green}{\cancel 5}}\times\frac{\color{green}{\cancel{5}}}{3}\times\frac{11}{\color{red}{\cancel 4}}. Commented Jun 19, 2022 at 19:20 • They are canceling in their heads. \require{cancel}$$\frac {\color{red}{\cancel{16}}^4}{\color{green}{\cancel 5}}\times\frac{\color{green}{\cancel{5}}}{3}\times\frac{11}{\color{red}{\cancel 4}}= \frac 41\times \frac 1 3\times \frac {11}1$. Commented Jun 19, 2022 at 19:26

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: $$\frac{16}5 \times \frac 53 \times \frac{11}4=\dfrac{16}{4}\times\dfrac{5}{5}\times\dfrac{11}{3}=\dfrac{4}{1}\times\dfrac{1}{1}\times\dfrac{11}{3}=\dfrac{4}{1}\times\dfrac{1}{3}\times\dfrac{11}{1}$$

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 $$\displaystyle\frac{4 \times 11 \times 1 \times 1}{3 \times 1 \times 1}$$.

• So as a general rule if a denominator and numerator have the same actual digit they cancel each other out? Commented Jun 19, 2022 at 19:28
• It’s not just a digit. Any number. If they both had 11, we would cancel. If they both had $\pi$, we would cancel. If they both had 96.3257, we would cancel. Any real number will do. Commented Jun 19, 2022 at 19:29
• So if I need a numerator to cancel out a denominator I can split it up (if possible) like you did with the 16? So it's about balance? Commented Jun 19, 2022 at 19:32
• Yes. it IS about balance. For eg., (38x5)/19 would be (19x2x5)/19 so cancel 19. You get (2x5)/1 which is 10/1 = 10. Commented Jun 19, 2022 at 19:32
• Great, I'm getting somewhere, what do you mean by what you said here " Let a, b be two real numbers. Then, for any real c ≠ 0, " Commented Jun 19, 2022 at 19:35

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.

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 $$\frac{16\times5\times11}{5\times3\times4} = \frac{16\times1\times11}{1\times3\times4} = \frac{4\times1\times11}{1\times3\times1}$$.

• Thanks for the reply Greg, So the demoninator 5 somehow cancels out the middle numerator 5? What is the logic behind that? Commented Jun 19, 2022 at 19:19